Sec 2 G1 Maths Tutor Bukit Timah | Answer Sense Checks

Did you know? A quick mathematical prediction can stop a wrong answer before the calculator is touched. A 20% discount cannot make a positive price go up, a probability cannot exceed one, and doubling both side lengths of a rectangle cannot merely double its area. These simple constraints allow a Secondary 2 G1 learner to test whether an answer could be right, not merely whether the arithmetic looked neat.

Quick answer: five sense-checks before calculating

Ask five questions: What kind of quantity is the answer? (length, area, probability, number of items); what sign or interval is possible? (positive, negative, zero, between zero and one); should the value rise or fall?; what rough magnitude would be reasonable?; and what independent calculation could reconstruct the original givens? Then calculate exactly and compare the result with the prediction. A mismatch signals an error worth locating, not permission to change a number by guesswork.

This guide is for parents exploring a Secondary 2 G1 Mathematics tutor in Bukit Timah. It differs from an article about decimal-place rounding or significant figures: this one teaches reasonableness constraints before arithmetic and independent checks afterward. MOE’s Secondary Two G1 course includes signed numbers, ratios, equations, rates, graphs, mensuration and simple probability. The published 2027 SEC G1 Mathematics K110 syllabus supports estimation and contextual reasoning for that graduating cohort; the current school assessment scope remains authoritative for a pupil in Secondary Two in 2026.

Choose a useful sense-check

The sixty-second reasoning test
Before, during and after solving
Signed-number and magnitude predictions
Fractions and decimals within sensible bounds
Percentages and discounts: predict direction
Ratio sharing and constant-rate checks
Algebra and independent inverse checks
Linear graph directions and output ranges
Geometry area, perimeter and dimensional checks
Statistics and probability plausibility
What each mismatch can tell a tutor
Short parent practice
Frequently asked questions
Official references

A sixty-second check with three impossible answers

A learner says a 25% discount on $80 produces a final price of $100; a bag with three red and seven blue counters gives P(red)=3/2; and a 4-cm by 5-cm rectangle has area 20 cm. Ask which answers can be rejected without redoing the full arithmetic. All three fail an independent constraint: a discount should reduce payment, probability must be between zero and one, and area needs square units. The first two also fail useful magnitude bounds.

A simple before–during–after routine

Before: predict the quantity, sign and size

For a ratio sharing fifty objects, the individual shares cannot be negative or exceed fifty if the groups partition that whole. For a direct proportional rate, doubling the number of identical units should double the total. For a signed-number sum, a rough number-line movement may predict the sign. A prediction does not have to be a precise decimal; it must follow logically from the story.

During: choose the method that matches the relation

A sensible estimate cannot replace an equation or a valid fraction calculation. After making a bound, solve with standard G1 techniques appropriate to the school task. Write intermediate steps clearly enough to identify where the model or arithmetic may have changed, especially around negative brackets, different units and percentage bases.

After: use a different path where available

Substitute an x-value back into its original equation, reconstruct a ratio’s total and proportions, convert measurement units backwards, or compare a fraction with a decimal bound. Simply repeating identical calculator keys may repeat the same error. The goal is an independent check connected with the original givens and requested output.

216 original sense-checking clinics

The following examples are a topic bank rather than compulsory daily homework. They vary signs, fractions, percentage directions, ratio parts, graphs, geometry and probabilities. Cover each worked answer, make the initial prediction, calculate and explain whether the result fits. All cases are original mathematical teaching illustrations, not official examination questions or actual commercial offers.

Signed-number and magnitude predictions

This group is about signed-number and magnitude predictions. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

1. Signed calculation 1: choose the answer’s direction

Question: Before calculating exactly, determine whether (-20)+(-18)−1 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -20 is the starting signed position, adding -18 changes it, and subtracting positive 1 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -20+(-18)=-38, then subtract 1 to obtain -39, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-20)+(-18)+1 adds the last quantity instead of subtracting it; the answer would be -37, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -39+1=-38. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 3, the answer should be -41; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

2. Signed calculation 2: choose the answer’s direction

Question: Before calculating exactly, determine whether (-13)+(-7)−4 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -13 is the starting signed position, adding -7 changes it, and subtracting positive 4 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -13+(-7)=-20, then subtract 4 to obtain -24, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-13)+(-7)+4 adds the last quantity instead of subtracting it; the answer would be -16, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -24+4=-20. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 6, the answer should be -26; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

3. Signed calculation 3: choose the answer’s direction

Question: Before calculating exactly, determine whether (-6)+(4)−7 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -6 is the starting signed position, adding 4 changes it, and subtracting positive 7 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -6+(4)=-2, then subtract 7 to obtain -9, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-6)+(4)+7 adds the last quantity instead of subtracting it; the answer would be 5, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -9+7=-2. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 9, the answer should be -11; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

4. Signed calculation 4: choose the answer’s direction

Question: Before calculating exactly, determine whether (1)+(15)−10 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 1 is the starting signed position, adding 15 changes it, and subtracting positive 10 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 1+(15)=16, then subtract 10 to obtain 6, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (1)+(15)+10 adds the last quantity instead of subtracting it; the answer would be 26, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 6+10=16. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 12, the answer should be 4; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

5. Signed calculation 5: choose the answer’s direction

Question: Before calculating exactly, determine whether (8)+(-11)−13 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 8 is the starting signed position, adding -11 changes it, and subtracting positive 13 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 8+(-11)=-3, then subtract 13 to obtain -16, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (8)+(-11)+13 adds the last quantity instead of subtracting it; the answer would be 10, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -16+13=-3. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 15, the answer should be -18; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

6. Signed calculation 6: choose the answer’s direction

Question: Before calculating exactly, determine whether (15)+(0)−16 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 15 is the starting signed position, adding 0 changes it, and subtracting positive 16 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 15+(0)=15, then subtract 16 to obtain -1, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (15)+(0)+16 adds the last quantity instead of subtracting it; the answer would be 31, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -1+16=15. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 18, the answer should be -3; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

7. Signed calculation 7: choose the answer’s direction

Question: Before calculating exactly, determine whether (-19)+(11)−3 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -19 is the starting signed position, adding 11 changes it, and subtracting positive 3 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -19+(11)=-8, then subtract 3 to obtain -11, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-19)+(11)+3 adds the last quantity instead of subtracting it; the answer would be -5, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -11+3=-8. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 5, the answer should be -13; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

8. Signed calculation 8: choose the answer’s direction

Question: Before calculating exactly, determine whether (-12)+(-15)−6 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -12 is the starting signed position, adding -15 changes it, and subtracting positive 6 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -12+(-15)=-27, then subtract 6 to obtain -33, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-12)+(-15)+6 adds the last quantity instead of subtracting it; the answer would be -21, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -33+6=-27. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 8, the answer should be -35; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

9. Signed calculation 9: choose the answer’s direction

Question: Before calculating exactly, determine whether (-5)+(-4)−9 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -5 is the starting signed position, adding -4 changes it, and subtracting positive 9 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -5+(-4)=-9, then subtract 9 to obtain -18, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-5)+(-4)+9 adds the last quantity instead of subtracting it; the answer would be 0, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -18+9=-9. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 11, the answer should be -20; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

10. Signed calculation 10: choose the answer’s direction

Question: Before calculating exactly, determine whether (2)+(7)−12 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 2 is the starting signed position, adding 7 changes it, and subtracting positive 12 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 2+(7)=9, then subtract 12 to obtain -3, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (2)+(7)+12 adds the last quantity instead of subtracting it; the answer would be 21, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -3+12=9. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 14, the answer should be -5; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

11. Signed calculation 11: choose the answer’s direction

Question: Before calculating exactly, determine whether (9)+(18)−15 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 9 is the starting signed position, adding 18 changes it, and subtracting positive 15 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 9+(18)=27, then subtract 15 to obtain 12, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (9)+(18)+15 adds the last quantity instead of subtracting it; the answer would be 42, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 12+15=27. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 17, the answer should be 10; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

12. Signed calculation 12: choose the answer’s direction

Question: Before calculating exactly, determine whether (16)+(-8)−2 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 16 is the starting signed position, adding -8 changes it, and subtracting positive 2 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 16+(-8)=8, then subtract 2 to obtain 6, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (16)+(-8)+2 adds the last quantity instead of subtracting it; the answer would be 10, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 6+2=8. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 4, the answer should be 4; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

13. Signed calculation 13: choose the answer’s direction

Question: Before calculating exactly, determine whether (-18)+(3)−5 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -18 is the starting signed position, adding 3 changes it, and subtracting positive 5 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -18+(3)=-15, then subtract 5 to obtain -20, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-18)+(3)+5 adds the last quantity instead of subtracting it; the answer would be -10, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -20+5=-15. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 7, the answer should be -22; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

14. Signed calculation 14: choose the answer’s direction

Question: Before calculating exactly, determine whether (-11)+(14)−8 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -11 is the starting signed position, adding 14 changes it, and subtracting positive 8 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -11+(14)=3, then subtract 8 to obtain -5, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-11)+(14)+8 adds the last quantity instead of subtracting it; the answer would be 11, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -5+8=3. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 10, the answer should be -7; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

15. Signed calculation 15: choose the answer’s direction

Question: Before calculating exactly, determine whether (-4)+(-12)−11 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -4 is the starting signed position, adding -12 changes it, and subtracting positive 11 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -4+(-12)=-16, then subtract 11 to obtain -27, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-4)+(-12)+11 adds the last quantity instead of subtracting it; the answer would be -5, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -27+11=-16. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 13, the answer should be -29; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

16. Signed calculation 16: choose the answer’s direction

Question: Before calculating exactly, determine whether (3)+(-1)−14 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 3 is the starting signed position, adding -1 changes it, and subtracting positive 14 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 3+(-1)=2, then subtract 14 to obtain -12, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (3)+(-1)+14 adds the last quantity instead of subtracting it; the answer would be 16, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -12+14=2. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 16, the answer should be -14; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

17. Signed calculation 17: choose the answer’s direction

Question: Before calculating exactly, determine whether (10)+(10)−1 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 10 is the starting signed position, adding 10 changes it, and subtracting positive 1 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 10+(10)=20, then subtract 1 to obtain 19, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (10)+(10)+1 adds the last quantity instead of subtracting it; the answer would be 21, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 19+1=20. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 3, the answer should be 17; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

18. Signed calculation 18: choose the answer’s direction

Question: Before calculating exactly, determine whether (17)+(-16)−4 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 17 is the starting signed position, adding -16 changes it, and subtracting positive 4 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 17+(-16)=1, then subtract 4 to obtain -3, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (17)+(-16)+4 adds the last quantity instead of subtracting it; the answer would be 5, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -3+4=1. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 6, the answer should be -5; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

19. Signed calculation 19: choose the answer’s direction

Question: Before calculating exactly, determine whether (-17)+(-5)−7 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -17 is the starting signed position, adding -5 changes it, and subtracting positive 7 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -17+(-5)=-22, then subtract 7 to obtain -29, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-17)+(-5)+7 adds the last quantity instead of subtracting it; the answer would be -15, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -29+7=-22. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 9, the answer should be -31; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

20. Signed calculation 20: choose the answer’s direction

Question: Before calculating exactly, determine whether (-10)+(6)−10 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -10 is the starting signed position, adding 6 changes it, and subtracting positive 10 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -10+(6)=-4, then subtract 10 to obtain -14, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-10)+(6)+10 adds the last quantity instead of subtracting it; the answer would be 6, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -14+10=-4. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 12, the answer should be -16; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

21. Signed calculation 21: choose the answer’s direction

Question: Before calculating exactly, determine whether (-3)+(17)−13 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -3 is the starting signed position, adding 17 changes it, and subtracting positive 13 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -3+(17)=14, then subtract 13 to obtain 1, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-3)+(17)+13 adds the last quantity instead of subtracting it; the answer would be 27, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 1+13=14. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 15, the answer should be -1; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

22. Signed calculation 22: choose the answer’s direction

Question: Before calculating exactly, determine whether (4)+(-9)−16 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 4 is the starting signed position, adding -9 changes it, and subtracting positive 16 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 4+(-9)=-5, then subtract 16 to obtain -21, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (4)+(-9)+16 adds the last quantity instead of subtracting it; the answer would be 11, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -21+16=-5. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 18, the answer should be -23; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

23. Signed calculation 23: choose the answer’s direction

Question: Before calculating exactly, determine whether (11)+(2)−3 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 11 is the starting signed position, adding 2 changes it, and subtracting positive 3 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 11+(2)=13, then subtract 3 to obtain 10, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (11)+(2)+3 adds the last quantity instead of subtracting it; the answer would be 16, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 10+3=13. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 5, the answer should be 8; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

24. Signed calculation 24: choose the answer’s direction

Question: Before calculating exactly, determine whether (18)+(13)−6 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: 18 is the starting signed position, adding 13 changes it, and subtracting positive 6 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine 18+(13)=31, then subtract 6 to obtain 25, which is positive. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (18)+(13)+6 adds the last quantity instead of subtracting it; the answer would be 37, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: 25+6=31. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 8, the answer should be 23; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

25. Signed calculation 25: choose the answer’s direction

Question: Before calculating exactly, determine whether (-16)+(-13)−9 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -16 is the starting signed position, adding -13 changes it, and subtracting positive 9 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -16+(-13)=-29, then subtract 9 to obtain -38, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-16)+(-13)+9 adds the last quantity instead of subtracting it; the answer would be -20, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -38+9=-29. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 11, the answer should be -40; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

26. Signed calculation 26: choose the answer’s direction

Question: Before calculating exactly, determine whether (-9)+(-2)−12 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -9 is the starting signed position, adding -2 changes it, and subtracting positive 12 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -9+(-2)=-11, then subtract 12 to obtain -23, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-9)+(-2)+12 adds the last quantity instead of subtracting it; the answer would be 1, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -23+12=-11. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 14, the answer should be -25; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

27. Signed calculation 27: choose the answer’s direction

Question: Before calculating exactly, determine whether (-2)+(9)−15 should be positive, negative or zero. Then find the exact result.

Predict first: Think of the signed quantities as movements on a number line: -2 is the starting signed position, adding 9 changes it, and subtracting positive 15 shifts it further left. A rough bound is enough to reject an answer whose sign makes no sense. The exact arithmetic should confirm that prediction, not replace it.

Exact working: First combine -2+(9)=7, then subtract 15 to obtain -8, which is negative. If an earlier rough prediction disagreed, inspect which signed move you visualised incorrectly rather than changing the final number by guesswork.

Tempting wrong line: Writing (-2)+(9)+15 adds the last quantity instead of subtracting it; the answer would be 22, a different operation. A second risk is interpreting a negative addend as a positive because its magnitude is written without brackets.

Independent check: Reverse the final operation: -8+15=7. The recovered intermediate value matches the first two terms. This check is different from simply retyping the entire original string.

Transfer: If the last quantity became 17, the answer should be -10; explain why the additional subtraction moves two units left. Predict the new sign before calculating.

Fractions and decimals within sensible bounds

This group is about fractions and decimals within sensible bounds. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

28. Fraction bound 1: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/2+1/4 is below one, equal to one or above one; then calculate exactly.

Predict first: Both given fractions are positive and less than one, so their sum must exceed either individual fraction while remaining below two. This bound prevents implausibly small or extremely large results. To decide the finer comparison with one, compare their complementary parts or use common units; a rough comparison is a hypothesis, not a guarantee.

Exact working: A common denominator 8 gives 4/8+2/8=6/8. Simplifying yields 3/4, or approximately 0.75. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (1+1)/(2+4); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 3/4 can be checked by multiplication back into denominator 8. Also compare its approximate decimal 0.75 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/4 and becomes 1. Predict the direction before recomputing the result.

29. Fraction bound 2: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/4+1/2 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 8 gives 2/8+4/8=6/8. Simplifying yields 3/4, or approximately 0.75. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (1+1)/(4+2); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/2 and becomes 5/4. Predict the direction before recomputing the result.

30. Fraction bound 3: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 3/5+15/20 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 60/100+75/100=135/100. Simplifying yields 27/20, or approximately 1.35. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (3+15)/(5+20); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 27/20 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.35 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 15 to 16. The exact sum increases by 1/20 and becomes 7/5. Predict the direction before recomputing the result.

31. Fraction bound 4: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 3/8+4/10 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 80 gives 30/80+32/80=62/80. Simplifying yields 31/40, or approximately 0.775. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (3+4)/(8+10); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 31/40 can be checked by multiplication back into denominator 80. Also compare its approximate decimal 0.775 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 4 to 5. The exact sum increases by 1/10 and becomes 7/8. Predict the direction before recomputing the result.

32. Fraction bound 5: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 4/10+1/8 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 80 gives 32/80+10/80=42/80. Simplifying yields 21/40, or approximately 0.525. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (4+1)/(10+8); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 21/40 can be checked by multiplication back into denominator 80. Also compare its approximate decimal 0.525 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/8 and becomes 13/20. Predict the direction before recomputing the result.

33. Fraction bound 6: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 16/20+4/5 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 80/100+80/100=160/100. Simplifying yields 8/5, or approximately 1.6. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (16+4)/(20+5); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 8/5 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.6 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 4 to 5. The exact sum increases by 1/5 and becomes 9/5. Predict the direction before recomputing the result.

34. Fraction bound 7: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/2+1/4 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/4 and becomes 1. Predict the direction before recomputing the result.

35. Fraction bound 8: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/4+1/2 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/2 and becomes 5/4. Predict the direction before recomputing the result.

36. Fraction bound 9: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/5+19/20 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 20/100+95/100=115/100. Simplifying yields 23/20, or approximately 1.15. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (1+19)/(5+20); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 23/20 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.15 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 19 to 20. The exact sum increases by 1/20 and becomes 6/5. Predict the direction before recomputing the result.

37. Fraction bound 10: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 7/8+1/10 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 80 gives 70/80+8/80=78/80. Simplifying yields 39/40, or approximately 0.975. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (7+1)/(8+10); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 39/40 can be checked by multiplication back into denominator 80. Also compare its approximate decimal 0.975 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/10 and becomes 43/40. Predict the direction before recomputing the result.

38. Fraction bound 11: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 4/10+1/8 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/8 and becomes 13/20. Predict the direction before recomputing the result.

39. Fraction bound 12: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 15/20+2/5 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 75/100+40/100=115/100. Simplifying yields 23/20, or approximately 1.15. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (15+2)/(20+5); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Transfer: Increase only the second numerator from 2 to 3. The exact sum increases by 1/5 and becomes 27/20. Predict the direction before recomputing the result.

40. Fraction bound 13: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/2+1/4 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/4 and becomes 1. Predict the direction before recomputing the result.

41. Fraction bound 14: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/4+1/2 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/2 and becomes 5/4. Predict the direction before recomputing the result.

42. Fraction bound 15: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 3/5+4/20 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 60/100+20/100=80/100. Simplifying yields 4/5, or approximately 0.8. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (3+4)/(5+20); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 4/5 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 0.8 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 4 to 5. The exact sum increases by 1/20 and becomes 17/20. Predict the direction before recomputing the result.

43. Fraction bound 16: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 4/8+7/10 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 80 gives 40/80+56/80=96/80. Simplifying yields 6/5, or approximately 1.2. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (4+7)/(8+10); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 6/5 can be checked by multiplication back into denominator 80. Also compare its approximate decimal 1.2 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 7 to 8. The exact sum increases by 1/10 and becomes 13/10. Predict the direction before recomputing the result.

44. Fraction bound 17: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 4/10+1/8 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/8 and becomes 13/20. Predict the direction before recomputing the result.

45. Fraction bound 18: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 14/20+4/5 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 70/100+80/100=150/100. Simplifying yields 3/2, or approximately 1.5. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (14+4)/(20+5); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 3/2 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.5 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 4 to 5. The exact sum increases by 1/5 and becomes 17/10. Predict the direction before recomputing the result.

46. Fraction bound 19: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/2+1/4 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/4 and becomes 1. Predict the direction before recomputing the result.

47. Fraction bound 20: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/4+1/2 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/2 and becomes 5/4. Predict the direction before recomputing the result.

48. Fraction bound 21: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/5+8/20 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 20/100+40/100=60/100. Simplifying yields 3/5, or approximately 0.6. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (1+8)/(5+20); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 3/5 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 0.6 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 8 to 9. The exact sum increases by 1/20 and becomes 13/20. Predict the direction before recomputing the result.

49. Fraction bound 22: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/8+4/10 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 80 gives 10/80+32/80=42/80. Simplifying yields 21/40, or approximately 0.525. This is less than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (1+4)/(8+10); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Transfer: Increase only the second numerator from 4 to 5. The exact sum increases by 1/10 and becomes 5/8. Predict the direction before recomputing the result.

50. Fraction bound 23: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 4/10+1/8 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/8 and becomes 13/20. Predict the direction before recomputing the result.

51. Fraction bound 24: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 13/20+2/5 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 65/100+40/100=105/100. Simplifying yields 21/20, or approximately 1.05. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (13+2)/(20+5); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 21/20 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.05 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 2 to 3. The exact sum increases by 1/5 and becomes 5/4. Predict the direction before recomputing the result.

52. Fraction bound 25: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/2+1/4 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/4 and becomes 1. Predict the direction before recomputing the result.

53. Fraction bound 26: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 1/4+1/2 is below one, equal to one or above one; then calculate exactly.

Transfer: Increase only the second numerator from 1 to 2. The exact sum increases by 1/2 and becomes 5/4. Predict the direction before recomputing the result.

54. Fraction bound 27: is the result below one?

Question: Without doing the complete fraction conversion, decide whether 3/5+12/20 is below one, equal to one or above one; then calculate exactly.

Exact working: A common denominator 100 gives 60/100+60/100=120/100. Simplifying yields 6/5, or approximately 1.2. This is greater than one. The two fractions contribute distinct positive amounts, never negative amounts.

Tempting wrong line: A tempting but invalid shortcut is (3+12)/(5+20); adding denominators changes the size of the units being counted. If a pupil’s answer is smaller than either positive starting fraction, it already violates a sound bound.

Independent check: The exact fraction 6/5 can be checked by multiplication back into denominator 100. Also compare its approximate decimal 1.2 against zero and two; those limits follow from the original quantities.

Transfer: Increase only the second numerator from 12 to 13. The exact sum increases by 1/20 and becomes 5/4. Predict the direction before recomputing the result.

Percentages and discounts: predict direction

This group is about percentages and discounts: predict direction. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

55. Discount sense-check 1: can payment exceed the original?

Question: A fictional item originally costs $160 and receives a straightforward 5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $160, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=5%×$160=$8. Remaining percentage is 95%, so payment=95%×$160=$152. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $8 with the amount paid $152 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 8+152=160. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $304, still below the new original $320.

56. Discount sense-check 2: can payment exceed the original?

Question: A fictional item originally costs $220 and receives a straightforward 10% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $220, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=10%×$220=$22. Remaining percentage is 90%, so payment=90%×$220=$198. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $22 with the amount paid $198 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 22+198=220. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $396, still below the new original $440.

57. Discount sense-check 3: can payment exceed the original?

Question: A fictional item originally costs $280 and receives a straightforward 12.5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $280, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=12.5%×$280=$35. Remaining percentage is 87.5%, so payment=87.5%×$280=$245. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $35 with the amount paid $245 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 35+245=280. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $490, still below the new original $560.

58. Discount sense-check 4: can payment exceed the original?

Question: A fictional item originally costs $340 and receives a straightforward 15% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $340, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=15%×$340=$51. Remaining percentage is 85%, so payment=85%×$340=$289. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $51 with the amount paid $289 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 51+289=340. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $578, still below the new original $680.

59. Discount sense-check 5: can payment exceed the original?

Question: A fictional item originally costs $400 and receives a straightforward 20% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $400, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=20%×$400=$80. Remaining percentage is 80%, so payment=80%×$400=$320. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $80 with the amount paid $320 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 80+320=400. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $640, still below the new original $800.

60. Discount sense-check 6: can payment exceed the original?

Question: A fictional item originally costs $460 and receives a straightforward 25% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $460, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=25%×$460=$115. Remaining percentage is 75%, so payment=75%×$460=$345. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $115 with the amount paid $345 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 115+345=460. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $690, still below the new original $920.

61. Discount sense-check 7: can payment exceed the original?

Question: A fictional item originally costs $520 and receives a straightforward 30% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $520, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=30%×$520=$156. Remaining percentage is 70%, so payment=70%×$520=$364. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $156 with the amount paid $364 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 156+364=520. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $728, still below the new original $1040.

62. Discount sense-check 8: can payment exceed the original?

Question: A fictional item originally costs $580 and receives a straightforward 40% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $580, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=40%×$580=$232. Remaining percentage is 60%, so payment=60%×$580=$348. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $232 with the amount paid $348 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 232+348=580. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $696, still below the new original $1160.

63. Discount sense-check 9: can payment exceed the original?

Question: A fictional item originally costs $640 and receives a straightforward 50% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $640, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=50%×$640=$320. Remaining percentage is 50%, so payment=50%×$640=$320. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $320 with the amount paid $320 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 320+320=640. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $640, still below the new original $1280.

64. Discount sense-check 10: can payment exceed the original?

Question: A fictional item originally costs $700 and receives a straightforward 60% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $700, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=60%×$700=$420. Remaining percentage is 40%, so payment=40%×$700=$280. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $420 with the amount paid $280 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 420+280=700. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $560, still below the new original $1400.

65. Discount sense-check 11: can payment exceed the original?

Question: A fictional item originally costs $760 and receives a straightforward 75% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $760, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=75%×$760=$570. Remaining percentage is 25%, so payment=25%×$760=$190. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $570 with the amount paid $190 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 570+190=760. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $380, still below the new original $1520.

66. Discount sense-check 12: can payment exceed the original?

Question: A fictional item originally costs $820 and receives a straightforward 80% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $820, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=80%×$820=$656. Remaining percentage is 20%, so payment=20%×$820=$164. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $656 with the amount paid $164 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 656+164=820. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $328, still below the new original $1640.

67. Discount sense-check 13: can payment exceed the original?

Question: A fictional item originally costs $880 and receives a straightforward 5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $880, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=5%×$880=$44. Remaining percentage is 95%, so payment=95%×$880=$836. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $44 with the amount paid $836 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 44+836=880. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $1672, still below the new original $1760.

68. Discount sense-check 14: can payment exceed the original?

Question: A fictional item originally costs $940 and receives a straightforward 10% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $940, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=10%×$940=$94. Remaining percentage is 90%, so payment=90%×$940=$846. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $94 with the amount paid $846 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 94+846=940. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $1692, still below the new original $1880.

69. Discount sense-check 15: can payment exceed the original?

Question: A fictional item originally costs $180 and receives a straightforward 12.5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $180, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=12.5%×$180=$22.5. Remaining percentage is 87.5%, so payment=87.5%×$180=$157.5. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $22.5 with the amount paid $157.5 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 22.5+157.5=180. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $315, still below the new original $360.

70. Discount sense-check 16: can payment exceed the original?

Question: A fictional item originally costs $240 and receives a straightforward 15% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $240, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=15%×$240=$36. Remaining percentage is 85%, so payment=85%×$240=$204. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $36 with the amount paid $204 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 36+204=240. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $408, still below the new original $480.

71. Discount sense-check 17: can payment exceed the original?

Question: A fictional item originally costs $300 and receives a straightforward 20% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $300, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=20%×$300=$60. Remaining percentage is 80%, so payment=80%×$300=$240. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $60 with the amount paid $240 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 60+240=300. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $480, still below the new original $600.

72. Discount sense-check 18: can payment exceed the original?

Question: A fictional item originally costs $360 and receives a straightforward 25% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $360, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=25%×$360=$90. Remaining percentage is 75%, so payment=75%×$360=$270. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $90 with the amount paid $270 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 90+270=360. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $540, still below the new original $720.

73. Discount sense-check 19: can payment exceed the original?

Question: A fictional item originally costs $420 and receives a straightforward 30% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $420, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=30%×$420=$126. Remaining percentage is 70%, so payment=70%×$420=$294. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $126 with the amount paid $294 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 126+294=420. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $588, still below the new original $840.

74. Discount sense-check 20: can payment exceed the original?

Question: A fictional item originally costs $480 and receives a straightforward 40% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $480, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=40%×$480=$192. Remaining percentage is 60%, so payment=60%×$480=$288. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $192 with the amount paid $288 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 192+288=480. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $576, still below the new original $960.

75. Discount sense-check 21: can payment exceed the original?

Question: A fictional item originally costs $540 and receives a straightforward 50% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $540, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=50%×$540=$270. Remaining percentage is 50%, so payment=50%×$540=$270. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $270 with the amount paid $270 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 270+270=540. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $540, still below the new original $1080.

76. Discount sense-check 22: can payment exceed the original?

Question: A fictional item originally costs $600 and receives a straightforward 60% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $600, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=60%×$600=$360. Remaining percentage is 40%, so payment=40%×$600=$240. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $360 with the amount paid $240 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 360+240=600. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $480, still below the new original $1200.

77. Discount sense-check 23: can payment exceed the original?

Question: A fictional item originally costs $660 and receives a straightforward 75% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $660, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=75%×$660=$495. Remaining percentage is 25%, so payment=25%×$660=$165. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $495 with the amount paid $165 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 495+165=660. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $330, still below the new original $1320.

78. Discount sense-check 24: can payment exceed the original?

Question: A fictional item originally costs $720 and receives a straightforward 80% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $720, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=80%×$720=$576. Remaining percentage is 20%, so payment=20%×$720=$144. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $576 with the amount paid $144 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 576+144=720. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $288, still below the new original $1440.

79. Discount sense-check 25: can payment exceed the original?

Question: A fictional item originally costs $780 and receives a straightforward 5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $780, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=5%×$780=$39. Remaining percentage is 95%, so payment=95%×$780=$741. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $39 with the amount paid $741 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 39+741=780. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $1482, still below the new original $1560.

80. Discount sense-check 26: can payment exceed the original?

Question: A fictional item originally costs $840 and receives a straightforward 10% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $840, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=10%×$840=$84. Remaining percentage is 90%, so payment=90%×$840=$756. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $84 with the amount paid $756 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 84+756=840. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $1512, still below the new original $1680.

81. Discount sense-check 27: can payment exceed the original?

Question: A fictional item originally costs $900 and receives a straightforward 12.5% discount. Without calculating exactly, bound the amount paid; then find both the saving and payment.

Predict first: A discount between zero and one hundred percent reduces the positive price. Therefore the payment is positive and below $900, while the saving is also positive and below the original. If the calculated final payment exceeds the original, the method has changed a discount into an increase.

Exact working: Saving=12.5%×$900=$112.5. Remaining percentage is 87.5%, so payment=87.5%×$900=$787.5. Subtracting savings from the original gives the same payment.

Tempting wrong line: Confusing the saving $112.5 with the amount paid $787.5 is an answer-form error. Adding the discount instead of removing it also breaks the direction bound. A reverse calculation based on the paid price is a different kind of task.

Independent check: Savings plus payment equals the original: 112.5+787.5=900. This part–whole check catches a misplaced percentage base or an arithmetic slip.

Transfer: If the same percentage applied to twice the original price, both savings and payment would double. The new payment is $1575, still below the new original $1800.

Ratio sharing and constant-rate checks

This group is about ratio sharing and constant-rate checks. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

82. Ratio-share bound 1: both shares must fit the total

Question: Two hypothetical groups divide 9 items in ratio 1:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 1 equal ratio parts, the second 2. The second has more parts. Both shares are nonnegative and together form all 9 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 3. Each part has 9/3=3 items. Thus the shares are 3 and 6 items. Their ratio is 1:2 as required.

Tempting wrong line: A child might divide 9 by only 1 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 9 but still violate the required ratio.

Independent check: Check both independently: 3+6=9, and 3:6 reduces to 1:2 after dividing by 3. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 18. The shares must double to 6 and 12; the ratio itself stays unchanged.

83. Ratio-share bound 2: both shares must fit the total

Question: Two hypothetical groups divide 56 items in ratio 2:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 2 equal ratio parts, the second 5. The second has more parts. Both shares are nonnegative and together form all 56 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 7. Each part has 56/7=8 items. Thus the shares are 16 and 40 items. Their ratio is 2:5 as required.

Tempting wrong line: A child might divide 56 by only 2 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 56 but still violate the required ratio.

Independent check: Check both independently: 16+40=56, and 16:40 reduces to 2:5 after dividing by 8. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 112. The shares must double to 32 and 80; the ratio itself stays unchanged.

84. Ratio-share bound 3: both shares must fit the total

Question: Two hypothetical groups divide 143 items in ratio 3:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 3 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 143 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 11. Each part has 143/11=13 items. Thus the shares are 39 and 104 items. Their ratio is 3:8 as required.

Tempting wrong line: A child might divide 143 by only 3 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 143 but still violate the required ratio.

Independent check: Check both independently: 39+104=143, and 39:104 reduces to 3:8 after dividing by 13. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 286. The shares must double to 78 and 208; the ratio itself stays unchanged.

85. Ratio-share bound 4: both shares must fit the total

Question: Two hypothetical groups divide 24 items in ratio 4:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 4 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 24 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 6. Each part has 24/6=4 items. Thus the shares are 16 and 8 items. Their ratio is 4:2 as required.

Tempting wrong line: A child might divide 24 by only 4 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 24 but still violate the required ratio.

Independent check: Check both independently: 16+8=24, and 16:8 reduces to 4:2 after dividing by 4. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 48. The shares must double to 32 and 16; the ratio itself stays unchanged.

86. Ratio-share bound 5: both shares must fit the total

Question: Two hypothetical groups divide 90 items in ratio 5:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 5 equal ratio parts, the second 5. Both have the same number of parts. Both shares are nonnegative and together form all 90 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 10. Each part has 90/10=9 items. Thus the shares are 45 and 45 items. Their ratio is 5:5 as required.

Tempting wrong line: A child might divide 90 by only 5 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 90 but still violate the required ratio.

Independent check: Check both independently: 45+45=90, and 45:45 reduces to 5:5 after dividing by 9. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 180. The shares must double to 90 and 90; the ratio itself stays unchanged.

87. Ratio-share bound 6: both shares must fit the total

Question: Two hypothetical groups divide 196 items in ratio 6:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 6 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 196 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 14. Each part has 196/14=14 items. Thus the shares are 84 and 112 items. Their ratio is 6:8 as required.

Tempting wrong line: A child might divide 196 by only 6 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 196 but still violate the required ratio.

Independent check: Check both independently: 84+112=196, and 84:112 reduces to 6:8 after dividing by 14. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 392. The shares must double to 168 and 224; the ratio itself stays unchanged.

88. Ratio-share bound 7: both shares must fit the total

Question: Two hypothetical groups divide 45 items in ratio 7:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 7 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 45 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 9. Each part has 45/9=5 items. Thus the shares are 35 and 10 items. Their ratio is 7:2 as required.

Tempting wrong line: A child might divide 45 by only 7 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 45 but still violate the required ratio.

Independent check: Check both independently: 35+10=45, and 35:10 reduces to 7:2 after dividing by 5. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 90. The shares must double to 70 and 20; the ratio itself stays unchanged.

89. Ratio-share bound 8: both shares must fit the total

Question: Two hypothetical groups divide 60 items in ratio 1:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 1 equal ratio parts, the second 5. The second has more parts. Both shares are nonnegative and together form all 60 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 6. Each part has 60/6=10 items. Thus the shares are 10 and 50 items. Their ratio is 1:5 as required.

Tempting wrong line: A child might divide 60 by only 1 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 60 but still violate the required ratio.

Independent check: Check both independently: 10+50=60, and 10:50 reduces to 1:5 after dividing by 10. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 120. The shares must double to 20 and 100; the ratio itself stays unchanged.

90. Ratio-share bound 9: both shares must fit the total

Question: Two hypothetical groups divide 150 items in ratio 2:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 2 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 150 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 10. Each part has 150/10=15 items. Thus the shares are 30 and 120 items. Their ratio is 2:8 as required.

Tempting wrong line: A child might divide 150 by only 2 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 150 but still violate the required ratio.

Independent check: Check both independently: 30+120=150, and 30:120 reduces to 2:8 after dividing by 15. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 300. The shares must double to 60 and 240; the ratio itself stays unchanged.

91. Ratio-share bound 10: both shares must fit the total

Question: Two hypothetical groups divide 30 items in ratio 3:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 3 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 30 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 5. Each part has 30/5=6 items. Thus the shares are 18 and 12 items. Their ratio is 3:2 as required.

Tempting wrong line: A child might divide 30 by only 3 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 30 but still violate the required ratio.

Independent check: Check both independently: 18+12=30, and 18:12 reduces to 3:2 after dividing by 6. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 60. The shares must double to 36 and 24; the ratio itself stays unchanged.

92. Ratio-share bound 11: both shares must fit the total

Question: Two hypothetical groups divide 99 items in ratio 4:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 4 equal ratio parts, the second 5. The second has more parts. Both shares are nonnegative and together form all 99 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 9. Each part has 99/9=11 items. Thus the shares are 44 and 55 items. Their ratio is 4:5 as required.

Tempting wrong line: A child might divide 99 by only 4 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 99 but still violate the required ratio.

Independent check: Check both independently: 44+55=99, and 44:55 reduces to 4:5 after dividing by 11. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 198. The shares must double to 88 and 110; the ratio itself stays unchanged.

93. Ratio-share bound 12: both shares must fit the total

Question: Two hypothetical groups divide 208 items in ratio 5:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 5 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 208 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 13. Each part has 208/13=16 items. Thus the shares are 80 and 128 items. Their ratio is 5:8 as required.

Tempting wrong line: A child might divide 208 by only 5 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 208 but still violate the required ratio.

Independent check: Check both independently: 80+128=208, and 80:128 reduces to 5:8 after dividing by 16. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 416. The shares must double to 160 and 256; the ratio itself stays unchanged.

94. Ratio-share bound 13: both shares must fit the total

Question: Two hypothetical groups divide 56 items in ratio 6:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 6 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 56 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 8. Each part has 56/8=7 items. Thus the shares are 42 and 14 items. Their ratio is 6:2 as required.

Tempting wrong line: A child might divide 56 by only 6 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 56 but still violate the required ratio.

Independent check: Check both independently: 42+14=56, and 42:14 reduces to 6:2 after dividing by 7. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 112. The shares must double to 84 and 28; the ratio itself stays unchanged.

95. Ratio-share bound 14: both shares must fit the total

Question: Two hypothetical groups divide 144 items in ratio 7:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 7 equal ratio parts, the second 5. The first has more parts. Both shares are nonnegative and together form all 144 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 12. Each part has 144/12=12 items. Thus the shares are 84 and 60 items. Their ratio is 7:5 as required.

Tempting wrong line: A child might divide 144 by only 7 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 144 but still violate the required ratio.

Independent check: Check both independently: 84+60=144, and 84:60 reduces to 7:5 after dividing by 12. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 288. The shares must double to 168 and 120; the ratio itself stays unchanged.

96. Ratio-share bound 15: both shares must fit the total

Question: Two hypothetical groups divide 27 items in ratio 1:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 1 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 27 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 9. Each part has 27/9=3 items. Thus the shares are 3 and 24 items. Their ratio is 1:8 as required.

Tempting wrong line: A child might divide 27 by only 1 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 27 but still violate the required ratio.

Independent check: Check both independently: 3+24=27, and 3:24 reduces to 1:8 after dividing by 3. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 54. The shares must double to 6 and 48; the ratio itself stays unchanged.

97. Ratio-share bound 16: both shares must fit the total

Question: Two hypothetical groups divide 32 items in ratio 2:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 2 equal ratio parts, the second 2. Both have the same number of parts. Both shares are nonnegative and together form all 32 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 4. Each part has 32/4=8 items. Thus the shares are 16 and 16 items. Their ratio is 2:2 as required.

Tempting wrong line: A child might divide 32 by only 2 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 32 but still violate the required ratio.

Independent check: Check both independently: 16+16=32, and 16:16 reduces to 2:2 after dividing by 8. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 64. The shares must double to 32 and 32; the ratio itself stays unchanged.

98. Ratio-share bound 17: both shares must fit the total

Question: Two hypothetical groups divide 104 items in ratio 3:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 3 equal ratio parts, the second 5. The second has more parts. Both shares are nonnegative and together form all 104 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 8. Each part has 104/8=13 items. Thus the shares are 39 and 65 items. Their ratio is 3:5 as required.

Tempting wrong line: A child might divide 104 by only 3 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 104 but still violate the required ratio.

Independent check: Check both independently: 39+65=104, and 39:65 reduces to 3:5 after dividing by 13. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 208. The shares must double to 78 and 130; the ratio itself stays unchanged.

99. Ratio-share bound 18: both shares must fit the total

Question: Two hypothetical groups divide 48 items in ratio 4:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 4 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 48 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 12. Each part has 48/12=4 items. Thus the shares are 16 and 32 items. Their ratio is 4:8 as required.

Tempting wrong line: A child might divide 48 by only 4 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 48 but still violate the required ratio.

Independent check: Check both independently: 16+32=48, and 16:32 reduces to 4:8 after dividing by 4. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 96. The shares must double to 32 and 64; the ratio itself stays unchanged.

100. Ratio-share bound 19: both shares must fit the total

Question: Two hypothetical groups divide 63 items in ratio 5:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 5 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 63 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 7. Each part has 63/7=9 items. Thus the shares are 45 and 18 items. Their ratio is 5:2 as required.

Tempting wrong line: A child might divide 63 by only 5 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 63 but still violate the required ratio.

Independent check: Check both independently: 45+18=63, and 45:18 reduces to 5:2 after dividing by 9. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 126. The shares must double to 90 and 36; the ratio itself stays unchanged.

101. Ratio-share bound 20: both shares must fit the total

Question: Two hypothetical groups divide 154 items in ratio 6:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 6 equal ratio parts, the second 5. The first has more parts. Both shares are nonnegative and together form all 154 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 11. Each part has 154/11=14 items. Thus the shares are 84 and 70 items. Their ratio is 6:5 as required.

Tempting wrong line: A child might divide 154 by only 6 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 154 but still violate the required ratio.

Independent check: Check both independently: 84+70=154, and 84:70 reduces to 6:5 after dividing by 14. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 308. The shares must double to 168 and 140; the ratio itself stays unchanged.

102. Ratio-share bound 21: both shares must fit the total

Question: Two hypothetical groups divide 75 items in ratio 7:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 7 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 75 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 15. Each part has 75/15=5 items. Thus the shares are 35 and 40 items. Their ratio is 7:8 as required.

Tempting wrong line: A child might divide 75 by only 7 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 75 but still violate the required ratio.

Independent check: Check both independently: 35+40=75, and 35:40 reduces to 7:8 after dividing by 5. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 150. The shares must double to 70 and 80; the ratio itself stays unchanged.

103. Ratio-share bound 22: both shares must fit the total

Question: Two hypothetical groups divide 30 items in ratio 1:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 1 equal ratio parts, the second 2. The second has more parts. Both shares are nonnegative and together form all 30 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 3. Each part has 30/3=10 items. Thus the shares are 10 and 20 items. Their ratio is 1:2 as required.

Tempting wrong line: A child might divide 30 by only 1 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 30 but still violate the required ratio.

Independent check: Check both independently: 10+20=30, and 10:20 reduces to 1:2 after dividing by 10. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 60. The shares must double to 20 and 40; the ratio itself stays unchanged.

104. Ratio-share bound 23: both shares must fit the total

Question: Two hypothetical groups divide 105 items in ratio 2:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 2 equal ratio parts, the second 5. The second has more parts. Both shares are nonnegative and together form all 105 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 7. Each part has 105/7=15 items. Thus the shares are 30 and 75 items. Their ratio is 2:5 as required.

Tempting wrong line: A child might divide 105 by only 2 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 105 but still violate the required ratio.

Independent check: Check both independently: 30+75=105, and 30:75 reduces to 2:5 after dividing by 15. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 210. The shares must double to 60 and 150; the ratio itself stays unchanged.

105. Ratio-share bound 24: both shares must fit the total

Question: Two hypothetical groups divide 66 items in ratio 3:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 3 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 66 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 11. Each part has 66/11=6 items. Thus the shares are 18 and 48 items. Their ratio is 3:8 as required.

Tempting wrong line: A child might divide 66 by only 3 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 66 but still violate the required ratio.

Independent check: Check both independently: 18+48=66, and 18:48 reduces to 3:8 after dividing by 6. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 132. The shares must double to 36 and 96; the ratio itself stays unchanged.

106. Ratio-share bound 25: both shares must fit the total

Question: Two hypothetical groups divide 66 items in ratio 4:2. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 4 equal ratio parts, the second 2. The first has more parts. Both shares are nonnegative and together form all 66 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 6. Each part has 66/6=11 items. Thus the shares are 44 and 22 items. Their ratio is 4:2 as required.

Tempting wrong line: A child might divide 66 by only 4 or 2, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 66 but still violate the required ratio.

Independent check: Check both independently: 44+22=66, and 44:22 reduces to 4:2 after dividing by 11. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 132. The shares must double to 88 and 44; the ratio itself stays unchanged.

107. Ratio-share bound 26: both shares must fit the total

Question: Two hypothetical groups divide 160 items in ratio 5:5. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 5 equal ratio parts, the second 5. Both have the same number of parts. Both shares are nonnegative and together form all 160 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 10. Each part has 160/10=16 items. Thus the shares are 80 and 80 items. Their ratio is 5:5 as required.

Tempting wrong line: A child might divide 160 by only 5 or 5, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 160 but still violate the required ratio.

Independent check: Check both independently: 80+80=160, and 80:80 reduces to 5:5 after dividing by 16. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 320. The shares must double to 160 and 160; the ratio itself stays unchanged.

108. Ratio-share bound 27: both shares must fit the total

Question: Two hypothetical groups divide 98 items in ratio 6:8. Before working out the exact shares, predict which is larger and explain why neither share can exceed the total.

Predict first: The first group has 6 equal ratio parts, the second 8. The second has more parts. Both shares are nonnegative and together form all 98 items. These constraints are useful before any division.

Exact working: The total number of equal parts is 14. Each part has 98/14=7 items. Thus the shares are 42 and 56 items. Their ratio is 6:8 as required.

Tempting wrong line: A child might divide 98 by only 6 or 8, using a part-to-part number instead of the combined number of blocks. A plausible-looking pair can total 98 but still violate the required ratio.

Independent check: Check both independently: 42+56=98, and 42:56 reduces to 6:8 after dividing by 7. The larger share must belong to the larger part count.

Transfer: Imagine the ratio parts remain fixed but the whole doubles to 196. The shares must double to 84 and 112; the ratio itself stays unchanged.

Algebra and independent inverse checks

This group is about algebra and independent inverse checks. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

109. Equation check 1: solve and reverse

Question: A rule is y=-5x−15, and the measured output in an abstract table is y=25. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient -5 gives the direction: a larger x lowers y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation 25=-5x+(-15) gives -5x=40. Divide by -5 to obtain x=-8. A forward substitution restores -5(-8)+(-15)=25.

Tempting wrong line: Dividing 25 by -5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-8, y=25; at x=-7, y=20. Difference 20-25=-5, the original coefficient.

Transfer: Replace the specified output by 20. The recovered input must be x=-7, one larger. Explain this from the sign of -5 and the unchanged constant.

110. Equation check 2: solve and reverse

Question: A rule is y=-3x−12, and the measured output in an abstract table is y=-9. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient -3 gives the direction: a larger x lowers y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation -9=-3x+(-12) gives -3x=3. Divide by -3 to obtain x=-1. A forward substitution restores -3(-1)+(-12)=-9.

Tempting wrong line: Dividing -9 by -3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-1, y=-9; at x=0, y=-12. Difference -12–9=-3, the original coefficient.

Transfer: Replace the specified output by -12. The recovered input must be x=0, one larger. Explain this from the sign of -3 and the unchanged constant.

111. Equation check 3: solve and reverse

Question: A rule is y=-2x−9, and the measured output in an abstract table is y=-21. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient -2 gives the direction: a larger x lowers y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation -21=-2x+(-9) gives -2x=-12. Divide by -2 to obtain x=6. A forward substitution restores -2(6)+(-9)=-21.

Tempting wrong line: Dividing -21 by -2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=6, y=-21; at x=7, y=-23. Difference -23–21=-2, the original coefficient.

Transfer: Replace the specified output by -23. The recovered input must be x=7, one larger. Explain this from the sign of -2 and the unchanged constant.

112. Equation check 4: solve and reverse

Question: A rule is y=-1x−6, and the measured output in an abstract table is y=-2. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient -1 gives the direction: a larger x lowers y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation -2=-1x+(-6) gives -1x=4. Divide by -1 to obtain x=-4. A forward substitution restores -1(-4)+(-6)=-2.

Tempting wrong line: Dividing -2 by -1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-4, y=-2; at x=-3, y=-3. Difference -3–2=-1, the original coefficient.

Transfer: Replace the specified output by -3. The recovered input must be x=-3, one larger. Explain this from the sign of -1 and the unchanged constant.

113. Equation check 5: solve and reverse

Question: A rule is y=1x−3, and the measured output in an abstract table is y=0. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient 1 gives the direction: a larger x raises y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation 0=1x+(-3) gives 1x=3. Divide by 1 to obtain x=3. A forward substitution restores 1(3)+(-3)=0.

Tempting wrong line: Dividing 0 by 1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=3, y=0; at x=4, y=1. Difference 1-0=1, the original coefficient.

Transfer: Replace the specified output by 1. The recovered input must be x=4, one larger. Explain this from the sign of 1 and the unchanged constant.

114. Equation check 6: solve and reverse

Question: A rule is y=2x+0, and the measured output in an abstract table is y=-14. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient 2 gives the direction: a larger x raises y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation -14=2x+(0) gives 2x=-14. Divide by 2 to obtain x=-7. A forward substitution restores 2(-7)+(0)=-14.

Tempting wrong line: Dividing -14 by 2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-7, y=-14; at x=-6, y=-12. Difference -12–14=2, the original coefficient.

Transfer: Replace the specified output by -12. The recovered input must be x=-6, one larger. Explain this from the sign of 2 and the unchanged constant.

115. Equation check 7: solve and reverse

Question: A rule is y=3x+3, and the measured output in an abstract table is y=3. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient 3 gives the direction: a larger x raises y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation 3=3x+(3) gives 3x=0. Divide by 3 to obtain x=0. A forward substitution restores 3(0)+(3)=3.

Tempting wrong line: Dividing 3 by 3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=0, y=3; at x=1, y=6. Difference 6-3=3, the original coefficient.

Transfer: Replace the specified output by 6. The recovered input must be x=1, one larger. Explain this from the sign of 3 and the unchanged constant.

116. Equation check 8: solve and reverse

Question: A rule is y=4x+6, and the measured output in an abstract table is y=34. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient 4 gives the direction: a larger x raises y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation 34=4x+(6) gives 4x=28. Divide by 4 to obtain x=7. A forward substitution restores 4(7)+(6)=34.

Tempting wrong line: Dividing 34 by 4 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=7, y=34; at x=8, y=38. Difference 38-34=4, the original coefficient.

Transfer: Replace the specified output by 38. The recovered input must be x=8, one larger. Explain this from the sign of 4 and the unchanged constant.

117. Equation check 9: solve and reverse

Question: A rule is y=5x+9, and the measured output in an abstract table is y=-6. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Predict first: The sign of the coefficient 5 gives the direction: a larger x raises y. The equation’s output contains both a repeated x-term and one constant. First remove the constant while preserving equality on both sides, then divide by the nonzero coefficient.

Exact working: The equation -6=5x+(9) gives 5x=-15. Divide by 5 to obtain x=-3. A forward substitution restores 5(-3)+(9)=-6.

Tempting wrong line: Dividing -6 by 5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-3, y=-6; at x=-2, y=-1. Difference -1–6=5, the original coefficient.

Transfer: Replace the specified output by -1. The recovered input must be x=-2, one larger. Explain this from the sign of 5 and the unchanged constant.

118. Equation check 10: solve and reverse

Question: A rule is y=-5x+12, and the measured output in an abstract table is y=-8. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -8=-5x+(12) gives -5x=-20. Divide by -5 to obtain x=4. A forward substitution restores -5(4)+(12)=-8.

Tempting wrong line: Dividing -8 by -5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=4, y=-8; at x=5, y=-13. Difference -13–8=-5, the original coefficient.

Transfer: Replace the specified output by -13. The recovered input must be x=5, one larger. Explain this from the sign of -5 and the unchanged constant.

119. Equation check 11: solve and reverse

Question: A rule is y=-3x+15, and the measured output in an abstract table is y=33. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 33=-3x+(15) gives -3x=18. Divide by -3 to obtain x=-6. A forward substitution restores -3(-6)+(15)=33.

Tempting wrong line: Dividing 33 by -3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-6, y=33; at x=-5, y=30. Difference 30-33=-3, the original coefficient.

Transfer: Replace the specified output by 30. The recovered input must be x=-5, one larger. Explain this from the sign of -3 and the unchanged constant.

120. Equation check 12: solve and reverse

Question: A rule is y=-2x−13, and the measured output in an abstract table is y=-15. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -15=-2x+(-13) gives -2x=-2. Divide by -2 to obtain x=1. A forward substitution restores -2(1)+(-13)=-15.

Tempting wrong line: Dividing -15 by -2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=1, y=-15; at x=2, y=-17. Difference -17–15=-2, the original coefficient.

Transfer: Replace the specified output by -17. The recovered input must be x=2, one larger. Explain this from the sign of -2 and the unchanged constant.

121. Equation check 13: solve and reverse

Question: A rule is y=-1x−10, and the measured output in an abstract table is y=-18. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -18=-1x+(-10) gives -1x=-8. Divide by -1 to obtain x=8. A forward substitution restores -1(8)+(-10)=-18.

Tempting wrong line: Dividing -18 by -1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=8, y=-18; at x=9, y=-19. Difference -19–18=-1, the original coefficient.

Transfer: Replace the specified output by -19. The recovered input must be x=9, one larger. Explain this from the sign of -1 and the unchanged constant.

122. Equation check 14: solve and reverse

Question: A rule is y=1x−7, and the measured output in an abstract table is y=-9. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -9=1x+(-7) gives 1x=-2. Divide by 1 to obtain x=-2. A forward substitution restores 1(-2)+(-7)=-9.

Tempting wrong line: Dividing -9 by 1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-2, y=-9; at x=-1, y=-8. Difference -8–9=1, the original coefficient.

Transfer: Replace the specified output by -8. The recovered input must be x=-1, one larger. Explain this from the sign of 1 and the unchanged constant.

123. Equation check 15: solve and reverse

Question: A rule is y=2x−4, and the measured output in an abstract table is y=6. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 6=2x+(-4) gives 2x=10. Divide by 2 to obtain x=5. A forward substitution restores 2(5)+(-4)=6.

Tempting wrong line: Dividing 6 by 2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=5, y=6; at x=6, y=8. Difference 8-6=2, the original coefficient.

Transfer: Replace the specified output by 8. The recovered input must be x=6, one larger. Explain this from the sign of 2 and the unchanged constant.

124. Equation check 16: solve and reverse

Question: A rule is y=3x−1, and the measured output in an abstract table is y=-16. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -16=3x+(-1) gives 3x=-15. Divide by 3 to obtain x=-5. A forward substitution restores 3(-5)+(-1)=-16.

Tempting wrong line: Dividing -16 by 3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-5, y=-16; at x=-4, y=-13. Difference -13–16=3, the original coefficient.

Transfer: Replace the specified output by -13. The recovered input must be x=-4, one larger. Explain this from the sign of 3 and the unchanged constant.

125. Equation check 17: solve and reverse

Question: A rule is y=4x+2, and the measured output in an abstract table is y=10. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 10=4x+(2) gives 4x=8. Divide by 4 to obtain x=2. A forward substitution restores 4(2)+(2)=10.

Tempting wrong line: Dividing 10 by 4 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=2, y=10; at x=3, y=14. Difference 14-10=4, the original coefficient.

Transfer: Replace the specified output by 14. The recovered input must be x=3, one larger. Explain this from the sign of 4 and the unchanged constant.

126. Equation check 18: solve and reverse

Question: A rule is y=5x+5, and the measured output in an abstract table is y=-35. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -35=5x+(5) gives 5x=-40. Divide by 5 to obtain x=-8. A forward substitution restores 5(-8)+(5)=-35.

Tempting wrong line: Dividing -35 by 5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-8, y=-35; at x=-7, y=-30. Difference -30–35=5, the original coefficient.

Transfer: Replace the specified output by -30. The recovered input must be x=-7, one larger. Explain this from the sign of 5 and the unchanged constant.

127. Equation check 19: solve and reverse

Question: A rule is y=-5x+8, and the measured output in an abstract table is y=13. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 13=-5x+(8) gives -5x=5. Divide by -5 to obtain x=-1. A forward substitution restores -5(-1)+(8)=13.

Tempting wrong line: Dividing 13 by -5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-1, y=13; at x=0, y=8. Difference 8-13=-5, the original coefficient.

Transfer: Replace the specified output by 8. The recovered input must be x=0, one larger. Explain this from the sign of -5 and the unchanged constant.

128. Equation check 20: solve and reverse

Question: A rule is y=-3x+11, and the measured output in an abstract table is y=-7. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -7=-3x+(11) gives -3x=-18. Divide by -3 to obtain x=6. A forward substitution restores -3(6)+(11)=-7.

Tempting wrong line: Dividing -7 by -3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=6, y=-7; at x=7, y=-10. Difference -10–7=-3, the original coefficient.

Transfer: Replace the specified output by -10. The recovered input must be x=7, one larger. Explain this from the sign of -3 and the unchanged constant.

129. Equation check 21: solve and reverse

Question: A rule is y=-2x+14, and the measured output in an abstract table is y=22. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 22=-2x+(14) gives -2x=8. Divide by -2 to obtain x=-4. A forward substitution restores -2(-4)+(14)=22.

Tempting wrong line: Dividing 22 by -2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-4, y=22; at x=-3, y=20. Difference 20-22=-2, the original coefficient.

Transfer: Replace the specified output by 20. The recovered input must be x=-3, one larger. Explain this from the sign of -2 and the unchanged constant.

130. Equation check 22: solve and reverse

Question: A rule is y=-1x−14, and the measured output in an abstract table is y=-17. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -17=-1x+(-14) gives -1x=-3. Divide by -1 to obtain x=3. A forward substitution restores -1(3)+(-14)=-17.

Tempting wrong line: Dividing -17 by -1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=3, y=-17; at x=4, y=-18. Difference -18–17=-1, the original coefficient.

Transfer: Replace the specified output by -18. The recovered input must be x=4, one larger. Explain this from the sign of -1 and the unchanged constant.

131. Equation check 23: solve and reverse

Question: A rule is y=1x−11, and the measured output in an abstract table is y=-18. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -18=1x+(-11) gives 1x=-7. Divide by 1 to obtain x=-7. A forward substitution restores 1(-7)+(-11)=-18.

Tempting wrong line: Dividing -18 by 1 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-7, y=-18; at x=-6, y=-17. Difference -17–18=1, the original coefficient.

Transfer: Replace the specified output by -17. The recovered input must be x=-6, one larger. Explain this from the sign of 1 and the unchanged constant.

132. Equation check 24: solve and reverse

Question: A rule is y=2x−8, and the measured output in an abstract table is y=-8. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -8=2x+(-8) gives 2x=0. Divide by 2 to obtain x=0. A forward substitution restores 2(0)+(-8)=-8.

Tempting wrong line: Dividing -8 by 2 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=0, y=-8; at x=1, y=-6. Difference -6–8=2, the original coefficient.

Transfer: Replace the specified output by -6. The recovered input must be x=1, one larger. Explain this from the sign of 2 and the unchanged constant.

133. Equation check 25: solve and reverse

Question: A rule is y=3x−5, and the measured output in an abstract table is y=16. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 16=3x+(-5) gives 3x=21. Divide by 3 to obtain x=7. A forward substitution restores 3(7)+(-5)=16.

Tempting wrong line: Dividing 16 by 3 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=7, y=16; at x=8, y=19. Difference 19-16=3, the original coefficient.

Transfer: Replace the specified output by 19. The recovered input must be x=8, one larger. Explain this from the sign of 3 and the unchanged constant.

134. Equation check 26: solve and reverse

Question: A rule is y=4x−2, and the measured output in an abstract table is y=-14. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation -14=4x+(-2) gives 4x=-12. Divide by 4 to obtain x=-3. A forward substitution restores 4(-3)+(-2)=-14.

Tempting wrong line: Dividing -14 by 4 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=-3, y=-14; at x=-2, y=-10. Difference -10–14=4, the original coefficient.

Transfer: Replace the specified output by -10. The recovered input must be x=-2, one larger. Explain this from the sign of 4 and the unchanged constant.

135. Equation check 27: solve and reverse

Question: A rule is y=5x+1, and the measured output in an abstract table is y=21. Find x. Predict whether a one-unit increase in x would make y rise or fall.

Exact working: The equation 21=5x+(1) gives 5x=20. Divide by 5 to obtain x=4. A forward substitution restores 5(4)+(1)=21.

Tempting wrong line: Dividing 21 by 5 before removing the constant gives the wrong inverse order. Another error is ignoring a negative coefficient when dividing, changing the sign of the recovered x.

Independent check: Check the exact output and the step-change separately: at x=4, y=21; at x=5, y=26. Difference 26-21=5, the original coefficient.

Transfer: Replace the specified output by 26. The recovered input must be x=5, one larger. Explain this from the sign of 5 and the unchanged constant.

Linear graph directions and output ranges

This group is about linear graph directions and output ranges. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

136. Graph slope 1: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-3,-2) and (-1,-10). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate -1 exceeds -3, so it is to the right. The vertical output -10 is smaller than -2, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-10−-2)/(-1−-3)=-8/2=-4. The graph’s y-intercept is -14 in the underlying linear rule y=-4x+(-14), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference -8 as the gradient without dividing by the horizontal step 2.

Independent check: A prediction of negative slope matches the exact -4. Check with a one-unit x-step: the output must change by -4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (-1,-10). The line should have output -22. Predict whether this rises or falls before calculating.

137. Graph slope 2: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-2,-1) and (1,-10). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 1 exceeds -2, so it is to the right. The vertical output -10 is smaller than -1, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-10−-1)/(1−-2)=-9/3=-3. The graph’s y-intercept is -7 in the underlying linear rule y=-3x+(-7), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference -9 as the gradient without dividing by the horizontal step 3.

Independent check: A prediction of negative slope matches the exact -3. Check with a one-unit x-step: the output must change by -3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (1,-10). The line should have output -19. Predict whether this rises or falls before calculating.

138. Graph slope 3: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-1,2) and (3,-6). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 3 exceeds -1, so it is to the right. The vertical output -6 is smaller than 2, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-6−2)/(3−-1)=-8/4=-2. The graph’s y-intercept is 0 in the underlying linear rule y=-2x+(0), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -2. Check with a one-unit x-step: the output must change by -2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (3,-6). The line should have output -12. Predict whether this rises or falls before calculating.

139. Graph slope 4: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (0,7) and (5,2). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 5 exceeds 0, so it is to the right. The vertical output 2 is smaller than 7, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(2−7)/(5−0)=-5/5=-1. The graph’s y-intercept is 7 in the underlying linear rule y=-1x+(7), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference -5 as the gradient without dividing by the horizontal step 5.

Independent check: A prediction of negative slope matches the exact -1. Check with a one-unit x-step: the output must change by -1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (5,2). The line should have output -1. Predict whether this rises or falls before calculating.

140. Graph slope 5: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (1,15) and (3,17). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 3 exceeds 1, so it is to the right. The vertical output 17 is larger than 15, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(17−15)/(3−1)=2/2=1. The graph’s y-intercept is 14 in the underlying linear rule y=1x+(14), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference 2 as the gradient without dividing by the horizontal step 2.

Independent check: A prediction of positive slope matches the exact 1. Check with a one-unit x-step: the output must change by 1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (3,17). The line should have output 20. Predict whether this rises or falls before calculating.

141. Graph slope 6: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (2,-4) and (5,2). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 5 exceeds 2, so it is to the right. The vertical output 2 is larger than -4, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(2−-4)/(5−2)=6/3=2. The graph’s y-intercept is -8 in the underlying linear rule y=2x+(-8), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference 6 as the gradient without dividing by the horizontal step 3.

Independent check: A prediction of positive slope matches the exact 2. Check with a one-unit x-step: the output must change by 2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (5,2). The line should have output 8. Predict whether this rises or falls before calculating.

142. Graph slope 7: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (3,8) and (7,20). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 7 exceeds 3, so it is to the right. The vertical output 20 is larger than 8, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(20−8)/(7−3)=12/4=3. The graph’s y-intercept is -1 in the underlying linear rule y=3x+(-1), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference 12 as the gradient without dividing by the horizontal step 4.

Independent check: A prediction of positive slope matches the exact 3. Check with a one-unit x-step: the output must change by 3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (7,20). The line should have output 29. Predict whether this rises or falls before calculating.

143. Graph slope 8: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-3,-6) and (2,14). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 2 exceeds -3, so it is to the right. The vertical output 14 is larger than -6, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(14−-6)/(2−-3)=20/5=4. The graph’s y-intercept is 6 in the underlying linear rule y=4x+(6), although that intercept is not itself supplied as a plotted point.

Tempting wrong line: A common error is reversing one subtraction but not the other, which flips the sign of the gradient. Another is treating the output difference 20 as the gradient without dividing by the horizontal step 5.

Independent check: A prediction of positive slope matches the exact 4. Check with a one-unit x-step: the output must change by 4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (2,14). The line should have output 26. Predict whether this rises or falls before calculating.

144. Graph slope 9: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-2,21) and (0,13). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 0 exceeds -2, so it is to the right. The vertical output 13 is smaller than 21, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(13−21)/(0−-2)=-8/2=-4. The graph’s y-intercept is 13 in the underlying linear rule y=-4x+(13), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -4. Check with a one-unit x-step: the output must change by -4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (0,13). The line should have output 1. Predict whether this rises or falls before calculating.

145. Graph slope 10: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-1,-6) and (2,-15). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 2 exceeds -1, so it is to the right. The vertical output -15 is smaller than -6, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-15−-6)/(2−-1)=-9/3=-3. The graph’s y-intercept is -9 in the underlying linear rule y=-3x+(-9), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -3. Check with a one-unit x-step: the output must change by -3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (2,-15). The line should have output -24. Predict whether this rises or falls before calculating.

146. Graph slope 11: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (0,-2) and (4,-10). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 4 exceeds 0, so it is to the right. The vertical output -10 is smaller than -2, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-10−-2)/(4−0)=-8/4=-2. The graph’s y-intercept is -2 in the underlying linear rule y=-2x+(-2), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -2. Check with a one-unit x-step: the output must change by -2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (4,-10). The line should have output -16. Predict whether this rises or falls before calculating.

147. Graph slope 12: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (1,4) and (6,-1). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 6 exceeds 1, so it is to the right. The vertical output -1 is smaller than 4, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-1−4)/(6−1)=-5/5=-1. The graph’s y-intercept is 5 in the underlying linear rule y=-1x+(5), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -1. Check with a one-unit x-step: the output must change by -1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (6,-1). The line should have output -4. Predict whether this rises or falls before calculating.

148. Graph slope 13: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (2,14) and (4,16). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 4 exceeds 2, so it is to the right. The vertical output 16 is larger than 14, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(16−14)/(4−2)=2/2=1. The graph’s y-intercept is 12 in the underlying linear rule y=1x+(12), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 1. Check with a one-unit x-step: the output must change by 1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (4,16). The line should have output 19. Predict whether this rises or falls before calculating.

149. Graph slope 14: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (3,-4) and (6,2). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 6 exceeds 3, so it is to the right. The vertical output 2 is larger than -4, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(2−-4)/(6−3)=6/3=2. The graph’s y-intercept is -10 in the underlying linear rule y=2x+(-10), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 2. Check with a one-unit x-step: the output must change by 2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (6,2). The line should have output 8. Predict whether this rises or falls before calculating.

150. Graph slope 15: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-3,-12) and (1,0). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 1 exceeds -3, so it is to the right. The vertical output 0 is larger than -12, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(0−-12)/(1−-3)=12/4=3. The graph’s y-intercept is -3 in the underlying linear rule y=3x+(-3), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 3. Check with a one-unit x-step: the output must change by 3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (1,0). The line should have output 9. Predict whether this rises or falls before calculating.

151. Graph slope 16: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-2,-4) and (3,16). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 3 exceeds -2, so it is to the right. The vertical output 16 is larger than -4, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(16−-4)/(3−-2)=20/5=4. The graph’s y-intercept is 4 in the underlying linear rule y=4x+(4), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 4. Check with a one-unit x-step: the output must change by 4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (3,16). The line should have output 28. Predict whether this rises or falls before calculating.

152. Graph slope 17: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-1,15) and (1,7). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 1 exceeds -1, so it is to the right. The vertical output 7 is smaller than 15, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(7−15)/(1−-1)=-8/2=-4. The graph’s y-intercept is 11 in the underlying linear rule y=-4x+(11), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -4. Check with a one-unit x-step: the output must change by -4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (1,7). The line should have output -5. Predict whether this rises or falls before calculating.

153. Graph slope 18: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (0,-11) and (3,-20). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 3 exceeds 0, so it is to the right. The vertical output -20 is smaller than -11, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-20−-11)/(3−0)=-9/3=-3. The graph’s y-intercept is -11 in the underlying linear rule y=-3x+(-11), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -3. Check with a one-unit x-step: the output must change by -3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (3,-20). The line should have output -29. Predict whether this rises or falls before calculating.

154. Graph slope 19: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (1,-6) and (5,-14). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 5 exceeds 1, so it is to the right. The vertical output -14 is smaller than -6, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-14−-6)/(5−1)=-8/4=-2. The graph’s y-intercept is -4 in the underlying linear rule y=-2x+(-4), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -2. Check with a one-unit x-step: the output must change by -2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (5,-14). The line should have output -20. Predict whether this rises or falls before calculating.

155. Graph slope 20: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (2,1) and (7,-4). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 7 exceeds 2, so it is to the right. The vertical output -4 is smaller than 1, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-4−1)/(7−2)=-5/5=-1. The graph’s y-intercept is 3 in the underlying linear rule y=-1x+(3), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -1. Check with a one-unit x-step: the output must change by -1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (7,-4). The line should have output -7. Predict whether this rises or falls before calculating.

156. Graph slope 21: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (3,13) and (5,15). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 5 exceeds 3, so it is to the right. The vertical output 15 is larger than 13, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(15−13)/(5−3)=2/2=1. The graph’s y-intercept is 10 in the underlying linear rule y=1x+(10), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 1. Check with a one-unit x-step: the output must change by 1 y-units along the same line.

Transfer: Move another 3 x-units to the right from (5,15). The line should have output 18. Predict whether this rises or falls before calculating.

157. Graph slope 22: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-3,-18) and (0,-12). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 0 exceeds -3, so it is to the right. The vertical output -12 is larger than -18, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(-12−-18)/(0−-3)=6/3=2. The graph’s y-intercept is -12 in the underlying linear rule y=2x+(-12), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 2. Check with a one-unit x-step: the output must change by 2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (0,-12). The line should have output -6. Predict whether this rises or falls before calculating.

158. Graph slope 23: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-2,-11) and (2,1). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 2 exceeds -2, so it is to the right. The vertical output 1 is larger than -11, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(1−-11)/(2−-2)=12/4=3. The graph’s y-intercept is -5 in the underlying linear rule y=3x+(-5), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 3. Check with a one-unit x-step: the output must change by 3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (2,1). The line should have output 10. Predict whether this rises or falls before calculating.

159. Graph slope 24: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (-1,-2) and (4,18). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 4 exceeds -1, so it is to the right. The vertical output 18 is larger than -2, so the line rises to the right. This predicts a positive gradient before division.

Exact working: Rise/run=(18−-2)/(4−-1)=20/5=4. The graph’s y-intercept is 2 in the underlying linear rule y=4x+(2), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of positive slope matches the exact 4. Check with a one-unit x-step: the output must change by 4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (4,18). The line should have output 30. Predict whether this rises or falls before calculating.

160. Graph slope 25: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (0,9) and (2,1). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 2 exceeds 0, so it is to the right. The vertical output 1 is smaller than 9, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(1−9)/(2−0)=-8/2=-4. The graph’s y-intercept is 9 in the underlying linear rule y=-4x+(9), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -4. Check with a one-unit x-step: the output must change by -4 y-units along the same line.

Transfer: Move another 3 x-units to the right from (2,1). The line should have output -11. Predict whether this rises or falls before calculating.

161. Graph slope 26: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (1,-16) and (4,-25). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 4 exceeds 1, so it is to the right. The vertical output -25 is smaller than -16, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-25−-16)/(4−1)=-9/3=-3. The graph’s y-intercept is -13 in the underlying linear rule y=-3x+(-13), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -3. Check with a one-unit x-step: the output must change by -3 y-units along the same line.

Transfer: Move another 3 x-units to the right from (4,-25). The line should have output -34. Predict whether this rises or falls before calculating.

162. Graph slope 27: is the line rising or falling?

Question: A hypothetical exact straight-line graph contains points (2,-10) and (6,-18). Without calculating the full fraction, predict the slope sign, then find its gradient.

Predict first: The second x-coordinate 6 exceeds 2, so it is to the right. The vertical output -18 is smaller than -10, so the line falls to the right. This predicts a negative gradient before division.

Exact working: Rise/run=(-18−-10)/(6−2)=-8/4=-2. The graph’s y-intercept is -6 in the underlying linear rule y=-2x+(-6), although that intercept is not itself supplied as a plotted point.

Independent check: A prediction of negative slope matches the exact -2. Check with a one-unit x-step: the output must change by -2 y-units along the same line.

Transfer: Move another 3 x-units to the right from (6,-18). The line should have output -24. Predict whether this rises or falls before calculating.

Geometry area, perimeter and dimensional checks

This group is about geometry area, perimeter and dimensional checks. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

163. Area versus boundary 1: predict units and scale

Question: An imaginary rectangular board measures 3 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Predict first: Area counts two-dimensional square units, so its unit is cm² and two doubled lengths quadruple it. Perimeter adds linear edges, so its unit is cm and all doubled lengths double it. This dimensional prediction is a strong sense-check against using the wrong formula.

Exact working: The area is 3×2=6 cm². The perimeter is 2(3+2)=10 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 6 cm as area or 10 cm² as perimeter is a unit error. If the learner calculates 3+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 24 cm² and new perimeter 20 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 9 cm², while perimeter rises by two centimetres to 12 cm. This contrast tests understanding rather than a memorised fourfold slogan.

164. Area versus boundary 2: predict units and scale

Question: An imaginary rectangular board measures 6 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 6×7=42 cm². The perimeter is 2(6+7)=26 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 42 cm as area or 26 cm² as perimeter is a unit error. If the learner calculates 6+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 168 cm² and new perimeter 52 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 48 cm², while perimeter rises by two centimetres to 28 cm. This contrast tests understanding rather than a memorised fourfold slogan.

165. Area versus boundary 3: predict units and scale

Question: An imaginary rectangular board measures 9 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 9×12=108 cm². The perimeter is 2(9+12)=42 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 108 cm as area or 42 cm² as perimeter is a unit error. If the learner calculates 9+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 432 cm² and new perimeter 84 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 117 cm², while perimeter rises by two centimetres to 44 cm. This contrast tests understanding rather than a memorised fourfold slogan.

166. Area versus boundary 4: predict units and scale

Question: An imaginary rectangular board measures 12 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 12×2=24 cm². The perimeter is 2(12+2)=28 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 24 cm as area or 28 cm² as perimeter is a unit error. If the learner calculates 12+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 96 cm² and new perimeter 56 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 36 cm², while perimeter rises by two centimetres to 30 cm. This contrast tests understanding rather than a memorised fourfold slogan.

167. Area versus boundary 5: predict units and scale

Question: An imaginary rectangular board measures 15 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 15×7=105 cm². The perimeter is 2(15+7)=44 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 105 cm as area or 44 cm² as perimeter is a unit error. If the learner calculates 15+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 420 cm² and new perimeter 88 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 120 cm², while perimeter rises by two centimetres to 46 cm. This contrast tests understanding rather than a memorised fourfold slogan.

168. Area versus boundary 6: predict units and scale

Question: An imaginary rectangular board measures 18 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 18×12=216 cm². The perimeter is 2(18+12)=60 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 216 cm as area or 60 cm² as perimeter is a unit error. If the learner calculates 18+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 864 cm² and new perimeter 120 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 234 cm², while perimeter rises by two centimetres to 62 cm. This contrast tests understanding rather than a memorised fourfold slogan.

169. Area versus boundary 7: predict units and scale

Question: An imaginary rectangular board measures 21 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 21×2=42 cm². The perimeter is 2(21+2)=46 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 42 cm as area or 46 cm² as perimeter is a unit error. If the learner calculates 21+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 168 cm² and new perimeter 92 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 63 cm², while perimeter rises by two centimetres to 48 cm. This contrast tests understanding rather than a memorised fourfold slogan.

170. Area versus boundary 8: predict units and scale

Question: An imaginary rectangular board measures 5 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 5×7=35 cm². The perimeter is 2(5+7)=24 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 35 cm as area or 24 cm² as perimeter is a unit error. If the learner calculates 5+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 140 cm² and new perimeter 48 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 40 cm², while perimeter rises by two centimetres to 26 cm. This contrast tests understanding rather than a memorised fourfold slogan.

171. Area versus boundary 9: predict units and scale

Question: An imaginary rectangular board measures 8 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 8×12=96 cm². The perimeter is 2(8+12)=40 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 96 cm as area or 40 cm² as perimeter is a unit error. If the learner calculates 8+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 384 cm² and new perimeter 80 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 104 cm², while perimeter rises by two centimetres to 42 cm. This contrast tests understanding rather than a memorised fourfold slogan.

172. Area versus boundary 10: predict units and scale

Question: An imaginary rectangular board measures 11 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 11×2=22 cm². The perimeter is 2(11+2)=26 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 22 cm as area or 26 cm² as perimeter is a unit error. If the learner calculates 11+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 88 cm² and new perimeter 52 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 33 cm², while perimeter rises by two centimetres to 28 cm. This contrast tests understanding rather than a memorised fourfold slogan.

173. Area versus boundary 11: predict units and scale

Question: An imaginary rectangular board measures 14 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 14×7=98 cm². The perimeter is 2(14+7)=42 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 98 cm as area or 42 cm² as perimeter is a unit error. If the learner calculates 14+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 392 cm² and new perimeter 84 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 112 cm², while perimeter rises by two centimetres to 44 cm. This contrast tests understanding rather than a memorised fourfold slogan.

174. Area versus boundary 12: predict units and scale

Question: An imaginary rectangular board measures 17 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 17×12=204 cm². The perimeter is 2(17+12)=58 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 204 cm as area or 58 cm² as perimeter is a unit error. If the learner calculates 17+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 816 cm² and new perimeter 116 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 221 cm², while perimeter rises by two centimetres to 60 cm. This contrast tests understanding rather than a memorised fourfold slogan.

175. Area versus boundary 13: predict units and scale

Question: An imaginary rectangular board measures 20 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 20×2=40 cm². The perimeter is 2(20+2)=44 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 40 cm as area or 44 cm² as perimeter is a unit error. If the learner calculates 20+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 160 cm² and new perimeter 88 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 60 cm², while perimeter rises by two centimetres to 46 cm. This contrast tests understanding rather than a memorised fourfold slogan.

176. Area versus boundary 14: predict units and scale

Question: An imaginary rectangular board measures 4 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 4×7=28 cm². The perimeter is 2(4+7)=22 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 28 cm as area or 22 cm² as perimeter is a unit error. If the learner calculates 4+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 112 cm² and new perimeter 44 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 32 cm², while perimeter rises by two centimetres to 24 cm. This contrast tests understanding rather than a memorised fourfold slogan.

177. Area versus boundary 15: predict units and scale

Question: An imaginary rectangular board measures 7 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 7×12=84 cm². The perimeter is 2(7+12)=38 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 84 cm as area or 38 cm² as perimeter is a unit error. If the learner calculates 7+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 336 cm² and new perimeter 76 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 91 cm², while perimeter rises by two centimetres to 40 cm. This contrast tests understanding rather than a memorised fourfold slogan.

178. Area versus boundary 16: predict units and scale

Question: An imaginary rectangular board measures 10 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 10×2=20 cm². The perimeter is 2(10+2)=24 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 20 cm as area or 24 cm² as perimeter is a unit error. If the learner calculates 10+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 80 cm² and new perimeter 48 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 30 cm², while perimeter rises by two centimetres to 26 cm. This contrast tests understanding rather than a memorised fourfold slogan.

179. Area versus boundary 17: predict units and scale

Question: An imaginary rectangular board measures 13 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 13×7=91 cm². The perimeter is 2(13+7)=40 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 91 cm as area or 40 cm² as perimeter is a unit error. If the learner calculates 13+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 364 cm² and new perimeter 80 cm, matching the earlier direction and scale predictions.

180. Area versus boundary 18: predict units and scale

Question: An imaginary rectangular board measures 16 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 16×12=192 cm². The perimeter is 2(16+12)=56 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 192 cm as area or 56 cm² as perimeter is a unit error. If the learner calculates 16+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 768 cm² and new perimeter 112 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 208 cm², while perimeter rises by two centimetres to 58 cm. This contrast tests understanding rather than a memorised fourfold slogan.

181. Area versus boundary 19: predict units and scale

Question: An imaginary rectangular board measures 19 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 19×2=38 cm². The perimeter is 2(19+2)=42 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 38 cm as area or 42 cm² as perimeter is a unit error. If the learner calculates 19+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 152 cm² and new perimeter 84 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 57 cm², while perimeter rises by two centimetres to 44 cm. This contrast tests understanding rather than a memorised fourfold slogan.

182. Area versus boundary 20: predict units and scale

Question: An imaginary rectangular board measures 3 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 3×7=21 cm². The perimeter is 2(3+7)=20 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 21 cm as area or 20 cm² as perimeter is a unit error. If the learner calculates 3+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 84 cm² and new perimeter 40 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 24 cm², while perimeter rises by two centimetres to 22 cm. This contrast tests understanding rather than a memorised fourfold slogan.

183. Area versus boundary 21: predict units and scale

Question: An imaginary rectangular board measures 6 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 6×12=72 cm². The perimeter is 2(6+12)=36 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 72 cm as area or 36 cm² as perimeter is a unit error. If the learner calculates 6+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 288 cm² and new perimeter 72 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 78 cm², while perimeter rises by two centimetres to 38 cm. This contrast tests understanding rather than a memorised fourfold slogan.

184. Area versus boundary 22: predict units and scale

Question: An imaginary rectangular board measures 9 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 9×2=18 cm². The perimeter is 2(9+2)=22 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 18 cm as area or 22 cm² as perimeter is a unit error. If the learner calculates 9+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 72 cm² and new perimeter 44 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 27 cm², while perimeter rises by two centimetres to 24 cm. This contrast tests understanding rather than a memorised fourfold slogan.

185. Area versus boundary 23: predict units and scale

Question: An imaginary rectangular board measures 12 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 12×7=84 cm². The perimeter is 2(12+7)=38 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 84 cm as area or 38 cm² as perimeter is a unit error. If the learner calculates 12+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 336 cm² and new perimeter 76 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 96 cm², while perimeter rises by two centimetres to 40 cm. This contrast tests understanding rather than a memorised fourfold slogan.

186. Area versus boundary 24: predict units and scale

Question: An imaginary rectangular board measures 15 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 15×12=180 cm². The perimeter is 2(15+12)=54 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 180 cm as area or 54 cm² as perimeter is a unit error. If the learner calculates 15+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 720 cm² and new perimeter 108 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 195 cm², while perimeter rises by two centimetres to 56 cm. This contrast tests understanding rather than a memorised fourfold slogan.

187. Area versus boundary 25: predict units and scale

Question: An imaginary rectangular board measures 18 cm by 2 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 18×2=36 cm². The perimeter is 2(18+2)=40 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 36 cm as area or 40 cm² as perimeter is a unit error. If the learner calculates 18+2 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 144 cm² and new perimeter 80 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 54 cm², while perimeter rises by two centimetres to 42 cm. This contrast tests understanding rather than a memorised fourfold slogan.

188. Area versus boundary 26: predict units and scale

Question: An imaginary rectangular board measures 21 cm by 7 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 21×7=147 cm². The perimeter is 2(21+7)=56 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 147 cm as area or 56 cm² as perimeter is a unit error. If the learner calculates 21+7 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 588 cm² and new perimeter 112 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 168 cm², while perimeter rises by two centimetres to 58 cm. This contrast tests understanding rather than a memorised fourfold slogan.

189. Area versus boundary 27: predict units and scale

Question: An imaginary rectangular board measures 5 cm by 12 cm. Before calculating, name the units for area and perimeter and predict how each changes if both lengths double.

Exact working: The area is 5×12=60 cm². The perimeter is 2(5+12)=34 cm. They may have similar-looking numbers but describe different physical quantities.

Tempting wrong line: Reporting 60 cm as area or 34 cm² as perimeter is a unit error. If the learner calculates 5+12 for full perimeter, two opposite sides have been forgotten. The unit and shape can expose those first incorrect steps.

Independent check: After doubling the two side lengths, the new area is 240 cm² and new perimeter 68 cm, matching the earlier direction and scale predictions.

Transfer: Instead of doubling both dimensions, increase width by one centimetre only. New area is 65 cm², while perimeter rises by two centimetres to 36 cm. This contrast tests understanding rather than a memorised fourfold slogan.

Statistics and probability plausibility

This group is about statistics and probability plausibility. Choose a small number of examples matching the current school chapter; explain the sign, size or unit constraint before performing exact arithmetic. If a correct-looking answer contradicts that constraint, identify the first incorrect assumption or line. An independent changed-number follow-up is better evidence than immediately copying the same solution twice.

190. Chance bound 1: an event cannot exceed one

Question: A hypothetical bag holds 1 red, 1 blue and 1 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Predict first: For one event in a valid probability model, the chance lies between zero and one. Because the bag has some red counters and also non-red counters, both red and not red are possible but neither is certain. Their probabilities must sum to one.

Exact working: Total counters=1+1+1=3. Thus P(red)=1/3 (about 0.333333), and P(not red)=2/3 (about 0.666667).

Tempting wrong line: Using only the number of distinct colour names (three) as denominator ignores their unequal counts. A calculated probability larger than one indicates that the favourable count or full sample size was misread.

Independent check: The favourable and complementary counts add 1+2=3; the probabilities add 1/3+2/3=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 3/5. The numerator and denominator both change, so adding 2/3 to the old chance is not valid.

191. Chance bound 2: an event cannot exceed one

Question: A hypothetical bag holds 4 red, 6 blue and 8 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=4+6+8=18. Thus P(red)=2/9 (about 0.222222), and P(not red)=7/9 (about 0.777778).

Independent check: The favourable and complementary counts add 4+14=18; the probabilities add 4/18+14/18=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 3/10. The numerator and denominator both change, so adding 2/18 to the old chance is not valid.

192. Chance bound 3: an event cannot exceed one

Question: A hypothetical bag holds 7 red, 11 blue and 3 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=7+11+3=21. Thus P(red)=1/3 (about 0.333333), and P(not red)=2/3 (about 0.666667).

Independent check: The favourable and complementary counts add 7+14=21; the probabilities add 7/21+14/21=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 9/23. The numerator and denominator both change, so adding 2/21 to the old chance is not valid.

193. Chance bound 4: an event cannot exceed one

Question: A hypothetical bag holds 10 red, 1 blue and 10 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=10+1+10=21. Thus P(red)=10/21 (about 0.47619), and P(not red)=11/21 (about 0.52381).

Independent check: The favourable and complementary counts add 10+11=21; the probabilities add 10/21+11/21=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 12/23. The numerator and denominator both change, so adding 2/21 to the old chance is not valid.

194. Chance bound 5: an event cannot exceed one

Question: A hypothetical bag holds 13 red, 6 blue and 5 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=13+6+5=24. Thus P(red)=13/24 (about 0.541667), and P(not red)=11/24 (about 0.458333).

Independent check: The favourable and complementary counts add 13+11=24; the probabilities add 13/24+11/24=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 15/26. The numerator and denominator both change, so adding 2/24 to the old chance is not valid.

195. Chance bound 6: an event cannot exceed one

Question: A hypothetical bag holds 3 red, 11 blue and 12 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=3+11+12=26. Thus P(red)=3/26 (about 0.115385), and P(not red)=23/26 (about 0.884615).

Independent check: The favourable and complementary counts add 3+23=26; the probabilities add 3/26+23/26=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 5/28. The numerator and denominator both change, so adding 2/26 to the old chance is not valid.

196. Chance bound 7: an event cannot exceed one

Question: A hypothetical bag holds 6 red, 1 blue and 7 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=6+1+7=14. Thus P(red)=3/7 (about 0.428571), and P(not red)=4/7 (about 0.571429).

Independent check: The favourable and complementary counts add 6+8=14; the probabilities add 6/14+8/14=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 1/2. The numerator and denominator both change, so adding 2/14 to the old chance is not valid.

197. Chance bound 8: an event cannot exceed one

Question: A hypothetical bag holds 9 red, 6 blue and 2 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=9+6+2=17. Thus P(red)=9/17 (about 0.529412), and P(not red)=8/17 (about 0.470588).

Independent check: The favourable and complementary counts add 9+8=17; the probabilities add 9/17+8/17=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 11/19. The numerator and denominator both change, so adding 2/17 to the old chance is not valid.

198. Chance bound 9: an event cannot exceed one

Question: A hypothetical bag holds 12 red, 11 blue and 9 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=12+11+9=32. Thus P(red)=3/8 (about 0.375), and P(not red)=5/8 (about 0.625).

Independent check: The favourable and complementary counts add 12+20=32; the probabilities add 12/32+20/32=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 7/17. The numerator and denominator both change, so adding 2/32 to the old chance is not valid.

199. Chance bound 10: an event cannot exceed one

Question: A hypothetical bag holds 2 red, 1 blue and 4 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=2+1+4=7. Thus P(red)=2/7 (about 0.285714), and P(not red)=5/7 (about 0.714286).

Independent check: The favourable and complementary counts add 2+5=7; the probabilities add 2/7+5/7=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 4/9. The numerator and denominator both change, so adding 2/7 to the old chance is not valid.

200. Chance bound 11: an event cannot exceed one

Question: A hypothetical bag holds 5 red, 6 blue and 11 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=5+6+11=22. Thus P(red)=5/22 (about 0.227273), and P(not red)=17/22 (about 0.772727).

Independent check: The favourable and complementary counts add 5+17=22; the probabilities add 5/22+17/22=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 7/24. The numerator and denominator both change, so adding 2/22 to the old chance is not valid.

201. Chance bound 12: an event cannot exceed one

Question: A hypothetical bag holds 8 red, 11 blue and 6 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=8+11+6=25. Thus P(red)=8/25 (about 0.32), and P(not red)=17/25 (about 0.68).

Independent check: The favourable and complementary counts add 8+17=25; the probabilities add 8/25+17/25=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 10/27. The numerator and denominator both change, so adding 2/25 to the old chance is not valid.

202. Chance bound 13: an event cannot exceed one

Question: A hypothetical bag holds 11 red, 1 blue and 1 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=11+1+1=13. Thus P(red)=11/13 (about 0.846154), and P(not red)=2/13 (about 0.153846).

Independent check: The favourable and complementary counts add 11+2=13; the probabilities add 11/13+2/13=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 13/15. The numerator and denominator both change, so adding 2/13 to the old chance is not valid.

203. Chance bound 14: an event cannot exceed one

Question: A hypothetical bag holds 1 red, 6 blue and 8 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=1+6+8=15. Thus P(red)=1/15 (about 0.066667), and P(not red)=14/15 (about 0.933333).

Independent check: The favourable and complementary counts add 1+14=15; the probabilities add 1/15+14/15=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 3/17. The numerator and denominator both change, so adding 2/15 to the old chance is not valid.

204. Chance bound 15: an event cannot exceed one

Question: A hypothetical bag holds 4 red, 11 blue and 3 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=4+11+3=18. Thus P(red)=2/9 (about 0.222222), and P(not red)=7/9 (about 0.777778).

Independent check: The favourable and complementary counts add 4+14=18; the probabilities add 4/18+14/18=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 3/10. The numerator and denominator both change, so adding 2/18 to the old chance is not valid.

205. Chance bound 16: an event cannot exceed one

Question: A hypothetical bag holds 7 red, 1 blue and 10 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=7+1+10=18. Thus P(red)=7/18 (about 0.388889), and P(not red)=11/18 (about 0.611111).

Independent check: The favourable and complementary counts add 7+11=18; the probabilities add 7/18+11/18=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 9/20. The numerator and denominator both change, so adding 2/18 to the old chance is not valid.

206. Chance bound 17: an event cannot exceed one

Question: A hypothetical bag holds 10 red, 6 blue and 5 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=10+6+5=21. Thus P(red)=10/21 (about 0.47619), and P(not red)=11/21 (about 0.52381).

Independent check: The favourable and complementary counts add 10+11=21; the probabilities add 10/21+11/21=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 12/23. The numerator and denominator both change, so adding 2/21 to the old chance is not valid.

207. Chance bound 18: an event cannot exceed one

Question: A hypothetical bag holds 13 red, 11 blue and 12 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=13+11+12=36. Thus P(red)=13/36 (about 0.361111), and P(not red)=23/36 (about 0.638889).

Independent check: The favourable and complementary counts add 13+23=36; the probabilities add 13/36+23/36=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 15/38. The numerator and denominator both change, so adding 2/36 to the old chance is not valid.

208. Chance bound 19: an event cannot exceed one

Question: A hypothetical bag holds 3 red, 1 blue and 7 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=3+1+7=11. Thus P(red)=3/11 (about 0.272727), and P(not red)=8/11 (about 0.727273).

Independent check: The favourable and complementary counts add 3+8=11; the probabilities add 3/11+8/11=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 5/13. The numerator and denominator both change, so adding 2/11 to the old chance is not valid.

209. Chance bound 20: an event cannot exceed one

Question: A hypothetical bag holds 6 red, 6 blue and 2 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=6+6+2=14. Thus P(red)=3/7 (about 0.428571), and P(not red)=4/7 (about 0.571429).

Independent check: The favourable and complementary counts add 6+8=14; the probabilities add 6/14+8/14=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 1/2. The numerator and denominator both change, so adding 2/14 to the old chance is not valid.

210. Chance bound 21: an event cannot exceed one

Question: A hypothetical bag holds 9 red, 11 blue and 9 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=9+11+9=29. Thus P(red)=9/29 (about 0.310345), and P(not red)=20/29 (about 0.689655).

Independent check: The favourable and complementary counts add 9+20=29; the probabilities add 9/29+20/29=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 11/31. The numerator and denominator both change, so adding 2/29 to the old chance is not valid.

211. Chance bound 22: an event cannot exceed one

Question: A hypothetical bag holds 12 red, 1 blue and 4 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=12+1+4=17. Thus P(red)=12/17 (about 0.705882), and P(not red)=5/17 (about 0.294118).

Independent check: The favourable and complementary counts add 12+5=17; the probabilities add 12/17+5/17=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 14/19. The numerator and denominator both change, so adding 2/17 to the old chance is not valid.

212. Chance bound 23: an event cannot exceed one

Question: A hypothetical bag holds 2 red, 6 blue and 11 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=2+6+11=19. Thus P(red)=2/19 (about 0.105263), and P(not red)=17/19 (about 0.894737).

Independent check: The favourable and complementary counts add 2+17=19; the probabilities add 2/19+17/19=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 4/21. The numerator and denominator both change, so adding 2/19 to the old chance is not valid.

213. Chance bound 24: an event cannot exceed one

Question: A hypothetical bag holds 5 red, 11 blue and 6 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=5+11+6=22. Thus P(red)=5/22 (about 0.227273), and P(not red)=17/22 (about 0.772727).

Independent check: The favourable and complementary counts add 5+17=22; the probabilities add 5/22+17/22=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 7/24. The numerator and denominator both change, so adding 2/22 to the old chance is not valid.

214. Chance bound 25: an event cannot exceed one

Question: A hypothetical bag holds 8 red, 1 blue and 1 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=8+1+1=10. Thus P(red)=4/5 (about 0.8), and P(not red)=1/5 (about 0.2).

Independent check: The favourable and complementary counts add 8+2=10; the probabilities add 8/10+2/10=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 5/6. The numerator and denominator both change, so adding 2/10 to the old chance is not valid.

215. Chance bound 26: an event cannot exceed one

Question: A hypothetical bag holds 11 red, 6 blue and 8 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=11+6+8=25. Thus P(red)=11/25 (about 0.44), and P(not red)=14/25 (about 0.56).

Independent check: The favourable and complementary counts add 11+14=25; the probabilities add 11/25+14/25=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 13/27. The numerator and denominator both change, so adding 2/25 to the old chance is not valid.

216. Chance bound 27: an event cannot exceed one

Question: A hypothetical bag holds 1 red, 11 blue and 3 green identical counters, thoroughly mixed for one equally likely draw. Predict the valid interval for P(red) and P(not red), then find both.

Exact working: Total counters=1+11+3=15. Thus P(red)=1/15 (about 0.066667), and P(not red)=14/15 (about 0.933333).

Independent check: The favourable and complementary counts add 1+14=15; the probabilities add 1/15+14/15=1. No counter is counted in both events.

Transfer: If two extra red counters were added, the probability of red becomes 3/17. The numerator and denominator both change, so adding 2/15 to the old chance is not valid.

The first mismatch is a better teaching target than a low score

A result can fail because the wrong quantity was modelled, a negative sign was lost, a unit conversion used the wrong dimension, or a valid numerical answer was attached to the wrong label. These are different errors. A tutor can ask what the pupil expected before calculating, then compare that expectation with the actual method. If the expectation was itself mistaken, repair the concept. If the expectation was sound but the computation failed, focus on the first arithmetic or notation slip.

The point is not to tell pupils that they must always know the answer beforehand. A genuinely unfamiliar problem may not yield a tight prediction immediately. But asking whether the output should be within zero and one, smaller than a total, or measured in cm² can be enough to detect a major issue before more work is done. Do not treat every non-integer or negative result as an error; judge those features in the actual mathematical context.

How a three-pupil Mathematics lesson can use these checks

The immutable eduKateSG Clementi three-pupil Mathematics reference describes 90-minute tutorials near Sixth Avenue MRT. In an illustrative Secondary 2 G1 lesson one pupil could need number-line signs, another might need area units and another might need comparing a ratio result with its stated whole. The tutor can model one general checking habit but assign different changed-number checks after identifying each first error. The programme reference is not evidence of current class availability or a guaranteed grade outcome.

First 10 minutes: predict the units, sign or bound of three short tasks without a calculator.
Next 20 minutes: teach one school-relevant concept and connect its formula to a reasonableness check.
Next 25 minutes: practise individual examples with first-error feedback.
Next 15 minutes: compare two methods of checking an answer independently.
Next 10 minutes: solve a mixed unseen question and state what would make its output impossible.
Final 10 minutes: choose one suitable delayed retest rather than another long worksheet.

Five short home checks without excessive homework

Day one: predict the sign of a short signed-number expression, then calculate.
Day two: compare a positive fraction sum against zero, one and two before simplifying.
Day three: predict that a straightforward percentage discount lowers the amount paid.
Day four: choose the correct area or perimeter unit before writing a formula.
Day five: check a probability or ratio share against the whole and explain the outcome.

The parent can ask, ‘Would that answer be bigger or smaller than the starting number?’ or ‘What unit should it have?’ These are practical questions even if the parent does not recall every algebraic formula. Keep the child’s school assignments and CCA commitments in mind. If a five-minute check reveals a fundamental misunderstanding, bring that exact first line to the tutor for targeted teaching.

Frequently asked questions

Is a rough estimate enough as a final answer?

Only if the question permits estimation. Otherwise use the prediction to guide and check the required exact or appropriately rounded calculation.

Can a negative answer ever be correct?

Yes, for signed-number differences, graph coordinates, temperatures and other quantities where negatives are permitted. A physical count or area usually cannot be negative.

What is the quickest geometry sense-check?

Name whether the question asks for a length, area or volume and predict the appropriate unit: cm, cm² or cm³.

What if two different methods produce different results?

Check the first step where they express different quantities, assumptions or operations; do not choose one result merely because it looks more familiar.

Is this the same as significant-figure rounding?

No. Rounding is about numerical presentation and approximation accuracy. Here the priority is whether the proposed result is logically plausible before and after calculating.

How can a tutor verify that sense-checking is independent?

Give an unfamiliar changed problem and ask the pupil to state a sign, interval, unit or direction bound before using the worked answer.

Official syllabus and reader notes

Reader note: All amounts, hypothetical groups and measurement problems are original mathematical examples. They do not represent live prices, examination predictions or real pupil performance. The 2027 G1 SEC syllabus belongs to its graduating cohort; current school work and the student’s eventual examination-year guidance take precedence. Programme availability, fees and outcomes must be confirmed directly.

For a focused learning discussion, bring one answer that seemed implausible before recalculating to eduKate Singapore, and ask which check would reveal the misunderstanding earliest.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Sense-Checking Answers Before Calculating

Quick answer: five sense-checks before calculating

Choose a useful sense-check

A sixty-second check with three impossible answers

A simple before–during–after routine

Before: predict the quantity, sign and size

During: choose the method that matches the relation

After: use a different path where available

216 original sense-checking clinics

Signed-number and magnitude predictions

1. Signed calculation 1: choose the answer’s direction

2. Signed calculation 2: choose the answer’s direction

3. Signed calculation 3: choose the answer’s direction

4. Signed calculation 4: choose the answer’s direction

5. Signed calculation 5: choose the answer’s direction

6. Signed calculation 6: choose the answer’s direction

7. Signed calculation 7: choose the answer’s direction

8. Signed calculation 8: choose the answer’s direction

9. Signed calculation 9: choose the answer’s direction

10. Signed calculation 10: choose the answer’s direction

11. Signed calculation 11: choose the answer’s direction

12. Signed calculation 12: choose the answer’s direction

13. Signed calculation 13: choose the answer’s direction

14. Signed calculation 14: choose the answer’s direction

15. Signed calculation 15: choose the answer’s direction

16. Signed calculation 16: choose the answer’s direction

17. Signed calculation 17: choose the answer’s direction

18. Signed calculation 18: choose the answer’s direction

19. Signed calculation 19: choose the answer’s direction

20. Signed calculation 20: choose the answer’s direction

21. Signed calculation 21: choose the answer’s direction

22. Signed calculation 22: choose the answer’s direction

23. Signed calculation 23: choose the answer’s direction

24. Signed calculation 24: choose the answer’s direction

25. Signed calculation 25: choose the answer’s direction

26. Signed calculation 26: choose the answer’s direction

27. Signed calculation 27: choose the answer’s direction

Fractions and decimals within sensible bounds

28. Fraction bound 1: is the result below one?

29. Fraction bound 2: is the result below one?

30. Fraction bound 3: is the result below one?

31. Fraction bound 4: is the result below one?

32. Fraction bound 5: is the result below one?

33. Fraction bound 6: is the result below one?

34. Fraction bound 7: is the result below one?

35. Fraction bound 8: is the result below one?

36. Fraction bound 9: is the result below one?

37. Fraction bound 10: is the result below one?

38. Fraction bound 11: is the result below one?

39. Fraction bound 12: is the result below one?

40. Fraction bound 13: is the result below one?

41. Fraction bound 14: is the result below one?

42. Fraction bound 15: is the result below one?

43. Fraction bound 16: is the result below one?

44. Fraction bound 17: is the result below one?

45. Fraction bound 18: is the result below one?

46. Fraction bound 19: is the result below one?

47. Fraction bound 20: is the result below one?

48. Fraction bound 21: is the result below one?

49. Fraction bound 22: is the result below one?

50. Fraction bound 23: is the result below one?

51. Fraction bound 24: is the result below one?

52. Fraction bound 25: is the result below one?

53. Fraction bound 26: is the result below one?

54. Fraction bound 27: is the result below one?

Percentages and discounts: predict direction

55. Discount sense-check 1: can payment exceed the original?

56. Discount sense-check 2: can payment exceed the original?

57. Discount sense-check 3: can payment exceed the original?

58. Discount sense-check 4: can payment exceed the original?

59. Discount sense-check 5: can payment exceed the original?

60. Discount sense-check 6: can payment exceed the original?

61. Discount sense-check 7: can payment exceed the original?

62. Discount sense-check 8: can payment exceed the original?

63. Discount sense-check 9: can payment exceed the original?

64. Discount sense-check 10: can payment exceed the original?

65. Discount sense-check 11: can payment exceed the original?

66. Discount sense-check 12: can payment exceed the original?

67. Discount sense-check 13: can payment exceed the original?

68. Discount sense-check 14: can payment exceed the original?