Sec 2 G1 Maths Tutor Bukit Timah

Did you know? ‘Two out of five’, ‘two to five’ and ‘two per five minutes’ can look similar on a worksheet but mean different things. The first may be a fraction of a whole, the second a ratio of one group to another, and the third a rate with time in the denominator. For a Secondary 2 G1 Mathematics tutor in Bukit Timah, helping a pupil choose the correct comparison is often more important than adding another page of easy division.

This Secondary 2 G1 Mathematics guide explains the differences between ratio, rate and percentage using familiar examples. Ratio compares labelled quantities such as red:blue=2:3. A rate may compare unlike quantities, such as 12 kilometres per hour. A percentage compares an amount with a reference base expressed per hundred; 20% of $80 is $16, but a price that rises from $80 to $100 has a 25% increase. The denominator tells the story.

Families choosing Secondary 2 G1 Maths tuition near Bukit Timah can use this as a chapter-connection guide. MOE’s 2020 Secondary Two G1 programme includes direct and inverse proportion, rates and average speed. The published 2027 G1 SEC Mathematics K110 graduating syllabus spans ratios, percentages and rates, including mathematical applications in everyday contexts. We distinguish these from a 2026 Secondary Two school’s immediate topic sequence, which should remain the teaching priority.

Quick answer: which comparison are you making?

Ratio: A:B=2:3 compares A with B; it does not automatically say A is two thirds of their total.
Part-to-whole: if only A and B make the whole, A is 2/(2+3)=2/5 of that whole.
Rate: 60 km/h is distance per hour, with different quantities and a compound unit.
Percentage: 2/5=40% expresses a fraction of a specified whole per hundred.
Percentage change: (new−old)/old×100% uses the old amount as base; percentage points are the arithmetic difference between two percentage rates.

Find the correct denominator

One-minute diagnostic
Why the denominator changes meaning
Ratio: one part compared with another
Part-to-whole fractions and percentages
Unit rates: money, mass and time
Percentage of an actual reference base
Percentage change versus percentage points
Comparing groups with unequal sample sizes
Direct and inverse proportional models
Choose the representation from the wording
Four high-impact mistakes
A three-pupil lesson
Frequently asked questions
Official sources

Three quick comparisons to test

Ask the pupil these questions about two groups of two and three counters: ‘What is A:B?’, ‘What fraction of all five counters is A?’, and ‘What percent of all five is A?’ The answers are 2:3, 2/5 and 40%. Next ask a fictional journey of six kilometres over two hours: the rate is 3 km/h, not a three-to-one share of a physical group. A child who simply writes the same number for every question needs a lesson on naming the whole and the units.

Start with the label and the whole

A numerator only makes sense beside its denominator. Six successes in twenty attempts gives relative frequency 6/20, while six successful pupils compared with fourteen unsuccessful pupils gives a part-to-part ratio 6:14. Six kilometres over two hours gives 3 km/h. These different outputs are not competing answers to one question: they respond to different comparisons. Before using a calculator, read what the question asks for and underline the reference group, elapsed interval or base quantity.

230 original ratio, rate and percentage clinics

Each group below uses original fictional amounts and asks the pupil to state which comparison is being made. The worked reasoning includes a check, a common denominator mistake and a changed context. Do not complete all questions in one sitting; choose several that match the school’s current G1 strands and revisit them later without headings or worked examples.

Ratio: one part compared with another

In this section, the learner should identify a reference quantity before applying a formula. Some questions ask for part-to-part; others for part-to-whole, rate per unit, an old-base change or an ideal proportional relationship. A correct number in the wrong unit or with the wrong denominator is not yet a correct interpretation.

1. Ratio comparison 1: part to part

Question. Two imaginary groups contain 2 and 4 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

What is being compared? A ratio compares the first amount with the second in the specified order. It is not automatically a fraction of the combined whole. The total is a separate number, 6, against which each share can be divided.

Worked answer. The first-to-second ratio is 2:4, simplifying by 2 to 1:2. Reversing labels gives 2:1. Their fractions of the whole are 1/3 and 2/3. The same numeral 1 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 1/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/3+2/3=1. The actual counts sum to 6. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 4 and 8, the ratio stays 1:2 while the total becomes 12. The percentage share of each group does not change when both scale equally.

2. Ratio comparison 2: part to part

Question. Two imaginary groups contain 18 and 45 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 18:45, simplifying by 9 to 2:5. Reversing labels gives 5:2. Their fractions of the whole are 2/7 and 5/7. The same numeral 2 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 2/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 2/7+5/7=1. The actual counts sum to 63. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 36 and 90, the ratio stays 2:5 while the total becomes 126. The percentage share of each group does not change when both scale equally.

3. Ratio comparison 3: part to part

Question. Two imaginary groups contain 9 and 24 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 9:24, simplifying by 3 to 3:8. Reversing labels gives 8:3. Their fractions of the whole are 3/11 and 8/11. The same numeral 3 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 3/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 3/11+8/11=1. The actual counts sum to 33. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 18 and 48, the ratio stays 3:8 while the total becomes 66. The percentage share of each group does not change when both scale equally.

4. Ratio comparison 4: part to part

Question. Two imaginary groups contain 40 and 20 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 40:20. Dividing both terms by 10 gives 4:2, and reducing their remaining common factor 2 gives the lowest terms 2:1. Reversing the labels gives 1:2. Their fractions of the whole are 2/3 and 1/3. The same numeral 4 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 4/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 2/3+1/3=1. The actual counts sum to 60. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 80 and 40, the ratio stays 2:1 while the total becomes 120. The percentage share of each group does not change when both scale equally.

5. Ratio comparison 5: part to part

Question. Two imaginary groups contain 20 and 20 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 20:20. Dividing both terms by 4 gives 5:5, and reducing their remaining common factor 5 gives the lowest terms 1:1. Reversing the labels gives 1:1. Their fractions of the whole are 1/2 and 1/2. The same numeral 5 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 5/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/2+1/2=1. The actual counts sum to 40. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 40 and 40, the ratio stays 1:1 while the total becomes 80. The percentage share of each group does not change when both scale equally.

6. Ratio comparison 6: part to part

Question. Two imaginary groups contain 66 and 88 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 66:88. Dividing both terms by 11 gives 6:8, and reducing their remaining common factor 2 gives the lowest terms 3:4. Reversing the labels gives 4:3. Their fractions of the whole are 3/7 and 4/7. The same numeral 6 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 6/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 3/7+4/7=1. The actual counts sum to 154. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 132 and 176, the ratio stays 3:4 while the total becomes 308. The percentage share of each group does not change when both scale equally.

7. Ratio comparison 7: part to part

Question. Two imaginary groups contain 35 and 10 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 35:10, simplifying by 5 to 7:2. Reversing labels gives 2:7. Their fractions of the whole are 7/9 and 2/9. The same numeral 7 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 7/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 7/9+2/9=1. The actual counts sum to 45. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 70 and 20, the ratio stays 7:2 while the total becomes 90. The percentage share of each group does not change when both scale equally.

8. Ratio comparison 8: part to part

Question. Two imaginary groups contain 96 and 60 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 96:60, simplifying by 12 to 8:5. Reversing labels gives 5:8. Their fractions of the whole are 8/13 and 5/13. The same numeral 8 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 8/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 8/13+5/13=1. The actual counts sum to 156. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 192 and 120, the ratio stays 8:5 while the total becomes 312. The percentage share of each group does not change when both scale equally.

9. Ratio comparison 9: part to part

Question. Two imaginary groups contain 6 and 48 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 6:48, simplifying by 6 to 1:8. Reversing labels gives 8:1. Their fractions of the whole are 1/9 and 8/9. The same numeral 1 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 1/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/9+8/9=1. The actual counts sum to 54. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 12 and 96, the ratio stays 1:8 while the total becomes 108. The percentage share of each group does not change when both scale equally.

10. Ratio comparison 10: part to part

Question. Two imaginary groups contain 26 and 26 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 26:26. Dividing both terms by 13 gives 2:2, and reducing their remaining common factor 2 gives the lowest terms 1:1. Reversing the labels gives 1:1. Their fractions of the whole are 1/2 and 1/2. The same numeral 2 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 2/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/2+1/2=1. The actual counts sum to 52. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 52 and 52, the ratio stays 1:1 while the total becomes 104. The percentage share of each group does not change when both scale equally.

11. Ratio comparison 11: part to part

Question. Two imaginary groups contain 21 and 35 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 21:35, simplifying by 7 to 3:5. Reversing labels gives 5:3. Their fractions of the whole are 3/8 and 5/8. The same numeral 3 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 3/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 3/8+5/8=1. The actual counts sum to 56. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 42 and 70, the ratio stays 3:5 while the total becomes 112. The percentage share of each group does not change when both scale equally.

12. Ratio comparison 12: part to part

Question. Two imaginary groups contain 56 and 112 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 56:112. Dividing both terms by 14 gives 4:8, and reducing their remaining common factor 4 gives the lowest terms 1:2. Reversing the labels gives 2:1. Their fractions of the whole are 1/3 and 2/3. The same numeral 4 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 4/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/3+2/3=1. The actual counts sum to 168. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 112 and 224, the ratio stays 1:2 while the total becomes 336. The percentage share of each group does not change when both scale equally.

13. Ratio comparison 13: part to part

Question. Two imaginary groups contain 40 and 16 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 40:16, simplifying by 8 to 5:2. Reversing labels gives 2:5. Their fractions of the whole are 5/7 and 2/7. The same numeral 5 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 5/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 5/7+2/7=1. The actual counts sum to 56. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 80 and 32, the ratio stays 5:2 while the total becomes 112. The percentage share of each group does not change when both scale equally.

14. Ratio comparison 14: part to part

Question. Two imaginary groups contain 12 and 10 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 12:10, simplifying by 2 to 6:5. Reversing labels gives 5:6. Their fractions of the whole are 6/11 and 5/11. The same numeral 6 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 6/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 6/11+5/11=1. The actual counts sum to 22. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 24 and 20, the ratio stays 6:5 while the total becomes 44. The percentage share of each group does not change when both scale equally.

15. Ratio comparison 15: part to part

Question. Two imaginary groups contain 63 and 72 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 63:72, simplifying by 9 to 7:8. Reversing labels gives 8:7. Their fractions of the whole are 7/15 and 8/15. The same numeral 7 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 7/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 7/15+8/15=1. The actual counts sum to 135. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 126 and 144, the ratio stays 7:8 while the total becomes 270. The percentage share of each group does not change when both scale equally.

16. Ratio comparison 16: part to part

Question. Two imaginary groups contain 24 and 6 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 24:6. Dividing both terms by 3 gives 8:2, and reducing their remaining common factor 2 gives the lowest terms 4:1. Reversing the labels gives 1:4. Their fractions of the whole are 4/5 and 1/5. The same numeral 8 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 8/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 4/5+1/5=1. The actual counts sum to 30. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 48 and 12, the ratio stays 4:1 while the total becomes 60. The percentage share of each group does not change when both scale equally.

17. Ratio comparison 17: part to part

Question. Two imaginary groups contain 10 and 50 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 10:50, simplifying by 10 to 1:5. Reversing labels gives 5:1. Their fractions of the whole are 1/6 and 5/6. The same numeral 1 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 1/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/6+5/6=1. The actual counts sum to 60. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 20 and 100, the ratio stays 1:5 while the total becomes 120. The percentage share of each group does not change when both scale equally.

18. Ratio comparison 18: part to part

Question. Two imaginary groups contain 8 and 32 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 8:32. Dividing both terms by 4 gives 2:8, and reducing their remaining common factor 2 gives the lowest terms 1:4. Reversing the labels gives 4:1. Their fractions of the whole are 1/5 and 4/5. The same numeral 2 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 2/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/5+4/5=1. The actual counts sum to 40. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 16 and 64, the ratio stays 1:4 while the total becomes 80. The percentage share of each group does not change when both scale equally.

19. Ratio comparison 19: part to part

Question. Two imaginary groups contain 33 and 22 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 33:22, simplifying by 11 to 3:2. Reversing labels gives 2:3. Their fractions of the whole are 3/5 and 2/5. The same numeral 3 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 3/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 3/5+2/5=1. The actual counts sum to 55. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 66 and 44, the ratio stays 3:2 while the total becomes 110. The percentage share of each group does not change when both scale equally.

20. Ratio comparison 20: part to part

Question. Two imaginary groups contain 20 and 25 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 20:25, simplifying by 5 to 4:5. Reversing labels gives 5:4. Their fractions of the whole are 4/9 and 5/9. The same numeral 4 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 4/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 4/9+5/9=1. The actual counts sum to 45. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 40 and 50, the ratio stays 4:5 while the total becomes 90. The percentage share of each group does not change when both scale equally.

21. Ratio comparison 21: part to part

Question. Two imaginary groups contain 60 and 96 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 60:96, simplifying by 12 to 5:8. Reversing labels gives 8:5. Their fractions of the whole are 5/13 and 8/13. The same numeral 5 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 5/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 5/13+8/13=1. The actual counts sum to 156. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 120 and 192, the ratio stays 5:8 while the total becomes 312. The percentage share of each group does not change when both scale equally.

22. Ratio comparison 22: part to part

Question. Two imaginary groups contain 36 and 12 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 36:12. Dividing both terms by 6 gives 6:2, and reducing their remaining common factor 2 gives the lowest terms 3:1. Reversing the labels gives 1:3. Their fractions of the whole are 3/4 and 1/4. The same numeral 6 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 6/2, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 3/4+1/4=1. The actual counts sum to 48. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 72 and 24, the ratio stays 3:1 while the total becomes 96. The percentage share of each group does not change when both scale equally.

23. Ratio comparison 23: part to part

Question. Two imaginary groups contain 91 and 65 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 91:65, simplifying by 13 to 7:5. Reversing labels gives 5:7. Their fractions of the whole are 7/12 and 5/12. The same numeral 7 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 7/5, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 7/12+5/12=1. The actual counts sum to 156. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 182 and 130, the ratio stays 7:5 while the total becomes 312. The percentage share of each group does not change when both scale equally.

24. Ratio comparison 24: part to part

Question. Two imaginary groups contain 56 and 56 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 56:56. Dividing both terms by 7 gives 8:8, and reducing their remaining common factor 8 gives the lowest terms 1:1. Reversing the labels gives 1:1. Their fractions of the whole are 1/2 and 1/2. The same numeral 8 appears in one ratio entry but is not necessarily an absolute number of objects.

Common denominator mistake. A pupil may write the first group’s part-of-total fraction as 8/8, which compares it with the other group rather than the whole. Another may reverse the labels while still copying the same ratio order.

Independent check. The shares add to 1/2+1/2=1. The actual counts sum to 112. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 112 and 112, the ratio stays 1:1 while the total becomes 224. The percentage share of each group does not change when both scale equally.

25. Ratio comparison 25: part to part

Question. Two imaginary groups contain 14 and 28 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 14:28, simplifying by 14 to 1:2. Reversing labels gives 2:1. Their fractions of the whole are 1/3 and 2/3. The same numeral 1 appears in one ratio entry but is not necessarily an absolute number of objects.

Independent check. The shares add to 1/3+2/3=1. The actual counts sum to 42. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 28 and 56, the ratio stays 1:2 while the total becomes 84. The percentage share of each group does not change when both scale equally.

26. Ratio comparison 26: part to part

Question. Two imaginary groups contain 16 and 40 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 16:40, simplifying by 8 to 2:5. Reversing labels gives 5:2. Their fractions of the whole are 2/7 and 5/7. The same numeral 2 appears in one ratio entry but is not necessarily an absolute number of objects.

Independent check. The shares add to 2/7+5/7=1. The actual counts sum to 56. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 32 and 80, the ratio stays 2:5 while the total becomes 112. The percentage share of each group does not change when both scale equally.

27. Ratio comparison 27: part to part

Question. Two imaginary groups contain 6 and 16 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 6:16, simplifying by 2 to 3:8. Reversing labels gives 8:3. Their fractions of the whole are 3/11 and 8/11. The same numeral 3 appears in one ratio entry but is not necessarily an absolute number of objects.

Independent check. The shares add to 3/11+8/11=1. The actual counts sum to 22. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 12 and 32, the ratio stays 3:8 while the total becomes 44. The percentage share of each group does not change when both scale equally.

28. Ratio comparison 28: part to part

Question. Two imaginary groups contain 36 and 18 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 36:18. Dividing both terms by 9 gives 4:2, and reducing their remaining common factor 2 gives the lowest terms 2:1. Reversing the labels gives 1:2. Their fractions of the whole are 2/3 and 1/3. The same numeral 4 appears in one ratio entry but is not necessarily an absolute number of objects.

Independent check. The shares add to 2/3+1/3=1. The actual counts sum to 54. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 72 and 36, the ratio stays 2:1 while the total becomes 108. The percentage share of each group does not change when both scale equally.

29. Ratio comparison 29: part to part

Question. Two imaginary groups contain 15 and 15 objects respectively. Give the first-to-second ratio, reverse it, and find each group’s fraction of the combined total.

Worked answer. The first-to-second ratio is 15:15. Dividing both terms by 3 gives 5:5, and reducing their remaining common factor 5 gives the lowest terms 1:1. Reversing the labels gives 1:1. Their fractions of the whole are 1/2 and 1/2. The same numeral 5 appears in one ratio entry but is not necessarily an absolute number of objects.

Independent check. The shares add to 1/2+1/2=1. The actual counts sum to 30. Both share-fraction and raw-count checks should reconstruct the same two-group partition.

Changed comparison. If each group size were doubled to 30 and 30, the ratio stays 1:1 while the total becomes 60. The percentage share of each group does not change when both scale equally.

Part-to-whole fractions and percentages

30. Part-to-whole comparison 1

Question. A hypothetical box contains 3 red and 3 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 6 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 3:3=1:1. Red fraction of the whole is 3/6=1/2. Red percentage is (3/6)×100%=50% (displayed to a practical number of digits). Its exact value is the fraction 1/2 of 100%.

Common denominator mistake. Writing 1:1 as 1/1 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/2. Add red plus blue shares: (1+1)/(2)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 6 and 6, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

31. Part-to-whole comparison 2

Question. A hypothetical box contains 20 red and 60 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 80 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 20:60, equivalent to 2:6 and simplified fully to 1:3. Red fraction of the whole is 20/80=1/4. Red percentage is (20/80)×100%=25% (displayed to a practical number of digits). Its exact value is the fraction 1/4 of 100%.

Common denominator mistake. Writing 2:6 as 2/6 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 3/4. Add red plus blue shares: (2+6)/(8)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 40 and 120, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

32. Part-to-whole comparison 3

Question. A hypothetical box contains 51 red and 51 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 102 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 51:51, equivalent to 3:3 and simplified fully to 1:1. Red fraction of the whole is 51/102=1/2. Red percentage is (51/102)×100%=50% (displayed to a practical number of digits). Its exact value is the fraction 1/2 of 100%.

Common denominator mistake. Writing 3:3 as 3/3 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/2. Add red plus blue shares: (3+3)/(6)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 102 and 102, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

33. Part-to-whole comparison 4

Question. A hypothetical box contains 36 red and 72 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 108 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 36:72, equivalent to 4:8 and simplified fully to 1:2. Red fraction of the whole is 36/108=1/3. Red percentage is (36/108)×100%=33.33333% (displayed to a practical number of digits). Its exact value is the fraction 1/3 of 100%.

Common denominator mistake. Writing 4:8 as 4/8 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 2/3. Add red plus blue shares: (4+8)/(12)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 72 and 144, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

34. Part-to-whole comparison 5

Question. A hypothetical box contains 80 red and 80 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 160 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 80:80, equivalent to 5:5 and simplified fully to 1:1. Red fraction of the whole is 80/160=1/2. Red percentage is (80/160)×100%=50% (displayed to a practical number of digits). Its exact value is the fraction 1/2 of 100%.

Common denominator mistake. Writing 5:5 as 5/5 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/2. Add red plus blue shares: (5+5)/(10)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 160 and 160, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

35. Part-to-whole comparison 6

Question. A hypothetical box contains 48 red and 16 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 64 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 48:16, equivalent to 6:2 and simplified fully to 3:1. Red fraction of the whole is 48/64=3/4. Red percentage is (48/64)×100%=75% (displayed to a practical number of digits). Its exact value is the fraction 3/4 of 100%.

Common denominator mistake. Writing 6:2 as 6/2 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/4. Add red plus blue shares: (6+2)/(8)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 96 and 32, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

36. Part-to-whole comparison 7

Question. A hypothetical box contains 105 red and 105 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 210 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 105:105, equivalent to 7:7 and simplified fully to 1:1. Red fraction of the whole is 105/210=1/2. Red percentage is (105/210)×100%=50% (displayed to a practical number of digits). Its exact value is the fraction 1/2 of 100%.

Common denominator mistake. Writing 7:7 as 7/7 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/2. Add red plus blue shares: (7+7)/(14)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 210 and 210, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

37. Part-to-whole comparison 8

Question. A hypothetical box contains 56 red and 28 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 84 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 56:28, equivalent to 8:4 and simplified fully to 2:1. Red fraction of the whole is 56/84=2/3. Red percentage is (56/84)×100%=66.66667% (displayed to a practical number of digits). Its exact value is the fraction 2/3 of 100%.

Common denominator mistake. Writing 8:4 as 8/4 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/3. Add red plus blue shares: (8+4)/(12)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 112 and 56, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

38. Part-to-whole comparison 9

Question. A hypothetical box contains 126 red and 14 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 140 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 126:14=9:1. Red fraction of the whole is 126/140=9/10. Red percentage is (126/140)×100%=90% (displayed to a practical number of digits). Its exact value is the fraction 9/10 of 100%.

Common denominator mistake. Writing 9:1 as 9/1 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/10. Add red plus blue shares: (9+1)/(10)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 252 and 28, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

39. Part-to-whole comparison 10

Question. A hypothetical box contains 6 red and 36 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 42 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 6:36=1:6. Red fraction of the whole is 6/42=1/7. Red percentage is (6/42)×100%=14.28571% (displayed to a practical number of digits). Its exact value is the fraction 1/7 of 100%.

Common denominator mistake. Writing 1:6 as 1/6 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 6/7. Add red plus blue shares: (1+6)/(7)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 12 and 72, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

40. Part-to-whole comparison 11

Question. A hypothetical box contains 26 red and 39 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 65 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 26:39=2:3. Red fraction of the whole is 26/65=2/5. Red percentage is (26/65)×100%=40% (displayed to a practical number of digits). Its exact value is the fraction 2/5 of 100%.

Common denominator mistake. Writing 2:3 as 2/3 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 3/5. Add red plus blue shares: (2+3)/(5)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 52 and 78, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

41. Part-to-whole comparison 12

Question. A hypothetical box contains 15 red and 40 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 55 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 15:40=3:8. Red fraction of the whole is 15/55=3/11. Red percentage is (15/55)×100%=27.27273% (displayed to a practical number of digits). Its exact value is the fraction 3/11 of 100%.

Common denominator mistake. Writing 3:8 as 3/8 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 8/11. Add red plus blue shares: (3+8)/(11)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 30 and 80, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

42. Part-to-whole comparison 13

Question. A hypothetical box contains 48 red and 60 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 48:60=4:5. Red fraction of the whole is 48/108=4/9. Red percentage is (48/108)×100%=44.44444% (displayed to a practical number of digits). Its exact value is the fraction 4/9 of 100%.

Common denominator mistake. Writing 4:5 as 4/5 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 5/9. Add red plus blue shares: (4+5)/(9)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 96 and 120, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

43. Part-to-whole comparison 14

Question. A hypothetical box contains 20 red and 8 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 28 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 20:8=5:2. Red fraction of the whole is 20/28=5/7. Red percentage is (20/28)×100%=71.42857% (displayed to a practical number of digits). Its exact value is the fraction 5/7 of 100%.

Common denominator mistake. Writing 5:2 as 5/2 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 2/7. Add red plus blue shares: (5+2)/(7)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 40 and 16, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

44. Part-to-whole comparison 15

Question. A hypothetical box contains 66 red and 77 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 143 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 66:77=6:7. Red fraction of the whole is 66/143=6/13. Red percentage is (66/143)×100%=46.15385% (displayed to a practical number of digits). Its exact value is the fraction 6/13 of 100%.

Common denominator mistake. Writing 6:7 as 6/7 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 7/13. Add red plus blue shares: (6+7)/(13)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 132 and 154, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

45. Part-to-whole comparison 16

Question. A hypothetical box contains 21 red and 12 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 33 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 21:12=7:4. Red fraction of the whole is 21/33=7/11. Red percentage is (21/33)×100%=63.63636% (displayed to a practical number of digits). Its exact value is the fraction 7/11 of 100%.

Common denominator mistake. Writing 7:4 as 7/4 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 4/11. Add red plus blue shares: (7+4)/(11)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 42 and 24, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

46. Part-to-whole comparison 17

Question. A hypothetical box contains 80 red and 10 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 90 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 80:10=8:1. Red fraction of the whole is 80/90=8/9. Red percentage is (80/90)×100%=88.88889% (displayed to a practical number of digits). Its exact value is the fraction 8/9 of 100%.

Common denominator mistake. Writing 8:1 as 8/1 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/9. Add red plus blue shares: (8+1)/(9)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 160 and 20, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

47. Part-to-whole comparison 18

Question. A hypothetical box contains 153 red and 102 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 255 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 153:102, equivalent to 9:6 and simplified fully to 3:2. Red fraction of the whole is 153/255=3/5. Red percentage is (153/255)×100%=60% (displayed to a practical number of digits). Its exact value is the fraction 3/5 of 100%.

Common denominator mistake. Writing 9:6 as 9/6 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 2/5. Add red plus blue shares: (9+6)/(15)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 306 and 204, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

48. Part-to-whole comparison 19

Question. A hypothetical box contains 9 red and 27 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 36 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 9:27=1:3. Red fraction of the whole is 9/36=1/4. Red percentage is (9/36)×100%=25% (displayed to a practical number of digits). Its exact value is the fraction 1/4 of 100%.

Common denominator mistake. Writing 1:3 as 1/3 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 3/4. Add red plus blue shares: (1+3)/(4)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 18 and 54, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

49. Part-to-whole comparison 20

Question. A hypothetical box contains 32 red and 128 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 32:128, equivalent to 2:8 and simplified fully to 1:4. Red fraction of the whole is 32/160=1/5. Red percentage is (32/160)×100%=20% (displayed to a practical number of digits). Its exact value is the fraction 1/5 of 100%.

Common denominator mistake. Writing 2:8 as 2/8 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 4/5. Add red plus blue shares: (2+8)/(10)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 64 and 256, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

50. Part-to-whole comparison 21

Question. A hypothetical box contains 24 red and 40 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 24:40=3:5. Red fraction of the whole is 24/64=3/8. Red percentage is (24/64)×100%=37.5% (displayed to a practical number of digits). Its exact value is the fraction 3/8 of 100%.

Common denominator mistake. Writing 3:5 as 3/5 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 5/8. Add red plus blue shares: (3+5)/(8)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 48 and 80, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

51. Part-to-whole comparison 22

Question. A hypothetical box contains 60 red and 30 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 60:30, equivalent to 4:2 and simplified fully to 2:1. Red fraction of the whole is 60/90=2/3. Red percentage is (60/90)×100%=66.66667% (displayed to a practical number of digits). Its exact value is the fraction 2/3 of 100%.

Common denominator mistake. Writing 4:2 as 4/2 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/3. Add red plus blue shares: (4+2)/(6)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 120 and 60, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

52. Part-to-whole comparison 23

Question. A hypothetical box contains 35 red and 49 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 35:49=5:7. Red fraction of the whole is 35/84=5/12. Red percentage is (35/84)×100%=41.66667% (displayed to a practical number of digits). Its exact value is the fraction 5/12 of 100%.

Common denominator mistake. Writing 5:7 as 5/7 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 7/12. Add red plus blue shares: (5+7)/(12)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 70 and 98, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

53. Part-to-whole comparison 24

Question. A hypothetical box contains 84 red and 56 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

Worked answer. Red:blue is 84:56, equivalent to 6:4 and simplified fully to 3:2. Red fraction of the whole is 84/140=3/5. Red percentage is (84/140)×100%=60% (displayed to a practical number of digits). Its exact value is the fraction 3/5 of 100%.

Common denominator mistake. Writing 6:4 as 6/4 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 2/5. Add red plus blue shares: (6+4)/(10)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 168 and 112, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

54. Part-to-whole comparison 25

Question. A hypothetical box contains 42 red and 6 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 48 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 42:6=7:1. Red fraction of the whole is 42/48=7/8. Red percentage is (42/48)×100%=87.5% (displayed to a practical number of digits). Its exact value is the fraction 7/8 of 100%.

Common denominator mistake. Writing 7:1 as 7/1 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/8. Add red plus blue shares: (7+1)/(8)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 84 and 12, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

55. Part-to-whole comparison 26

Question. A hypothetical box contains 104 red and 78 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 182 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 104:78, equivalent to 8:6 and simplified fully to 4:3. Red fraction of the whole is 104/182=4/7. Red percentage is (104/182)×100%=57.14286% (displayed to a practical number of digits). Its exact value is the fraction 4/7 of 100%.

Common denominator mistake. Writing 8:6 as 8/6 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 3/7. Add red plus blue shares: (8+6)/(14)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 208 and 156, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

56. Part-to-whole comparison 27

Question. A hypothetical box contains 45 red and 15 blue counters. A parent asks for the ratio red:blue, the fraction red out of the whole, and the percentage red. Calculate all three distinctly.

What is being compared? The ratio compares two subgroups. The fraction uses the entire total 60 as denominator. The percentage is that part-of-whole fraction expressed per hundred. These are three representations of the same data but answer three different wordings.

Worked answer. Red:blue is 45:15, equivalent to 9:3 and simplified fully to 3:1. Red fraction of the whole is 45/60=3/4. Red percentage is (45/60)×100%=75% (displayed to a practical number of digits). Its exact value is the fraction 3/4 of 100%.

Common denominator mistake. Writing 9:3 as 9/3 of the whole ignores the blue group in the denominator. A percentage can also be given with too much claimed precision if the original fraction’s decimal expansion repeats; keep the fraction as the exact statement.

Independent check. The blue fraction is 1/4. Add red plus blue shares: (9+3)/(12)=1. The displayed rounded percentages may not add to exactly 100.000000% unless their precision is sufficient; the exact fractions do.

Changed comparison. Change the box total by repeating every counter group twice. Raw counts double to 90 and 30, while the ratio, exact fractions and percentages remain unchanged. Explain why scale and composition are different ideas.

Unit rates: money, mass and time

57. Price per item is a rate 1

Question. A fictitious set of 2 identical supplies costs $4 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 5 items. Is this a ratio or a rate question?

What is being compared? A unit rate compares different kinds of quantity: money divided by number of items. It may be written $/item. A ratio of costs of two purchases would compare quantities with the same unit and would be a different question.

Worked answer. Unit cost is $4/2 items = $2 per item. The new purchase has 5 items, costing 5×$2 = $10. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $4, they have assumed each extra item costs one dollar. If they compare 4:2 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 4/2=2 dollars per item and 10/5=2 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $2 per item remains, but total becomes 5+2n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

58. Price per item is a rate 2

Question. A fictitious set of 7 identical supplies costs $35 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 10 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $35/7 items = $5 per item. The new purchase has 10 items, costing 10×$5 = $50. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $35, they have assumed each extra item costs one dollar. If they compare 35:7 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 35/7=5 dollars per item and 50/10=5 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $5 per item remains, but total becomes 5+5n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

59. Price per item is a rate 3

Question. A fictitious set of 12 identical supplies costs $96 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 15 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $96/12 items = $8 per item. The new purchase has 15 items, costing 15×$8 = $120. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $96, they have assumed each extra item costs one dollar. If they compare 96:12 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 96/12=8 dollars per item and 120/15=8 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $8 per item remains, but total becomes 5+8n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

60. Price per item is a rate 4

Question. A fictitious set of 17 identical supplies costs $187 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 20 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $187/17 items = $11 per item. The new purchase has 20 items, costing 20×$11 = $220. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $187, they have assumed each extra item costs one dollar. If they compare 187:17 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 187/17=11 dollars per item and 220/20=11 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $11 per item remains, but total becomes 5+11n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

61. Price per item is a rate 5

Question. A fictitious set of 22 identical supplies costs $308 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 25 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $308/22 items = $14 per item. The new purchase has 25 items, costing 25×$14 = $350. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $308, they have assumed each extra item costs one dollar. If they compare 308:22 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 308/22=14 dollars per item and 350/25=14 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $14 per item remains, but total becomes 5+14n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

62. Price per item is a rate 6

Question. A fictitious set of 4 identical supplies costs $68 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 7 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $68/4 items = $17 per item. The new purchase has 7 items, costing 7×$17 = $119. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $68, they have assumed each extra item costs one dollar. If they compare 68:4 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 68/4=17 dollars per item and 119/7=17 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $17 per item remains, but total becomes 5+17n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

63. Price per item is a rate 7

Question. A fictitious set of 9 identical supplies costs $180 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 12 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $180/9 items = $20 per item. The new purchase has 12 items, costing 12×$20 = $240. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $180, they have assumed each extra item costs one dollar. If they compare 180:9 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 180/9=20 dollars per item and 240/12=20 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $20 per item remains, but total becomes 5+20n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

64. Price per item is a rate 8

Question. A fictitious set of 14 identical supplies costs $56 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 17 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $56/14 items = $4 per item. The new purchase has 17 items, costing 17×$4 = $68. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $56, they have assumed each extra item costs one dollar. If they compare 56:14 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 56/14=4 dollars per item and 68/17=4 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $4 per item remains, but total becomes 5+4n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

65. Price per item is a rate 9

Question. A fictitious set of 19 identical supplies costs $133 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 22 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $133/19 items = $7 per item. The new purchase has 22 items, costing 22×$7 = $154. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $133, they have assumed each extra item costs one dollar. If they compare 133:19 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 133/19=7 dollars per item and 154/22=7 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $7 per item remains, but total becomes 5+7n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

66. Price per item is a rate 10

Question. A fictitious set of 24 identical supplies costs $240 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 27 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $240/24 items = $10 per item. The new purchase has 27 items, costing 27×$10 = $270. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $240, they have assumed each extra item costs one dollar. If they compare 240:24 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 240/24=10 dollars per item and 270/27=10 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $10 per item remains, but total becomes 5+10n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

67. Price per item is a rate 11

Question. A fictitious set of 6 identical supplies costs $78 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 9 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $78/6 items = $13 per item. The new purchase has 9 items, costing 9×$13 = $117. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $78, they have assumed each extra item costs one dollar. If they compare 78:6 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 78/6=13 dollars per item and 117/9=13 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $13 per item remains, but total becomes 5+13n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

68. Price per item is a rate 12

Question. A fictitious set of 11 identical supplies costs $176 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 14 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $176/11 items = $16 per item. The new purchase has 14 items, costing 14×$16 = $224. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $176, they have assumed each extra item costs one dollar. If they compare 176:11 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 176/11=16 dollars per item and 224/14=16 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $16 per item remains, but total becomes 5+16n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

69. Price per item is a rate 13

Question. A fictitious set of 16 identical supplies costs $304 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 19 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $304/16 items = $19 per item. The new purchase has 19 items, costing 19×$19 = $361. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $304, they have assumed each extra item costs one dollar. If they compare 304:16 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 304/16=19 dollars per item and 361/19=19 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $19 per item remains, but total becomes 5+19n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

70. Price per item is a rate 14

Question. A fictitious set of 21 identical supplies costs $63 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 24 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $63/21 items = $3 per item. The new purchase has 24 items, costing 24×$3 = $72. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $63, they have assumed each extra item costs one dollar. If they compare 63:21 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 63/21=3 dollars per item and 72/24=3 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $3 per item remains, but total becomes 5+3n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

71. Price per item is a rate 15

Question. A fictitious set of 3 identical supplies costs $18 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 6 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $18/3 items = $6 per item. The new purchase has 6 items, costing 6×$6 = $36. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $18, they have assumed each extra item costs one dollar. If they compare 18:3 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 18/3=6 dollars per item and 36/6=6 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $6 per item remains, but total becomes 5+6n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

72. Price per item is a rate 16

Question. A fictitious set of 8 identical supplies costs $72 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 11 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $72/8 items = $9 per item. The new purchase has 11 items, costing 11×$9 = $99. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $72, they have assumed each extra item costs one dollar. If they compare 72:8 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 72/8=9 dollars per item and 99/11=9 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $9 per item remains, but total becomes 5+9n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

73. Price per item is a rate 17

Question. A fictitious set of 13 identical supplies costs $156 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 16 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $156/13 items = $12 per item. The new purchase has 16 items, costing 16×$12 = $192. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $156, they have assumed each extra item costs one dollar. If they compare 156:13 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 156/13=12 dollars per item and 192/16=12 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $12 per item remains, but total becomes 5+12n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

74. Price per item is a rate 18

Question. A fictitious set of 18 identical supplies costs $270 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 21 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $270/18 items = $15 per item. The new purchase has 21 items, costing 21×$15 = $315. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $270, they have assumed each extra item costs one dollar. If they compare 270:18 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 270/18=15 dollars per item and 315/21=15 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $15 per item remains, but total becomes 5+15n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

75. Price per item is a rate 19

Question. A fictitious set of 23 identical supplies costs $414 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 26 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $414/23 items = $18 per item. The new purchase has 26 items, costing 26×$18 = $468. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $414, they have assumed each extra item costs one dollar. If they compare 414:23 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 414/23=18 dollars per item and 468/26=18 dollars per item. Units and the fixed-rate assumption match.

Changed comparison. Introduce a hypothetical one-off delivery fee $5. The rate $18 per item remains, but total becomes 5+18n instead of pure direct proportion. Ask which part of the earlier proportional reasoning no longer holds.

76. Price per item is a rate 20

Question. A fictitious set of 5 identical supplies costs $10 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 8 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $10/5 items = $2 per item. The new purchase has 8 items, costing 8×$2 = $16. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $10, they have assumed each extra item costs one dollar. If they compare 10:5 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 10/5=2 dollars per item and 16/8=2 dollars per item. Units and the fixed-rate assumption match.

77. Price per item is a rate 21

Question. A fictitious set of 10 identical supplies costs $50 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 13 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $50/10 items = $5 per item. The new purchase has 13 items, costing 13×$5 = $65. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $50, they have assumed each extra item costs one dollar. If they compare 50:10 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 50/10=5 dollars per item and 65/13=5 dollars per item. Units and the fixed-rate assumption match.

78. Price per item is a rate 22

Question. A fictitious set of 15 identical supplies costs $120 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 18 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $120/15 items = $8 per item. The new purchase has 18 items, costing 18×$8 = $144. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $120, they have assumed each extra item costs one dollar. If they compare 120:15 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 120/15=8 dollars per item and 144/18=8 dollars per item. Units and the fixed-rate assumption match.

79. Price per item is a rate 23

Question. A fictitious set of 20 identical supplies costs $220 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 23 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $220/20 items = $11 per item. The new purchase has 23 items, costing 23×$11 = $253. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $220, they have assumed each extra item costs one dollar. If they compare 220:20 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 220/20=11 dollars per item and 253/23=11 dollars per item. Units and the fixed-rate assumption match.

80. Price per item is a rate 24

Question. A fictitious set of 2 identical supplies costs $28 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 5 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $28/2 items = $14 per item. The new purchase has 5 items, costing 5×$14 = $70. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $28, they have assumed each extra item costs one dollar. If they compare 28:2 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 28/2=14 dollars per item and 70/5=14 dollars per item. Units and the fixed-rate assumption match.

81. Price per item is a rate 25

Question. A fictitious set of 7 identical supplies costs $119 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 10 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $119/7 items = $17 per item. The new purchase has 10 items, costing 10×$17 = $170. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $119, they have assumed each extra item costs one dollar. If they compare 119:7 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 119/7=17 dollars per item and 170/10=17 dollars per item. Units and the fixed-rate assumption match.

82. Price per item is a rate 26

Question. A fictitious set of 12 identical supplies costs $240 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 15 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $240/12 items = $20 per item. The new purchase has 15 items, costing 15×$20 = $300. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $240, they have assumed each extra item costs one dollar. If they compare 240:12 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 240/12=20 dollars per item and 300/15=20 dollars per item. Units and the fixed-rate assumption match.

83. Price per item is a rate 27

Question. A fictitious set of 17 identical supplies costs $68 altogether, with no fixed fee and a constant unit price. Find dollars per item, then the price of 20 items. Is this a ratio or a rate question?

Worked answer. Unit cost is $68/17 items = $4 per item. The new purchase has 20 items, costing 20×$4 = $80. The constant unit price makes cost directly proportional to quantity in this ideal model.

Common denominator mistake. If the learner simply adds three dollars to the original $68, they have assumed each extra item costs one dollar. If they compare 68:17 as a two-group part-to-whole ratio, they lose the compound unit.

Independent check. Check both cost/quantity quotients: 68/17=4 dollars per item and 80/20=4 dollars per item. Units and the fixed-rate assumption match.

Percentage of an actual reference base

84. Percentage depends on the stated whole 1

Question. A made-up amount of $100 is treated as 100%. Find 5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 5%=5/100. The problem defines $100 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $100×(5/100)=$5. The remaining share is 95% of the original, or $95. The same result follows by subtracting 5 from 100.

Common denominator mistake. A pupil may calculate 5% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 5 dollars instead of the computed fraction of $100.

Independent check. Reconstruct the whole: $5+$95=$100. These amounts correspond to shares 5% and 95%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $200, the 5% share becomes $10. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

85. Percentage depends on the stated whole 2

Question. A made-up amount of $120 is treated as 100%. Find 10% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 10%=10/100. The problem defines $120 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $120×(10/100)=$12. The remaining share is 90% of the original, or $108. The same result follows by subtracting 12 from 120.

Common denominator mistake. A pupil may calculate 10% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 10 dollars instead of the computed fraction of $120.

Independent check. Reconstruct the whole: $12+$108=$120. These amounts correspond to shares 10% and 90%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $240, the 10% share becomes $24. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

86. Percentage depends on the stated whole 3

Question. A made-up amount of $140 is treated as 100%. Find 12.5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 12.5%=12.5/100. The problem defines $140 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $140×(12.5/100)=$17.5. The remaining share is 87.5% of the original, or $122.5. The same result follows by subtracting 17.5 from 140.

Common denominator mistake. A pupil may calculate 12.5% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 12.5 dollars instead of the computed fraction of $140.

Independent check. Reconstruct the whole: $17.5+$122.5=$140. These amounts correspond to shares 12.5% and 87.5%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $280, the 12.5% share becomes $35. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

87. Percentage depends on the stated whole 4

Question. A made-up amount of $160 is treated as 100%. Find 15% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 15%=15/100. The problem defines $160 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $160×(15/100)=$24. The remaining share is 85% of the original, or $136. The same result follows by subtracting 24 from 160.

Common denominator mistake. A pupil may calculate 15% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 15 dollars instead of the computed fraction of $160.

Independent check. Reconstruct the whole: $24+$136=$160. These amounts correspond to shares 15% and 85%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $320, the 15% share becomes $48. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

88. Percentage depends on the stated whole 5

Question. A made-up amount of $180 is treated as 100%. Find 20% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 20%=20/100. The problem defines $180 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $180×(20/100)=$36. The remaining share is 80% of the original, or $144. The same result follows by subtracting 36 from 180.

Common denominator mistake. A pupil may calculate 20% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 20 dollars instead of the computed fraction of $180.

Independent check. Reconstruct the whole: $36+$144=$180. These amounts correspond to shares 20% and 80%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $360, the 20% share becomes $72. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

89. Percentage depends on the stated whole 6

Question. A made-up amount of $200 is treated as 100%. Find 25% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 25%=25/100. The problem defines $200 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $200×(25/100)=$50. The remaining share is 75% of the original, or $150. The same result follows by subtracting 50 from 200.

Common denominator mistake. A pupil may calculate 25% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 25 dollars instead of the computed fraction of $200.

Independent check. Reconstruct the whole: $50+$150=$200. These amounts correspond to shares 25% and 75%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $400, the 25% share becomes $100. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

90. Percentage depends on the stated whole 7

Question. A made-up amount of $220 is treated as 100%. Find 30% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 30%=30/100. The problem defines $220 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $220×(30/100)=$66. The remaining share is 70% of the original, or $154. The same result follows by subtracting 66 from 220.

Common denominator mistake. A pupil may calculate 30% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 30 dollars instead of the computed fraction of $220.

Independent check. Reconstruct the whole: $66+$154=$220. These amounts correspond to shares 30% and 70%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $440, the 30% share becomes $132. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

91. Percentage depends on the stated whole 8

Question. A made-up amount of $240 is treated as 100%. Find 40% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 40%=40/100. The problem defines $240 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $240×(40/100)=$96. The remaining share is 60% of the original, or $144. The same result follows by subtracting 96 from 240.

Common denominator mistake. A pupil may calculate 40% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 40 dollars instead of the computed fraction of $240.

Independent check. Reconstruct the whole: $96+$144=$240. These amounts correspond to shares 40% and 60%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $480, the 40% share becomes $192. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

92. Percentage depends on the stated whole 9

Question. A made-up amount of $260 is treated as 100%. Find 50% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 50%=50/100. The problem defines $260 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $260×(50/100)=$130. The remaining share is 50% of the original, or $130. The same result follows by subtracting 130 from 260.

Common denominator mistake. A pupil may calculate 50% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 50 dollars instead of the computed fraction of $260.

Independent check. Reconstruct the whole: $130+$130=$260. These amounts correspond to shares 50% and 50%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $520, the 50% share becomes $260. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

93. Percentage depends on the stated whole 10

Question. A made-up amount of $280 is treated as 100%. Find 60% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 60%=60/100. The problem defines $280 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $280×(60/100)=$168. The remaining share is 40% of the original, or $112. The same result follows by subtracting 168 from 280.

Common denominator mistake. A pupil may calculate 60% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 60 dollars instead of the computed fraction of $280.

Independent check. Reconstruct the whole: $168+$112=$280. These amounts correspond to shares 60% and 40%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $560, the 60% share becomes $336. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

94. Percentage depends on the stated whole 11

Question. A made-up amount of $300 is treated as 100%. Find 75% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 75%=75/100. The problem defines $300 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $300×(75/100)=$225. The remaining share is 25% of the original, or $75. The same result follows by subtracting 225 from 300.

Common denominator mistake. A pupil may calculate 75% of the reduced remainder when the question actually asks for a share of the original. That changes the base. Another may subtract 75 dollars instead of the computed fraction of $300.

Independent check. Reconstruct the whole: $225+$75=$300. These amounts correspond to shares 75% and 25%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $600, the 75% share becomes $450. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

95. Percentage depends on the stated whole 12

Question. A made-up amount of $320 is treated as 100%. Find 5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 5%=5/100. The problem defines $320 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $320×(5/100)=$16. The remaining share is 95% of the original, or $304. The same result follows by subtracting 16 from 320.

Independent check. Reconstruct the whole: $16+$304=$320. These amounts correspond to shares 5% and 95%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $640, the 5% share becomes $32. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

96. Percentage depends on the stated whole 13

Question. A made-up amount of $340 is treated as 100%. Find 10% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 10%=10/100. The problem defines $340 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $340×(10/100)=$34. The remaining share is 90% of the original, or $306. The same result follows by subtracting 34 from 340.

Independent check. Reconstruct the whole: $34+$306=$340. These amounts correspond to shares 10% and 90%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $680, the 10% share becomes $68. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

97. Percentage depends on the stated whole 14

Question. A made-up amount of $360 is treated as 100%. Find 12.5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 12.5%=12.5/100. The problem defines $360 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $360×(12.5/100)=$45. The remaining share is 87.5% of the original, or $315. The same result follows by subtracting 45 from 360.

Independent check. Reconstruct the whole: $45+$315=$360. These amounts correspond to shares 12.5% and 87.5%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $720, the 12.5% share becomes $90. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

98. Percentage depends on the stated whole 15

Question. A made-up amount of $380 is treated as 100%. Find 15% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 15%=15/100. The problem defines $380 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $380×(15/100)=$57. The remaining share is 85% of the original, or $323. The same result follows by subtracting 57 from 380.

Independent check. Reconstruct the whole: $57+$323=$380. These amounts correspond to shares 15% and 85%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $760, the 15% share becomes $114. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

99. Percentage depends on the stated whole 16

Question. A made-up amount of $400 is treated as 100%. Find 20% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 20%=20/100. The problem defines $400 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $400×(20/100)=$80. The remaining share is 80% of the original, or $320. The same result follows by subtracting 80 from 400.

Independent check. Reconstruct the whole: $80+$320=$400. These amounts correspond to shares 20% and 80%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $800, the 20% share becomes $160. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

100. Percentage depends on the stated whole 17

Question. A made-up amount of $420 is treated as 100%. Find 25% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 25%=25/100. The problem defines $420 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $420×(25/100)=$105. The remaining share is 75% of the original, or $315. The same result follows by subtracting 105 from 420.

Independent check. Reconstruct the whole: $105+$315=$420. These amounts correspond to shares 25% and 75%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $840, the 25% share becomes $210. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

101. Percentage depends on the stated whole 18

Question. A made-up amount of $440 is treated as 100%. Find 30% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 30%=30/100. The problem defines $440 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $440×(30/100)=$132. The remaining share is 70% of the original, or $308. The same result follows by subtracting 132 from 440.

Independent check. Reconstruct the whole: $132+$308=$440. These amounts correspond to shares 30% and 70%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $880, the 30% share becomes $264. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

102. Percentage depends on the stated whole 19

Question. A made-up amount of $460 is treated as 100%. Find 40% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 40%=40/100. The problem defines $460 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $460×(40/100)=$184. The remaining share is 60% of the original, or $276. The same result follows by subtracting 184 from 460.

Independent check. Reconstruct the whole: $184+$276=$460. These amounts correspond to shares 40% and 60%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $920, the 40% share becomes $368. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

103. Percentage depends on the stated whole 20

Question. A made-up amount of $480 is treated as 100%. Find 50% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 50%=50/100. The problem defines $480 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $480×(50/100)=$240. The remaining share is 50% of the original, or $240. The same result follows by subtracting 240 from 480.

Independent check. Reconstruct the whole: $240+$240=$480. These amounts correspond to shares 50% and 50%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $960, the 50% share becomes $480. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

104. Percentage depends on the stated whole 21

Question. A made-up amount of $500 is treated as 100%. Find 60% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 60%=60/100. The problem defines $500 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $500×(60/100)=$300. The remaining share is 40% of the original, or $200. The same result follows by subtracting 300 from 500.

Independent check. Reconstruct the whole: $300+$200=$500. These amounts correspond to shares 60% and 40%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1000, the 60% share becomes $600. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

105. Percentage depends on the stated whole 22

Question. A made-up amount of $520 is treated as 100%. Find 75% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 75%=75/100. The problem defines $520 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $520×(75/100)=$390. The remaining share is 25% of the original, or $130. The same result follows by subtracting 390 from 520.

Independent check. Reconstruct the whole: $390+$130=$520. These amounts correspond to shares 75% and 25%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1040, the 75% share becomes $780. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

106. Percentage depends on the stated whole 23

Question. A made-up amount of $540 is treated as 100%. Find 5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 5%=5/100. The problem defines $540 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $540×(5/100)=$27. The remaining share is 95% of the original, or $513. The same result follows by subtracting 27 from 540.

Independent check. Reconstruct the whole: $27+$513=$540. These amounts correspond to shares 5% and 95%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1080, the 5% share becomes $54. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

107. Percentage depends on the stated whole 24

Question. A made-up amount of $560 is treated as 100%. Find 10% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 10%=10/100. The problem defines $560 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $560×(10/100)=$56. The remaining share is 90% of the original, or $504. The same result follows by subtracting 56 from 560.

Independent check. Reconstruct the whole: $56+$504=$560. These amounts correspond to shares 10% and 90%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1120, the 10% share becomes $112. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

108. Percentage depends on the stated whole 25

Question. A made-up amount of $580 is treated as 100%. Find 12.5% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 12.5%=12.5/100. The problem defines $580 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $580×(12.5/100)=$72.5. The remaining share is 87.5% of the original, or $507.5. The same result follows by subtracting 72.5 from 580.

Independent check. Reconstruct the whole: $72.5+$507.5=$580. These amounts correspond to shares 12.5% and 87.5%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1160, the 12.5% share becomes $145. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

109. Percentage depends on the stated whole 26

Question. A made-up amount of $600 is treated as 100%. Find 15% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 15%=15/100. The problem defines $600 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $600×(15/100)=$90. The remaining share is 85% of the original, or $510. The same result follows by subtracting 90 from 600.

Independent check. Reconstruct the whole: $90+$510=$600. These amounts correspond to shares 15% and 85%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1200, the 15% share becomes $180. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

110. Percentage depends on the stated whole 27

Question. A made-up amount of $620 is treated as 100%. Find 20% of the amount and the remainder after removing that share. Say explicitly which amount is the reference whole.

What is being compared? A percentage is not an extra physical unit or a fixed dollar saving. It is a fraction of a chosen base: 20%=20/100. The problem defines $620 as that base, so a percentage calculation must begin there.

Worked answer. The specified share is $620×(20/100)=$124. The remaining share is 80% of the original, or $496. The same result follows by subtracting 124 from 620.

Independent check. Reconstruct the whole: $124+$496=$620. These amounts correspond to shares 20% and 80%, totaling 100%.

Changed comparison. If the stated reference amount doubles to $1240, the 20% share becomes $248. The percentage stays fixed but the actual amount changes, proving that ‘percent’ needs a base.

Percentage change versus percentage points

111. Percentage points are not percent change 1

Question. A hypothetical participation rate rises from 5% to 7%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

What is being compared? A percentage-point difference subtracts the two printed percentages directly. A relative percent change divides the change by the old percentage as the reference value. They have different denominators and describe different aspects of the same change.

Worked answer. The percentage-point increase is 7−5=2 percentage points. Relative increase from the old rate is (2/5)×100%=40%. If the rate were to fall back from 7% to 5%, the relative decrease would be 28.57143%, not necessarily 40%.

Common denominator mistake. Reporting a 2% relative increase simply because the rate changed by 2 printed points misuses the base. Similarly, using new rate 7% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 5% + 2 percentage points = 7%. The relative growth factor is 7/5, which equals 1+(2/5).

Changed comparison. Suppose another fictional report rose from 5% to 6%. The point change would be 1 and the relative increase 20%. Say which of the two measures a headline actually uses.

112. Percentage points are not percent change 2

Question. A hypothetical participation rate rises from 12% to 17%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 17−12=5 percentage points. Relative increase from the old rate is (5/12)×100%=41.66667%. If the rate were to fall back from 17% to 12%, the relative decrease would be 29.41176%, not necessarily 41.66667%.

Common denominator mistake. Reporting a 5% relative increase simply because the rate changed by 5 printed points misuses the base. Similarly, using new rate 17% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 12% + 5 percentage points = 17%. The relative growth factor is 17/12, which equals 1+(5/12).

Changed comparison. Suppose another fictional report rose from 12% to 14.5%. The point change would be 2.5 and the relative increase 20.83333%. Say which of the two measures a headline actually uses.

113. Percentage points are not percent change 3

Question. A hypothetical participation rate rises from 19% to 27%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 27−19=8 percentage points. Relative increase from the old rate is (8/19)×100%=42.10526%. If the rate were to fall back from 27% to 19%, the relative decrease would be 29.62963%, not necessarily 42.10526%.

Common denominator mistake. Reporting a 8% relative increase simply because the rate changed by 8 printed points misuses the base. Similarly, using new rate 27% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 19% + 8 percentage points = 27%. The relative growth factor is 27/19, which equals 1+(8/19).

Changed comparison. Suppose another fictional report rose from 19% to 23%. The point change would be 4 and the relative increase 21.05263%. Say which of the two measures a headline actually uses.

114. Percentage points are not percent change 4

Question. A hypothetical participation rate rises from 26% to 37%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 37−26=11 percentage points. Relative increase from the old rate is (11/26)×100%=42.30769%. If the rate were to fall back from 37% to 26%, the relative decrease would be 29.72973%, not necessarily 42.30769%.

Common denominator mistake. Reporting a 11% relative increase simply because the rate changed by 11 printed points misuses the base. Similarly, using new rate 37% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 26% + 11 percentage points = 37%. The relative growth factor is 37/26, which equals 1+(11/26).

Changed comparison. Suppose another fictional report rose from 26% to 31.5%. The point change would be 5.5 and the relative increase 21.15385%. Say which of the two measures a headline actually uses.

115. Percentage points are not percent change 5

Question. A hypothetical participation rate rises from 33% to 47%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 47−33=14 percentage points. Relative increase from the old rate is (14/33)×100%=42.42424%. If the rate were to fall back from 47% to 33%, the relative decrease would be 29.78723%, not necessarily 42.42424%.

Common denominator mistake. Reporting a 14% relative increase simply because the rate changed by 14 printed points misuses the base. Similarly, using new rate 47% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 33% + 14 percentage points = 47%. The relative growth factor is 47/33, which equals 1+(14/33).

Changed comparison. Suppose another fictional report rose from 33% to 40%. The point change would be 7 and the relative increase 21.21212%. Say which of the two measures a headline actually uses.

116. Percentage points are not percent change 6

Question. A hypothetical participation rate rises from 40% to 57%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 57−40=17 percentage points. Relative increase from the old rate is (17/40)×100%=42.5%. If the rate were to fall back from 57% to 40%, the relative decrease would be 29.82456%, not necessarily 42.5%.

Common denominator mistake. Reporting a 17% relative increase simply because the rate changed by 17 printed points misuses the base. Similarly, using new rate 57% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 40% + 17 percentage points = 57%. The relative growth factor is 57/40, which equals 1+(17/40).

Changed comparison. Suppose another fictional report rose from 40% to 48.5%. The point change would be 8.5 and the relative increase 21.25%. Say which of the two measures a headline actually uses.

117. Percentage points are not percent change 7

Question. A hypothetical participation rate rises from 47% to 51%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 51−47=4 percentage points. Relative increase from the old rate is (4/47)×100%=8.510638%. If the rate were to fall back from 51% to 47%, the relative decrease would be 7.843137%, not necessarily 8.510638%.

Common denominator mistake. Reporting a 4% relative increase simply because the rate changed by 4 printed points misuses the base. Similarly, using new rate 51% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 47% + 4 percentage points = 51%. The relative growth factor is 51/47, which equals 1+(4/47).

Changed comparison. Suppose another fictional report rose from 47% to 49%. The point change would be 2 and the relative increase 4.255319%. Say which of the two measures a headline actually uses.

118. Percentage points are not percent change 8

Question. A hypothetical participation rate rises from 54% to 61%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 61−54=7 percentage points. Relative increase from the old rate is (7/54)×100%=12.96296%. If the rate were to fall back from 61% to 54%, the relative decrease would be 11.47541%, not necessarily 12.96296%.

Common denominator mistake. Reporting a 7% relative increase simply because the rate changed by 7 printed points misuses the base. Similarly, using new rate 61% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 54% + 7 percentage points = 61%. The relative growth factor is 61/54, which equals 1+(7/54).

Changed comparison. Suppose another fictional report rose from 54% to 57.5%. The point change would be 3.5 and the relative increase 6.481481%. Say which of the two measures a headline actually uses.

119. Percentage points are not percent change 9

Question. A hypothetical participation rate rises from 61% to 71%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 71−61=10 percentage points. Relative increase from the old rate is (10/61)×100%=16.39344%. If the rate were to fall back from 71% to 61%, the relative decrease would be 14.08451%, not necessarily 16.39344%.

Common denominator mistake. Reporting a 10% relative increase simply because the rate changed by 10 printed points misuses the base. Similarly, using new rate 71% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 61% + 10 percentage points = 71%. The relative growth factor is 71/61, which equals 1+(10/61).

Changed comparison. Suppose another fictional report rose from 61% to 66%. The point change would be 5 and the relative increase 8.196721%. Say which of the two measures a headline actually uses.

120. Percentage points are not percent change 10

Question. A hypothetical participation rate rises from 68% to 81%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 81−68=13 percentage points. Relative increase from the old rate is (13/68)×100%=19.11765%. If the rate were to fall back from 81% to 68%, the relative decrease would be 16.04938%, not necessarily 19.11765%.

Common denominator mistake. Reporting a 13% relative increase simply because the rate changed by 13 printed points misuses the base. Similarly, using new rate 81% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 68% + 13 percentage points = 81%. The relative growth factor is 81/68, which equals 1+(13/68).

Changed comparison. Suppose another fictional report rose from 68% to 74.5%. The point change would be 6.5 and the relative increase 9.558824%. Say which of the two measures a headline actually uses.

121. Percentage points are not percent change 11

Question. A hypothetical participation rate rises from 10% to 26%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 26−10=16 percentage points. Relative increase from the old rate is (16/10)×100%=160%. If the rate were to fall back from 26% to 10%, the relative decrease would be 61.53846%, not necessarily 160%.

Common denominator mistake. Reporting a 16% relative increase simply because the rate changed by 16 printed points misuses the base. Similarly, using new rate 26% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 10% + 16 percentage points = 26%. The relative growth factor is 26/10, which equals 1+(16/10).

Changed comparison. Suppose another fictional report rose from 10% to 18%. The point change would be 8 and the relative increase 80%. Say which of the two measures a headline actually uses.

122. Percentage points are not percent change 12

Question. A hypothetical participation rate rises from 17% to 20%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 20−17=3 percentage points. Relative increase from the old rate is (3/17)×100%=17.64706%. If the rate were to fall back from 20% to 17%, the relative decrease would be 15%, not necessarily 17.64706%.

Common denominator mistake. Reporting a 3% relative increase simply because the rate changed by 3 printed points misuses the base. Similarly, using new rate 20% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 17% + 3 percentage points = 20%. The relative growth factor is 20/17, which equals 1+(3/17).

Changed comparison. Suppose another fictional report rose from 17% to 18.5%. The point change would be 1.5 and the relative increase 8.823529%. Say which of the two measures a headline actually uses.

123. Percentage points are not percent change 13

Question. A hypothetical participation rate rises from 24% to 30%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 30−24=6 percentage points. Relative increase from the old rate is (6/24)×100%=25%. If the rate were to fall back from 30% to 24%, the relative decrease would be 20%, not necessarily 25%.

Common denominator mistake. Reporting a 6% relative increase simply because the rate changed by 6 printed points misuses the base. Similarly, using new rate 30% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 24% + 6 percentage points = 30%. The relative growth factor is 30/24, which equals 1+(6/24).

Changed comparison. Suppose another fictional report rose from 24% to 27%. The point change would be 3 and the relative increase 12.5%. Say which of the two measures a headline actually uses.

124. Percentage points are not percent change 14

Question. A hypothetical participation rate rises from 31% to 40%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 40−31=9 percentage points. Relative increase from the old rate is (9/31)×100%=29.03226%. If the rate were to fall back from 40% to 31%, the relative decrease would be 22.5%, not necessarily 29.03226%.

Common denominator mistake. Reporting a 9% relative increase simply because the rate changed by 9 printed points misuses the base. Similarly, using new rate 40% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 31% + 9 percentage points = 40%. The relative growth factor is 40/31, which equals 1+(9/31).

Changed comparison. Suppose another fictional report rose from 31% to 35.5%. The point change would be 4.5 and the relative increase 14.51613%. Say which of the two measures a headline actually uses.

125. Percentage points are not percent change 15

Question. A hypothetical participation rate rises from 38% to 50%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 50−38=12 percentage points. Relative increase from the old rate is (12/38)×100%=31.57895%. If the rate were to fall back from 50% to 38%, the relative decrease would be 24%, not necessarily 31.57895%.

Common denominator mistake. Reporting a 12% relative increase simply because the rate changed by 12 printed points misuses the base. Similarly, using new rate 50% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 38% + 12 percentage points = 50%. The relative growth factor is 50/38, which equals 1+(12/38).

Changed comparison. Suppose another fictional report rose from 38% to 44%. The point change would be 6 and the relative increase 15.78947%. Say which of the two measures a headline actually uses.

126. Percentage points are not percent change 16

Question. A hypothetical participation rate rises from 45% to 60%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 60−45=15 percentage points. Relative increase from the old rate is (15/45)×100%=33.33333%. If the rate were to fall back from 60% to 45%, the relative decrease would be 25%, not necessarily 33.33333%.

Common denominator mistake. Reporting a 15% relative increase simply because the rate changed by 15 printed points misuses the base. Similarly, using new rate 60% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 45% + 15 percentage points = 60%. The relative growth factor is 60/45, which equals 1+(15/45).

Changed comparison. Suppose another fictional report rose from 45% to 52.5%. The point change would be 7.5 and the relative increase 16.66667%. Say which of the two measures a headline actually uses.

127. Percentage points are not percent change 17

Question. A hypothetical participation rate rises from 52% to 54%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 54−52=2 percentage points. Relative increase from the old rate is (2/52)×100%=3.846154%. If the rate were to fall back from 54% to 52%, the relative decrease would be 3.703704%, not necessarily 3.846154%.

Common denominator mistake. Reporting a 2% relative increase simply because the rate changed by 2 printed points misuses the base. Similarly, using new rate 54% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 52% + 2 percentage points = 54%. The relative growth factor is 54/52, which equals 1+(2/52).

Changed comparison. Suppose another fictional report rose from 52% to 53%. The point change would be 1 and the relative increase 1.923077%. Say which of the two measures a headline actually uses.

128. Percentage points are not percent change 18

Question. A hypothetical participation rate rises from 59% to 64%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 64−59=5 percentage points. Relative increase from the old rate is (5/59)×100%=8.474576%. If the rate were to fall back from 64% to 59%, the relative decrease would be 7.8125%, not necessarily 8.474576%.

Common denominator mistake. Reporting a 5% relative increase simply because the rate changed by 5 printed points misuses the base. Similarly, using new rate 64% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 59% + 5 percentage points = 64%. The relative growth factor is 64/59, which equals 1+(5/59).

Changed comparison. Suppose another fictional report rose from 59% to 61.5%. The point change would be 2.5 and the relative increase 4.237288%. Say which of the two measures a headline actually uses.

129. Percentage points are not percent change 19

Question. A hypothetical participation rate rises from 66% to 74%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 74−66=8 percentage points. Relative increase from the old rate is (8/66)×100%=12.12121%. If the rate were to fall back from 74% to 66%, the relative decrease would be 10.81081%, not necessarily 12.12121%.

Common denominator mistake. Reporting a 8% relative increase simply because the rate changed by 8 printed points misuses the base. Similarly, using new rate 74% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 66% + 8 percentage points = 74%. The relative growth factor is 74/66, which equals 1+(8/66).

Changed comparison. Suppose another fictional report rose from 66% to 70%. The point change would be 4 and the relative increase 6.060606%. Say which of the two measures a headline actually uses.

130. Percentage points are not percent change 20

Question. A hypothetical participation rate rises from 8% to 19%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 19−8=11 percentage points. Relative increase from the old rate is (11/8)×100%=137.5%. If the rate were to fall back from 19% to 8%, the relative decrease would be 57.89474%, not necessarily 137.5%.

Common denominator mistake. Reporting a 11% relative increase simply because the rate changed by 11 printed points misuses the base. Similarly, using new rate 19% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 8% + 11 percentage points = 19%. The relative growth factor is 19/8, which equals 1+(11/8).

Changed comparison. Suppose another fictional report rose from 8% to 13.5%. The point change would be 5.5 and the relative increase 68.75%. Say which of the two measures a headline actually uses.

131. Percentage points are not percent change 21

Question. A hypothetical participation rate rises from 15% to 29%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 29−15=14 percentage points. Relative increase from the old rate is (14/15)×100%=93.33333%. If the rate were to fall back from 29% to 15%, the relative decrease would be 48.27586%, not necessarily 93.33333%.

Common denominator mistake. Reporting a 14% relative increase simply because the rate changed by 14 printed points misuses the base. Similarly, using new rate 29% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 15% + 14 percentage points = 29%. The relative growth factor is 29/15, which equals 1+(14/15).

Changed comparison. Suppose another fictional report rose from 15% to 22%. The point change would be 7 and the relative increase 46.66667%. Say which of the two measures a headline actually uses.

132. Percentage points are not percent change 22

Question. A hypothetical participation rate rises from 22% to 39%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 39−22=17 percentage points. Relative increase from the old rate is (17/22)×100%=77.27273%. If the rate were to fall back from 39% to 22%, the relative decrease would be 43.58974%, not necessarily 77.27273%.

Common denominator mistake. Reporting a 17% relative increase simply because the rate changed by 17 printed points misuses the base. Similarly, using new rate 39% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 22% + 17 percentage points = 39%. The relative growth factor is 39/22, which equals 1+(17/22).

Changed comparison. Suppose another fictional report rose from 22% to 30.5%. The point change would be 8.5 and the relative increase 38.63636%. Say which of the two measures a headline actually uses.

133. Percentage points are not percent change 23

Question. A hypothetical participation rate rises from 29% to 33%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 33−29=4 percentage points. Relative increase from the old rate is (4/29)×100%=13.7931%. If the rate were to fall back from 33% to 29%, the relative decrease would be 12.12121%, not necessarily 13.7931%.

Common denominator mistake. Reporting a 4% relative increase simply because the rate changed by 4 printed points misuses the base. Similarly, using new rate 33% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 29% + 4 percentage points = 33%. The relative growth factor is 33/29, which equals 1+(4/29).

Changed comparison. Suppose another fictional report rose from 29% to 31%. The point change would be 2 and the relative increase 6.896552%. Say which of the two measures a headline actually uses.

134. Percentage points are not percent change 24

Question. A hypothetical participation rate rises from 36% to 43%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 43−36=7 percentage points. Relative increase from the old rate is (7/36)×100%=19.44444%. If the rate were to fall back from 43% to 36%, the relative decrease would be 16.27907%, not necessarily 19.44444%.

Common denominator mistake. Reporting a 7% relative increase simply because the rate changed by 7 printed points misuses the base. Similarly, using new rate 43% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 36% + 7 percentage points = 43%. The relative growth factor is 43/36, which equals 1+(7/36).

Changed comparison. Suppose another fictional report rose from 36% to 39.5%. The point change would be 3.5 and the relative increase 9.722222%. Say which of the two measures a headline actually uses.

135. Percentage points are not percent change 25

Question. A hypothetical participation rate rises from 43% to 53%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 53−43=10 percentage points. Relative increase from the old rate is (10/43)×100%=23.25581%. If the rate were to fall back from 53% to 43%, the relative decrease would be 18.86792%, not necessarily 23.25581%.

Common denominator mistake. Reporting a 10% relative increase simply because the rate changed by 10 printed points misuses the base. Similarly, using new rate 53% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 43% + 10 percentage points = 53%. The relative growth factor is 53/43, which equals 1+(10/43).

Changed comparison. Suppose another fictional report rose from 43% to 48%. The point change would be 5 and the relative increase 11.62791%. Say which of the two measures a headline actually uses.

136. Percentage points are not percent change 26

Question. A hypothetical participation rate rises from 50% to 63%. Find the increase in percentage points and the relative percentage increase compared with the old participation rate.

Worked answer. The percentage-point increase is 63−50=13 percentage points. Relative increase from the old rate is (13/50)×100%=26%. If the rate were to fall back from 63% to 50%, the relative decrease would be 20.63492%, not necessarily 26%.

Common denominator mistake. Reporting a 13% relative increase simply because the rate changed by 13 printed points misuses the base. Similarly, using new rate 63% as denominator for the original increase answers a different question.

Independent check. Check the original rate plus the point increase: 50% + 13 percentage points = 63%. The relative growth factor is 63/50, which equals 1+(13/50).

Changed comparison. Suppose another fictional report rose from 50% to 56.5%. The point change would be 6.5 and the relative increase 13%. Say which of the two measures a headline actually uses.

Comparing groups with unequal sample sizes

137. Compare unequal group sizes 1

Question. Two fictional survey groups report 3 favourable responses out of 20 and 3 favourable responses out of 35, respectively. Compare the count and the share, and say why those can give different impressions.

What is being compared? The numerator gives the count, while the denominator is the total eligible responses in each group. Comparing raw counts answers ‘which group observed more favourable responses?’ Comparing rates or percentages answers ‘which share of its own group was favourable?’ The two questions are distinct.

Worked answer. Group A has count 3 and percentage 15%. Group B has count 3 and percentage 8.571429%. Count comparison: tied. Share comparison: A has the larger percentage, with exact shares 3/20 and 3/35.

Common denominator mistake. A pupil may choose the larger raw count without checking total group sizes, or compare rounded percentages so aggressively that a small real difference is hidden. Keep the exact fractions to support the conclusion before display rounding.

Independent check. For Group A, 3/20×100≈15%; for B, 3/35×100≈8.571429%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 6 out of 40. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

138. Compare unequal group sizes 2

Question. Two fictional survey groups report 6 favourable responses out of 25 and 10 favourable responses out of 44, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 6 and percentage 24%. Group B has count 10 and percentage 22.72727%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 6/25 and 5/22.

Independent check. For Group A, 6/25×100≈24%; for B, 10/44×100≈22.72727%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 12 out of 50. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

139. Compare unequal group sizes 3

Question. Two fictional survey groups report 9 favourable responses out of 30 and 17 favourable responses out of 53, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 9 and percentage 30%. Group B has count 17 and percentage 32.07547%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 3/10 and 17/53.

Independent check. For Group A, 9/30×100≈30%; for B, 17/53×100≈32.07547%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 18 out of 60. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

140. Compare unequal group sizes 4

Question. Two fictional survey groups report 12 favourable responses out of 35 and 24 favourable responses out of 62, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 12 and percentage 34.28571%. Group B has count 24 and percentage 38.70968%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 12/35 and 12/31.

Independent check. For Group A, 12/35×100≈34.28571%; for B, 24/62×100≈38.70968%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 24 out of 70. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

141. Compare unequal group sizes 5

Question. Two fictional survey groups report 15 favourable responses out of 40 and 6 favourable responses out of 71, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 15 and percentage 37.5%. Group B has count 6 and percentage 8.450704%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 3/8 and 6/71.

Independent check. For Group A, 15/40×100≈37.5%; for B, 6/71×100≈8.450704%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 30 out of 80. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

142. Compare unequal group sizes 6

Question. Two fictional survey groups report 18 favourable responses out of 45 and 13 favourable responses out of 80, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 18 and percentage 40%. Group B has count 13 and percentage 16.25%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 2/5 and 13/80.

Independent check. For Group A, 18/45×100≈40%; for B, 13/80×100≈16.25%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 36 out of 90. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

143. Compare unequal group sizes 7

Question. Two fictional survey groups report 21 favourable responses out of 50 and 20 favourable responses out of 89, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 21 and percentage 42%. Group B has count 20 and percentage 22.47191%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 21/50 and 20/89.

Independent check. For Group A, 21/50×100≈42%; for B, 20/89×100≈22.47191%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 42 out of 100. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

144. Compare unequal group sizes 8

Question. Two fictional survey groups report 5 favourable responses out of 55 and 27 favourable responses out of 98, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 5 and percentage 9.090909%. Group B has count 27 and percentage 27.55102%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 1/11 and 27/98.

Independent check. For Group A, 5/55×100≈9.090909%; for B, 27/98×100≈27.55102%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 10 out of 110. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

145. Compare unequal group sizes 9

Question. Two fictional survey groups report 8 favourable responses out of 60 and 9 favourable responses out of 107, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 8 and percentage 13.33333%. Group B has count 9 and percentage 8.411215%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 2/15 and 9/107.

Independent check. For Group A, 8/60×100≈13.33333%; for B, 9/107×100≈8.411215%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 16 out of 120. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

146. Compare unequal group sizes 10

Question. Two fictional survey groups report 11 favourable responses out of 65 and 16 favourable responses out of 116, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 11 and percentage 16.92308%. Group B has count 16 and percentage 13.7931%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 11/65 and 4/29.

Independent check. For Group A, 11/65×100≈16.92308%; for B, 16/116×100≈13.7931%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 22 out of 130. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

147. Compare unequal group sizes 11

Question. Two fictional survey groups report 14 favourable responses out of 70 and 23 favourable responses out of 125, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 14 and percentage 20%. Group B has count 23 and percentage 18.4%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 1/5 and 23/125.

Independent check. For Group A, 14/70×100≈20%; for B, 23/125×100≈18.4%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 28 out of 140. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

148. Compare unequal group sizes 12

Question. Two fictional survey groups report 17 favourable responses out of 75 and 5 favourable responses out of 134, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 17 and percentage 22.66667%. Group B has count 5 and percentage 3.731343%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 17/75 and 5/134.

Independent check. For Group A, 17/75×100≈22.66667%; for B, 5/134×100≈3.731343%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 34 out of 150. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

149. Compare unequal group sizes 13

Question. Two fictional survey groups report 20 favourable responses out of 80 and 12 favourable responses out of 143, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 20 and percentage 25%. Group B has count 12 and percentage 8.391608%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 1/4 and 12/143.

Independent check. For Group A, 20/80×100≈25%; for B, 12/143×100≈8.391608%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 40 out of 160. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

150. Compare unequal group sizes 14

Question. Two fictional survey groups report 4 favourable responses out of 85 and 19 favourable responses out of 41, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 4 and percentage 4.705882%. Group B has count 19 and percentage 46.34146%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 4/85 and 19/41.

Independent check. For Group A, 4/85×100≈4.705882%; for B, 19/41×100≈46.34146%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 8 out of 170. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

151. Compare unequal group sizes 15

Question. Two fictional survey groups report 7 favourable responses out of 90 and 26 favourable responses out of 50, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 7 and percentage 7.777778%. Group B has count 26 and percentage 52%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 7/90 and 13/25.

Independent check. For Group A, 7/90×100≈7.777778%; for B, 26/50×100≈52%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 14 out of 180. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

152. Compare unequal group sizes 16

Question. Two fictional survey groups report 10 favourable responses out of 95 and 8 favourable responses out of 59, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 10 and percentage 10.52632%. Group B has count 8 and percentage 13.55932%. Count comparison: A has the larger count. Share comparison: B has the larger percentage, with exact shares 2/19 and 8/59.

Independent check. For Group A, 10/95×100≈10.52632%; for B, 8/59×100≈13.55932%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 20 out of 190. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

153. Compare unequal group sizes 17

Question. Two fictional survey groups report 13 favourable responses out of 100 and 15 favourable responses out of 68, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 13 and percentage 13%. Group B has count 15 and percentage 22.05882%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 13/100 and 15/68.

Independent check. For Group A, 13/100×100≈13%; for B, 15/68×100≈22.05882%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 26 out of 200. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

154. Compare unequal group sizes 18

Question. Two fictional survey groups report 16 favourable responses out of 24 and 22 favourable responses out of 77, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 16 and percentage 66.66667%. Group B has count 22 and percentage 28.57143%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 2/3 and 2/7.

Independent check. For Group A, 16/24×100≈66.66667%; for B, 22/77×100≈28.57143%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 32 out of 48. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

155. Compare unequal group sizes 19

Question. Two fictional survey groups report 19 favourable responses out of 29 and 4 favourable responses out of 86, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 19 and percentage 65.51724%. Group B has count 4 and percentage 4.651163%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 19/29 and 2/43.

Independent check. For Group A, 19/29×100≈65.51724%; for B, 4/86×100≈4.651163%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 38 out of 58. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

156. Compare unequal group sizes 20

Question. Two fictional survey groups report 3 favourable responses out of 34 and 11 favourable responses out of 95, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 3 and percentage 8.823529%. Group B has count 11 and percentage 11.57895%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 3/34 and 11/95.

Independent check. For Group A, 3/34×100≈8.823529%; for B, 11/95×100≈11.57895%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 6 out of 68. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

157. Compare unequal group sizes 21

Question. Two fictional survey groups report 6 favourable responses out of 39 and 18 favourable responses out of 104, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 6 and percentage 15.38462%. Group B has count 18 and percentage 17.30769%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 2/13 and 9/52.

Independent check. For Group A, 6/39×100≈15.38462%; for B, 18/104×100≈17.30769%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 12 out of 78. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

158. Compare unequal group sizes 22

Question. Two fictional survey groups report 9 favourable responses out of 44 and 25 favourable responses out of 113, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 9 and percentage 20.45455%. Group B has count 25 and percentage 22.12389%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 9/44 and 25/113.

Independent check. For Group A, 9/44×100≈20.45455%; for B, 25/113×100≈22.12389%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 18 out of 88. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

159. Compare unequal group sizes 23

Question. Two fictional survey groups report 12 favourable responses out of 49 and 7 favourable responses out of 122, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 12 and percentage 24.4898%. Group B has count 7 and percentage 5.737705%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 12/49 and 7/122.

Independent check. For Group A, 12/49×100≈24.4898%; for B, 7/122×100≈5.737705%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 24 out of 98. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

160. Compare unequal group sizes 24

Question. Two fictional survey groups report 15 favourable responses out of 54 and 14 favourable responses out of 131, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 15 and percentage 27.77778%. Group B has count 14 and percentage 10.68702%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 5/18 and 14/131.

Independent check. For Group A, 15/54×100≈27.77778%; for B, 14/131×100≈10.68702%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 30 out of 108. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

161. Compare unequal group sizes 25

Question. Two fictional survey groups report 18 favourable responses out of 59 and 21 favourable responses out of 140, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 18 and percentage 30.50847%. Group B has count 21 and percentage 15%. Count comparison: B has the larger count. Share comparison: A has the larger percentage, with exact shares 18/59 and 3/20.

Independent check. For Group A, 18/59×100≈30.50847%; for B, 21/140×100≈15%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 36 out of 118. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

162. Compare unequal group sizes 26

Question. Two fictional survey groups report 21 favourable responses out of 64 and 3 favourable responses out of 38, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 21 and percentage 32.8125%. Group B has count 3 and percentage 7.894737%. Count comparison: A has the larger count. Share comparison: A has the larger percentage, with exact shares 21/64 and 3/38.

Independent check. For Group A, 21/64×100≈32.8125%; for B, 3/38×100≈7.894737%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 42 out of 128. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

163. Compare unequal group sizes 27

Question. Two fictional survey groups report 5 favourable responses out of 69 and 10 favourable responses out of 47, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 5 and percentage 7.246377%. Group B has count 10 and percentage 21.2766%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 5/69 and 10/47.

Independent check. For Group A, 5/69×100≈7.246377%; for B, 10/47×100≈21.2766%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 10 out of 138. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

164. Compare unequal group sizes 28

Question. Two fictional survey groups report 8 favourable responses out of 74 and 17 favourable responses out of 56, respectively. Compare the count and the share, and say why those can give different impressions.

Worked answer. Group A has count 8 and percentage 10.81081%. Group B has count 17 and percentage 30.35714%. Count comparison: B has the larger count. Share comparison: B has the larger percentage, with exact shares 4/37 and 17/56.

Independent check. For Group A, 8/74×100≈10.81081%; for B, 17/56×100≈30.35714%. Each favourable count is no greater than its own sample total, so the fractions are valid shares.

Changed comparison. Double the size of Group A hypothetically by duplicating its observations: 16 out of 148. Its percentage remains the same even though the count doubles. Explain how this exposes the difference between count and rate.

Direct and inverse proportional models

165. Inverse proportion with a fixed task 1

Question. In an idealised model, 2 equally productive workers complete one perfectly divisible fixed task using 60 hours each. If 4 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 120 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 2×60=120 worker-hours. The doubled workforce needs 120/4=30 hours. The product workers×time stays constant even as each individual factor changes.

Common denominator mistake. A student might double the time because they see twice as many workers, importing a direct-proportion rule into an inverse relationship. Real projects may not obey perfect divisibility; the answer is conditional on the explicitly idealised assumptions.

Independent check. Check 4×30≈120 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 60 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 60 hours. Explain which quantity was held constant in each scenario before choosing a formula.

166. Inverse proportion with a fixed task 2

Question. In an idealised model, 3 equally productive workers complete one perfectly divisible fixed task using 44 hours each. If 6 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 132 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 3×44=132 worker-hours. The doubled workforce needs 132/6=22 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 6×22≈132 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 44 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 44 hours. Explain which quantity was held constant in each scenario before choosing a formula.

167. Inverse proportion with a fixed task 3

Question. In an idealised model, 4 equally productive workers complete one perfectly divisible fixed task using 36 hours each. If 8 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 144 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 4×36=144 worker-hours. The doubled workforce needs 144/8=18 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 8×18≈144 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 36 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 36 hours. Explain which quantity was held constant in each scenario before choosing a formula.

168. Inverse proportion with a fixed task 4

Question. In an idealised model, 5 equally productive workers complete one perfectly divisible fixed task using 31.2 hours each. If 10 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 156 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 5×31.2=156 worker-hours. The doubled workforce needs 156/10=15.6 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 10×15.6≈156 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 31.2 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 31.2 hours. Explain which quantity was held constant in each scenario before choosing a formula.

169. Inverse proportion with a fixed task 5

Question. In an idealised model, 6 equally productive workers complete one perfectly divisible fixed task using 28 hours each. If 12 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 168 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 6×28=168 worker-hours. The doubled workforce needs 168/12=14 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 12×14≈168 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 28 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 28 hours. Explain which quantity was held constant in each scenario before choosing a formula.

170. Inverse proportion with a fixed task 6

Question. In an idealised model, 8 equally productive workers complete one perfectly divisible fixed task using 22.5 hours each. If 16 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 180 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 8×22.5=180 worker-hours. The doubled workforce needs 180/16=11.25 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 16×11.25≈180 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 22.5 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 22.5 hours. Explain which quantity was held constant in each scenario before choosing a formula.

171. Inverse proportion with a fixed task 7

Question. In an idealised model, 10 equally productive workers complete one perfectly divisible fixed task using 19.2 hours each. If 20 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 192 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 10×19.2=192 worker-hours. The doubled workforce needs 192/20=9.6 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 20×9.6≈192 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 19.2 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 19.2 hours. Explain which quantity was held constant in each scenario before choosing a formula.

172. Inverse proportion with a fixed task 8

Question. In an idealised model, 12 equally productive workers complete one perfectly divisible fixed task using 17 hours each. If 24 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 204 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 12×17=204 worker-hours. The doubled workforce needs 204/24=8.5 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 24×8.5≈204 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 17 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 17 hours. Explain which quantity was held constant in each scenario before choosing a formula.

173. Inverse proportion with a fixed task 9

Question. In an idealised model, 2 equally productive workers complete one perfectly divisible fixed task using 108 hours each. If 4 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 216 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 2×108=216 worker-hours. The doubled workforce needs 216/4=54 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 4×54≈216 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 108 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 108 hours. Explain which quantity was held constant in each scenario before choosing a formula.

174. Inverse proportion with a fixed task 10

Question. In an idealised model, 3 equally productive workers complete one perfectly divisible fixed task using 76 hours each. If 6 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 228 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 3×76=228 worker-hours. The doubled workforce needs 228/6=38 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 6×38≈228 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 76 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 76 hours. Explain which quantity was held constant in each scenario before choosing a formula.

175. Inverse proportion with a fixed task 11

Question. In an idealised model, 4 equally productive workers complete one perfectly divisible fixed task using 60 hours each. If 8 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 240 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 4×60=240 worker-hours. The doubled workforce needs 240/8=30 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 8×30≈240 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 60 hours.

176. Inverse proportion with a fixed task 12

Question. In an idealised model, 5 equally productive workers complete one perfectly divisible fixed task using 50.4 hours each. If 10 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 252 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 5×50.4=252 worker-hours. The doubled workforce needs 252/10=25.2 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 10×25.2≈252 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 50.4 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 50.4 hours. Explain which quantity was held constant in each scenario before choosing a formula.

177. Inverse proportion with a fixed task 13

Question. In an idealised model, 6 equally productive workers complete one perfectly divisible fixed task using 44 hours each. If 12 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 264 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 6×44=264 worker-hours. The doubled workforce needs 264/12=22 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 12×22≈264 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 44 hours.

178. Inverse proportion with a fixed task 14

Question. In an idealised model, 8 equally productive workers complete one perfectly divisible fixed task using 34.5 hours each. If 16 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 276 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 8×34.5=276 worker-hours. The doubled workforce needs 276/16=17.25 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 16×17.25≈276 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 34.5 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 34.5 hours. Explain which quantity was held constant in each scenario before choosing a formula.

179. Inverse proportion with a fixed task 15

Question. In an idealised model, 10 equally productive workers complete one perfectly divisible fixed task using 28.8 hours each. If 20 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 288 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 10×28.8=288 worker-hours. The doubled workforce needs 288/20=14.4 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 20×14.4≈288 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 28.8 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 28.8 hours. Explain which quantity was held constant in each scenario before choosing a formula.

180. Inverse proportion with a fixed task 16

Question. In an idealised model, 12 equally productive workers complete one perfectly divisible fixed task using 25 hours each. If 24 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 300 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 12×25=300 worker-hours. The doubled workforce needs 300/24=12.5 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 24×12.5≈300 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 25 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 25 hours. Explain which quantity was held constant in each scenario before choosing a formula.

181. Inverse proportion with a fixed task 17

Question. In an idealised model, 2 equally productive workers complete one perfectly divisible fixed task using 156 hours each. If 4 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 312 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 2×156=312 worker-hours. The doubled workforce needs 312/4=78 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 4×78≈312 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 156 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 156 hours. Explain which quantity was held constant in each scenario before choosing a formula.

182. Inverse proportion with a fixed task 18

Question. In an idealised model, 3 equally productive workers complete one perfectly divisible fixed task using 108 hours each. If 6 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 324 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 3×108=324 worker-hours. The doubled workforce needs 324/6=54 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 6×54≈324 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 108 hours.

183. Inverse proportion with a fixed task 19

Question. In an idealised model, 4 equally productive workers complete one perfectly divisible fixed task using 84 hours each. If 8 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 336 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 4×84=336 worker-hours. The doubled workforce needs 336/8=42 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 8×42≈336 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 84 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 84 hours. Explain which quantity was held constant in each scenario before choosing a formula.

184. Inverse proportion with a fixed task 20

Question. In an idealised model, 5 equally productive workers complete one perfectly divisible fixed task using 69.6 hours each. If 10 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 348 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 5×69.6=348 worker-hours. The doubled workforce needs 348/10=34.8 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 10×34.8≈348 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 69.6 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 69.6 hours. Explain which quantity was held constant in each scenario before choosing a formula.

185. Inverse proportion with a fixed task 21

Question. In an idealised model, 6 equally productive workers complete one perfectly divisible fixed task using 60 hours each. If 12 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 360 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 6×60=360 worker-hours. The doubled workforce needs 360/12=30 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 12×30≈360 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 60 hours.

186. Inverse proportion with a fixed task 22

Question. In an idealised model, 8 equally productive workers complete one perfectly divisible fixed task using 46.5 hours each. If 16 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 372 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 8×46.5=372 worker-hours. The doubled workforce needs 372/16=23.25 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 16×23.25≈372 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 46.5 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 46.5 hours. Explain which quantity was held constant in each scenario before choosing a formula.

187. Inverse proportion with a fixed task 23

Question. In an idealised model, 10 equally productive workers complete one perfectly divisible fixed task using 38.4 hours each. If 20 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 384 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 10×38.4=384 worker-hours. The doubled workforce needs 384/20=19.2 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 20×19.2≈384 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 38.4 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 38.4 hours. Explain which quantity was held constant in each scenario before choosing a formula.

188. Inverse proportion with a fixed task 24

Question. In an idealised model, 12 equally productive workers complete one perfectly divisible fixed task using 33 hours each. If 24 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 396 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 12×33=396 worker-hours. The doubled workforce needs 396/24=16.5 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 24×16.5≈396 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 33 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 33 hours. Explain which quantity was held constant in each scenario before choosing a formula.

189. Inverse proportion with a fixed task 25

Question. In an idealised model, 2 equally productive workers complete one perfectly divisible fixed task using 204 hours each. If 4 identical workers share exactly the same task under the same assumptions, how much time does it take? Is this direct or inverse proportion?

What is being compared? The amount of work is fixed at 408 worker-hours in this theoretical model. Doubling workers halves the required time when every worker contributes equally and there are no coordination or setup losses. This is inverse rather than direct proportion.

Worked answer. The first arrangement has 2×204=408 worker-hours. The doubled workforce needs 408/4=102 hours. The product workers×time stays constant even as each individual factor changes.

Independent check. Check 4×102≈408 worker-hours, allowing for display rounding. The new time is exactly one half the original unrounded 204 hours.

Changed comparison. If the fixed task became twice as much work with the doubled workers, time would return to the original 204 hours. Explain which quantity was held constant in each scenario before choosing a formula.

Choose the representation from the wording

190. Ratio without a total

Question. Red:blue=2:3. Can you state an exact red count without a total or either group count?

What is being compared? Identify the comparison’s numerator, denominator, reference whole and units before choosing a representation. A pair of quantities with identical dimensions may form a ratio; a different-dimension quotient is a rate; a fraction of a stated base may be expressed as a percentage.

Worked answer. No. The ratio gives only relative parts; any positive common scale fits.

Common denominator mistake. Calling two items red automatically treats a unit part as one physical object.

Independent check. Try 2:3 and 20:30 as distinct actual groups with the same ratio.

Changed comparison. Build a changed context that uses the same numbers but asks a different question. A learner who can name the new denominator and units before calculating has understood the distinction rather than memorising one formula.

191. Rate with different units

Question. A journey covers 120 km in two hours. Is 120:2 a dimensionless red:blue-style ratio?

Worked answer. No. It is distance divided by time, giving 60 km/h.

Common denominator mistake. Dropping units hides what the quotient measures.

Independent check. 60 km/h×2h=120 km.

192. Percentage of an unknown whole

Question. A pie chart says 30% choose A but gives no sample size. Can exact count be found?

Worked answer. No. The absolute count is 0.3 times the unknown total.

Common denominator mistake. Calling the count thirty invents a hundred-person sample.

Independent check. If N=20 then six; if N=100 then thirty.

193. Share and rate can look like the same fraction

Question. Six of ten pupils prefer an activity. Is 6/10 a count, a ratio or a share?

Worked answer. It is a part-to-whole fraction and relative frequency 0.6; a separate ratio of prefer:not prefer is 6:4=3:2.

Common denominator mistake. Using the same denominator ten for a part-to-part ratio changes the question.

Independent check. Count four not-prefer to form 6:4.

194. Ratio can be written with the same units

Question. Three litres of juice and five litres of water. State juice:water.

Worked answer. 3:5, a dimensionless comparison of like volume units.

Common denominator mistake. Calling 3 L:5 min a dimensionless juice ratio would compare unlike quantities.

Independent check. Both use litres, so units cancel in a comparison.

195. Unit conversion before rate

Question. A trip is 600 m in two minutes. Find average m/s.

Worked answer. Two minutes is 120 s; speed=600/120=5 m/s.

Common denominator mistake. Dividing 600/2 gives 300 m/min, not m/s.

Independent check. 5m/s×120s=600m.

196. Unit conversion before ratio

Question. One length is 50 cm, another 1 m. Find first:second.

Worked answer. Convert one metre to 100 cm; ratio 50:100=1:2.

Common denominator mistake. Raw 50:1 compares inconsistent units.

Independent check. Convert both to metres: 0.5:1=1:2.

197. Percentage change uses old base

Question. A price goes $80 to $100. Find percentage increase.

Worked answer. Increase $20 relative to $80; 20/80=25%.

Common denominator mistake. Using 20/100=20% chooses new amount as base.

Independent check. 80×1.25=100.

198. Percentage decrease can differ on reversal

Question. A price falls $100 to $80. Find percentage decrease.

Worked answer. Decrease $20 relative to $100; 20%.

Common denominator mistake. Assuming reverse percentage must equal earlier 25% increase.

Independent check. 100×0.8=80.

199. Percentage point comparison

Question. A survey rate rises from 20% to 30%. Find point and relative increase.

Worked answer. 10 percentage points, but (10/20)×100=50% relative increase.

Common denominator mistake. Calling both ten percent confuses point difference with old-base comparison.

Independent check. 20%×1.5=30%.

200. Average speed versus instantaneous speed

Question. A hypothetical vehicle covers 100 km in two hours. Must it travel constantly at 50 km/h?

Worked answer. No. Fifty is average speed; instantaneous speed could vary within the interval.

Common denominator mistake. Treating distance/time as a guarantee about every instant.

Independent check. Intermediate journey may have different speeds and rest periods.

201. Rate vs percentage

Question. A subscription costs $5 per week and a separate report states 5% growth. Do these two ‘fives’ have the same meaning?

Worked answer. No. $5/week is a compound rate with units; 5% is a proportion relative to a reference quantity.

Common denominator mistake. The repeated numeral does not make the quantities comparable.

Independent check. Ask for denominator and units.

202. Direct proportion test

Question. Does C=5n+3 describe direct proportion between number of items and cost?

Worked answer. No. A nonzero intercept is a fixed charge, so C/n is not constant over positive n.

Common denominator mistake. Every increasing straight-line model is not a direct proportion.

Independent check. At n=0, C=3 instead of zero.

203. Inverse proportion test

Question. If workers double and time halves for a fixed perfect task, what stays constant?

Worked answer. Workers×time, the product, not workers/time.

Common denominator mistake. Calling it direct because the story contains a doubling factor.

Independent check. 2×10=4×5.

204. Different total sizes

Question. One group has 20 favourable of 40, another 30 of 100. Which share is higher?

Worked answer. First 50%, second 30%, even though second has larger favourable count.

Common denominator mistake. Comparing counts alone answers the wrong quantity.

Independent check. Convert each to its own percentage.

205. Same ratio, different sizes

Question. Are 2:3 and 8:12 equivalent?

Worked answer. Yes, each simplifies to 2:3, but actual amounts may differ.

Common denominator mistake. Assuming ratios being equal forces raw counts to be equal.

Independent check. 8/12=2/3 as first-to-second quotient.

206. Part-to-whole and odds

Question. A game has two winning and three losing equally likely outcomes. What are win:lose odds and P(win)?

Worked answer. Win:lose ratio 2:3; probability of winning 2/5.

Common denominator mistake. Writing probability 2/3 confuses part-to-part with part-to-whole.

Independent check. Favourable plus unfavourable is five.

207. Density as compound rate

Question. An ideal uniform material has mass 18 g and volume 6 cm³. Find mass per volume.

Worked answer. 3 g/cm³.

Common denominator mistake. Calling three grams without cubic-centimetre denominator loses the unit.

Independent check. 3 g/cm³×6cm³=18g.

208. Percentage of another quantity may exceed 100

Question. Quantity A=15 and B=10. What percent is A of B?

Worked answer. 15/10×100=150%.

Common denominator mistake. Assuming percentages cannot exceed 100 in every comparison.

Independent check. 150% of 10=15.

209. A share of a fixed whole cannot exceed 100

Question. Can 150% be one category’s share of a complete one-choice survey?

Worked answer. No. Its proportion of that same whole cannot exceed one.

Common denominator mistake. Confusing percentage comparison with percentage share.

Independent check. Valid mutually exclusive survey shares sum to 100%.

210. Rates can be negative in signed contexts

Question. Temperature falls by 5 °C over 2 min. What is the average rate of temperature change?

Worked answer. −2.5 °C/min.

Common denominator mistake. Reporting +2.5 because only absolute magnitude was used.

Independent check. Final minus initial divided by elapsed time is negative.

211. Favourable count vs share

Question. A class has 18 of 30 voting A. Give count, fraction and percentage.

Worked answer. Count 18; share 18/30=3/5; percentage 60%.

Common denominator mistake. A fraction and a raw count are different answer types.

Independent check. 3/5×30=18.

212. Percent is per hundred

Question. What does 12.5% mean as a fraction?

Worked answer. 12.5/100=1/8.

Common denominator mistake. Writing 12.5 as though it were the probability or fraction itself.

Independent check. One eighth of 80 is 10.

213. Compare two unit costs

Question. Four identical pens cost $6; five of a different pen cost $9. Which has lower unit price?

Worked answer. First $1.50 each, second $1.80 each; first cheaper.

Common denominator mistake. Comparing totals $6 and $9 without counts is insufficient as a per-unit comparison.

Independent check. Four×1.50=6, five×1.80=9.

214. Percent of a number and percentage change are different

Question. What is 20% of $80, and what is the % increase from $80 to $100?

Worked answer. The part is $16; the increase is $20/$80=25%.

Common denominator mistake. Using twenty percent to report the change simply because the increase is $20.

Independent check. Different quantities require different denominators.

215. One value without a denominator

Question. A report says ‘twelve successes’. Can a success rate be determined?

Worked answer. No. The number of attempts is missing.

Common denominator mistake. Inventing a 12% rate assumes 100 attempts.

Independent check. With 20 attempts rate60%, with 40 rate30%.

216. Ratios need ordered labels

Question. If A:B=3:2, what is B:A?

Worked answer. 2:3.

Common denominator mistake. Repeating 3:2 after reversing labels changes meaning.

Independent check. An example A=6 B=4 verifies both.

217. Proportion is a relationship, not a single outcome

Question. If two quantities are directly proportional, what does doubling one do to the other?

Worked answer. It doubles the other under the stated constant proportionality model.

Common denominator mistake. Adding two rather than scaling by two is different.

Independent check. If y=3x, x=2 gives6 and x=4 gives12.

218. A graph gradient needs compound units

Question. A graph uses y=litres and x=minutes, and rises twelve litres in four minutes. What is slope?

Worked answer. 3 L/min.

Common denominator mistake. Calling it 3 L because the denominator was omitted.

Independent check. 3×4=12 L over 4 min.

219. A mixed-units ratio needs conversion

Question. Compare 500 mL to 1.5 L as a ratio.

Worked answer. 1.5L=1500mL; 500:1500=1:3.

Common denominator mistake. 500:1.5 ignores unit compatibility.

Independent check. Both in litres 0.5:1.5=1:3.

220. Percentage points need two rates

Question. Does the change from 40 people to 50 people automatically represent ten percentage points?

Worked answer. No. Percentage points require percentages relative to specified totals; the difference in counts is ten people.

Common denominator mistake. Using a raw count difference as a rate change invents denominators.

Independent check. If totals are each 100, rates change 40%→50%; different totals yield different rates.

221. Rate change vs total price

Question. A fixed-fee service charges $5 plus $3/visit. What is the total after four visits?

Worked answer. $17. The per-visit rate is $3/visit; the fixed fee is $5.

Common denominator mistake. Calling $17/4=$4.25 the service’s stated marginal rate confuses average cost per visit with marginal per-visit charge.

Independent check. 5+3×4=17.

222. Average cost versus marginal rate

Question. Using $5+$3 per visit, what is average cost per visit after four visits?

Worked answer. Total $17/4=$4.25 per visit average; marginal fee remains $3 for an extra visit.

Common denominator mistake. Using these two rates interchangeably hides the once-only charge.

Independent check. After fifth visit total $20, $3 higher.

223. A same-unit ratio is dimensionless

Question. A length A=4m and B=6m. What is A:B?

Worked answer. 4:6=2:3 with metres cancelling.

Common denominator mistake. Writing 2:3 m as though the ratio were a physical length is imprecise.

Independent check. 4/6=2/3.

224. Scientific percentage and density answer different questions

Question. A sample contains 12 g of material in 20 g total. What is mass percentage?

Worked answer. 12/20×100=60% by mass, not 0.6 g/cm³.

Common denominator mistake. Conflating percent composition with density needs a volume measure.

Independent check. 12+8=20 g if remaining material is eight.

225. An inverse proportional model can fail physically

Question. Ten machines are assumed identical and perfectly parallel. Does doubling them always halve real project time?

Worked answer. Not necessarily; real setup and coordination can limit the inverse model. The school result is conditional on stated assumptions.

Common denominator mistake. Treating idealised inverse proportion as universal physical law overstates evidence.

Independent check. Name what is held constant and why.

226. Comparing probabilities uses fractions of sample space

Question. Event A has 3/8 and event B 2/5 probabilities. Which is larger?

Worked answer. 2/5=0.4 exceeds 3/8=0.375.

Common denominator mistake. Comparing only numerators three versus two ignores denominators.

Independent check. Convert to denominator 40: 15/40 vs16/40.

227. Same percentage, different base

Question. 25% of 80 and 25% of 120 are what?

Worked answer. 20 and 30. Same fraction of different wholes.

Common denominator mistake. Claiming the results equal because the printed percentage matches.

Independent check. 20/80=30/120=1/4.

228. A percentage decrease after increase

Question. A positive price rises 20%, then falls 20%. Does it return to original?

Worked answer. No. It becomes 1.2×0.8=0.96 of original, a net 4% reduction.

Common denominator mistake. Assuming opposite percentage labels cancel ignores changed base.

Independent check. Starting 100 gives120 then96.

229. A total increase isn’t automatically a rate increase

Question. A school sees twenty more event attendees, but no original count given. Can percentage growth be found?

Worked answer. No. Need original attendee count to divide the increase by it.

Common denominator mistake. Calling twenty people 20% growth assumes old count 100.

Independent check. If original 200, growth10%; if100, growth20%.

230. Choice of operation comes from the noun

Question. A problem asks ‘how much per hour’ from a total and duration. Which quotient?

Worked answer. Total amount divided by elapsed hours, giving units per hour.

Common denominator mistake. Adding totals or forming an unlabeled ratio doesn’t answer rate.

Independent check. Multiply computed rate by time to reconstruct amount.

Four comparison mistakes that deserve separate repairs

The first ratio part is taken as a fraction of the whole

In A:B=2:3, A is two parts while B is three. The combined amount is five parts, so A is 2/5, not 2/3 of the whole. Draw five equal boxes, shade two for A and three for B, and read the resulting ratio and fraction separately. The first mistaken denominator is the teaching target.

Different units are silently cancelled

A distance of 120 km in two hours is 60 km/h, not simply the unitless ratio 60. Its numerical quotient is accompanied by a compound unit. On a graph, vertical change divided by horizontal change similarly produces the appropriate measurement per input unit. The pupil should write the units before dividing.

The wrong amount is treated as 100%

A 20% discount on $100 gives a $20 saving and a $80 payment. To reverse this process, $80 is 80% of the original, so the original is $80/0.8=$100. Adding 20% of $80 uses the wrong base. Ask the learner to label the original as 100% in every reverse-percentage story.

A percentage-point difference is called percentage growth

If a rate rises from 20% to 30%, the point increase is ten percentage points. The relative increase is (30−20)/20×100%=50%. These are both correct calculations but answer different questions. The tutor should ask for the exact wording rather than accepting an unlabeled ‘ten percent’.

How a three-pupil tutorial could connect these ideas

The eduKateSG Clementi small-group Maths programme describes a ninety-minute three-pupil format near Sixth Avenue MRT. A G1 lesson can allow one child to rebuild part-to-whole ratio fractions, another to choose units in rate problems and a third to compare percentages with different bases. A shared theme is useful, but individual first errors and changed-number retests may differ. The format does not establish current class availability or guarantee results.

An illustrative lesson might begin with ten minutes identifying denominator and unit labels, twenty minutes using a five-box ratio model, twenty-five minutes of independent rate and percentage tasks, fifteen minutes of percentage-point comparisons, ten minutes of mixed questions without topic headings, and ten minutes explaining how each solution reconstructs the given quantities.

A five-day parent–student comparison routine

Day one: write ratio A:B and A’s fraction of total for two concrete groups.
Day two: calculate one unit price and label it $/item.
Day three: find one percentage part using a known base and check the remainder.
Day four: distinguish percentage-point change from old-base percentage increase.
Day five: compare counts and percentages across two sample groups of different sizes.

Frequently asked questions

Is a ratio the same as a fraction of the whole?

Not automatically. A:B compares two labelled amounts; A/(A+B) gives the first amount’s fraction of the two-group whole.

Does a rate always need time?

No. Many rates compare unlike quantities: dollars per item, grams per cubic centimetre and litres per minute are examples.

Why can a percentage exceed 100%?

An amount may be greater than its reference, such as 150 being 150% of 100. But a single category’s share of its own nonoverlapping whole cannot exceed 100%.

What are percentage points?

They are the direct difference between two rates expressed as percentages, such as 30%−20%=10 percentage points.

What is the difference between direct and inverse proportion?

In direct proportion one quantity is a fixed multiplier of the other; in inverse proportion their product is fixed under the stated model.

How should a tutor diagnose confusion quickly?

Present the same raw numbers in a ratio, a fraction-of-total and a unit-rate question. Ask the learner to name the comparison and units before calculating each result.

Official references and scope note

Reader note: All prices, worker examples, travel rates and survey counts are fictional mathematical models. A direct or inverse relationship holds only under its stated assumptions. These are not live price quotes or experimental observations. Current tuition availability, syllabus coverage, tutor placement and fees require confirmation directly, and no grade improvement is guaranteed.

Bring a marked question where the pupil used the wrong denominator to eduKate Singapore and ask how a tutor will teach the difference between part-to-part, part-to-whole and per-unit rate.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Ratio, Rate and Percentage: What Is Being Compared?