Did you know? A phone plan with the lowest starting fee may cost more than a pricier plan once data usage grows. The reason is mathematical: an included allowance changes which cost rule applies. The question is not merely ‘Which monthly price is smallest?’ It is ‘Which bill is smallest for this particular amount of data?’
For a Secondary 4 G3 Mathematics student in Bukit Timah, the useful route is to define the units, write a cost expression for every valid interval, identify where two charges become equal, and then check which option is lower on either side. That is what a break-even graph tells the reader.
Every plan, price and consumption figure in this guide is fictional, not a real Singapore telecommunications offer. The scenario is designed to connect linear graphs, simultaneous equations, inequalities, rates, percentages and interpretation in the published 2027 SEC G3 Mathematics K310 syllabus.
Start by defining the variables and the billing assumptions
Let x denote gigabytes of mobile data used in one month, with x ≥ 0. Let a monthly bill be measured in Singapore dollars. The unit of a cost-graph gradient is therefore Singapore dollars per gigabyte, not simply dollars or gigabytes.
Assume all invented prices are final GST-inclusive amounts. Included data reset at the beginning of each month. There are no promotions, joining fees, handset payments, roaming charges, rollover allowances or speed restrictions in the initial model.
We also initially assume fractional extra gigabytes are billed proportionally: if the rate is S$3/GB, half an excess gigabyte costs S$1.50. This is a stated classroom assumption, not a claim about how a particular company bills its subscribers.
Three invented options, each with its own meaning
Plan A: S$18 per month includes 4 GB, and additional use beyond that allowance costs S$3 per GB. Plan B: S$30 per month includes 10 GB, and additional use beyond ten costs S$2 per GB.
Plan C: a fixed S$6 administration charge plus S$4 per GB used, with no included free amount. This plan has the lowest starting bill but the highest per-GB gradient.
The three plans must be compared at the same amount of usage. Comparing A at 2 GB with B at 10 GB would not tell us which option costs less for a single student’s needs.
Write Plan A correctly using two intervals
For usage between 0 and 4 GB inclusive, A(x) = 18. The student has not exceeded the included allowance, so the bill remains flat at S$18.
For x > 4, the extra amount is x−4 GB, and A(x) = 18 + 3(x−4) = 3x+6. This higher-usage expression is valid only beyond the four-gigabyte allowance.
At x = 4, the two expressions both give S$18, which means the two pieces of the graph meet. A common mistake is writing 18+3x and charging for the first four gigabytes a second time.
Plan B has a different break point
For 0 ≤ x ≤ 10, B(x) = 30. The monthly bill remains S$30 while the usage falls within the ten-gigabyte allowance.
When x > 10, the excess is x−10, and B(x) = 30 + 2(x−10) = 2x+10. The line rises by S$2 for each additional GB over the threshold.
Students should check the boundary x = 10: 30+2(10−10) = 30. An incorrectly written formula that produces a sudden jump at exactly ten would contradict the stated continuous-billing conditions.
Plan C is one straight-line relationship
Plan C has C(x) = 6+4x for x ≥ 0. Its y-intercept is S$6 because that is what the user pays even when no data is used. Its gradient is S$4/GB.
At 2 GB, Plan C costs 6+8 = S$14. At 3 GB it costs S$18, and at 4 GB it costs S$22.
The difference between intercept and gradient is essential. The intercept gives the cost at zero usage. The gradient describes how the bill changes as usage increases. Neither alone is always sufficient to compare total costs.
A first table: zero, two and four gigabytes
At 0 GB, the final charges are A S$18, B S$30 and C S$6. At 2 GB, they are A S$18, B S$30 and C S$14. At 4 GB, they are A S$18, B S$30 and C S$22.
Plan C is the cheapest at zero and two, but Plan A is cheapest at four. This makes it clear why advertising the smallest monthly basic charge does not settle the comparison for every user.
The entries must be checked against the correct branch. For example, A(2) is S$18, not 3(2)+6 = S$12, because the latter expression applies only above four GB.
First genuine intersection: Plans A and C
While Plan A is in its flat 0–4 GB region, compare its S$18 charge with Plan C’s 6+4x. Set 18=6+4x, giving 12=4x and thus x=3 GB.
Substitute the candidate into both actual plans: A(3) = S$18 and C(3) = 6+4(3) = S$18. The valid intersection point is (3,18), where x measures gigabytes and the other coordinate measures dollars.
To determine which plan is cheaper nearby, use the table: C wins at 2 GB, while A wins at 4 GB. A break-even point is not the same as a universal recommendation.
An algebraic intersection outside an interval is not valid
Suppose we compare Plan A’s overage expression 3x+6 with Plan C’s 4x+6. Setting them equal gives x=0. But at zero data usage, Plan A’s actual charge is S$18, not S$6.
The line 3x+6 is an extension of a piece of the price curve beyond the interval where that piece is valid. The algebra can be correct while its application to the story is wrong.
Did you know? A piecewise graph has a domain condition attached to each rule. It is not enough to solve two formulas; the result must lie where both formulas are legitimate.
Second genuine intersection: Plans A and B
Between 4 and 10 GB, Plan A costs 3x+6 while Plan B still costs S$30. To find their tie, solve 3x+6=30. Subtract six and divide by three to get x=8 GB.
Check at eight: Plan A is 18+3(8−4) = S$30, while Plan B is S$30. At seven, A costs S$27, lower than B. At nine, A costs S$33, higher than B.
The intersection is (8,30) on the cost-versus-usage graph. It lies inside the interval where A’s overage formula and B’s flat formula both apply, so it is meaningful.
Compare the plans when use is above ten gigabytes
For x > 10, the Plan A bill is 3x+6 and the Plan B bill is 2x+10. Their difference is A(x)−B(x) = x−4.
Since x>10 implies x−4 is positive, A is more expensive than B throughout this high-use interval. We do not need to evaluate every possible usage value separately.
The mathematical extensions of those two lines meet at x=4, but that is outside this high-use comparison. Stating the interval is part of giving a valid conclusion.
The complete cheapest-price ranges
Under these exact fictional assumptions, Plan C is cheapest for 0 ≤ x < 3 GB. At x=3, Plans A and C tie at S$18.
Plan A is cheapest for 3 < x < 8 GB. At x=8, Plans A and B tie at S$30. Plan B is cheapest for x>8 GB.
This result is a lower-cost envelope across three piecewise graphs. It is a statement about the final modelled monthly bill, not about real network speed, service quality or a provider’s current availability.
Worked example: twelve gigabytes in one month
Plan A includes four GB and therefore charges on eight additional GB: 18+3(8) = S$42. Plan B includes ten GB and charges on two additional GB: 30+2(2) = S$34.
Plan C costs 6+4(12) = S$54. Therefore the lowest cost at 12 GB is Plan B, with a saving of S$8 against A and S$20 against C.
Check the overage quantities separately from the payment calculation. Charging A on all twelve GB would ignore the first four included gigabytes.
Worked example: a much higher usage value
At 15 GB, Plan A costs 18+3(15−4) = S$51. Plan B costs 30+2(15−10) = S$40, while Plan C costs 6+4(15) = S$66.
This confirms the inequality argument that B remains below A above ten GB, and it illustrates why a large fixed allowance can matter more than the smallest subscription starting fee.
However, a real plan could have a hard data cap or a separate speed policy. Our invented model ignores those features, so the result should not be transferred uncritically to any actual operator.
Draw the graph with the correct corners
Use monthly usage in GB for the horizontal axis and bill in S$ for the vertical axis. Plan A is horizontal at 18 through x=4, then slopes upward by 3 for each GB.
Plan B is horizontal at 30 through x=10, then slopes upward by 2. Plan C starts at S$6 when x=0 and rises continuously at a gradient of 4.
Mark both genuine intersections, (3,18) and (8,30). Do not accidentally draw Plan A’s 3x+6 rule back to zero usage, because that creates a false low starting price.
The gradient has units: Singapore dollars per gigabyte
Plan A’s flat section has gradient zero, but it still has a nonzero monthly bill. Beyond the allowance its gradient is S$3/GB. Plan B’s corresponding gradients are zero and S$2/GB.
Plan C has gradient S$4/GB for every x in the model. These numbers tell the reader how a small increase in usage changes the total charge within each interval.
The line’s gradient is not the same as the total bill. Saying ‘Plan B costs S$2’ because its overage slope is two would be a units-and-interpretation error.
An S$32 budget becomes an inequality
Suppose an invented household places a maximum of S$32 per month on data use. On Plan A’s overage segment, write 18 + 3(x−4) ≤ 32, giving 3(x−4) ≤ 14 and x ≤ 8⅔ GB while x is above four.
For Plan B, usage through ten GB costs S$30. Above ten, solve 30 + 2(x−10) ≤ 32, giving x ≤ 11 GB. For Plan C, solve 6 + 4x ≤ 32, giving x ≤ 6.5 GB.
Thus under the same continuous-billing assumptions Plan B allows the greatest usage without crossing S$32: eleven gigabytes. A student should explain the unit and check that the chosen formula’s interval includes this value.
An inequality answers more than one intersection point
The equation A(x)=B(x) finds a tie at eight gigabytes. The inequality A(x)<B(x), with valid formulas, instead asks when A is cheaper than B. Those are different questions and the inequality describes a set of usage values.
Between four and ten GB, compare 3x+6 with 30. Solving 3x+6 < 30 gives x < 8; combined with the segment requirement, the range is 4 < x < 8.
For usage at or below four, A costs S$18 and B costs S$30, so A is cheaper there too. Putting both intervals together shows A costs less than B for 0 ≤ x < 8, with equality at eight.
Percentages need a named comparison base
At twelve gigabytes, Plan A costs S$42 and Plan B costs S$34, so the saving from using B instead of A is S$8. Relative to the S$42 Plan A cost, the saving percentage is 8/42 × 100% ≈ 19.0%.
If a question instead asks how much more expensive A is compared with B, the comparison base is S$34. The corresponding percentage is 8/34 × 100% ≈ 23.5%.
Both answers use the same S$8 difference and are mathematically correct for different questions. Write what represents the original 100% before touching the percentage key.
Six months of changing usage can change the winner
Imagine monthly data use of 2, 5, 7, 9, 11 and 14 GB across six months, with each plan’s allowance resetting monthly. Do not add the gigabytes first and pretend the total belongs to one monthly allowance.
The six Plan A bills are S$18, S$21, S$27, S$33, S$39 and S$48, totaling S$186. Plan B bills are S$30, S$30, S$30, S$30, S$32 and S$38, totaling S$190.
Plan C bills are S$14, S$26, S$34, S$42, S$50 and S$62, totaling S$228. Plan A has the lowest six-month total in this particular pattern, even though B wins several high-usage months.
Why the cheapest high-use month does not decide the full year
At fourteen gigabytes, B costs S$38 and A costs S$48. Plan B wins that month by S$10. But A wins in some low-usage months because its fixed charge is S$12 below B’s basic fee.
Across the six-month example, Plan A is S$4 cheaper than B: S$186 versus S$190. A real user’s pattern may differ; neither line graph alone predicts future consumption.
This is why the comparison period must be named. ‘B is cheapest at 14 GB’ and ‘A has the lowest total across these six months’ can both be true. A student’s answer should not silently switch between them.
Average monthly payment and cost per gigabyte are different
Across the six-month example, Plan A’s average bill is 186/6 = S$31 per month. Plan B averages about S$31.67 per month, while Plan C averages S$38 per month.
The denominator is six because the question asks about months. If the question instead asked for a price per GB, its denominator would involve an appropriate quantity of data and its interpretation would be different.
Did you know? The word ‘average’ does not completely define a calculation. ‘Average money paid each month’, ‘average price per GB’ and ‘average data used per day’ are separate quantities.
An entire year at the same twelve-gigabyte usage
Change the assumption. Suppose usage is fixed at 12 GB in each of twelve months, with no change in monthly rates. Under our earlier calculations, the respective monthly bills are A S$42, B S$34 and C S$54.
Multiplying by twelve gives annual totals of S$504, S$408 and S$648. In this pattern, B saves S$96 compared with A over the year.
This result differs from the six-month mixed-usage example because the usage pattern changed. Neither is a forecast or a claim about current Singapore providers.
A one-time joining charge can reverse a short decision
Now imagine a different fictional condition: Plan B has an additional one-time S$20 joining charge. For the first month at 12 GB, B requires S$34 + S$20 = S$54, while A’s bill remains S$42.
For twelve identical months, however, B’s total is 12×34 + 20 = S$428, still below A’s S$504. The joining fee is paid once, not twelve times.
The lesson is about scope and frequency. A fixed one-time cost is not the same as a monthly basic fee. A tutor should ask the student which costs occur once and which recur before the algebra starts.
A data cap changes the permitted domain
Imagine a separate fictional plan that permits no data usage beyond 20 GB. Its mathematical usage domain is 0 ≤ x ≤ 20, rather than all nonnegative x.
A formula could output a number at x=25, but the plan as described does not permit that situation. The calculated number would fall outside the domain of the model.
This same principle applies to a water tank that cannot exceed its capacity or a school event limited to a stated number of participants. The equation may be valid algebraically but its real-world conditions still matter.
Charging for each started extra gigabyte creates steps
Our initial plans charge fractional gigabytes proportionally. Suppose a different version of Plan A charges an entire extra gigabyte whenever any excess part is used.
At 4.2 GB, only 0.2 GB exceeds the original allowance. Proportional billing would cost 18 + 3(0.2) = S$18.60. Charging for one started whole overage GB would instead produce S$21.
The second arrangement is a stepwise function, not the continuous line used earlier. The question must specify how overage is billed; a student should never introduce a rounding rule without support.
Use an estimate to catch a bad billing expression
At fifteen gigabytes on Plan A, excess data is eleven gigabytes. Eleven times S$3 is S$33, and adding the S$18 basic fee suggests a bill of S$51.
If a student’s calculator returns S$63, perhaps they applied the variable rate to all fifteen GB. If it returns S$33, perhaps they forgot the monthly basic fee.
Both are concrete first-error diagnoses. Rather than saying ‘careless’, name the missing allowance subtraction or fixed charge and practise one changed example.
The plotted graph should match a cost table
Check x = 3, 4, 8, 10 and 12 GB. Plan A’s costs are 18, 18, 30, 36 and 42. Plan B’s costs are 30, 30, 30, 30 and 34.
Plan C’s corresponding costs are 18, 22, 38, 46 and 54. These five rows capture the two important ties and both included-allowance thresholds.
If the sketch shows B increasing before ten gigabytes, it contradicts the table. If A appears to cost only S$6 at zero usage, the student has plotted the wrong branch. A table and a graph should support the same model.
A straight-line extension can look plausible but still be wrong
For x>4, the Plan A expression is 3x+6. If extended backwards to x=0 it gives S$6, a number that looks like a legitimate intercept. But Plan A’s actual zero-usage bill is S$18.
A mathematical line segment and its infinite straight-line extension are different objects when a tariff has thresholds. The correct drawing begins with the flat portion, then joins the overage portion at x=4.
Students may know how to sketch y=mx+c perfectly yet use the wrong interval in a word problem. That is a modelling error rather than a basic graph-drawing error.
Which plan has a smaller gradient?
Above ten gigabytes, Plan B’s gradient S$2/GB is below Plan A’s S$3/GB and Plan C’s S$4/GB. It therefore adds the smallest amount to the bill for each further gigabyte in that region.
That does not justify saying B is cheapest at x=0, because the monthly fixed fees and initial allowances create different vertical positions. The graph needs both its gradient and its starting conditions.
A concise explanation would be: ‘Beyond 10 GB, B has the lowest total of the three under the stated formulas, and its cost also rises least rapidly.’
The fastest way to solve a changed-price challenge
Suppose Plan A keeps its S$18 basic charge and S$3 excess rate, but the included allowance grows from four to five GB. Its overage expression becomes 18+3(x−5) = 3x+3 for x>5.
Plan B still costs S$30 up to 10 GB. Their valid intersection now solves 3x+3=30, giving x=9 GB.
The threshold shifts from eight to nine because the improved allowance delays A’s extra charge. A learner who can explain that shift has understood the piecewise structure, not merely memorised a single answer.
AO1, AO2 and AO3 in one price-comparison task
Applying an algebraic technique to 3x+6=30 is a standard mathematical step. Recognising that this comparison is valid only between four and ten gigabytes is a modelling decision.
Explaining which plan is cheaper on either side of x=8 and verifying both original charges is mathematical communication. The published K310 syllabus emphasises techniques, problem solving and reasoning; one case can exercise all three.
A correct isolated ‘x=8’ answer is useful, but a final statement of ‘8 GB, at S$30 for both plans’ gives the result its proper context and units.
A diagnostic for parents reviewing a marked paper
Ask what the child believed the x-axis represented. Did they subtract the included allowance? Did they apply the correct billing interval? Did they use the same consumption for every plan? Did they compare one month or several?
If the first error is selecting the wrong plan branch, more simultaneous-equation drills may not help. If the branch is correct but the equation is solved with a sign mistake, shorter algebra repair may be appropriate.
Once the mistake is repaired, change a plan threshold or a basic fee and ask the student to solve without the model in sight.
A seven-day practice sequence
Day 1: identify included allowances and fixed monthly charges. Day 2: write piecewise expressions for Plans A and B. Day 3: calculate a table at several usage values and sketch the graphs.
Day 4: solve the meaningful intersections and explain both sides. Day 5: practise an S$32 budget inequality. Day 6: total several months of different usage. Day 7: change a tariff condition and independently rebuild the model.
Keep a dated error log using the labels allowance subtraction, interval, fixed fee, marginal rate, percentage base, comparison period and domain. The next practice question should target the earliest mistake.
Frequently asked questions about mobile-tariff Mathematics
Does the lowest starting fee always win? No, as Plan C demonstrates. When do A and C tie? At 3 GB, both costing S$18. When do A and B tie? At 8 GB, both costing S$30.
Why is Plan A’s excess x−4? Four gigabytes are included before overage begins. Can a lower slope guarantee lower total cost? Not without comparing fixed charges and the relevant interval.
Is the model a current Singapore product recommendation? No. All three plans are invented mathematical examples. Does K310 include this kind of skill? Yes: linear functions, graphs, equations, inequalities, rates, percentages and real-world financial modelling.
A complete conclusion that a reader can actually use
Under the simplified charges and continuous-billing convention, Plan C costs least below three gigabytes, Plan A between three and eight, and Plan B above eight. Ties occur at exactly three and eight gigabytes.
The conclusion is supported by valid piecewise rules and checked intersection values. It does not assert that the same option has better coverage, speed, reliability or other features not provided by the problem.
In an examination-style answer, state the usage intervals, total monthly charges at the ties, and the assumptions. That is clearer than declaring one plan ‘best’ with no conditions.
Small-group Mathematics lessons in Bukit Timah
The original eduKateSG small-group Mathematics reference remains unchanged. It describes premium tutorials with a maximum of three pupils, lasting 1.5 hours weekly, with careful explanations and guided corrections. The stated tuition venue is 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. This guide does not claim another branch or guarantee a vacant seat.
One learner may solve the intersection correctly but choose the wrong interval. Another may omit a fixed fee; a third may confuse a six-month total with one monthly allowance. A tutor can diagnose these as separate first errors and retest each one using changed numbers.
Enquire about a parent–student eduKate Singapore Mathematics consultation and bring a recent unaided school script with the student’s actual examination year.
National examination guidance and official references
The official 2027 SEC G3 subject list identifies Mathematics code K310, with 4052 the earlier GCE O-Level Mathematics reference. The 2027 K310 syllabus addresses graphs, rates, percentages and real-world applications. A 2026 O-Level candidate must follow their applicable 4052 syllabus and exam instructions rather than treat the 2027 paper as identical.
- SEAB: Official 2027 SEC G3 Mathematics K310 syllabus (PDF)
- SEAB: 2027 G3 subjects and codes
- eduKateSG: Original Mathematics small-group programme
- Linear Graphs, Gradient and Intercepts
- School Canteen Break-Even Case Study
- Cylinder Tank Flow and Time Graphs
- Library Survey and Honest Charts
All plan names, amounts, usage values and restrictions are fictional classroom teaching data. No actual telecom product is recommended and no exam question or grade outcome is promised.