Did you know? The bicycle rental with the lowest starting fee does not have to remain cheapest during a longer ride. The amount paid may depend on a fixed fee, free included minutes and a different rate for extra time. Once a student separates those conditions, the question becomes a practical problem involving straight-line graphs, equations and inequalities.
For Secondary 4 G3 Mathematics learners in Bukit Timah, the quickest reliable method is to put every duration in minutes, use the correct charge formula for that interval, solve where two bills become equal, and check the resulting time against each pricing rule’s valid range. One algebraic intersection outside that range is not a real-world break-even point.
The three plans below and all rides are invented for teaching. They are not actual Singapore bicycle rental offers, local route descriptions or recommendations. The 2027 SEC G3 Mathematics K310 syllabus contains the underlying skills of rates, unit conversion, linear functions, inequalities, percentage comparison and interpretation of contextual problems.
Define the variable and the assumptions
Let t represent elapsed rental time in minutes, with t≥0. Let the bill be measured in Singapore dollars. We initially assume that the clock runs continuously for one session and that every fraction of an extra minute is billed proportionally.
There are no membership fees, deposits, discounts, damage charges, mandatory equipment charges or additional taxes in the original model. Each invented price is a final charge under these stated conditions.
The meaning of a rate matters. S$0.20 per minute is not S$0.20 per hour, and an included thirty-minute allowance does not mean thirty minutes of all future sessions can be accumulated.
Plan A: thirty minutes included
Plan A costs S$8 for any rental up to and including thirty minutes. If 0≤t≤30, its cost expression is A(t)=8.
Beyond thirty minutes, each extra minute costs S$0.20. Thus for t>30, A(t)=8+0.20(t−30). Simplifying gives A(t)=0.20t+2, but that expanded expression applies only to the overage interval.
At t=30 both branches give S$8. A student who writes 8+0.20t at t=30 has charged for included minutes a second time. The phrase ‘additional minutes’ is part of the model, not a decorative adjective.
Plan B: a smaller starting fee but a rate from minute one
Plan B charges S$5 plus S$0.25 for every elapsed minute. There is no free included period, so B(t)=5+0.25t for t≥0.
At ten minutes, B costs 5+2.50 = S$7.50. At twenty minutes it costs 5+5 = S$10. The slope is S$0.25 per minute and the vertical intercept is the S$5 base fee.
The simplest error here is to use only 0.25t and forget the five-dollar charge. Checking the cost at t=0 helps reveal that omission immediately.
Plan C: a ninety-minute included period
Plan C costs S$18 for any rental through ninety minutes. If 0≤t≤90, then C(t)=18.
Beyond ninety minutes, each additional minute costs S$0.10, so for t>90, C(t)=18+0.10(t−90). The equivalent expanded overage expression is C(t)=0.10t+9.
The two branches meet at t=90 with a charge of S$18. That consistency check confirms there is no invented sudden jump in the proportional-billing model.
Which plan costs least at the beginning?
At t=0, the modelled charges are A S$8, B S$5 and C S$18. Plan B has the smallest starting charge.
At t=10 minutes, A still costs S$8. B costs S$7.50, and C still costs S$18. B remains cheapest.
At t=20 minutes, A costs S$8, B costs S$10 and C S$18. Plan A has become cheapest. The table shows a crossing must have occurred between ten and twenty minutes.
Break-even point one: A compared with B
While Plan A is in its included thirty-minute period, its price is constant at S$8. Set 8 = 5+0.25t to find when B catches up.
Subtract five to get 3=0.25t. Divide by 0.25 and obtain t=12 minutes. Both plans cost S$8, and twelve minutes falls inside the required Plan A interval.
This is the first genuine intersection. It can be written as (12,8) if time is on the horizontal axis and cost on the vertical axis. At ten minutes B wins; at twenty A wins.
Do not accept an intersection outside its formula’s interval
A student might compare Plan A’s expanded overage line 0.20t+2 with Plan B’s line 0.25t+5 and obtain a negative-time intersection.
Set 0.20t+2 = 0.25t+5. This gives −3 = 0.05t and therefore t=−60 minutes.
That is not a permitted rental time. In addition, Plan A’s expression 0.20t+2 is only valid for t>30. An algebraic solution outside the interval is not a real tie under the described tariffs.
Break-even point two: A compared with C
Between thirty and ninety minutes, Plan A charges 0.20t+2 while Plan C remains at S$18. Solve 0.20t+2=18.
Subtract two to obtain 0.20t=16, then t=80 minutes. Check with the original formulas: A costs 8+0.20(80−30)=18, and C costs 18.
At seventy minutes, Plan A costs S$16, below C. At ninety minutes, A costs S$20 while C costs S$18. Thus the lowest cost changes at the valid eighty-minute intersection.
Why Plan C stays below A beyond ninety minutes
When t>90, Plan A costs 0.20t+2, and Plan C costs 0.10t+9. Subtract the two charges to find A(t)−C(t)=0.10t−7.
For every t>90, 0.10t−7 is positive. Thus Plan A costs more than C throughout this overage interval.
Solving the two extended lines as equal would give t=70, which is outside the shared t>90 interval. That mathematical extension does not create a third valid break-even.
The complete cheapest-price ranges
Under the original pricing rules, Plan B is cheapest for 0≤t<12 minutes. Plan A and B tie at twelve minutes.
Plan A is cheapest for 12
This is a comparison across three piecewise graphs. It does not say which actual bicycle operator has the best service or which bicycle is safest. The result is limited to the fictional charges supplied.
Worked rental: sixty-eight minutes
Suppose a rider rents for 68 minutes. Plan A’s excess is 68−30=38 minutes. Its cost is 8+0.20×38 = S$15.60.
Plan B costs 5+0.25×68 = S$22, while Plan C remains S$18 because the rental is shorter than ninety minutes.
Plan A is cheapest, saving S$2.40 against C and S$6.40 against B. These are differences between the final bills for the same duration.
Worked rental: two hours
Two hours equals 120 minutes, not the number two in a per-minute formula. Plan A costs 8+0.20×90 = S$26.
Plan B costs 5+0.25×120 = S$35. Plan C costs 18+0.10×30 = S$21.
Plan C has the lowest modelled charge for that duration, consistent with the rule that it becomes cheapest beyond eighty minutes. Converting hours into minutes before calculating is essential.
Build an honest graph from the three rules
Put elapsed rental time in minutes along the horizontal axis and cost in Singapore dollars along the vertical axis.
Plan A begins at (0,8), stays horizontal to (30,8), then rises by S$0.20 per additional minute. Plan B begins at (0,5) and rises by S$0.25 per minute. Plan C stays at S$18 until ninety minutes and then rises by S$0.10 per minute.
Mark the genuine intersections at (12,8) and (80,18). A correct graph includes the corners at thirty and ninety minutes rather than extending the overage rules through their included periods.
Gradient does not equal the total monthly or session cost
The Plan A graph has gradient zero up to thirty minutes, even though its bill is S$8. After thirty minutes it has gradient S$0.20 per minute.
Plan B has gradient S$0.25/min throughout. Plan C has gradient zero within ninety minutes and then S$0.10/min thereafter.
These gradients tell how the cost changes with one additional minute in the relevant interval. They do not include all the information needed to decide which final bill is smallest.
Check a table against the graph
For t=0,12,30,68,80,90 and 120 minutes, Plan A’s charges are S$8, S$8, S$8, S$15.60, S$18, S$20 and S$26.
At the same times, Plan B’s charges are S$5, S$8, S$12.50, S$22, S$25, S$27.50 and S$35. Plan C’s charges are S$18, S$18, S$18, S$18, S$18, S$18 and S$21.
The table confirms the intersections and thresholds. A graph that shows Plan C rising before ninety minutes or Plan A starting below S$8 contradicts the original pricing data.
An S$22 budget as an inequality
Suppose the rider can spend at most S$22 on one session. Under Plan A after thirty minutes, 8+0.20(t−30)≤22 gives t≤100 minutes.
Plan B requires 5+0.25t≤22, so t≤68 minutes. Plan C after ninety minutes requires 18+0.10(t−90)≤22, so t≤130 minutes.
Under the assumptions of a continuous charge, no separate time cap and one rental session, Plan C permits the longest time inside S$22: 130 minutes.
What if a different rule limits a rental to two hours?
Now add a separate hypothetical condition that the maximum permitted single rental duration is 120 minutes.
Although the Plan C budget inequality would allow 130 minutes, the contract now limits the permitted time to 120 minutes. At 120 minutes its charge is S$21, which is below S$22.
This demonstrates that a valid price calculation must also satisfy the full domain of the real-world model. The separate maximum does not change the earlier mathematical inequality; it adds another restriction.
Change the billing rule: charge every started extra minute
The original pricing model charged fractional extra minutes proportionally. Imagine a separate fictional condition that rounds any started overage minute up to a whole chargeable minute. A Plan A rental lasting 30.4 minutes would then be billed for one full additional minute.
The revised cost becomes S$8+S$0.20 = S$8.20. Under the proportional original rule, it would be 8+0.20(0.4)=S$8.08. Both calculations are correct under different stated conventions.
The student must not introduce an extra rounding rule just because it seems realistic. A problem’s wording determines whether the tariff graph is continuous or stepwise. A change in the billing rule is a change in the mathematical model.
A 10% discount: which price does the percentage affect?
Suppose a fictional promotion reduces the entire Plan A bill for a 68-minute rental by 10%. The undiscounted bill was S$15.60, so the new amount is 0.90×15.60 = S$14.04.
If the offer instead reduces only the S$8 basic fee, the result is 0.90×8 + 0.20(68−30) = 7.20+7.60 = S$14.80.
Both expressions use ten percent, but they act on different bases. A student should underline ‘entire final bill’ or ‘basic fee only’ before calculating. A percentage without its defined reference quantity is incomplete.
A twenty-percent discount after the ride
Take a separate hypothetical 120-minute Plan C rental, which originally costs S$21. Suppose a clearly stated promotion reduces the total bill by 20% without changing the duration or rental conditions.
The amount saved is 0.20×21 = S$4.20, giving a final bill of S$16.80. The discount does not create extra included rental minutes; it changes only the final amount payable.
Do not subtract twenty percent from the number of minutes when the question says the price has been reduced. Units can reveal the error: a percentage of money remains money, while a percentage of time remains time.
Turn a rental duration into a clock time
Imagine a short fictional outing beginning at 16:15 and lasting 32 minutes. Add thirty-two minutes to obtain an end time of 16:47.
If a fifteen-minute stop or delay is added to that elapsed rental time, the finish becomes 17:02. The new duration is 47 minutes, not 32 plus 0.15 hours.
Clock time uses sixty minutes per hour. A decimal such as 0.5 hour is thirty minutes, not fifty minutes. Translate between clock notation and numerical duration before calculating a per-minute charge.
A different fictional cycling route gives a speed exercise
Consider a three-leg imaginary ride: 1.5 km at 12 km/h, then 2 km at 10 km/h, then 1.5 km at 12 km/h. These are mathematical distances and rates, not a real route or safety recommendation.
Each outer 1.5 km section takes 1.5/12 hour = 0.125 hour = 7.5 minutes. The middle 2 km takes 2/10 hour = 0.2 hour = 12 minutes.
The total distance is 5 km and moving time is 7.5+12+7.5 = 27 minutes. This illustrates how three different speeds and lengths combine; merely averaging 12, 10 and 12 would not produce the overall average speed.
Five minutes of rest changes the whole-ride average
Suppose the rider pauses for five minutes during the imaginary trip, with the rental clock continuing. Total elapsed time is now 27+5 = 32 minutes.
Average speed for the entire elapsed journey equals total distance divided by total elapsed time: 5/(32/60) = 9.375 km/h, or 9.38 km/h to three significant figures.
If the question asks for average speed while moving, exclude the rest and calculate 5/(27/60) ≈ 11.1 km/h. Both numbers are mathematically valid for their specified intervals. The wording tells us which one to report.
The 32-minute ride and its rental cost
A 32-minute ride exceeds Plan A’s thirty-minute allowance by two minutes. Its Plan A charge is 8+0.20(2) = S$8.40.
Plan B would charge 5+0.25(32) = S$13, while Plan C would still charge S$18. Plan A has the lowest final bill for this fictional 32-minute journey.
The journey distance does not appear in any of the original price formulas because they depend on elapsed time rather than kilometres. A cost-per-kilometre comparison is possible but requires a separate step after the bill is known.
Cost per kilometre is not the contractual rate
The 32-minute, five-kilometre journey costs S$8.40 with Plan A. Dividing by five gives S$1.68 per kilometre for that particular trip.
Using the other two modelled bills, Plan B would cost 13/5 = S$2.60/km, and Plan C would cost 18/5 = S$3.60/km for the same trip.
These are derived cost-per-distance ratios, not advertised per-kilometre rental rates. Under a time-based tariff, a slower journey may cost more per kilometre even if the total route length remains unchanged.
The effect of a fifteen-minute delay on the bill
Suppose the 32-minute ride is delayed by fifteen minutes while the bicycle remains rented. The revised duration is 47 minutes.
Plan A now charges 8+0.20(47−30) = 8+3.40 = S$11.40. Plan B charges 5+0.25(47) = S$16.75, and Plan C remains S$18.
The student should not invent a separate late-return penalty; none was specified. The increase comes entirely from the continuing per-minute rental clock under the fictional tariff rules.
Two separate bicycles for two riders
Imagine two people each rent their own bicycle for the same 68-minute interval, with the fictional price applied independently to each bicycle.
Plan A costs S$15.60 per bicycle, so both rentals together cost 2×15.60 = S$31.20. The factor two multiplies the entire final charge for each rental.
Do not multiply only the overage fee while leaving the basic charge paid once, unless the problem expressly describes a shared booking discount. Two independent rentals are two complete tariff calculations.
Two separate sessions versus one continuous session
Suppose a person rents under Plan C for 120 minutes, returns the bicycle, and later starts a completely new 120-minute session on the same fictional terms. Each session costs S$21, so the combined outlay is S$42.
By contrast, if a continuous 240-minute rental were allowed and billed as one session, Plan C would cost 18+0.10(240−90)=S$33.
This variation depends on when the included allowance resets. It illustrates why an event occurring once per session must not be combined with a continuous-time model without reading the wording.
A duration cap can invalidate the continuous-session example
Now suppose the fictional rental agreement separately limits each session to no more than 120 minutes. Under this rule, one continuous 240-minute hire is not permitted.
Its algebraic S$33 expression is still a calculation using the original cost formula, but it lies outside the revised rental domain. Two separate permitted 120-minute sessions could still total S$42 if rebooking is explicitly allowed.
A useful student answer distinguishes the amount a formula calculates from the situations the stated terms actually permit. That same domain habit applies to physical capacity and ticket-count problems.
Six different sessions and the total cost of each plan
Imagine six separate rental sessions lasting 10, 20, 40, 70, 90 and 120 minutes, with every session starting a fresh included-time allowance. We use the original proportional-billing model.
Plan A bills are S$8, S$8, S$10, S$16, S$20 and S$26, totaling S$88. Plan B bills are S$7.50, S$10, S$15, S$22.50, S$27.50 and S$35, totaling S$117.50.
Plan C bills are S$18, S$18, S$18, S$18, S$18 and S$21, totaling S$111. The lowest aggregate here is Plan A, even though Plan C is cheapest for the longest session individually.
Why the cheapest long ride does not determine the whole period
At 120 minutes, Plan C costs S$21 while Plan A costs S$26, a five-dollar advantage for C. But the six-session total gives A S$88 versus C S$111.
Short sessions benefit from A’s smaller basic charge than C’s S$18. Those savings accumulate, so the best total across a mixture of durations can differ from the best price for one long ride.
A complete answer should name the comparison period and the set of durations. Saying ‘Plan C is always cheapest’ would contradict the short-session rows even under the fictional assumptions.
Mean charge across the six sessions
The total Plan A spending in the six-session example is S$88. Its mean charge per session is 88/6 ≈ S$14.67.
Plan B averages 117.50/6 ≈ S$19.58, while Plan C averages 111/6 = S$18.50. These are per-session averages for this particular list of durations.
Do not divide by the combined number of minutes if the question asks for mean cost per session. That would instead produce a different ratio, dollars per minute across the group of sessions.
A percentage saving needs a named base
For a two-hour rental, Plan A costs S$26 while Plan C costs S$21. Choosing C instead of A saves S$5.
Relative to A’s price, the saving is 5/26×100% ≈ 19.2%. If asked how much more expensive A is relative to C, use the C price as the base: 5/21×100% ≈ 23.8%.
The same absolute five-dollar difference gives different percentage results because the original base changes. State what represents 100% before dividing.
A quick reasonableness estimate
For Plan A at 120 minutes, the included thirty minutes leave ninety minutes of overage. At twenty cents per minute, that is S$18; adding the S$8 charge gives S$26.
A calculator answer of S$44 might indicate all 120 minutes were charged as excess. An answer of S$18 might indicate the basic fee was omitted.
These are different first mistakes requiring different corrections. An estimate is useful precisely because it checks the structure of a calculation, not merely the last digit.
A change to Plan A’s included time
Suppose a separate fictional promotion extends Plan A’s included time from thirty to forty minutes, while keeping its S$8 basic fee and S$0.20 per excess minute.
At a 68-minute rental, the revised overage is 68−40 = 28 minutes. The new bill is 8+0.20(28) = S$13.60, compared with the original S$15.60.
A child who continues subtracting thirty minutes has memorised an old rule instead of reading the changed condition. This is an effective independent retest of the idea behind a piecewise tariff.
Find a new tie after that promotion
Under the revised forty-minute allowance, Plan A costs S$8 through t=40, then costs 8+0.20(t−40)=0.20t above forty.
Plan C remains S$18 through ninety minutes. Their cost equality in the interval 40
Check at exactly ninety: revised A = 8+0.20(50) = S$18, and C = S$18. The tie moved from eighty minutes to ninety, consistent with the extra ten included minutes.
A seven-day targeted practice sequence
Day 1: identify the fixed charge and the per-minute rate. Day 2: write Plan A and C as piecewise functions. Day 3: verify the two genuine intersections and check their intervals.
Day 4: solve a budget inequality. Day 5: convert minutes and hours for a distance–time problem. Day 6: compare an entire series of rental sessions and calculate mean charge. Day 7: change one allowance and solve the model again without viewing the original answer.
Keep a dated error log with categories wrong overage amount, missing base fee, invalid interval, hours-to-minutes conversion, incorrect denominator, percentage base and domain restriction.
Parent questions about rental-rate Mathematics
Why does Plan B win at ten minutes? It costs S$7.50 compared with Plan A’s S$8. Why does Plan A win at 68 minutes? Its final charge of S$15.60 is below B’s S$22 and C’s S$18.
Which plan wins at 120 minutes? Plan C at S$21. Does an included period mean the rental is free? No. A nonzero basic charge is still payable. Why can a correct algebraic intersection be invalid? It may be outside the interval where the expression applies.
Are these actual rental offers? No, all figures are fictional teaching data. What exam skills are involved? Rates, time, linear functions, inequalities, percentages and mathematical interpretation in the G3 Mathematics K310 content.
Secondary 4 G3 Mathematics tuition in Bukit Timah
The original eduKateSG Mathematics small-group teaching reference remains unchanged. It describes premium three-pupil Mathematics tutorials with 1.5-hour weekly lessons, individual feedback, guided corrections and independent practice. The stated teaching venue is 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. This article does not advertise another branch or guarantee an available seat.
One learner may confuse time billed with time included, another may solve an intersection outside the applicable interval, and a third may calculate average speed using moving time when the question asks for elapsed time. A useful small-group tutor identifies the first faulty step in each learner’s work and checks the repair using changed numbers.
Enquire about an eduKate Singapore parent–student Mathematics consultation with the learner’s current school script and actual examination year.
Official national examination route and continued study
The first Singapore–Cambridge SEC sitting takes place in 2027, when G3 Mathematics is subject code K310. The official 2027 syllabus includes rates, time, linear functions, inequalities and contextual applications. Candidates sitting 2026 GCE O-Level Mathematics must instead follow the applicable 4052 documents and exam instructions.
- SEAB: 2027 SEC G3 Mathematics K310 syllabus (PDF)
- SEAB: 2027 G3 school-candidate subject list
- eduKateSG: original Mathematics small-group reference (unchanged)
- Earlier: Speed–Time Graphs and Average Speed
- Earlier: Mobile Data Plans and Break-Even Graphs
- Robotics Route: Vectors and Bearings
- School Hall Seating and Capacity
- Recycling Audit and Statistics
All plans, rates, rental times and journeys are original classroom examples, not real bicycle-hire offers or directions for a real road. No examination question or grade is promised.