These ten modelling challenges require interpreting the data before writing equations. The first five rework classical patterns; the others were composed for this collection with new settings and data, without claiming that their mathematical patterns have never appeared before. They use linear or quadratic equations and one linear system. Choose an unknown, state its domain and justify every term before opening the solution. A root must also make sense in the original problem.
- A boat journey with a hidden stop
- A job completed in three stages
- Equal selling prices, unequal gains and losses
- Two identical dilution steps
- A garden whose side lengths are unknown
- Two alternative theatre layouts
- Lost packets and added headers
- A robot keeping its energy reserve
- An invoice with two unknown percentages
- A sensor panel: boundary and interior
1. A boat journey with a hidden stop
Reworked classic
A boat travels 18 km along a river and returns to its starting point. The current is constantly 3 km/h and the boat maintains the same speed relative to the water in both directions. It stops for 30 minutes between legs. Total elapsed time is 5 hours. Find its still-water speed and both travel times.
Hint
Remove the stop from the total time. Do not average the two speeds arithmetically.
Solution and check
Let v be still-water speed, with v>3 so upstream travel is possible. Ground speeds are v+3 and v−3. Each travel time is distance divided by speed. Subtract the half-hour stop and clear the positive denominator v²−9:
18/(v+3) + 18/(v−3) = 5−1/2 = 9/2 36v/(v²−9) = 9/2 v²−8v−9 = 0 (v−9)(v+1) = 0 v=9; v=−1
Reject −1. The speed is 9 km/h. Downstream takes 18/12=1.5 hours, upstream 18/6=3 hours. Including the stop gives 1.5+3+0.5=5. The average speed over the full journey is not v.
2. A job completed in three stages
Reworked classic
Machines A and B work at constant rates, and B alone needs 4 hours longer than A for a batch. A works alone for 2 hours, B alone for 4 hours, then both together for 2 hours, completing exactly one batch. Their output adds without interference. How long would each need alone?
Hint
If A needs x hours, its hourly output is 1/x of a batch.
Solution and check
Let x>0 be A’s solo time; B needs x+4. A works 4 hours in total, B 6, although their last 2 hours overlap. Add fractions of a batch, not elapsed durations, then multiply by x(x+4):
2/x + 4/(x+4) + 2[1/x+1/(x+4)] = 1 4/x + 6/(x+4) = 1 4(x+4)+6x=x(x+4) x²−6x−16=0 (x−8)(x+2)=0 x=8; x=−2
Reject −2. A needs 8 hours, B 12. The stages produce 1/4, 1/3 and 5/12 of a batch, summing to 1. The batch remains unfinished before the joint stage.
3. Equal selling prices, unequal gains and losses
Reworked classic
Two objects sell for 240 euros each. The first earns a percentage profit on its purchase cost; the second loses the same percentage of its own purchase cost. The combined loss is 20 euros. Find the percentage and both purchase costs.
Hint
The percentages have different bases. Write the rate as a fraction p.
Solution and check
We require 0<p<1. Revenue is 480, so total cost is 500. A selling price equal to cost·(1+p) implies cost=price/(1+p); for a loss use 1−p. Do not multiply the selling prices by those factors.
240/(1+p) + 240/(1−p) = 500 480/(1−p²)=500 p²=1/25 p=1/5; p=−1/5
Choose p=1/5=20%; a negative p reverses the specified roles. Costs are 200 and 300 euros. Profit 40 minus loss 60 gives the required net loss 20.
4. Two identical dilution steps
Reworked classic
A tank contains 100 litres of solution, including 20 litres of dissolved substance. Assume additive volumes and perfect mixing. Remove x litres of solution, replace them with x litres of pure water and mix. Repeat exactly once. The final concentration is 9.8% by volume. Find x.
Hint
Each removal takes the same fraction of the substance currently present, not the same amount.
Solution and check
The domain is 0≤x≤100. Volume returns to 100 litres after each replacement. Each step preserves a fraction 1−x/100 of the substance. Apply that factor twice; 9.8% of 100 litres means 9.8 litres.
20(1−x/100)² = 9.8 (1−x/100)²=49/100 1−x/100=±7/10 x=30; x=170
Reject 170 as physically impossible. Replace 30 litres each time. The first step leaves 14 litres of substance; the second removes 30% of 14, or 4.2, leaving 9.8.
5. A garden whose side lengths are unknown
Reworked classic
A rectangular garden has perimeter 92 m and diagonal 34 m. A uniform-width path runs inside all four edges, leaving a central rectangular lawn of area 240 m². Find the garden dimensions and path width.
Hint
First find the sum and product of the side lengths.
Solution and check
Let the sides be a,b. Perimeter gives a+b=46 and Pythagoras gives a²+b²=1156. Expand (a+b)² to obtain ab. The sides are roots of z²−(a+b)z+ab=0. Only then introduce path width x: subtract 2x from each dimension.
a+b=46; a²+b²=34²=1156 2ab=46²−1156=960 ab=480 z²−46z+480=(z−30)(z−16)=0 (30−2x)(16−2x)=240 4x²−92x+240=0 x²−23x+60=(x−3)(x−20)=0 x=3; x=20
Sides are 30 m and 16 m. Require 0<x<8, so reject 20 and take 3 m. The lawn is 24×10=240; the original perimeter is 92 and diagonal √1156=34. These are two successive quadratics, not a fourth-degree equation.
6. Two alternative theatre layouts
Original setting
A theatre has r rows of c seats. One proposed change removes 2 rows and adds 4 seats to every remaining row, increasing capacity by 16. Another proposal, also relative to the original layout, adds 2 rows and removes 3 seats per row, reducing capacity by 10. Find the original layout and capacity.
Hint
Expand both capacities: the products rc cancel.
Solution and check
r,c are integers with r>2,c>3. Both proposals start from the original capacity rc. After expanding, cancel rc and add the resulting linear equations to eliminate c:
(r−2)(c+4)=rc+16 (r+2)(c−3)=rc−10 4r−2c=24 −3r+2c=−4 r=20; c=28 rc=560
There are 20 rows of 28 seats: 560 seats. The alternatives give 18×32=576 and 22×25=550. Applying the second change to the first would be a different problem.
7. Lost packets and added headers
Original setting
A sender prepares n packets, each containing n−5 payload bytes. Eight packets are lost. The receiver saves all payloads of the remaining packets and adds a 3-byte header to each. The total saved, including headers, is 144 bytes less than the sender’s original payload total. Find n and both totals.
Hint
The receiver has n−8 packets, each occupying n−2 bytes.
Solution and check
n is an integer greater than 8. The sender prepares n(n−5) bytes; the receiver stores (n−8)(n−2). Their difference is 144. Expanding cancels the quadratic terms:
n(n−5) − (n−8)(n−2) = 144 n²−5n − (n²−10n+16) = 144 5n−16=144 n=32
n=32. The sender has 864 bytes and the receiver 720. Alternatively, lost payload is 8·27=216 and added headers are 24·3=72, giving a net loss of 144.
8. A robot keeping its energy reserve
Original setting
A robot starts with 725 energy units. Outbound leg j uses exactly 2j+3 units, for j=1,…,n. It then returns directly to base at a cost of 5n units. There are no other costs or recharges. On return, exactly 20% of its initial energy remains. How many outbound legs did it complete?
Hint
Sum the varying leg costs rather than multiplying by the final leg’s cost.
Solution and check
n is a positive integer. The reserve is 145, so expenditure is 580. Outbound costs are 5,7,…,2n+3. Using 1+…+n=n(n+1)/2 gives n²+4n outbound, plus 5n for returning:
725−0.20·725=580 Σⱼ₌₁ⁿ(2j+3)+5n=580 2·n(n+1)/2+3n+5n=580 n²+9n−580=0 (n−20)(n+29)=0 n=20; n=−29
Reject −29. There are 20 legs. Outbound cost is 20(5+43)/2=480, return cost 100; 145 remains. Before returning the robot has 245 units, sufficient to reach base.
9. An invoice with two unknown percentages
Original setting
A service costs 500 euros before adjustments. Apply an increase of x%, then a discount of (x−10)% to the increased price, then a fixed 20-euro delivery fee not subject to discount. The total is 560. Both percentages are positive and the discount is below 100%. Find x.
Hint
x is measured in percentage points; the increase factor is 1+x/100.
Solution and check
The domain is 10<x<110. Before delivery the price is 540. The two factors multiply because the discount applies after the increase. Keep the quadratic cross-term:
500(1+x/100)[1−(x−10)/100]+20=560 (1+x/100)(1.10−x/100)=1.08 1.10+0.001x−0.0001x²=1.08 x²−10x−200=0 (x−20)(x+10)=0 x=20; x=−10
Choose x=20; −10 violates the domain. Increase: 500→600; 10% discount: 600→540; delivery: 560. The service itself rose by 8%, not 10%.
10. A sensor panel: boundary and interior
Original setting
A full rectangular grid has r rows and c columns of sensors. Its boundary contains 74 sensors, counting each corner once. Every strictly interior sensor is connected to one of 22 full 12-input modules, without repetitions or omissions. Find the grid dimensions and total sensor count.
Hint
Convert modules to an interior count and avoid double-counting corners.
Solution and check
r,c are integers greater than 2. The interior count is 22·12=264. Boundary count is 2c+2(r−2)=74. The interior is (r−2)(c−2). Obtain r+c and rc, then solve the quadratic having r and c as roots:
2r+2c−4=74
(r−2)(c−2)=264
r+c=39
rc−2(r+c)+4=264
rc=338
z²−39z+338=(z−13)(z−26)=0
{r,c}={13,26}The grid is 13×26, or rotated, with 338 sensors. Interior 11×24=264, boundary 338−264=74. Both roots are needed: they are the two dimensions, not rival answers.