AQA A-level Mathematics: Problem Solving, Accuracy and Modelling (7357) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover outcomes OT2.1 to OT2.7 and OT3.1 to OT3.5 of the AQA A-level Mathematics (7357) specification, for first teaching 2017. Use g = 9.8 m s⁻² where needed. Learn the ideas in the study guide or the revision notes. See also the course hub, the printable checklist and the free diagnostics.

Questions

1. A school wants to decide how many photocopiers to buy. Describe, in order, the four stages of the problem-solving cycle, giving a specific action at each stage for this problem. [4]

2. Plan P costs £8 per month plus 5p per minute of calls. Plan Q costs £14 per month plus 2p per minute.

(a) Write an expression for the monthly cost of each plan for m minutes. [2] (b) Find the number of minutes for which the two plans cost the same. [1] (c) A customer uses 150 minutes a month. Which plan is cheaper, and by how much? [2]

3. Let f(x) = x³ + 4x − 7. Show that f(x) = 0 has a root which is 1.3 to 1 decimal place. [3]

4. A plot measures 16.4 m by 7.2 m, each to 1 decimal place.

(a) Find the lower and upper bounds for its area. [2] (b) Give the area to a justified degree of accuracy, with a reason. [2]

5. A hall is 20 m long, 12 m wide and 5 m high. One tin of paint covers 40 m² for one coat. Estimate the tins needed for two coats on the four walls. State your assumption and whether it makes the estimate too high or too low. [4]

6. A box of mass 4 kg is held at rest on a smooth slope inclined at 20° to the horizontal by a light string parallel to the slope.

(a) Draw and label a force diagram. [2] (b) Find the tension in the string. [2] (c) Find the normal reaction. [1]

7. A tank holds 800 litres. A model for the volume V litres left t minutes after the outlet opens is V = 800 − 24t.

(a) State two assumptions of this model. [2] (b) Find when the model says the tank is empty. [1] (c) At t = 20 the measured volume is 395 litres. Compare this with the model. [1] (d) A second model is V = 800e^(−kt), where V = 560 at t = 10. Find k to 3 significant figures, and decide which model fits the measurement at t = 20 better. [3]

8. A parcel of mass 3 kg is released from rest at the top of a straight chute 4 m long, inclined at 30° to the horizontal.

(a) Model the parcel as a particle on a smooth chute. Find its speed at the bottom. [3] (b) The measured speed is 5.4 m s⁻¹. Say what this suggests about the model, then find the coefficient of friction needed to match it. [3]

9. A van makes a 360 km journey at a constant speed of v km h⁻¹. The running cost is £(30 + 0.00003v³) per hour, which includes the driver’s pay.

(a) Show that the total cost is C = 10 800/v + 0.0108v². [2] (b) Find the speed that minimises C, and justify that it is a minimum. [4] (c) The road has a speed limit of 70 km h⁻¹. State the cheapest allowed speed and its cost, and compare it with the answer to part (b). [2]

10. A model for the number of subscribers N thousand, t months after a service launches, is N = 12e^(0.15t). The market contains at most 400 thousand potential subscribers.

(a) Find when the model gives N = 100. [2] (b) Use t = 36 to show the model is unsuitable for long-term prediction. [2] (c) At t = 24 the true value is 250. A refined model is N = 400 − 388e^(−kt). Find k to 3 significant figures, then predict N when t = 36. [3]

11. Weekly sales t weeks after launch are modelled by S = 5e^(0.4t) (model A) or S = 40 + 12t (model B).

(a) Let g(t) = 5e^(0.4t) − 40 − 12t. Show that g(t) = 0 has a root between 8 and 8.5. [2] (b) Show that g(t) = 0 can be written t = ln((40 + 12t)/5)/0.4. Starting from t₀ = 8, find t₁ and t₂ to 3 decimal places. [3] (c) Show that the root is 8.33 to 2 decimal places. [2] (d) Interpret the root in context, and comment on the models at 20 weeks. [1]

Answers

1. Specify: decide exactly what is wanted, for example the smallest number of copiers so that no job waits more than 10 minutes [1]. Collect: gather copies needed per day, copier speed, opening hours and cost [1]. Process and represent: set demand against capacity and calculate the smallest number that meets it [1]. Interpret: state the number in context, check it against budget and queue times, and repeat the cycle with changed assumptions if it is unaffordable or implausible [1]. Examiner insight: Answers are more likely to gain credit when each stage names an action specific to the copiers; generic phrases such as “do the maths” say little.

2. (a) P = 8 + 0.05m (pounds) [1]; Q = 14 + 0.02m [1]. (b) 8 + 0.05m = 14 + 0.02m, so 0.03m = 6 and m = 200 minutes [1]. (c) At 150 minutes: P = £15.50 and Q = £17.00 [1]. Plan P is cheaper, by £1.50 [1]. Examiner insight: Mixing pence and pounds is a common slip; convert once at the start.

3. f(1.25) = 1.953125 + 5 − 7 = −0.047 [1]. f(1.3) = 2.197 + 5.2 − 7 = 0.397 [1]. f is continuous and changes sign, so a root lies in 1.25 < x < 1.3; every number in this interval rounds to 1.3 to 1 d.p. [1] Examiner insight: A “show that” answer is written for a reader who has not been told the result, so give both values, the sign change, and the conclusion.

4. (a) Lower: 16.35 × 7.15 = 116.9025 [1]. Upper: 16.45 × 7.25 = 119.2625 [1]. (b) Both bounds round to 120 at 2 significant figures [1], so the area is 120 m² to 2 significant figures; the product 118.08 would suggest more precision than the data has [1]. Examiner insight: Show the bounds you use for each length; a bare final answer leaves nothing to credit if it is wrong.

5. Wall area = 2(20 + 12) × 5 = 320 m² [1]. Two coats: 640 m² [1]. Tins = 640 ÷ 40 = 16 [1]. Assumption: doors and windows are ignored, so the wall area is too large and the estimate is too high [1]. Examiner insight: An estimate question rewards the stated assumption and its direction of error, not just the number.

6. (a) Three forces: weight 4g = 39.2 N vertically down, normal reaction R perpendicular to the slope, tension T up the slope [1]. The 20° marked between the weight and the perpendicular to the slope, or between the slope and the horizontal [1]. (b) Along the slope: T = 4g sin 20° [1] = 13.4 N [1]. (c) R = 4g cos 20° = 36.8 N [1]. Examiner insight: Label each force and mark the angle; a labelled diagram helps you catch sin and cos swapped.

7. (a) Water leaves at a constant rate [1]; nothing enters the tank, and the 800 litres is the starting volume [1]. (b) 800 − 24t = 0, so t = 33.3 minutes [1]. (c) The model gives 320 litres at t = 20, but 395 was measured: the model underestimates the volume by 75 litres, so a constant outflow rate is not realistic [1]. (d) 560 = 800e^(−10k), so k = ln(800/560)/10 [1]; k = 0.0357 [1]. At t = 20, V = 800e^(−0.714) = 392, which is much closer to 395 than 320 is, so the exponential model fits better [1]. Examiner insight: Comparing two models needs a number from each, not just the word “better”.

8. (a) Along the chute: a = g sin 30° = 4.9 m s⁻² [1]. v² = 2(4.9)(4) = 39.2 [1]. v = 6.26 m s⁻¹ [1]. (b) The model predicts a higher speed than measured, which is consistent with friction being ignored [1]. Required acceleration: a = 5.4²/(2 × 4) = 3.645 [1]. With friction, a = g(sin 30° − μ cos 30°), so 3.645 = 4.9 − 8.487μ and μ = 0.148 [1]. Examiner insight: The mass of the parcel cancels.

9. (a) Time = 360/v hours [1]. C = (360/v)(30 + 0.00003v³) = 10 800/v + 0.0108v² [1]. (b) dC/dv = −10 800/v² + 0.0216v = 0 [1], so v³ = 500 000 [1] and v = 79.4 km h⁻¹ [1]. d²C/dv² = 21 600/v³ + 0.0216 > 0 for v > 0, so this is a minimum [1]. (c) C decreases for 0 < v < 79.4, so the cheapest allowed speed is the limit, 70 km h⁻¹ [1], with C = 154.29 + 52.92 = £207 compared with £204 at 79.4 km h⁻¹ [1]. Examiner insight: The unconstrained optimum lies outside the allowed range, so the answer in context is the boundary value, and the reason should be stated.

10. (a) 12e^(0.15t) = 100, so t = ln(100/12)/0.15 [1] = 14.1 months [1]. (b) At t = 36, N = 12e^(5.4) ≈ 2660 thousand [1], which is far above the market limit of 400 thousand, so unlimited growth is not a valid assumption in the long term [1]. (c) 400 − 388e^(−24k) = 250, so e^(−24k) = 150/388 [1], giving k = 0.0396 [1]. At t = 36, N = 400 − 388e^(−1.426) = 307 thousand [1]. Examiner insight: Quote the figure that breaks the assumption, here 2660 against 400.

11. (a) g(8) = 5e^(3.2) − 40 − 96 = −13.3 and g(8.5) = 5e^(3.4) − 40 − 102 = 7.82 [1]. g is continuous and changes sign, so there is a root between 8 and 8.5 [1]. (b) 5e^(0.4t) = 40 + 12t, so e^(0.4t) = (40 + 12t)/5 and t = ln((40 + 12t)/5)/0.4 [1]. t₁ = 8.258 [1], t₂ = 8.314 [1]. (c) g(8.325) = −0.208 and g(8.335) = 0.232 [1]. The sign change means the root lies in 8.325 < t < 8.335, which rounds to 8.33 [1]. (d) The models predict equal weekly sales after about 8.3 weeks, and model A predicts more after that; at 20 weeks model A gives about 14 900 and model B gives 280, so data must decide which to trust [1]. Examiner insight: Use two values of g to either side of the root for the accuracy argument. The iteration values alone do not prove the second decimal place.

Where marks are usually lost

• Writing generic stage names in a cycle question without tying them to the situation.
• Saying “the model is unrealistic” without naming an assumption and its effect.
• Quoting a raw product with more figures than the given data supports.
• Showing a sign change at the interval ends without saying the function is continuous.
• Ignoring a restriction on the allowed range.
• Omitting a force or an angle from a force diagram.
• Giving an optimisation value with no units or meaning.
• Claiming an iteration has converged after only one or two steps.

Next steps

• Go back over the revision notes and the study guide.
• Use the course hub and printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: AQA A-level Mathematics 7357, sections 3.1.2 (OT2) and 3.1.3 (OT3), published by AQA, for first teaching 2017.