Ratio, Proportion & Rates of Change

A car costing £8,000 depreciates by 10% in a year. What is it worth after one year?

• A£7,200✓
• B£7,900
• C£8,800
• D£800

Answer: 10% of £8,000 = £800. £8,000 − £800 = £7,200 (or 0.9 × £8,000).

£500 is invested at 4% simple interest per year. How much interest is earned in 3 years?

• A£540
• B£600
• C£60✓
• D£20

Answer: Simple interest = 0.04 × 500 × 3 = £60.

y is directly proportional to x, and y = 12 when x = 4. Find y when x = 7.

• A19
• B15
• C21✓
• D28

Answer: y = kx; 12 = k×4 so k = 3. When x = 7, y = 3 × 7 = 21.

Increase £80 by 15%.

• A£92✓
• B£104
• C£95
• D£68

Answer: 15% of 80 = 12. £80 + £12 = £92 (or 1.15 × 80).

A price of £60 includes 20% VAT. What was the price before VAT?

• A£40
• B£50✓
• C£72
• D£48

Answer: £60 is 120% of the original. Original = 60 ÷ 1.2 = £50.

y is directly proportional to x². When x = 4, y = 48. Work out the value of y when x = 7.

• A21
• B84
• C147✓
• D588

Answer: y = kx², so 48 = k × 16 and k = 3. Then y = 3 × 7² = 3 × 49 = 147. Using y = kx instead would give a constant of 12 and miss the squaring.

y is inversely proportional to √x. When x = 9, y = 12. Work out the value of y when x = 36.

• A6✓
• B24
• C3
• D1

Answer: y = k/√x, so k = y√x = 12 × 3 = 36. When x = 36, √x = 6, so y = 36 ÷ 6 = 6. Quadrupling x only doubles √x, so y halves rather than quartering.

£4,500 is invested at 3.2% compound interest per year. Work out the value of the investment after 6 years, to the nearest penny.

• A£5,364.00
• B£936.14
• C£5,267.58
• D£5,436.14✓

Answer: Each year the amount is multiplied by 1.032, so after 6 years the value is 4500 × 1.032⁶ = £5,436.14. Working out 6 × 3.2% of £4,500 would give simple interest, which is less.

A machine bought for £18,000 loses 15% of its value each year. Work out its value after 4 years, to the nearest pound.

• A£7,200
• B£9,396✓
• C£11,054
• D£31,482

Answer: Each year the value is multiplied by 0.85, so after 4 years it is 18000 × 0.85⁴ = £9,396. Taking 4 × 15% = 60% off in one go removes too much, because each year's loss is 15% of a smaller amount.

A velocity–time graph for a cyclist has three straight sections: the velocity rises steadily from 0 to 8 m/s during the first 5 seconds, stays at 8 m/s for the next 12 seconds, then falls steadily to 0 over the final 4 seconds. Work out the total distance travelled.

• A96 m
• B116 m
• C168 m
• D132 m✓

Answer: Distance is the area under the graph: ½ × 5 × 8 = 20, plus 12 × 8 = 96, plus ½ × 4 × 8 = 16, giving 132 m. Treating the sloping sections as rectangles double-counts them.

On a velocity–time graph, a car's velocity increases in a straight line from 6 m/s at t = 3 s to 27 m/s at t = 10 s. Work out the acceleration over this section.

• A3 m/s²✓
• B2.1 m/s²
• C2.7 m/s²
• D21 m/s²

Answer: Acceleration is the gradient of a velocity–time graph: (27 − 6) ÷ (10 − 3) = 21 ÷ 7 = 3 m/s². The change in time is 7 s, not 10 s.

The depth of water, d cm, in a container is plotted against time, t seconds. A tangent drawn to the curve at t = 20 passes through the points (10, 12) and (30, 40). Estimate the rate at which the depth is increasing at t = 20 seconds.

• A28 cm/s
• B0.71 cm/s
• C1.4 cm/s✓
• D2.0 cm/s

Answer: The rate of change is the gradient of the tangent: (40 − 12) ÷ (30 − 10) = 28 ÷ 20 = 1.4 cm/s. Dividing the time change by the depth change inverts the gradient.