Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
- 2.A plan is drawn to a scale of 1 : 500. Work out how many metres are represented by 1 cm on the plan.
- 3.A car uses 6 litres of petrol to travel 90 km. Write the ratio of the distance travelled in km to the amount of petrol used in litres, in its simplest form.
- 4.A laptop bag has a mass of 2.4 kg. A school bag has a mass of 3.6 kg. Write the mass of the laptop bag as a fraction of the mass of the school bag, giving your answer in its simplest form.
- 5.A force of 20 N acts on an area of 4 m². Using pressure = force ÷ area, work out the pressure, in pascals.
- 6.Amelia uses 2 kg of flour to bake 5 cakes. Using the same recipe, work out how many cakes she can bake with 6 kg of flour.
- 7.A cyclist rides at a steady speed of 20 mph. Given that 1 mile ≈ 1.6 km, work out the cyclist's speed in km/h.
- 8.There are 84 passengers on a double-decker bus. 35 of the passengers are adults and the rest are children. Write the number of adults as a fraction of the number of children. Give your answer in its simplest form.
- 9.An amount of money is shared in the ratio 1:2:3. The largest share is £90 more than the smallest share. Work out the total amount that was shared.
- 10.A concrete block exerts a force of 45 N on the ground through a base with an area of 0.05 m². Work out the pressure on the ground, in N/m².
- 11.A plumber's charge, C pounds, for a job lasting h hours is shown on a straight-line graph. The line passes through the points (1, 45) and (3, 85). Work out what the gradient of this line represents, in context.
- 12.The number of biscuits, b, that each guest at a party receives is inversely proportional to the number of guests, g. Which equation could show this relationship?
- 13.The ratio of two lengths is 0.6 : 1.5. Write this ratio in its simplest whole-number form.
- 14.Write the ratio 4 : 9 in the form 1 : n.
- 15.A road sign 3 m tall casts a shadow 5 m long. At the same time, a nearby lamppost casts a shadow 10 m long. Work out the height of the lamppost.
Answer key
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
- (b) 5 m — A scale of 1 : 500 means 1 cm on the plan represents 500 cm in real life. Converting to metres: 500 ÷ 100 = 5 m. 500 m comes from forgetting to convert the 500 cm into metres at all. 50 m comes from dividing by 10 instead of 100 when converting centimetres to metres: 500 ÷ 10 = 50. 0.5 m comes from dividing by 1000 instead of 100, as if converting to kilometres instead of metres: 500 ÷ 1000 = 0.5.
- (a) 15:1 — Write the ratio distance : petrol using the numbers in the question: 90 : 6. Divide both parts by their highest common factor, 6, to give 15 : 1. (1:15 comes from writing the ratio the wrong way round, petrol : distance. 5:2 comes from dividing only the distance by 6 and leaving the 6 litres unchanged, giving 15:6, which does simplify to 5:2 — the error is the step before, because both parts of a ratio must be divided by the same number. 14:1 comes from subtracting at the first step instead of dividing, 90 − 6 = 84, and then writing 84 : 6 and dividing both parts by 6 to get 14 : 1; subtracting has no part in simplifying a ratio.)
- (a) 2/3 — Put the laptop bag's mass over the school bag's mass: 2.4/3.6. Multiply both numbers by 10 to clear the decimals: 24/36. Divide both by their highest common factor, 12: 24÷12 = 2, 36÷12 = 3, giving 2/3. (3/2 comes from writing the masses the wrong way round. 1/3 comes from finding the difference in the masses, 3.6 − 2.4 = 1.2 kg, and writing it as a fraction of the school bag's mass, 1.2/3.6. 2/5 comes from comparing the laptop bag's mass to the total mass of both bags, 2.4/6.)
- (a) 5 — Pressure = force ÷ area = 20 ÷ 4 = 5 pascals. Getting 80 comes from multiplying the force and area instead of dividing. Getting 16 comes from subtracting the area from the force (20 − 4 = 16) instead of dividing. Getting 24 comes from adding the force and area (20 + 4 = 24) instead of dividing.
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (c) 32 km/h — Method: to change mph into km/h, multiply by the number of kilometres in a mile. Working: 20 × 1.6 = 32 km/h. So the cyclist's speed is 32 km/h. Distractor 12.5 km/h comes from dividing by 1.6 instead of multiplying. Distractor 21.6 km/h comes from adding 1.6 instead of multiplying by it. Distractor 20 km/h comes from not converting the units at all.
- (c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
- (d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (b) 900 N/m² — Pressure = force ÷ area, so 45 ÷ 0.05 = 900 N/m². Working out 45 ÷ 5 = 9 misplaces the decimal point in the area, treating 0.05 m² as 5 m². Working out 45 × 0.05 = 2.25 multiplies force and area together instead of dividing. Working out 45 + 0.05 = 45.05 simply adds the two given numbers, which does not give a pressure at all. The pressure on the ground is 900 N/m².
- (a) The plumber charges £20 for each extra hour worked — The gradient is the change in C divided by the change in h: (85 − 45) ÷ (3 − 1) = 40 ÷ 2 = £20. On this graph the gradient represents the extra amount charged for each extra hour worked, so the answer is 'The plumber charges £20 for each extra hour worked'. Not dividing by the change in h gives 40, the option that reads 'charges £40 for each extra hour worked' — that is the total change in cost between the two points, not the rate per hour. Reading off the C-value of the first point, 45, gives the option using £45 — that is the cost of a job lasting 1 hour, not the rate. Treating the gradient as a flat total charge regardless of the time taken misunderstands what a straight-line graph through two different h-values shows: the cost does depend on h, so it cannot be a single fixed total.
- (d) b = k ÷ g — Inverse proportion means b × g stays constant (equal to k), so b = k ÷ g. 'b = kg' is the equation for direct proportion, not inverse. 'b = k + g' treats the relationship as additive, which is not proportion at all. 'b = g ÷ k' is direct proportion in disguise: rearranging it gives g = kb, so b still grows as g grows. For inverse proportion the g must be underneath the constant, not above it.
- (d) 2:5 — Multiply both parts by 10 to clear the decimals: 6 : 15. Divide both parts by their common factor 3: 2 : 5.
- (b) 1 : 2.25 — Method: to write a ratio in the form 1 : n, divide both parts by the first part. Working: 4 ÷ 4 = 1 and 9 ÷ 4 = 2.25, so 4 : 9 = 1 : 2.25. Working out 9 ÷ 4 = 2.25 correctly but then writing it as the first part gives 2.25 : 1, the two parts the wrong way round. Subtracting 9 − 4 = 5 gives 1 : 5, confusing the difference between the parts with the ratio. Multiplying 4 × 9 = 36 gives 1 : 36, confusing the product of the parts with the ratio.
- (c) 6.00 m — Method: the ratio of height to shadow length is the same for both objects. Working: road sign height ÷ shadow = 3 ÷ 5 = 0.6. Lamppost height = 0.6 × 10 = 6.00 m. Wrong options: 16.67 m comes from inverting the ratio, using shadow ÷ height instead of height ÷ shadow (10 × 5 ÷ 3); 8.00 m comes from adding the difference between the two shadow lengths to the road sign's height instead of scaling (3 + (10 − 5)); 1.50 m comes from multiplying by the ratio of the two shadow lengths the wrong way round (3 × 5 ÷ 10).
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