Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
- 2.Two numbers A and B are in the ratio 3:5. When 6 is added to B, the ratio of A to the new value of B is 1:2. Work out the value of A.
- 3.Tap A fills a swimming pool in 6 hours. Tap B pours water twice as fast as tap A. The time taken to fill the pool is inversely proportional to the rate of flow. Work out how long tap B takes to fill the pool.
- 4.Two numbers are in the ratio 3 : 7. Their sum is 60. Work out the positive difference between the two numbers.
- 5.A car travels for 3 hours at an average speed of 80 km/h and then for 1 hour at an average speed of 40 km/h. Work out the average speed for the whole journey.
- 6.Two mathematically similar water bottles have a volume scale factor of 8 from the smaller bottle to the larger bottle. Work out the surface area scale factor from the smaller bottle to the larger bottle.
- 7.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 8.A stall sells bottles of juice. Two bottles cost £3.00, five bottles cost £7.50 and eight bottles cost £12.00. The cost, C pounds, is in a fixed ratio to the number of bottles, n. Write down a formula for C in terms of n.
- 9.A recipe uses flour, sugar and butter in the ratio 8 : 3 : 5. Write the ratio of flour to the rest of the mixture (sugar and butter combined) in its simplest form.
- 10.A tangent to a curve at the point where x = 3 passes through the points (1, 4) and (5, 20). Work out the gradient of the tangent at x = 3, and hence the instantaneous rate of change of y with respect to x at that point.
- 11.It takes 8 painters 6 days to paint a fence. Working at the same rate, work out how many days 4 painters would take to paint the same fence.
- 12.Two scale plans of the same football pitch are drawn: Plan A has a scale of 1 : 1000 and Plan B has a scale of 1 : 200. A pitch length is 8 cm on Plan A. Work out the length of the same pitch on Plan B, in centimetres.
- 13.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 14.A recipe needs 1.2 kg of flour. Dan has 750 g of flour in his cupboard. Write the mass of flour the recipe needs as a fraction of the mass of flour Dan has. Give your answer in its simplest form.
- 15.A water tank has a capacity of 500 litres. It currently holds 350 litres of water. Write the amount of empty space in the tank as a fraction of the tank's capacity. Give your answer in its simplest form.
Answer key
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (a) 18 — Method: write both numbers with the same multiplier, turn the second ratio into an equation by cross-multiplying, solve for the multiplier and then build A from it. Working: let A = 3k and B = 5k, so 3k : (5k + 6) = 1 : 2; cross-multiplying gives 2 × 3k = 5k + 6, so 6k = 5k + 6 and k = 6; A = 3 × 6 = 18. Answer: 18, and the check works, because B = 30, B + 6 = 36 and 18:36 = 1:2. The distractors: 9 comes from reading the 6 as the difference between the two numbers — 5 − 3 = 2 parts, so one part is 3 and A is 3 × 3 — but the 6 is added to B, it is not the gap between A and B; 6 comes from solving A : (A + 6) = 1 : 2, adding the 6 to A instead of to B; 30 is the value of B, found from the correct multiplier but given in place of A.
- (a) 3 hours — Method: for a fixed pool the rate of flow multiplied by the time taken is constant, so multiplying the rate by a factor divides the time by that same factor. Working: tap B's rate is 2 times tap A's rate, so tap B's time is 6 ÷ 2 = 3 hours. Answer: 3 hours. The distractors: 12 hours comes from multiplying the time by 2 as well, which treats the time as directly proportional to the rate and has the faster tap taking longer; 4 hours comes from reading ‘twice as fast’ additively, as two hours quicker, and working out 6 − 2 instead of scaling the time by a factor of 2; 1.5 hours comes from applying the factor of 2 twice, halving 6 to 3 and then halving again.
- (d) 24 — Method: split 60 into 3 + 7 = 10 equal parts, find the value of one part, then use the difference in ratio parts. Working: 60 ÷ 10 = 6, so the numbers are 3 × 6 = 18 and 7 × 6 = 42, and their difference is 42 − 18 = 24. Answer: 24. 4 comes from finding the difference between the ratio numbers, 7 − 3, but forgetting to multiply by the value of one part. 60 comes from adding the two numbers back together instead of subtracting, which just repeats the given sum. 80 comes from dividing 60 by the first ratio number, 3, instead of by the total number of parts, 10, giving a part value of 20 and a difference of 7 × 20 − 3 × 20 = 80.
- (a) 70 km/h — Method: average speed for a whole journey is the total distance divided by the total time, not the mean of the separate speeds. Working: 3 × 80 = 240 km and 1 × 40 = 40 km, giving 280 km in 3 + 1 = 4 hours, so 280 ÷ 4 = 70. Answer: 70 km/h. The distractors: 60 km/h comes from taking the mean of 80 and 40, which would only be right if equal times were spent at each speed; 280 km/h is the total distance written with a speed unit, from forgetting to divide by the total time; 80 km/h comes from quoting the speed of the longer leg as the average for the whole journey.
- (b) 4 — Method: find the length scale factor by taking the cube root of the volume scale factor, then square it to get the area scale factor. Working: 8 = 2³, so the length scale factor is 2, and the area scale factor is 2² = 4. Answer: 4. 8 comes from using the volume scale factor itself as if it were the area scale factor. 64 comes from squaring the volume scale factor, 8² = 64, instead of first taking its cube root. 2 comes from correctly finding the length scale factor but then forgetting to square it.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (d) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (c) 40 cm — First find the real pitch length using Plan A's scale: 8 × 1000 = 8000 cm. Now find its length on Plan B using Plan B's scale: 8000 ÷ 200 = 40 cm. Assuming both plans use the same scale and leaving the length as 8 cm ignores that Plan B has a different scale. Using the ratio of the two scales upside down, 200 ÷ 1000 = 0.2, then 8 × 0.2 = 1.6, gives 1.6 cm — Plan B is drawn to a larger scale than Plan A, so lengths on it should be bigger, not smaller. Dividing by 2000 instead of 200 gives 4 cm, ten times too small.
- (b) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (d) 8/5 — Two masses can only be compared once they are in the same unit. Since 1 kg is 1000 g, the recipe needs 1200 g. The recipe's mass is being written as a fraction of Dan's mass, so 1200 goes on the top and 750 on the bottom, giving 1200/750. The highest common factor of the two is 150: 1200 ÷ 150 = 8 and 750 ÷ 150 = 5. The fraction is 8/5, which is greater than 1 because the recipe needs more flour than Dan has.
- (b) 3/10 — Work out the empty space: 500 − 350 = 150 litres. Form the fraction 150/500; both numbers share a factor of 50, so 150 ÷ 50 = 3 and 500 ÷ 50 = 10, giving 3/10. 7/10 comes from writing the fraction of the tank that is full (350/500), instead of the empty space. 1/2 comes from miscalculating 500 − 350 as 250 instead of 150. 3/7 comes from comparing the empty space with the water held (150/350), instead of with the tank's total capacity.
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