Printable · GCSE Higher · ages 14-16
Direct and inverse proportion worksheet — GCSE Higher
Fifteen questions on "direct and inverse proportion" — DfE statement R10. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Direct and inverse proportion worksheet — GCSE Higher
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- 1.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.
- 2.A journey takes 4 hours at an average speed of 60 km/h. The time taken is inversely proportional to the average speed. Work out how long the same journey takes at an average speed of 80 km/h.
- 3.A spring's extension is directly proportional to the force applied to it. A force of 5 N produces an extension of 12 mm. Work out the extension produced by a force of 20 N.
- 4.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.
- 5.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.
- 6.y is directly proportional to x. When x = 5, y = 18. Work out the value of y when x = 15.
- 7.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 8.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 9.Two numbers a and b are in the ratio a : b = 3 : 4. Given that a = 15, work out the value of b.
- 10.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 11.Work out the value of x when 4/x = 2/5.
- 12.A 750 g box of cereal costs £2.70. A 500 g box of the same cereal costs £1.95. Work out which box is better value, and its cost per 100 g.
- 13.y is inversely proportional to x. When x = 5, y = 8. Work out the value of y when x = 10.
- 14.A recipe for 8 muffins needs 200 g of sugar. Sam wants to make 20 muffins for a bake sale, and he already has 350 g of sugar. Work out how many more grams of sugar he needs to buy.
Answer key
- (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) 3 hours — Method: in inverse proportion the product of the two quantities is constant, and here that product is the distance. Working: 60 × 4 = 240 km, so at 80 km/h the time is 240 ÷ 80 = 3. Answer: 3 hours. The distractors: 5 hours 20 minutes comes from treating the relationship as direct, working out 4 × 80 ÷ 60; 2 hours 40 minutes comes from cutting the time by the fraction the speed rose by — the speed went up by one third, so the time was cut by one third — which is not how inverse proportion works; 4 hours comes from dividing the 240 km by the original speed of 60 km/h again instead of by the new speed.
- (c) 48 mm — Method: extension = k × force, where k = extension ÷ force. Working: k = 12 ÷ 5 = 2.4 mm per N. At 20 N: extension = 2.4 × 20 = 48 mm. Wrong options: 32 mm comes from adding the extension and force numbers instead of scaling (12 + 20); 3 mm comes from treating the relationship as inverse proportion (12 × 5 ÷ 20); 36 mm comes from using an incorrect scale factor of 3 between the forces instead of the correct factor of 4 (20 ÷ 5).
- (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.
- (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.
- (c) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
- (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (a) 10 — Method: two equal fractions can be rearranged by cross-multiplying, multiplying each numerator by the other denominator. Working: 4 × 5 = 2 × x, so 2x = 20 and x = 20 ÷ 2 = 10. Answer: 10. The distractors: 20 comes from cross-multiplying to 4 × 5 = 20 and stopping there, without dividing by the 2; 8 comes from multiplying the two numerators, 4 × 2; 2.5 comes from working only with the right-hand fraction, 5 ÷ 2, and ignoring the 4.
- (d) The 750 g box, at 36p per 100 g — Work out the cost per 100 g of each box. 750 g box: 270p ÷ 7.5 = 36p per 100 g. 500 g box: 195p ÷ 5 = 39p per 100 g. The lower cost per 100 g is the better value, so the 750 g box at 36p per 100 g is the answer. Choosing the 500 g box at 39p per 100 g gets the maths right but picks the higher unit price, not realising a smaller cost per 100 g is the better deal. Choosing the 500 g box because £1.95 is lower than £2.70 compares the total prices without allowing for the different pack sizes at all. Working out 270 ÷ 5 = 54p divides the 750 g box's price by the wrong number of hundred-grams (the 500 g box's), giving a rate that belongs to neither box. The 750 g box, at 36p per 100 g, is the better value.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
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