Printable · GCSE Foundation · ages 14-16
Proportion as equality of ratios worksheet — GCSE Foundation
Fifteen questions on "proportion as equality of ratios" — DfE statement R7. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Answer key: Proportion as equality of ratios worksheet — GCSE Foundation
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- (c) Yes — 700 ÷ 7 = 100 ml for the 1 part of concentrate — Method: add the ratio parts to find the total number of parts, divide the total volume by this, then use the ratio to find concentrate's share. Working: 1 + 6 = 7 parts. 700 ÷ 7 = 100 ml per part. Concentrate = 1 part = 100 ml, so Freya is correct. Wrong options: 'divide 700 by 6' uses only one of the ratio numbers instead of the total of 7 parts, giving about 117 ml; '600 ml is concentrate' swaps which ratio number belongs to the concentrate and which belongs to the water; 'half of 700 ml should be concentrate' ignores the ratio altogether and assumes an equal split.
- (c) 25 — Method: divide the larger number by its ratio part to find the value of one part, then multiply by the smaller number's ratio part. Working: 40 ÷ 8 = 5 (value of one part). Smaller number = 5 × 5 = 25. Wrong options: 64 comes from dividing by the smaller ratio part instead of the larger (40 ÷ 5 × 8); 45 comes from adding the value of one part onto 40 instead of scaling down (40 + 5); 35 comes from subtracting the value of one part from 40 (40 − 5) instead of multiplying it by the smaller ratio part.
- (b) 12 cm — Method: height:width = 3:2 means both lengths are built from parts of the same size, so write 3:2 = 18:x, find one part and multiply by the number of parts in the width. Working: the height is 3 parts and measures 18 cm, so one part is 18 ÷ 3 = 6 cm; the width is 2 parts, so 2 × 6 = 12 cm. Answer: 12 cm. The distractors: 27 cm comes from using the ratio the wrong way round, 18 ÷ 2 × 3, which makes the width longer than the height even though 2 is the smaller part; 6 cm is the value of one part, given as the width instead of being doubled; 17 cm comes from treating the ratio as a difference — 3 − 2 = 1, so 1 cm is taken off the height — but a ratio scales the lengths, it does not subtract from them.
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (a) 20% — Method: a percentage concentration compares the sugar with the whole solution, so add the two masses to get the mass of solution and then scale the ratio of sugar to solution to a denominator of 100. Working: the solution has a mass of 200 + 50 = 250 g; sugar:solution = 50:250, and scaling to per 100 gives 50 ÷ 250 × 100 = 20, so the ratio is 20:100. Answer: 20%. The distractors: 25% comes from comparing the sugar with the 200 g of water, 50:200, rather than with the whole solution; 80% is the percentage of the solution that is water, 200:250, which answers for the wrong part of the mixture; 0.2% comes from working out 50 ÷ 250 = 0.2 and writing that decimal down as a percentage without multiplying by 100.
- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (a) 160 g — Method: use the ratio 20:100 to find the mass of the whole solution from the mass of acid, then take the acid away to leave the water. Working: 20:100 = 40:m, and 40 ÷ 20 = 2, so m = 2 × 100 = 200 g of solution; the water is 200 − 40 = 160 g. Answer: 160 g. The distractors: 200 g is the mass of the whole solution, which is the middle step and includes the acid the question asks you to leave out; 8 g comes from working out 20% of 40 g, which treats the 40 g as the whole solution rather than as the 20% inside it; 10 g comes from reading the 40 g as the 80% that is water, giving a solution of 50 g and a difference of 50 − 40.
- (d) Yes, because 4 × 9 = 6 × 6 — Method: two ratios are equal when their cross-products are equal, so multiply the first part of each ratio by the second part of the other. Working: 4 × 9 = 36 and 6 × 6 = 36; the two products match, so the ratios are equal, and simplifying both to 2:3 shows the same thing. Answer: yes, because 4 × 9 = 6 × 6. The distractors: the reason that the number 6 appears in both ratios reaches the right verdict from a surface match, since a figure shared by two ratios says nothing about equivalence — 4:6 and 6:5 share a 6 and are not equal; the reason built on 9 − 6 and 6 − 4 compares the differences inside each ratio, 3 against 2, which is additive thinking and ends at a verdict of no; the reason built on 4 × 6 and 6 × 9 multiplies the two parts of each ratio together instead of across the pair, giving 24 against 54 and again a verdict of no.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (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.
- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (c) 500 g — Method: a concentration of 10% is the ratio 10:100, and the salt and the solution in the beaker must be in that same ratio, so write 10:100 = 50:m and scale. Working: 50 ÷ 10 = 5, so the salt is 5 times the 10 of the ratio; the solution must be 5 times the 100 of the ratio, giving 5 × 100 = 500 g. Answer: 500 g. The distractors: 5 g comes from working out 10% of 50 g, which treats the 50 g as the whole solution when it is the salt inside it; 450 g comes from scaling correctly and then taking the 50 g of salt away, which gives the mass of water rather than the mass of the whole solution; 5000 g comes from dividing by 0.01 instead of 0.1, that is from writing 10% as 0.01.
- (d) 450 g — Find the amount of rice per person: 300 ÷ 4 = 75 g. Multiply by the new number of people: 75 × 6 = 450 g. Giving 200 g swaps which number the rice is divided and multiplied by (300 ÷ 6 × 4 = 200), scaling the wrong way. Giving 180 g uses 4 + 6 = 10 as the base instead of the original 4 people (300 × 6 ÷ 10 = 180). Giving 400 g assumes each of the 2 extra people needs 300 ÷ 6 = 50 g on top of the original 300 g (300 + 2 × 50 = 400), rather than scaling the whole amount in proportion.
- (b) 250 g — Method: adding water changes the total mass but not the mass of salt, so find the salt, hold it fixed, use the new ratio to find the new total mass and subtract the mass already in the beaker. Working: 12:100 = x:500 gives 12 ÷ 100 × 500 = 60 g of salt; that 60 g must be 8% of the new mixture, so 8:100 = 60:y gives y = 60 ÷ 8 × 100 = 750 g; the water added is 750 − 500 = 250 g. Answer: 250 g. The distractors: 750 g is the mass of the diluted solution, given without taking away the 500 g that was in the beaker to start with; 60 g is the mass of salt, the quantity that stays the same, given instead of the mass of water; 20 g comes from treating the fall from 12% to 8% as 4% of the original 500 g, which measures a change in concentration as though it were a mass of water.
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