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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Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (b) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (c) 4 : 5 — Method: a formula of the form y = kx says that for every 1 of x there are k of y, so the ratio y : x is k : 1, which is then written with whole numbers and cancelled. Working: here k = 0.8, so y : x = 0.8 : 1; multiplying both parts by 10 gives 8 : 10, and dividing both parts by their highest common factor 2 gives 4 : 5. Answer: 4 : 5. The distractors: 5 : 4 comes from writing x : y, reversing the order asked for; 8 : 10 has the decimal cleared but the ratio left unsimplified, and the question asks for the simplest form; 8 : 1 comes from multiplying only the left-hand part by 10 and leaving the 1 unchanged, which changes the relationship instead of keeping it.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (b) 6 litres — Method: find the volume of the cuboid in cm³, then change cm³ into litres using 1 litre = 1000 cm³. Working: 30 × 20 × 10 = 6000 cm³, and 6000 ÷ 1000 = 6. Answer: 6 litres. The distractors: 60 litres comes from using 1 litre = 100 cm³; 600 litres comes from using 1 litre = 10 cm³; 0.6 litres comes from using 1 litre = 10 000 cm³.
- (b) 48 m² — Method: lengths are multiplied by the scale factor, but areas are multiplied by its square. Working: 1 cm on the plan stands for 200 cm = 2 m, so 1 cm² on the plan stands for 2 × 2 = 4 m², and 12 × 4 = 48. Answer: 48 m². The distractors: 24 m² comes from scaling the area by the length factor 2 instead of by its square; 2400 m² comes from multiplying the area by the scale 200 as though it were a length; 4800 m² comes from working in centimetres, 12 × 200² = 480 000 cm², and then dividing by 100 instead of by 10 000 to reach square metres.
- (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.
- (a) 7 — The average rate of change of y with respect to x over an interval is the change in y divided by the change in x between its two endpoints — the gradient of the chord joining them, not a rate at a single point. At x = 1: 2 × 1 = 2, so y = 1 + 2 = 3. At x = 4: 2 × 4 = 8, so y = 16 + 8 = 24. The change in y is 24 − 3 = 21 and the change in x is 4 − 1 = 3, so the average rate of change is 21 ÷ 3 = 7. Reporting the change in y, 21, on its own is not a rate of change, because it has not been divided by the 3 units of x over which it happened. Reporting 3 is not a rate either — 3 is the width of the interval, and also the value of y at x = 1, and neither of those measures how fast y is changing. Subtracting in the wrong order gives −21 ÷ 3 = −7, the wrong sign. The average rate of change of y with respect to x over the interval is 7.
- (d) 100 — Rearrange L = 0.6L + 40 by collecting the L terms on one side: L − 0.6L = 40, which gives 0.4L = 40, then L = 40 ÷ 0.4 = 100. Subtracting the other way round, 0.6L − L = 40, gives −0.4L = 40, then L = 40 ÷ (−0.4) = −100 — a sign error that flips the answer negative even though a long-run value here must be positive. Ignoring the 0.6L term completely and solving L = 40 directly gives 40, which throws away the recurrence's own multiplier. Dividing 40 by 0.6 instead of by the correct coefficient 0.4 gives 40 ÷ 0.6 ≈ 66.7, a slip that comes from dividing by the coefficient of L on the RIGHT of the original equation rather than by what is left once the L terms are collected on one side. Always collect the L terms first, then divide by whatever coefficient of L remains.
- (d) 400 — The area scale factor is the length scale factor squared: 20² = 400, so the real car's surface area is 400 times the model's. 20 comes from using the length scale factor itself, without squaring it. 8000 comes from cubing the length scale factor (20³), instead of squaring it — cubing is the rule for volume, not area. 40 comes from doubling the length scale factor (2 × 20), instead of squaring it.
- (a) 2 — Since y is inversely proportional to x², y = k/x². Using x = 2, y = 8: 2² = 4, so 8 = k ÷ 4, giving k = 8 × 4 = 32. The equation is y = 32/x². When x = 4: 4² = 16, so y = 32 ÷ 16 = 2. Treating the relationship as inversely proportional to x itself, rather than to x², gives k = 8 × 2 = 16 and then y = 16 ÷ 4 = 4, a different relationship. Using x instead of x² in the new calculation gives y = 32 ÷ 4 = 8, skipping the square. Multiplying by x² instead of dividing by it gives y = 32 × 16 = 512, the wrong operation for an inverse relationship. When x = 4, y = 2.
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
- (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) 9:15:21 — 6 : 10 : 14 simplifies to 3 : 5 : 7 (divide every part by 2). Multiplying every part of 3 : 5 : 7 by 3 gives 9 : 15 : 21, so 9 : 15 : 21 is equivalent to 6 : 10 : 14. Adding 2 to every part of 6 : 10 : 14 gives 8 : 12 : 16, which is not equivalent — ratios are equivalent when every part is multiplied by the same number, not when the same number is added to every part. Doubling only the first two parts, 6 × 2 = 12 and 10 × 2 = 20, but leaving the third part unchanged at 14, gives 12 : 20 : 14 — a scaling applied to two parts and not the third. Cancelling the first two parts correctly, 6 ÷ 2 = 3 and 10 ÷ 2 = 5, then treating the three numbers as a sequence and making the third part the sum of the first two, 3 + 5 = 8, gives 3 : 5 : 8 — the third part was never divided by 2 at all.
- (b) 5:8 — The mix has 5 parts sand and 3 parts cement, so 5 + 3 = 8 parts in total. Sand to total is 5 : 8, and since the highest common factor of 5 and 8 is 1, this is already in its simplest form. Giving 5 : 3 answers sand to cement, not sand to the total mix. Giving 3 : 8 is cement to total, the wrong part of the mix. Giving 8 : 5 has the total and the sand swapped round.
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