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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- (d) The candle's height decreases by 0.3 cm every minute. — A negative gradient means the quantity on the vertical axis decreases as the quantity on the horizontal axis increases. The size of the gradient, 0.3, gives the amount of decrease per minute.
- (c) 500 — Pressure = force ÷ area = 250 ÷ 0.5 = 500 pascals. Getting 125 comes from multiplying the force by the area instead of dividing (250 × 0.5 = 125). Getting 249.5 comes from subtracting the area from the force instead of dividing. Getting 50 comes from misreading the area as 5 m² instead of 0.5 m² and dividing 250 by 5.
- (d) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (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).
- (d) 1.5 — Method: the length scale factor is the square root of the area scale factor, not the area scale factor itself. Working: the area scale factor is 45 ÷ 20 = 2.25, and the square root of 2.25 is 1.5. Answer: 1.5. Nadia's answer, 2.25, is the AREA scale factor — she never took the square root to get back to the length scale factor. 4.5 comes from doubling the area scale factor instead of taking its square root. 0.67 comes from taking the square root in the wrong direction, finding the scale factor from the larger rug to the smaller rug instead of the other way round.
- (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.
- (b) 1500 m — Method: multiply the map length by the scale to get the real length in centimetres, then divide by 100 to change centimetres into metres. Working: 3 × 50 000 = 150 000 cm, and 150 000 ÷ 100 = 1500. Answer: 1500 m. The distractors: 15000 m comes from dividing the 150 000 cm by 10 instead of 100; 150 m comes from dividing by 1000, the conversion for kilometres; 50000 m comes from writing the scale itself as the answer and ignoring the 3 cm measured on the map.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
- (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.
- (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.
- (b) 2, the cost in pounds of each extra gigabyte — Method: the gradient is the change in cost divided by the change in data, so it is the cost of each extra gigabyte; the value where the line meets the vertical axis is the charge before any data is used, which is a different quantity. Working: from (0, 10) to (8, 26) the cost rises by 26 − 10 = 16 pounds while the data rises by 8 − 0 = 8 gigabytes, so the gradient is 16 ÷ 8 = 2, meaning each extra gigabyte costs £2. Answer: 2, the cost in pounds of each extra gigabyte. The distractors: '10, the cost in pounds of each extra gigabyte' reads the intercept as the gradient, but 10 is what the tariff costs when no data at all has been used; '3.25, the cost in pounds of each extra gigabyte' comes from 26 ÷ 8, treating the line as though it passed through the origin when it starts at 10; '2, the fixed monthly charge in pounds' has the gradient right but describes the intercept, and the fixed charge on this tariff is £10.
- (d) 60 km/h — Method: use the formula v = d ÷ t with the distance and time given. Working: 180 ÷ 3 = 60 km/h. So the average speed is 60 km/h. Distractor 540 km/h comes from multiplying the distance and time instead of dividing. Distractor 90 km/h comes from dividing by 2 instead of 3. Distractor 18 km/h comes from dividing by 10 instead of 3, a decimal-point slip.
- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
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