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) 1:200000 — '1 cm represents 2 km' means 1 cm on the map is 2 km in real life. Convert 2 km into centimetres, the same unit as the 1: 2 km = 2000 m = 200 000 cm. So the scale is 1 : 200 000. Converting only as far as metres gives 1 : 2000 — the conversion to centimetres was never finished. Losing a zero in the conversion gives 1 : 20 000, ten times too small. Writing 1 : 2 without converting units at all compares 1 cm to 2 km directly, which is not a valid ratio since the units do not match.
- (c) 3 — Method: for a tangent written in the form y = mx + c, the coefficient m is the gradient of the line, and the gradient of the tangent at its point of contact equals the curve's instantaneous rate of change there. Working: y = 3x − 2 has gradient 3, so the instantaneous rate of change of y with respect to x at x = 4 is 3. Reading the constant term as the rate instead of the coefficient of x gives −2, but −2 is only where the tangent crosses the y-axis, not a rate. Reading the x-coordinate of the point of contact as the rate gives 4, but 4 only tells you where on the curve the tangent touches, not how fast y is changing there. Substituting x = 4 into the tangent equation, 3 × 4 − 2 = 10, gives the y-coordinate of the point of contact, not the rate; a candidate who works out the height of the point instead of the gradient gives 10. Whenever a tangent is given as an equation, the rate of change is always the coefficient of x — do not let the constant term, the x-value or a substituted y-value stand in for 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.
- (a) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
- (c) 6.00 m — Method: the ratio of height to shadow length is the same for both objects. Working: road sign height ÷ shadow = 3 ÷ 5 = 0.6. Lamppost height = 0.6 × 10 = 6.00 m. Wrong options: 16.67 m comes from inverting the ratio, using shadow ÷ height instead of height ÷ shadow (10 × 5 ÷ 3); 8.00 m comes from adding the difference between the two shadow lengths to the road sign's height instead of scaling (3 + (10 − 5)); 1.50 m comes from multiplying by the ratio of the two shadow lengths the wrong way round (3 × 5 ÷ 10).
- (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.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (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.
- (c) 36 cm — A perimeter is built from lengths, so it is multiplied by the length scale factor and not by the square of it. The scale factor from the smaller rectangle to the larger one is 10 ÷ 5 = 2. The larger perimeter is therefore 18 × 2 = 36 cm.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (a) 4:3 — Multiply both parts by the lowest common denominator, 6: (2/3) × 6 = 4 and (1/2) × 6 = 3, giving the ratio 4 : 3, which is already in simplest form.
- (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.
- (b) 3:2 — Write both fractions over a common denominator of 4: 3/4 stays as 3/4, and 1/2 = 2/4. Comparing the numerators gives the ratio 3 : 2. Getting 2 : 3 swaps the two parts round. Getting 3 : 1 comes from using the numerator of the first fraction and the original numerator of the second fraction (1) without converting to a common denominator. Getting 2 : 1 comes from using only the denominators, 4 and 2, and simplifying those instead of the numerators.
- (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.
- (b) 4/3 — Work out the number of bananas: 3/4 of 16 = 12. Form the fraction 16/12 (apples over bananas); both numbers share a factor of 4, so 16 ÷ 4 = 4 and 12 ÷ 4 = 3, giving 4/3. 3/4 comes from giving the fraction that was already stated in the question (bananas as a fraction of apples), instead of inverting it. 4/1 comes from reading '3/4 of the apples' as 'the apples reduced by 3/4', giving only 4 bananas, then 16/4 = 4. 4/7 comes from comparing the apples with the total number of pieces of fruit (16/28), instead of with the bananas alone.
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