Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number 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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Number worksheet — GCSE Higher
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- 1.A café sells ice cream in three flavours: vanilla, mango and pistachio. Isla buys a cone with two scoops, using two different flavours, one for the bottom scoop and one for the top scoop. Work out how many different cones she could buy.
- 2.A cyclist rides at a steady speed of 8 metres per second. Work out this speed in kilometres per hour.
- 3.Robert is converting 0.999... (with the 9s recurring forever) into a fraction. He lets x = 0.999... . Multiplying by 10 gives 10x = 9.999... . Subtracting x from 10x gives 9x = 9, so x = 1. Which statement correctly explains this result?
- 4.n is a whole number and 2 ≤ n < 6. Write down all the possible values of n.
- 5.A vending machine sells 4 types of crisps, 5 types of chocolate bar and 2 types of drink. Work out how many different combinations of one crisp packet, one chocolate bar and one drink can be bought.
- 6.Grace works out 7 × 99 by writing 99 as 100 − 1. Use her method to work out 7 × 99.
- 7.A shop assistant says that 7.2 × 3.9 = 56.16. Work out an estimate for 7.2 × 3.9, by rounding each number to the nearest whole number, to show that the assistant’s answer cannot be correct.
- 8.A square tile has an area of 72 cm². Work out the exact perimeter of the tile, giving your answer in the form k√2 cm.
- 9.Which of these is written correctly in standard form?
- 10.A gardener has 42 tulip bulbs and 56 daffodil bulbs. She plants them in rows, with every row containing the same number of tulip bulbs and the same number of daffodil bulbs, and no bulbs left over. Work out the greatest number of rows she can plant.
- 11.Work out (−6) + (−4) × 3
- 12.Write 90 as a product of its prime factors.
- 13.Work out (4 × 10⁻³) × (2 × 10⁵). Give your answer in standard form.
- 14.A photograph uses 4 × 10⁶ bytes of storage. A memory card holds 3.2 × 10¹⁰ bytes. Work out how many of these photographs the card can hold. Give your answer in standard form.
- 15.A charity bake sale sells 187 cakes at £2.95 each. By rounding each number to 1 significant figure, work out an estimate for the total amount raised.
Answer key
- (a) 6 — Method: the two scoops sit in different places on the cone, so a cone is an ordered choice; the possibilities can be listed systematically or counted by multiplying the choices available at each stage. Working: there are 3 flavours for the bottom scoop, and once that flavour is used only 2 flavours remain for the top scoop, so there are 3 × 2 = 6 cones; listing them confirms this, since vanilla on the bottom allows mango or pistachio on top, mango on the bottom allows vanilla or pistachio, and pistachio on the bottom allows vanilla or mango. Answer: 6. The distractors: 3 comes from treating the two scoops as interchangeable, so that vanilla under mango and mango under vanilla are counted as one cone; 9 comes from allowing the same flavour to be used for both scoops, giving 3 × 3; 5 comes from adding the 3 choices for the bottom scoop to the 2 choices left for the top scoop instead of multiplying them.
- (a) 28.8 km/h — A compound unit is converted one part at a time. There are 3600 seconds in an hour, so in one hour the cyclist travels 8 × 3600 = 28 800 metres. There are 1000 metres in a kilometre, so 28 800 m = 28 800 ÷ 1000 = 28.8 km/h. 28 800 km/h leaves the distance in metres, 0.48 km/h converts the seconds to minutes rather than to hours, and 2.22 km/h divides by 3.6 instead of multiplying.
- (c) Exactly 1: the subtraction has no rounding at any step. — 10x − x removes the recurring part completely, because the digits after the decimal point in 10x and in x are identical from the tenths place onward, so they cancel exactly: 9.999... − 0.999... = 9.000... = 9. Nothing was rounded to reach 9x = 9, so x = 1 is an exact equality, not an approximation, and the statement that the value is exactly 1, with no rounding at any step, is the correct one. Calling it only approximately 1, on the ground that a recurring decimal can never reach a whole number, misunderstands what the subtraction has just shown: the recurring tail cancels completely, leaving no gap to approximate away. Claiming the method only works because the recurring digit is 9 is also wrong — the same subtraction cancels the recurring part for any repeating digit, not just 9; it is the choice of multiplier (10, matching the one-digit repeat) that makes the cancellation exact, not the digit itself. Saying 10x minus x gives 8.999... rather than 9 misreads the subtraction: 9.999... − 0.999... has no digit to borrow from, since every decimal digit in the two numbers matches, so the result is exactly 9, not 8.999... .
- (a) 2, 3, 4, 5 — Method: work out which whole numbers satisfy both parts of the inequality. Working: n ≥ 2 means n can be 2 or more; n < 6 means n must be less than 6, so 6 itself is not included. The whole numbers that fit both conditions are 2, 3, 4 and 5. Answer: 2, 3, 4, 5. 2, 3, 4, 5, 6 treats < 6 as ≤ 6 and wrongly includes 6. 3, 4, 5 treats ≥ 2 as > 2 and wrongly leaves out 2. 1, 2, 3, 4, 5 wrongly includes 1, which does not satisfy n ≥ 2.
- (a) 40 — Multiply the number of choices for each item: 4 × 5 × 2 = 40. 11 comes from adding the three numbers instead of multiplying them. 20 comes from multiplying only the crisps and chocolate bars, 4 × 5, and forgetting the drink. 10 comes from multiplying only the chocolate bars and drinks, 5 × 2, and forgetting the crisps.
- (c) 693 — Method: multiplying a bracket by a number multiplies every term inside it, so 7 × (100 − 1) = 7 × 100 − 7 × 1. Working: 7 × 100 = 700 and 7 × 1 = 7, so the calculation becomes 700 − 7 = 693. Answer: 693. The distractors: 699 comes from subtracting the 1 itself rather than 7 lots of it, giving 700 − 1 = 699; 707 comes from adding the second product instead of subtracting it, giving 700 + 7 = 707; 700 comes from rounding 99 up to 100 and then offering the estimate 7 × 100 as an exact value.
- (d) 28 — Method: round each number to the nearest whole number, then multiply the rounded numbers to get an estimate that can be compared with the assistant's answer. Working: 7.2 rounds to 7, and 3.9 rounds to 4, so the estimate is 7 × 4 = 28. Since 28 is much smaller than 56.16, the assistant's answer cannot be correct. 56 comes from rounding the assistant's answer to the nearest whole number, instead of rounding the two numbers being multiplied and then multiplying them. 35 comes from rounding both numbers correctly but then slipping in the seven times table, writing 7 × 5 = 35 in place of 7 × 4 = 28. 21 comes from rounding 3.9 down to 3 instead of 4, giving 7 × 3 = 21. Answer: 28.
- (b) 24√2 — The side length of the tile is √72. Since 72 = 36 × 2, √72 = √36 × √2 = 6√2 cm. A square has four equal sides, so the perimeter is 4 × 6√2 = 24√2 cm. Simplifying √72 by writing the perfect-square factor itself as the coefficient instead of its root, 36√2 instead of 6√2, and then multiplying by 4 lands on 144√2. Working out the correct side length, 6√2 cm, but then giving that as the final answer without multiplying by 4 for the perimeter gives 6√2. Doubling the side length instead of quadrupling it, as if the perimeter were 2 × 6√2 rather than 4 × 6√2, gives 12√2.
- (a) 7.5 × 10⁴ — Standard form is A × 10ⁿ, where A is between 1 and 10 (1 ≤ A < 10) and n is an integer — 7.5 × 10⁴ satisfies all of this. 12 × 10³ fails because 12 is not below 10. 0.5 × 10⁴ fails because 0.5 is not at least 1. 6.2 × 4⁵ fails because standard form always uses a power of 10, not a power of 4.
- (b) 14 — Method: the greatest number of identical rows is the highest common factor of the two bulb totals, found by taking every prime factor the two totals share. Working: 42 = 2 × 3 × 7 and 56 = 2 × 2 × 2 × 7, so the prime factors common to both are 2 and 7, giving a highest common factor of 2 × 7 = 14. 2 comes from taking only the common factor 2 and forgetting the common factor 7. 7 comes from taking only the common factor 7 and forgetting the common factor 2. 168 is the lowest common multiple of 42 and 56, not their highest common factor. Answer: 14.
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
- (a) 8 × 10² — Method: to multiply numbers written in standard form, multiply the coefficients and add the indices. Working: 4 × 2 = 8 for the coefficients, and −3 + 5 = 2 for the indices; 8 already lies between 1 and 10, so no adjustment is needed. Answer: 8 × 10². The distractors: 6 × 10² comes from adding the coefficients, 4 + 2, instead of multiplying them; 8 × 10⁸ comes from ignoring the minus sign and adding 3 + 5; 8 × 10⁻¹⁵ comes from multiplying the indices, −3 × 5, instead of adding them.
- (a) 8 × 10³ — Method: divide the capacity of the card by the size of one photograph, dividing the coefficients and subtracting the indices, then bring the coefficient back into the range 1 to 10. Working: 3.2 ÷ 4 = 0.8 and 10 − 6 = 4, which gives 0.8 × 10⁴; a coefficient of 0.8 is smaller than 1, so the decimal point moves one place to the right and the index falls by 1. Answer: 8 × 10³. The distractors: 8 × 10⁴ comes from correcting 0.8 to 8 without reducing the index, which makes the answer ten times too large; 1.28 × 10¹⁷ comes from multiplying the two numbers instead of dividing them, since 3.2 × 4 = 12.8 and 10 + 6 = 16; 8 × 10¹⁵ comes from dividing the coefficients but adding the indices instead of subtracting them.
- (a) £600 — Method: round the number of cakes and the price of each cake to 1 significant figure, then multiply the rounded values. Working: 187 rounds to 200, and £2.95 rounds to £3 (the digit after the first, 9, rounds the 2 up to 3), so the estimate is 200 × £3 = £600. £400 comes from rounding £2.95 down to £2 instead of up to £3, giving 200 × £2 = £400. £561 comes from rounding only the price and using the exact number of cakes, 187 × £3 = £561. £570 comes from rounding 187 to the nearest 10 as 190 instead of to 1 significant figure as 200, giving 190 × £3 = £570. Answer: £600.
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