Printable · GCSE Foundation · ages 14-16

Number worksheet — GCSE Foundation

Fifteen questions across the number statements at Foundation 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 Foundation

Maths · GCSE Foundation · Number · 15 questions

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1.Work out √(4 × 9)
(a)5
(b)18
(c)36
(d)6
2.Postage on a parcel is calculated as £4.60, correct to the nearest 20p. Which of these could not be the actual cost of the postage?
(a)£4.55
(b)£4.50
(c)£4.65
(d)£4.80
3.Freya types 4² + 3² into her calculator and writes down 49. Work out the correct value of 4² + 3².
(a)14
(b)25
(c)49
(d)12
4.On one night in Manchester the temperature at 3 am was 8 °C below zero and at 9 am it was 3 °C below zero. Write down the warmer of the two readings.
(a)−3 °C
(b)3 °C
(c)5 °C
(d)−8 °C
5.A recipe for one cake needs 2/3 of a cup of sugar. Priya has 3 1/2 cups of sugar. Work out how many complete cakes she can make.
(a)5.25
(b)5
(c)6
(d)7
6.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
(a)0.5, 0.55, 0.56, 0.6, 0.601
(b)0.5, 0.56, 0.55, 0.6, 0.601
(c)0.5, 0.6, 0.55, 0.56, 0.601
(d)0.601, 0.6, 0.56, 0.55, 0.5
7.A sponsored walk raised £350 for charity. 20% of the money raised is spent on equipment. Work out how much is spent on equipment.
(a)£35
(b)£280
(c)£70
(d)£17.50
8.Simplify x⁷ × x⁴, giving your answer as a single power of x.
(a)x¹¹
(b)11x
(c)x³
(d)x²⁸
9.Tickets for a fairground ride are sold in packs of 6. Tokens for the dodgems are sold in packs of 10. Yusuf wants to buy the smallest number of packs of each so that he ends up with the same number of ride tickets as dodgem tokens. Work out how many ride tickets that is.
(a)16
(b)30
(c)60
(d)2
10.A newspaper reports that 48,000 people attended a festival. The figure is correct to the nearest thousand. Which statement about the number of people who attended must be true?
(a)Fewer than 48,500 people attended
(b)More than 48,500 people attended
(c)Fewer than 47,500 people attended
(d)More than 48,000 people attended
11.A charity trek covers 830 miles over roughly 19 days. By rounding each number to 1 significant figure, work out an estimate for the number of miles walked per day.
(a)4 miles
(b)830 miles
(c)41.5 miles
(d)40 miles
12.Write 260,000 in standard form.
(a)2.6 × 10⁶
(b)26 × 10⁴
(c)2.6 × 10⁵
(d)0.26 × 10⁶
13.Two-digit numbers are made using the digits 1, 2, 3 and 4, and no digit may be used twice in the same number. Work out how many different two-digit numbers can be made.
(a)12
(b)6
(c)16
(d)8
14.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
(a)42.5 ≤ t < 47.5
(b)40 ≤ t < 50
(c)42.5 ≤ t ≤ 47.5
(d)44.5 ≤ t < 45.5
15.Grace buys three items costing £9.95, £19.90 and £4.99. Work out an estimate for the total cost, by rounding each price to the nearest pound.
(a)£34.84
(b)£35
(c)£32
(d)£40

Answer key

(d) 6 — Method: the square root of a product can be found either by multiplying first and then rooting, or by rooting each factor and multiplying the two roots together. Working: 4 × 9 = 36, and 6 × 6 = 36, so the root is 6; the same value comes from √4 × √9 = 2 × 3. Answer: 6. The distractors: 36 comes from multiplying inside the root and then leaving the root untaken; 5 comes from rooting each factor and adding the results, 2 + 3, instead of multiplying them; 18 comes from rooting the 4 only and leaving the 9 untouched, giving 2 × 9.
(d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
(b) 25 — Method: check the calculator answer by following the order of operations — each power is worked out before the addition. Working: 4² = 4 × 4 = 16 and 3² = 3 × 3 = 9, and 16 + 9 = 25. Answer: 25. The distractors: 49 is the value Freya wrote down and comes from adding first and then squaring, working out (4 + 3)² instead of 4² + 3²; 14 comes from doubling each number instead of squaring it, adding 8 and 6; 12 comes from multiplying 4 by 3 instead of squaring each number and adding the results.
(a) −3 °C — Method: write each reading as a signed temperature, then choose the one further to the right on a number line. Working: 8 °C below zero is −8 °C and 3 °C below zero is −3 °C. On a number line −3 lies to the right of −8, so it is the warmer reading. Answer: −3 °C. The distractors: −8 °C comes from ordering negatives by the size of their digits, treating −8 as the larger number; 3 °C has the right size but the sign dropped, and a reading of 3 °C is above zero rather than below it; 5 °C comes from working out the difference between the two readings instead of choosing one of them.
(b) 5 — Method: divide the total amount of sugar by the amount needed for one cake, then round down because a part-used amount of sugar cannot make an extra whole cake. Working: 3 1/2 ÷ 2/3 = 7/2 × 3/2 = 21/4 = 5.25; only 5 complete cakes can be made, since the leftover 0.25 of a portion is not enough for a 6th cake. Answer: 5. 5.25 gives the exact result of the division without rounding down to a whole number of cakes. 7 comes from multiplying 3.5 by 2 and ignoring the need to also divide by 3 as part of dividing by the fraction 2/3. 6 comes from rounding 5.25 up to the nearest whole number instead of down, wrongly assuming a 6th cake could be made from the leftover sugar.
(a) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
(c) £70 — Method: 20% of an amount is 20/100 of it; a reliable route is to find 10% by dividing by 10 and then double it. Working: 10% of £350 is £350 ÷ 10 = £35, and 20% is twice as much, £35 × 2 = £70. Answer: £70. The distractors: £35 comes from finding 10% and stopping there; £17.50 comes from reading 20% as one twentieth and working out £350 ÷ 20 = £17.50; £280 comes from taking 20% off the money raised rather than finding 20% of it, giving £350 ÷ 5 = £70 and then £350 − £70 = £280.
(a) x¹¹ — Method: when multiplying powers of the same base, add the indices. Working: 7 + 4 = 11, so x⁷ × x⁴ = x¹¹. x²⁸ comes from multiplying the indices, 7 × 4 = 28, instead of adding them. x³ comes from working out 7 − 4 = 3, which is the rule for dividing powers, not multiplying them. 11x comes from adding the indices to make 11 but then treating x as a coefficient instead of a power. Answer: x¹¹.
(b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
(a) Fewer than 48,500 people attended — Method: a figure given to the nearest thousand lies within half of 1,000, that is 500, of the figure printed, so the attendance is at least 47,500 and below 48,500. Working: 48,000 − 500 = 47,500 and 48,000 + 500 = 48,500, and an attendance of 48,500 would have been reported as 49,000, so every possible attendance is below 48,500. Answer: Fewer than 48,500 people attended. The distractors: more than 48,500 turns the upper limit of the range into a minimum; fewer than 47,500 uses the lower limit as though it were the upper one; more than 48,000 assumes the printed figure was rounded down, when it could just as well have been rounded up from a smaller attendance.
(d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
(c) 2.6 × 10⁵ — In standard form, A must satisfy 1 ≤ A < 10. Moving the decimal point in 260,000 to just after the 2 gives A = 2.6, and the decimal point moved 5 places, so 260,000 = 2.6 × 10⁵. A candidate who wrote 26 × 10⁴ used a value of A outside the required range, even though it has the same overall value. A candidate who wrote 2.6 × 10⁶ counted one place too many when moving the decimal point. A candidate who wrote 0.26 × 10⁶ used a value of A below 1, again outside the required range.
(a) 12 — Method: build the number one place at a time, listing systematically: fix the tens digit, then run through every units digit that is still available. Working: any of the 4 digits can go in the tens place, and once it has been used only 3 digits are left for the units place, so there are 4 × 3 = 12 numbers; listing the numbers that begin with 1 gives 12, 13 and 14, and each of the other three starting digits gives 3 numbers in the same way. Answer: 12. The distractors: 16 comes from working out 4 × 4, which allows a digit to be used twice; 8 comes from multiplying the 4 digits by the 2 places in the number instead of multiplying the choices available at each place; 6 comes from treating a number and its reverse as the same, counting only the unordered pairs of digits.
(a) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
(b) £35 — Method: round each price to the nearest pound, then add the rounded prices. Working: £9.95 rounds to £10, £19.90 rounds to £20 and £4.99 rounds to £5, and £10 + £20 + £5 gives the estimate. Answer: £35. The distractors: £40 comes from rounding each price up to the nearest £10 rather than to the nearest pound, giving £10 + £20 + £10; £32 comes from cutting the pence off each price instead of rounding it, giving £9 + £19 + £4; £34.84 is the exact total, worked out in full when the question asks for an estimate.

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