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
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- 1.Write 260,000 in standard form.
- 2.Amelia has 49 boxes of apples with 21 apples in each box. Work out an estimate for the total number of apples, by rounding each number to 1 significant figure.
- 3.Write 0.06 as a fraction in its simplest form.
- 4.Work out 3/7 × 14/9. Give your answer as a fraction in its simplest form.
- 5.Chloe wants to work out 3 1/4 − 1 2/3. She converts both mixed numbers to twelfths, then subtracts the whole numbers and the fraction parts separately, without checking whether she needs to exchange first. Work out the correct value of 3 1/4 − 1 2/3, giving your answer as a mixed number in its simplest form.
- 6.Work out 5/6 × 2/9, giving your answer in its simplest form.
- 7.A carpenter has a plank of wood 4.8 m long. She cuts off 3 pieces, each 0.9 m long, to make shelves. Work out the length of wood remaining.
- 8.Write 90 as a product of its prime factors.
- 9.A carton of orange juice holds 1.35 litres. Ruby pours the juice equally into 4 identical glasses. Work out how much juice is in each glass, giving your answer as a fraction of a litre in its simplest form.
- 10.Work out an estimate for 588 ÷ 31, by rounding each number to 1 significant figure.
- 11.A tank contains 120 litres of water. Water is drained out at a rate of 8 litres per minute for 6 minutes, and then a hose adds 15 litres. Work out how much water is left in the tank.
- 12.Write these numbers in order, starting with the smallest: 5, −8, 0, −3, 2
- 13.A box holds 140 pens. 25% of the pens are red. Work out how many of the pens are red.
- 14.Write 60 as a product of its prime factors, using index notation.
- 15.In a fruit basket the ratio of bananas to oranges is 2 : 5. What fraction of the fruit are oranges?
Answer key
- (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.
- (d) 1,000 — Method: round each number to 1 significant figure, then multiply the rounded values. Working: 49 rounds to 50 and 21 rounds to 20, and 50 × 20 = 1,000 because 5 × 2 = 10 and the two rounded numbers carry one zero each. Answer: 1,000. The distractors: 800 comes from rounding 49 down to 40 instead of to the nearest ten; 1,500 comes from rounding 21 up to 30 rather than down to 20; 1,029 is the exact product 49 × 21, worked out in full when the question asks for an estimate.
- (b) 3/50 — Method: write the decimal over the power of ten that matches the number of digits after the point, counting every digit including a zero, then divide the numerator and the denominator by their highest common factor. Working: 0.06 has two digits after the point, so it is 6 hundredths and can be written as 6/100; the highest common factor of 6 and 100 is 2, and 6 ÷ 2 = 3 with 100 ÷ 2 = 50. Answer: 3/50. The distractors: 3/5 comes from ignoring the zero straight after the point and converting 0.6 instead, giving 6/10, which cancels to 3/5; 3/500 comes from counting three decimal places instead of two and writing 6/1000, which cancels to 3/500; 1/6 comes from putting 1 over the digits after the point, as though 0.06 meant one sixth.
- (c) 2/3 — Method: multiply the numerators together and the denominators together, then divide both parts of the result by their highest common factor. Working: 3 × 14 = 42 and 7 × 9 = 63, giving 42/63; the highest common factor of 42 and 63 is 21, and 42 ÷ 21 = 2 with 63 ÷ 21 = 3. Answer: 2/3. The distractors: 17/16 comes from adding the numerators and adding the denominators, giving (3 + 14)/(7 + 9); 27/98 comes from turning the second fraction upside down and multiplying, which divides instead of multiplying and gives 3/7 × 9/14; 2/21 comes from cancelling the 7 into the 14 in the numerator but leaving the 7 in the denominator, giving 6/63.
- (d) 1 7/12 — Method: convert both mixed numbers to improper fractions with a common denominator, then subtract. Working: 3 1/4 = 39/12 and 1 2/3 = 20/12, so 39/12 − 20/12 = 19/12 = 1 7/12. Answer: 1 7/12. Chloe's method, subtracting whole numbers (3−1=2) and fraction parts (2/3−1/4=5/12) separately without exchanging, gives 2 5/12. 1 3/4 comes from converting 2/3 to twelfths incorrectly as 6/12 instead of 8/12, then subtracting. 8 comes from converting both mixed numbers to improper fractions correctly (13/4 and 5/3) but then subtracting numerators and denominators separately: (13−5)/(4−3) = 8/1.
- (b) 5/27 — Method: multiply the numerators together and the denominators together, then simplify. Working: (5 × 2)/(6 × 9) = 10/54 = 5/27. Answer: 5/27. 7/15 comes from adding the fractions instead of multiplying: (5+2)/(6+9) = 7/15. 15/4 comes from flipping the second fraction, as if dividing: (5 × 9)/(6 × 2) = 45/12 = 15/4. 5/3 comes from cancelling the two denominators against each other, dividing both 6 and 9 by 3 to leave 5/2 × 2/3 = 10/6 = 5/3; cancelling is only valid between a numerator and a denominator, never between two denominators.
- (b) 2.1 m — The three pieces use 3 × 0.9 = 2.7 m of wood. Remaining wood = 4.8 − 2.7 = 2.1 m. A candidate who miscounts and only subtracts 2 pieces instead of 3 gets 4.8 − 1.8 = 3.0 m. A candidate who adds instead of subtracting gets 4.8 + 2.7 = 7.5 m. A candidate who gives the length used instead of the length remaining gets 2.7 m.
- (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.
- (d) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (a) 20 — Method: round each number to 1 significant figure, then divide the rounded values. Working: 588 rounds to 600 (1 s.f.) and 31 rounds to 30 (1 s.f.). 600 ÷ 30 = 20. Answer: 20. 17 comes from cutting 588 down to 500, keeping the leading digit as it stands instead of rounding it up to 1 significant figure, 600, then dividing by the correctly rounded 30. 200 comes from misreading the rounded divisor 30 as 3, giving 600 ÷ 3 instead of 600 ÷ 30. 19 is the exact value of 588 ÷ 31 rounded to the nearest whole number, found without rounding the numbers first.
- (c) 87 litres — Work out how much water is drained: 8 × 6 = 48 litres. Subtract this from the starting amount: 120 − 48 = 72 litres. Then add the 15 litres from the hose: 72 + 15 = 87 litres. Subtracting the 15 litres instead of adding it, as though the hose also removed water, gives 120 − 48 − 15 = 57 litres. Stopping after the drain step, without adding the hose water back in, leaves the working at 72 litres. Adding the rate and the time instead of multiplying them, 8 + 6 = 14 litres drained, and then working from there gives 120 − 14 + 15 = 121 litres. So 87 litres of water is left in the tank.
- (d) −8, −3, 0, 2, 5 — Method: place the numbers on a number line and read them from left to right. Working: the negative numbers −8 and −3 come before 0, with −8 further left than −3 because it is further from zero in the negative direction; then 0, then the positive numbers 2 and 5 in increasing size. Answer: −8, −3, 0, 2, 5. −3, −8, 0, 2, 5 swaps −3 and −8, putting the negative number closer to zero first. 5, 2, 0, −3, −8 lists the numbers from largest to smallest instead of smallest to largest. −8, −3, 2, 0, 5 puts 2 before 0, treating positive numbers as if they always come before zero.
- (a) 35 — Method: 25% is 25/100, which cancels to 1/4, so finding 25% of an amount means dividing it by 4. Working: 25% = 25/100 = 1/4, and 140 ÷ 4 = 35. Answer: 35 pens. The distractors: 70 comes from halving instead of quartering, confusing 25% with 50%; 105 comes from working out the pens that are not red, which is 75% of 140, instead of the pens that are; 25 comes from ignoring the percent sign and reading the 25% as a count of 25 pens.
- (b) 2² × 3 × 5 — Repeatedly divide 60 by prime numbers: 60 ÷ 2 = 30, 30 ÷ 2 = 15, 15 ÷ 3 = 5, and 5 is itself prime. So 60 is 2 × 2 × 3 × 5, which in index notation is 2² × 3 × 5. Stopping the factor tree after only three divisions and writing 2 × 3 × 5 misses that the 2 divides in twice, and gives only 30, not 60. Squaring the 3 as well as the 2 gives 2² × 3² × 5, which comes to 180, far too big. Squaring the 5 instead of the 2 gives 2 × 3 × 5², which comes to 150, also too big. So 60 = 2² × 3 × 5.
- (d) 5/7 — Add the parts of the ratio to find the total: 2 + 5 = 7. Oranges are 5 of those 7 parts, so the fraction of the fruit that are oranges is 5/7. 2/7 comes from finding the fraction of bananas instead of oranges. 5/2 comes from writing the ratio itself as a fraction, without adding the parts to find the total. 7/5 comes from putting the total number of parts over the number of oranges instead of the number of oranges over the total.
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