Printable · GCSE Foundation · ages 14-16
The four operations and place value worksheet — GCSE Foundation
Fifteen questions on "the four operations and place value" — DfE statement N2. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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The four operations and place value worksheet — GCSE Foundation
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- 1.Jack buys 3 books, each costing £4.25, and pays with a £20 note. Work out how much change he receives.
- 2.In the number 3.472, work out the value of the digit 7.
- 3.Work out 0.6 × 0.04.
- 4.Work out 250 ÷ 1000.
- 5.Work out 3 + 4 × (−2).
- 6.A roll of ribbon is 8.4 m long. Ribbon is cut into pieces that are each 0.6 m long. Work out how many complete pieces can be cut from the roll.
- 7.Grace drinks 1/3 of a bottle of water in the morning and another 1/3 of the same bottle in the afternoon. Work out what fraction of the bottle she has drunk altogether.
- 8.Work out (2/3)² + (1/2)². Give your answer as a fraction.
- 9.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.
- 10.Write the mixed number 2 1/4 as an improper fraction.
- 11.Meera says that 0.7 ÷ 0.1 = 0.07. Work out the correct value of 0.7 ÷ 0.1.
- 12.Work out 2 3/4 − 1 5/6. Give your answer as a fraction in its simplest form.
- 13.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.
- 14.Work out 3/7 × 14/9. Give your answer as a fraction in its simplest form.
- 15.Work out 4368 ÷ 12.
Answer key
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (c) 0.07 — Each digit after the decimal point has a place value: the first digit is tenths, the second is hundredths, the third is thousandths. In 3.472, the 4 is in the tenths place and the 7 is in the hundredths place, so it is worth 0.07. Reading it as 7 ignores place value altogether, treating it as if it were a whole number. Reading it as 0.7 puts it one place too big, in the tenths place. Reading it as 0.007 puts it one place too small, in the thousandths place. The digit 7 in 3.472 is worth 0.07.
- (c) 0.024 — Multiply the digits ignoring the decimal points: 6 × 4 = 24. Count the total number of decimal places in the two numbers being multiplied: 0.6 has 1 decimal place and 0.04 has 2, giving 3 in total. Place the decimal point in 24 so that there are 3 digits after it: 0.024. Counting only 2 decimal places instead of 3 gives 0.24. Counting 4 decimal places instead of 3 gives 0.0024. Counting only 1 decimal place instead of 3 — in effect moving the point in just one of the two numbers, as if the calculation were 6 × 0.4 — gives 2.4. So 0.6 × 0.04 = 0.024.
- (d) 0.25 — Dividing by 1000 moves every digit three place-value columns, so 250 ÷ 1000 = 0.25. A candidate who divides by 100 instead of 1000 gets 2.5. A candidate who divides by 10,000 instead of 1000 gets 0.025. A candidate who divides by 10 instead of 1000 gets 25.
- (b) −5 — Using the order of operations, work out the multiplication first: 4 × (−2) = −8. Then 3 + (−8) = −5. A candidate who adds before multiplying gets (3 + 4) × (−2) = −14. A candidate who drops the negative sign on the multiplication gets 3 + 4 × 2 = 11. A candidate who works out the multiplication correctly but gives that as the final answer, forgetting to combine it with the 3, gets −8.
- (a) 14 — Multiply both numbers by 10 to clear the decimals: 8.4 becomes 84 and 0.6 becomes 6. Then divide: 84 ÷ 6 = 14, so 14 complete pieces can be cut. Scaling only the divisor by 10 and leaving the dividend as 8.4 gives 8.4 ÷ 6 = 1.4, which rounds down to 1 complete piece — the dividend was never converted. Scaling only the dividend by 10 and leaving the divisor as 0.6 gives 84 ÷ 0.6 = 140. Rounding the divisor from 0.6 to 0.7 before dividing, trading accuracy for a rounder number, gives 8.4 ÷ 0.7 = 12. So 14 complete pieces of ribbon can be cut.
- (c) 2/3 — Method: fractions with the same denominator are added by adding the numerators and leaving the denominator alone, because the parts are already the same size. Working: 1/3 + 1/3 has numerators 1 + 1 = 2 and the denominator stays as 3, giving 2/3. Answer: 2/3. The distractors: 2/6 comes from adding the denominators as well as the numerators, 1 + 1 over 3 + 3; 2/9 comes from adding the numerators but multiplying the denominators, 1 + 1 over 3 × 3; 1/9 comes from multiplying throughout instead of adding, 1 × 1 over 3 × 3.
- (c) 25/36 — Method: square a fraction by squaring its numerator and its denominator separately, then add the two results over a common denominator. Working: (2/3)² = 4/9 and (1/2)² = 1/4; the lowest common denominator of 9 and 4 is 36, so 4/9 = 16/36 and 1/4 = 9/36, and 16 + 9 = 25. Answer: 25/36. The distractors: 49/36 comes from adding the two fractions first and squaring the total, giving (7/6)²; 5/13 comes from squaring correctly but then adding the numerators and the denominators, as (4 + 1)/(9 + 4); 7/3 comes from doubling each fraction instead of squaring it, giving 4/3 + 1.
- (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.
- (b) 9/4 — Method: write the whole part as a fraction with the same denominator, then add the fraction part to it. Working: there are 4 quarters in 1 whole, so 2 wholes are 2 × 4 = 8 quarters; adding the 1 quarter that is already there gives 8 + 1 = 9 quarters over a denominator of 4. Answer: 9/4. The distractors: 3/4 comes from adding the whole number to the numerator, as 2 + 1, instead of multiplying it by the denominator first; 7/4 comes from multiplying correctly but then subtracting the numerator, as 2 × 4 − 1; 5/4 comes from multiplying the numerator by the denominator instead of the whole number, as 1 × 4 + 1.
- (a) 7 — Dividing by 0.1 is the same as multiplying by 10, so 0.7 ÷ 0.1 = 7. Meera's answer of 0.07 comes from dividing 0.7 by 10 instead of by 0.1, the wrong way round. A candidate who confuses 0.1 with 0.01 multiplies by 100 instead of 10 and gets 70. A candidate who thinks dividing by a number less than 1 does not change the value gets 0.7.
- (d) 11/12 — Convert both mixed numbers to improper fractions with a common denominator. 2 3/4 = 11/4, which is 33/12, and 1 5/6 = 11/6, which is 22/12. Subtracting, 33/12 − 22/12 gives 11/12, already in its simplest form. Forgetting to borrow, and instead subtracting the fraction parts the other way round to avoid a negative, 10/12 − 9/12 gives 1/12; adding that to the whole-number difference of 1 gives 13/12. Subtracting only the fraction parts, 9/12 − 10/12, and reporting just the size of that difference gives 1/12, which ignores the whole numbers altogether. Adding the two improper fractions instead of subtracting them, 33/12 + 22/12, gives 55/12. So 2 3/4 − 1 5/6 = 11/12.
- (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.
- (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.
- (b) 364 — Divide in stages using multiples of 12. 12 × 300 = 3600, leaving a remainder of 4368 − 3600 = 768. Then 12 × 64 = 768, so 4368 ÷ 12 = 300 + 64 = 364. Placing the decimal point as though dividing 436.8 by 12 gives 36.4. Transposing the last two digits of 364 gives 346. Working out 768 ÷ 12 as 4 instead of 64, losing the tens digit, and adding 300 + 4 gives 304. So 4368 ÷ 12 = 364.
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