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.
Answer key: Number worksheet — GCSE Foundation
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- (d) 600 — Method: the number of seats is the number of rows multiplied by the number of seats in each row, so round each number to 1 significant figure and then multiply the rounded values, which is quick because a product of two multiples of ten is found by multiplying the non-zero digits and attaching the zeros. Working: 21 rounds to 20 and 29 rounds to 30; 2 × 3 = 6, and 20 and 30 carry one zero each, so two zeros follow the 6. Answer: about 600 seats. The distractors: 50 comes from adding the two rounded numbers instead of multiplying them, 20 + 30; 60 comes from multiplying 20 by the 3 of 30 and forgetting the zero in 30; 6,000 comes from attaching three zeros to 2 × 3 when 20 and 30 provide only two between them.
- (c) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (d) 60 — Method: round each number to 1 significant figure and multiply; the estimate then shows whether the calculator answer is sensible. Working: 3.1 rounds to 3 and 19.6 rounds to 20, so the estimate is 3 × 20 = 60. Answer: 60. Hannah's 6.076 is about ten times too small, which is what happens when 19.6 is keyed in as 1.96. The distractors: 62 comes from rounding 19.6 only and leaving 3.1 as it stands, giving 3.1 × 20 = 62; 6 comes from trusting the calculator display rather than checking it against an estimate; 600 comes from rounding 19.6 to 200 instead of to 20, a place-value slip, giving 3 × 200 = 600.
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (a) 0.625 — Method: convert the fraction to a decimal so it can be compared properly with 0.6. Working: 5/8 = 0.625, and since 0.625 > 0.6, the larger value is 0.625. Answer: 0.625. 0.6 repeats Sam's incorrect claim, made by comparing single digits rather than full place value. 0.58 comes from converting 5/8 incorrectly, treating it as if it read 5 tenths and 8 hundredths. 0.85 comes from turning the fraction upside down and writing its digits straight after the decimal point, 8 then 5, instead of dividing.
- (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.
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (a) 187.5 g — Method: the smallest possible actual mass is half the rounding unit below the given value. Working: half of 25 g is 12.5 g, so the smallest possible mass is 200 − 12.5 = 187.5 g. Answer: 187.5 g. (175 g comes from subtracting the whole rounding unit, 25, instead of half of it. 200 g comes from giving the rounded value itself rather than the lower bound. 212.5 g comes from adding the half unit instead of subtracting it, giving the upper bound.)
- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (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.
- (d) 9 — List all the factors of 36 in pairs that multiply to give 36: 1 × 36, 2 × 18, 3 × 12, 4 × 9, and 6 × 6. This gives the factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 — nine factors in total, with 6 counted only once even though it appears in a pair with itself. Forgetting that 36 is itself a factor of 36 and leaving it off the list gives 8. Counting the number of factor pairs, five of them, rather than the number of individual factors gives 5. Treating the repeated pair 6 × 6 as two separate factors, 6 and 6 again, gives 10 instead of 9. So 36 has 9 factors.
- (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
- (b) £37 — One box costs £4 + £3 = £7. Five boxes cost 5 × £7 = £35. Adding the single £2 delivery fee gives £35 + £2 = £37. A candidate who added the £2 delivery fee to each box instead of once for the whole order worked out 5 × (£7 + £2) = 5 × £9 = £45. A candidate who forgot the £3 markup and used the shop's buying price worked out 5 × £4 + £2 = £22. A candidate who added the £3 markup only once, after multiplying the buying price by 5, worked out 5 × £4 + £3 + £2 = £25.
- (b) 18.5 ≤ T < 18.7 — Method: the error interval reaches half the rounding unit either side of the recorded value. Working: half of 0.2 is 0.1, so the interval runs from 18.6 − 0.1 to 18.6 + 0.1. Answer: 18.5 ≤ T < 18.7. (18.4 ≤ T < 18.8 comes from using the full rounding unit, 0.2, either side instead of half of it. 18.5 ≤ T ≤ 18.7 comes from including the upper bound with ≤ instead of excluding it with <. 18.6 ≤ T < 18.8 comes from treating the recorded value as the start of the interval and adding the whole rounding unit, 0.2, above it.)
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