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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Answer key: Number worksheet — GCSE Foundation
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- (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.
- (a) 25 — Reverse the operations in reverse order: undo the subtraction by adding 6, then undo the division by multiplying by 5. −1 + 6 = 5, so the number divided by 5 equals 5, and 5 × 5 = 25 — checking, 25 ÷ 5 − 6 = 5 − 6 = −1. A candidate who subtracted 6 again instead of adding worked out −1 − 6 = −7, then −7 × 5 = −35. A candidate who multiplied by 5 before undoing the subtraction, doing the inverse operations in the wrong order, worked out −1 × 5 = −5, then −5 + 6 = 1. A candidate who multiplied by 5 but forgot to undo the subtraction at all worked out −1 × 5 = −5 and stopped there.
- (c) 8 — Each of the 4 spinner outcomes can be paired with each of the 2 coin outcomes, so multiply: 4 × 2 = 8. 6 comes from adding the two numbers instead of multiplying them. 4 comes from using only the spinner outcomes and ignoring the coin. 2 comes from using only the coin outcomes and ignoring the spinner.
- (c) 3:4 — The jacket costs 3/7 of £84, which is 3 × (84 ÷ 7) = 3 × 12 = £36. The bag then costs the rest of the money, £84 − £36 = £48. The ratio of the jacket to the bag is 36:48, which simplifies to 3:4. Writing the fraction spent on the jacket, 3/7, directly as the ratio, without working out that the bag's share is the remaining 4/7, gives 3:7. Giving the ratio the wrong way round, bag to jacket instead of jacket to bag, gives 4:3. Assuming the jacket and bag cost the same, ignoring the fraction given, gives 1:1.
- (a) 36 — Method: for the highest common factor, take the LOWER power of each prime that appears in both numbers. Working: for 2, the lower power is 2² (from 108); for 3, the lower power is 3² (from 72), so the highest common factor is 2² × 3² = 4 × 9 = 36. 216 comes from taking the higher power of each prime instead, 2³ × 3³ = 8 × 27 = 216, which gives the lowest common multiple, not the highest common factor. 108 is simply one of the two numbers, not their highest common factor. 6 comes from multiplying the primes without any powers at all, 2 × 3 = 6. Answer: 36.
- (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (c) 12 — Method: division and multiplication have equal priority, so they are carried out in the order they are written, from left to right; a negative divided by a negative is positive. Working: (−36) ÷ (−6) = 6, and then 6 × 2 = 12. Answer: 12. The distractors: 3 comes from carrying out the multiplication first, (−6) × 2 = −12 followed by (−36) ÷ (−12) = 3; −12 comes from treating a negative divided by a negative as negative, giving −6 and then −6 × 2 = −12; 6 comes from stopping at the division and never carrying out the multiplication by 2.
- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
- (d) 51 is not prime, because 51 = 3 × 17. — Check 51 for small prime factors: 51 ÷ 3 = 17, and both 3 and 17 are themselves prime, so 51 = 3 × 17 and 51 is not a prime number — it has factors other than 1 and itself. Checking only 2, 3 and 5 and concluding wrongly that none of them divide 51 misses that 3 does divide it exactly, so the claim that 51 is prime because it avoids 2, 3 and 5 is false. Assuming any odd number must be prime ignores that 51 = 3 × 17 is a counterexample — plenty of odd numbers are not prime. Misreading 51 as the even number 52 leads to the false claim that it is divisible by 2; 51 itself is odd, and 2 is not one of its factors. So 51 is not prime, because 51 = 3 × 17.
- (c) Ben: 2 × 3 = 6, then 6 + 6 = 12 — Multiplication has priority over addition, so 2 × 3 = 6 is worked out first, then 6 + 6 = 12 — this is Ben's method. Amy adds 6 and 2 before multiplying: 6 + 2 = 8, then 8 × 3 = 24, breaking the priority rule. Chen multiplies the wrong pair of numbers, 6 and 2, instead of 2 and 3: 6 × 2 = 12, then 12 + 3 = 15. Dev applies the right order but slips when multiplying, using 5 instead of 6 for 2 × 3, so the final step becomes 6 + 5 = 11.
- (a) 49 — Method: 7² means 7 multiplied by itself. Working: 7 × 7 = 49. Answer: 49. 14 comes from working out 7 × 2, treating the power 2 as a number to multiply by rather than an instruction to multiply 7 by itself. 77 comes from writing the digit 7 twice side by side, treating the power as an instruction to repeat the digit rather than to multiply. 9 comes from working out 7 + 2, adding the base and the power instead of multiplying the base by itself.
- (c) 27 — Each of the 3 digits can be chosen independently for each of the 3 positions, so multiply: 3 × 3 × 3 = 27. 9 comes from multiplying only two of the three positions, 3 × 3, and forgetting the third. 6 comes from working out 3 × 2 × 1 = 6, which counts codes where digits do not repeat, but the question allows repeated digits. 3 comes from considering only one digit position.
- (a) 11.5 ≤ L < 12.5 — Rounding to the nearest centimetre means L can be up to half a centimetre below or above 12 before it would round to a different whole number. Half of 1 cm is 0.5 cm, so the lower bound is 12 − 0.5 = 11.5 and the upper bound is 12 + 0.5 = 12.5. A value exactly at the upper bound, 12.5, would round up to 13, not 12, so 12.5 itself is excluded, giving 11.5 ≤ L < 12.5. Writing 11.5 ≤ L ≤ 12.5 wrongly includes 12.5 on both ends. Writing 11 ≤ L < 13 uses a whole centimetre either side instead of half a centimetre. Writing 11.5 < L < 12.5 wrongly excludes the lower bound, which is a value that does round to 12.
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