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.
Non-calculator
Answer key: Number worksheet — GCSE Foundation
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- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (d) 1 — Method: any non-zero number raised to the power zero has the same value, which follows from dividing a power by itself. Working: 3² ÷ 3² subtracts the indices to give 3⁰, and the same division worked out directly is 9 ÷ 9 = 1, so 3⁰ = 1. Answer: 1. The distractors: 0 comes from reading the index as the value of the whole expression; 3 comes from treating a zero index as leaving the base unchanged; 1/3 comes from confusing a zero index with a negative index and taking the reciprocal of 3.
- (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
- (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.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (d) 13/20 — Method: write both fractions over a common denominator, then add the numerators. Working: 2/5 = 8/20 and 1/4 = 5/20, so 2/5 + 1/4 = 8/20 + 5/20 = 13/20, which is already in its simplest form. Answer: 13/20. 1/3 comes from adding the numerators and the denominators separately: (2+1)/(5+4) = 3/9 = 1/3. 13/40 comes from converting both fractions to twentieths correctly but then adding the denominators as well as the numerators: (8+5)/(20+20) = 13/40. 3/20 comes from adding the original numerators (2+1) but keeping them over the common denominator without converting them first: 3/20.
- (a) 4.5 × 10⁴ — 45,000 = 4.5 × 10,000 = 4.5 × 10⁴, with the coefficient between 1 and 10 as standard form requires. Writing 45 × 10³ keeps the coefficient too large — 45 is not between 1 and 10. Writing 4.5 × 10³ undercounts the places moved, giving only 4,500. Writing 4.5 × 10⁵ overcounts the places moved, giving 450,000.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (c) £80 — Method: a unit fraction acts as an operator, so finding 1/9 of a price means dividing that price by 9. Working: £720 ÷ 9 = £80. Answer: £80. The distractors: £6480 comes from multiplying by the denominator instead of dividing by it, giving £720 × 9 = £6480; £640 comes from working out what is left of the £720 once the case is paid for, £720 − £80, instead of the cost of the case itself; £72 comes from dividing by 10 instead of 9, treating one ninth as one tenth.
- (d) 11 mm — Method: for a square, the side length is the square root of the area. Working: 11 × 11 = 121, so the side length is 11 mm. 60.5 mm comes from working out 121 ÷ 2 = 60.5, halving the area instead of finding its square root. 242 mm comes from working out 121 × 2 = 242, doubling the area instead of finding its square root. 22 mm comes from working out 11 × 2 = 22, doubling the correct side length. Answer: 11 mm.
- (b) −4 — To subtract a negative number, add its positive equivalent: −7 − (−3) becomes −7 + 3. Work out −7 + 3 to get −4. Treating "− (−3)" as simply "−3" without flipping the sign gives the wrong working −7 − 3, which is −10. Ignoring the negative sign on −7 and just subtracting the values, 7 − 3, gives 4, which loses the sign of the starting number. Flipping the sign of both numbers, 7 + 3, gives 10, which changes more than the double negative allows. So −7 − (−3) = −4.
- (b) 1/2 — To multiply fractions, multiply the numerators together and multiply the denominators together: 2 × 3 = 6 and 3 × 4 = 12, giving 6/12, which simplifies to 1/2. Adding the fractions instead of multiplying them, using a common denominator of 12, gives 8/12 + 9/12 = 17/12. Dividing by 3/4 instead of multiplying by it, so multiplying by its reciprocal 4/3, gives 2/3 × 4/3 = 8/9. Multiplying only the numerators, 2 × 3 = 6, and keeping the first denominator, 3, unchanged gives 6/3 = 2.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
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