Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Algebra worksheet — GCSE Foundation
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- (a) x > 4 — Subtract 2x from both sides: 3x − 3 > 9. Add 3 to both sides: 3x > 12. Divide both sides by 3: x > 4. A candidate who subtracts 3 from 9 instead of adding gets 3x > 6, so x > 2. A candidate who divides correctly but wrongly flips the inequality (as if dividing by a negative) gets x < 4. A candidate who multiplies by 3 instead of dividing gets x > 36.
- (c) 7 — To reverse the rule, subtract the constant then divide by the coefficient: 30−2=28, then 28÷4=7, so n=7. A candidate who adds the constant instead of subtracting it, a sign error when rearranging, would compute (30+2)÷4=32÷4=8. A candidate who subtracts the constant correctly but then forgets to divide by the coefficient would stop at 30−2=28. A candidate who treats 4n+2 as a single term 6n, adding the coefficient and constant together, would compute 30÷6=5.
- (a) 30 US dollars — £24 is 6 times £4, and the same scale factor applies to US dollars, so multiply 5 by 6: 5 × 6 = 30, giving 30 US dollars. Multiplying £24 by 4/5, the pounds-per-dollar ratio, instead of by 5/4, the dollars-per-pound ratio, gives 24 × 4/5 = 19.2, applying the ratio the wrong way round and giving 19.20 US dollars. Adding the extra £20 straight onto the 5 as if £1 equalled 1 US dollar, 5 + 20 = 25, gives 25 US dollars. Multiplying £24 by 5 without first dividing by the original £4, 24 × 5 = 120, gives 120 US dollars.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
- (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.
- (a) (n − 5)/2 — Method: do the subtraction first, keep it together as a single bracket, then divide that whole bracket by 2. Working: 'subtract 5 from n' is (n − 5); 'divide the result by 2' means the whole bracket goes over 2, giving (n − 5)/2. Answer: (n − 5)/2. n/2 − 5 comes from dividing n by 2 first and only then subtracting 5, the wrong order. 2(n − 5) comes from multiplying by 2 instead of dividing. (5 − n)/2 comes from subtracting n from 5 instead of subtracting 5 from n, the wrong way round.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (c) 9 — Method: replace the letter with the value given for it, then work out the multiplication before the addition. Working: 5y means 5 × y, and 5 × 0 = 0, so the expression becomes 0 + 9, which is 9. Answer: 9. The distractors: 14 comes from treating 5 × 0 as 5, a common slip with zero, and then adding the 9; 0 comes from assuming that an expression containing a zero must itself be zero, so the + 9 is ignored; 45 comes from multiplying the whole expression by 5 instead of multiplying only y, giving 5 × (0 + 9).
- (b) P = 4s — A formula is a rule connecting two or more different quantities, so that one of them can be worked out once the others are known. P = 4s does exactly that: it gives P from s. The line with no equals sign is an expression; the line with an equals sign and a single unknown letter is an equation, because it is solved to find that letter; the line joined by a comparison sign is an inequality.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (a) 8 — Both points share the x-coordinate, so the distance between them is the difference between the y-coordinates: 3 − (−5) = 8. A candidate who mistakenly uses the equal x-coordinates instead of the y-coordinates gets 4 − 4 = 0. A candidate who adds the y-coordinates instead of subtracting them gets 3 + (−5) = −2. A candidate who reads off only point B's y-coordinate as the distance gets 3.
- (b) (−1, 6) — '6 units above the x-axis' gives a y-coordinate of 6, and the x-coordinate is given as −1, so F = (−1, 6). (6, −1) comes from swapping the x- and y-coordinates. (−1, −6) comes from treating 'above' as a negative direction instead of positive. (1, 6) comes from dropping the negative sign on the given x-coordinate.
- (d) (4, 8) — The other endpoint is found from 2 × midpoint − known endpoint: x = 2 × (−1) − (−6) = −2 + 6 = 4, y = 2 × 5 − 2 = 10 − 2 = 8, giving (4, 8). (−3.5, 3.5) comes from averaging the given endpoint and the midpoint as if they were the two endpoints of a segment, instead of working backwards from the midpoint. (5, 3) comes from working out (−1 − (−6), 5 − 2) instead of doubling the midpoint before subtracting. (4, 5) comes from correctly finding the x-coordinate but copying the midpoint's y-coordinate of 5 instead of doubling it.
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