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.
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Answer key: Algebra worksheet — GCSE Foundation
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- (a) x = (y + 9) / 2 — Method: undo the subtraction of 9 first, then undo the multiplication by 2. Working: y = 2x − 9, so adding 9 to both sides gives y + 9 = 2x, then dividing both sides by 2 gives x = (y + 9) / 2. The value x = (y − 9) / 2 comes from a sign error, keeping the −9 instead of moving it to +9. The value x = 2(y + 9) comes from multiplying by 2 instead of dividing. The value x = y / 2 + 9 comes from dividing by 2 before adding 9, the wrong order of operations, instead of dividing the whole bracket.
- (b) (−5, −2) — Both points have the same y-coordinate, so the midpoint lies on the same horizontal line: y = −2. The x-coordinate is the average of −9 and −1: (−9 + (−1)) ÷ 2 = −10 ÷ 2 = −5, giving (−5, −2). (−10, −2) comes from adding the x-coordinates but forgetting to divide by 2. (−4, −2) comes from a sign error on the second x-coordinate, treating −1 as +1: (−9 + 1) ÷ 2 = −4. (5, −2) comes from dropping the negative sign on the x-coordinate.
- (d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
- (a) x/5 — Read the decimal as a fraction first: 0.2 is 2 tenths, so the coefficient is 2/10. Dividing the numerator and the denominator by 2 cancels this to 1/5, and a coefficient of 1/5 in front of a letter is written x/5. Reading the digit 2 as 'a half' gives x/2; reading 0.2 as 1/20 gives x/20; dividing only the denominator by 2 leaves 2/5, which is the coefficient 0.4 and twice as large as it should be.
- (c) x = 2, y = 3 — Method: no letter cancels straight away, so make one letter the subject of one equation and substitute that expression into the other. Working: 2x + y = 7 gives y = 7 − 2x, so x + 2y = 8 becomes x + 2(7 − 2x) = 8, that is x + 14 − 4x = 8, so −3x = −6 and x = 2; substituting back gives y = 7 − 4 = 3. Answer: x = 2, y = 3, which checks in 2 + 6 = 8. The distractors: x = 3, y = 2 comes from adding the equations to x + y = 5 and then subtracting them the wrong way round, writing x − y = 1 instead of x − y = −1; x = 2, y = 5 comes from substituting x = 2 into 2x + y = 7 as 2 + y = 7, dropping the coefficient; x = −2, y = 11 comes from dividing −3x = −6 as though two minus signs gave a negative answer.
- (b) 3 + 2m — Method: add the fixed fee to the cost of the miles travelled. Working: the miles cost 2 × m = 2m pounds, and the fixed fee is £3, so the total is 3 + 2m. Answer: 3 + 2m. 2 + 3m comes from swapping which number is the fixed fee and which is the per-mile rate. 5 + m comes from adding the fixed fee and the rate together into one number, 3 + 2 = 5, before adding the miles, instead of multiplying the rate by the number of miles. 5m comes from multiplying that combined total by the miles, treating the whole £5 as if it were a per-mile rate.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (c) 4x − 3 — Subtract each term inside the bracket: 5x − (x + 3) = 5x − x − 3 = 4x − 3. Writing 4x + 3 comes from subtracting the x but not the 3, keeping its sign positive. Writing 6x − 3 comes from adding the x term, 5x + x = 6x, instead of subtracting it. Writing 5x − 3 comes from ignoring the x term inside the bracket and only subtracting the 3.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (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.
- (a) 5k — Method: adding a letter to itself repeatedly means counting how many of that letter you have, written as a coefficient in front of the letter. Working: k + k + k + k + k is five lots of k, written 5k. Answer: 5k. k⁵ comes from treating repeated addition as repeated multiplication (raising to a power) instead of counting copies. 5 + k comes from adding the count of terms (5) to a single k instead of multiplying. k/5 comes from dividing instead of counting how many k's there are.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (b) 23 — Term 1 is 2. Term 2 = 2 × 2 + 1 = 5. Term 3 = 2 × 5 + 1 = 11. Term 4 = 2 × 11 + 1 = 23. A candidate who doubles each term but forgets to add 1 gets 2, 4, 8, 16. A candidate who adds 1 before doubling at each step (the wrong order) gets 2, 6, 14, 30. A candidate who forgets to add 1 only on the final step gets 2 × 11 = 22.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (c) x = 2y + 10 — Method: undo the operations done to x in reverse order — add 5, then multiply by 2. Working: y = x/2 − 5, so y + 5 = x/2, so x = 2(y + 5) = 2y + 10. Answer: x = 2y + 10. x = 2y + 5 comes from multiplying only the x/2 term by 2 and forgetting to multiply the 5 as well. x = 2y − 10 comes from a sign error, subtracting 5 instead of adding it before multiplying by 2. x = (y + 5)/2 comes from dividing by 2 instead of multiplying, the wrong operation to undo a division.
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