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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- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (a) 27 — Method: let the smaller number be n, so the larger number is n + 9. Form the equation n + (n + 9) = 45. Working: simplify: 2n + 9 = 45, so 2n = 36, giving n = 18. The larger number is 18 + 9 = 27. Answer: 27. 18 is the smaller number, found correctly, but given as the answer to the question asking for the larger number. 22.5 comes from ignoring the "9 more" part and simply halving the total, 45 ÷ 2. 36 comes from a sign error when isolating the constant, treating the equation as 2n = 45 + 9 instead of 2n = 45 − 9.
- (d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
- (a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (a) x = 4 — Method: undo the addition of 4 first, then undo the multiplication by 6. Working: subtracting 4 from both sides gives 6x = 24, and dividing both sides by 6 gives x = 4. Checking: 6 × 4 + 4 = 28. Answer: x = 4. The distractors: x = 24 comes from stopping at 6x = 24 and writing 24 as the value of x; x = 18 comes from subtracting 6 from 24 instead of dividing by 6; x = 144 comes from multiplying 24 by 6 instead of dividing by 6.
- (c) 32 — Using the rule 3n + 4: the 3rd term is 3 × 3 + 4 = 13, and the 5th term is 3 × 5 + 4 = 19, so their sum is 13 + 19 = 32. Forgetting to add the 4 for the 3rd term, 3 × 3 = 9, and adding the correct 5th term, gives 9 + 19 = 28. Rounding the 19 up to 20 to make the addition easier and then forgetting to take the extra 1 back off, 13 + 20 = 33, gives 33. Using the rule 4n + 3 instead of 3n + 4 gives 4 × 3 + 3 = 15 and 4 × 5 + 3 = 23, summing to 38.
- (d) 21 − 3n — Method: find the common difference, then find the constant that fits the first term. Working: 15 − 18 = −3, 12 − 15 = −3, 9 − 12 = −3, so the terms decrease by 3 each time and the nth term has the form c − 3n. Substituting n = 1: c − 3(1) = 18, so c = 21. Answer: the nth term is 21 − 3n. The value 18 − 3n comes from using the first term as c directly, without adding back the difference that was removed. The value 3n − 21 comes from a sign error that flips the whole expression. The value −3n comes from using only the common difference and leaving out the constant.
- (c) 5x² — A term is one part of an expression, separated from the others by a + or − sign. The expression 5x² − 3x + 7 has three terms: 5x², −3x and 7. Distractor origins: 5x² − 3x combines two terms together, not a single term; −3x + 7 also combines two terms together; 5x² − 3x + 7 gives the whole expression, not a single term.
- (c) 15 and 9 — Method: write the two facts as two equations in the same pair of letters and add them, because the letter with opposite signs cancels. Working: with x the larger number and y the smaller, x + y = 24 and x − y = 6; adding gives 2x = 30, so x = 15, and substituting into x + y = 24 gives y = 9. Answer: 15 and 9, which add to 24 and differ by 6. The distractors: 18 and 6 come from halving 24 to 12 and then adding and subtracting the whole difference of 6 instead of half of it, which leaves a difference of 12; 15 and 21 come from finding the larger number correctly and then adding 6 to it instead of subtracting; 15 and 6 come from finding the larger number and then taking the 6 in the question to be the smaller number itself.
- (b) £14 — Method: read the fixed charge (the cost at 0 miles) and the rate (the cost per extra mile) from the graph, then use them to work out the cost for a distance beyond the part that is plotted. Working: the graph shows a fixed charge of £2 at 0 miles, and the cost rises by £2 for every extra mile, so for 6 miles the cost is £2 + (£2 × 6) = £2 + £12 = £14. Answer: £14. Distractor refutation: £12 comes from multiplying the rate by the distance and leaving out the £2 fixed charge. £8 comes from misreading the rate as £1 per mile instead of £2 per mile. £24 comes from adding the fixed charge to the rate first and then multiplying the total by the distance, instead of multiplying the rate by the distance and then adding the fixed charge.
- (b) x = 4 — Method: every point on a vertical line has the same x-coordinate, so its equation sets x equal to a number, and that number is the x-coordinate of any point known to be on the line. Working: the line is vertical and passes through (4, 7), whose x-coordinate is 4, so every point on the line has x-coordinate 4 and the equation is x = 4. Answer: x = 4. The distractors: y = 7 is the horizontal line through the same point, obtained by fixing the y-coordinate instead of the x-coordinate; x = 7 uses the right form but takes the y-coordinate of the point rather than its x-coordinate; y = 4x treats the 4 as a gradient and gives a sloping line through the origin, which does not pass through (4, 7) at all.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (b) 3p — p + p + p means three lots of p added together, and repeated addition of the same term is written as a coefficient: 3p. Writing p³ mistakes the repeated addition for repeated multiplication, as if the expression had been p × p × p. Writing 3 + p adds the number of terms (3) onto p as a separate constant, instead of writing 3 as a coefficient of p. Writing just p forgets to count the terms at all, as though repeating the same letter makes no difference. The simplified expression is 3p.
- (a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
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