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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- (b) 37 — The terms are 7, 13, 19, 25, 31, 37 — each found by adding 6 to the term before, so the 6th term is 37. Adding 6 six times to the first term instead of five times, 7 + 6 × 6 = 43, treats the first term as if it were before the sequence starts. Stopping one term early gives the 5th term, 31. Stopping two terms early gives the 4th term, 25.
- (d) V = lwh — V = lwh is a formula: it relates one quantity, the volume V, to others, the length, width and height, in a way that is true generally. A candidate who picks lwh has chosen the expression, not a full statement relating two quantities. A candidate who picks lwh = 60 has chosen an equation, since it is only true for particular values of l, w and h that multiply to 60. A candidate who picks 2(l + w) ≡ 2l + 2w has chosen an identity, mistaking the identity symbol ≡ for a sign that makes a statement a formula.
- (b) 64 — Method: a position-to-term rule is applied straight to the position number, so a term far along the sequence can be found without writing out the terms in between. Working: the rule multiplies the position by itself and the position asked for is 8, so the term is 8 × 8. Answer: 64. The distractors: 16 comes from multiplying the position by 2 instead of by itself; 81 comes from working out 9 × 9, one position too far along; 8 comes from writing down the position number itself as the term.
- (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.
- (a) y = 2 — Method: make x the subject of the simpler equation, substitute it into the other equation and then read off the letter the question asks for. Working: x − y = 1 gives x = y + 1, so 2x + 3y = 12 becomes 2(y + 1) + 3y = 12, that is 2y + 2 + 3y = 12, so 5y = 10 and y = 2. Answer: y = 2, and the matching value x = 3 checks in 2 × 3 + 3 × 2 = 12. The distractors: y = 3 comes from solving the pair correctly and then writing down the value of x; y = 2.2 comes from expanding 2(y + 1) as 2y + 1, which leaves 5y = 11; y = 10 comes from rearranging x − y = 1 as x = 1 − y, which turns the first equation into 2 + y = 12.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
- (c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
- (b) 9 — Add the two equations to eliminate y: (x + y) + (x − y) = 15 + 3, so 2x = 18, and x = 9 (then y = 15 − 9 = 6). A candidate who reports the value of y instead of x would give 6. A candidate who forgets to divide by 2 after adding would give 18. A candidate who simply subtracts the two totals (15 − 3) instead of adding the equations would give 12.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (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.
- (b) k − 5 — Method: the width is 5 less than the length, so subtract 5 from the length. Working: length − 5 = k − 5. Answer: k − 5. 5 − k comes from subtracting in the wrong order, taking the length away from 5 instead of the other way round. k + 5 comes from adding instead of subtracting, missing that the width is smaller than the length. 5k comes from multiplying the length by 5 instead of subtracting 5 from it.
- (b) −1 — Method: in the form y = mx + c the gradient m is the number multiplying x, and an x term written with no number in front of it has a coefficient of 1. Working: y = −x + 3 is the same equation as y = (−1)x + 3, so comparing it with y = mx + c gives m = −1 and c = 3. Answer: −1. The distractors: 1 comes from taking the coefficient as 1 and leaving the minus sign out of the answer; 3 comes from reading the constant as the gradient, confusing m with c; −3 comes from moving the minus sign across to the constant term and then reading the gradient off that term instead of the x term.
- (d) 28 cm — Method: rate = 12 cm ÷ 3 min = 4 cm per minute. Depth after 7 minutes = 4 × 7 = 28 cm. Distractor origins: 84 cm multiplies the given depth by 7 directly, without first finding the rate per minute (12 × 7 = 84); 24 cm simply doubles the given depth instead of scaling correctly by the ratio of times; 16 cm combines the numbers with subtraction and addition (12 − 3 + 7 = 16) instead of finding a rate.
- (a) 5x − 3 = 12 — 5x − 3 = 12 is an equation with exactly one solution: adding 3 and dividing by 5 gives x = 3, and no other value works. 5x − 3 = 5x − 3 is true for every value of x, since both sides are identical — it has infinitely many solutions, not one. 5x − 3 > 12 is an inequality: any value of x greater than 3 satisfies it, so it has a whole range of solutions, not a single one. 5x − 3 = 5x + 2 has no solution at all, since subtracting 5x from both sides leaves −3 = 2, which is never true. The equation with exactly one solution is 5x − 3 = 12.
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