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) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
- (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.
- (b) m = w + 8 — To make m the subject of w = m − 8, add 8 to both sides so the subtraction is undone: m = w + 8. Writing m = w − 8 forgets to change the operation at all — it just relabels the equation without moving the 8 across. Writing m = 8 − w swaps the order and puts the wrong sign on w, as if the equation had been w = 8 − m instead. Writing m = w/8 mistakes subtraction for a scaling relationship and divides by 8 instead of adding it. The correct rearrangement is m = w + 8.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (d) 50/3 — Method: call the smaller number x, write the larger number in terms of x and use the total. Working: the larger number is 2x, so x + 2x = 50, that is 3x = 50 and x = 50/3. Answer: 50/3, since 50/3 added to 100/3 makes 50 and 100/3 is twice 50/3. The distractors: 100/3 is the larger of the two numbers rather than the smaller one asked for; 25 comes from halving 50 and treating the two numbers as equal; 24 comes from reading 'twice the smaller number' as 'two more than the smaller number' and solving x + (x + 2) = 50.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
- (d) a/2 — Method: write both terms over the same denominator, subtract the numerators, then cancel the fraction down. Working: 6 is a multiple of 3, so a/3 is rewritten as 2a/6; the calculation becomes 5a/6 − 2a/6 = 3a/6, and dividing numerator and denominator by 3 gives a/2. Answer: a/2. The distractors: 4a/3 comes from subtracting the denominators as well as the numerators, giving (5 − 1)a over (6 − 3); 2a/3 comes from subtracting 1 from 5 without first rewriting a/3 as 2a/6, giving 4a/6; 7a/6 comes from adding the two fractions instead of subtracting them, giving 5a/6 + 2a/6.
- (a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
- (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
- (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (c) 1, 7, 13, 19 — Method: substitute n = 1, 2, 3, 4 into the rule 6n − 5 in turn. Working: n = 1: 6 − 5 = 1. n = 2: 12 − 5 = 7. n = 3: 18 − 5 = 13. n = 4: 24 − 5 = 19. Answer: 1, 7, 13, 19. 6, 12, 18, 24 comes from using 6n on its own, forgetting to subtract 5. 5, 11, 17, 23 comes from using the rule 6n − 1 instead of 6n − 5, a slip in the constant. 0, 6, 12, 18 comes from using 6(n − 1) instead of 6n − 5, effectively shifting every term one position along.
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (d) 68, and it is a formula — Substituting C = 20 gives F = 1.8 × 20 + 32 = 36 + 32 = 68. The statement F = 1.8C + 32 is a formula because it expresses a general relationship between F and C that holds for any value of C, not just C = 20. A candidate who gets the correct value but calls it an equation is confusing "has an equals sign" with "is an equation" — an equation is only true for particular value(s) of the unknown, whereas this relationship is used to calculate rather than to be solved. A candidate who forgets to multiply C by 1.8 and just adds 32 to 20 gets 52. A candidate who multiplies correctly but forgets to add 32 gets 36.
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