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) 50 — Method: in ab², only the b is squared, so square b first, then multiply by a. Working: b² = 5² = 25, then a × b² = 2 × 25 = 50. Answer: 50. 100 comes from squaring the product ab instead of just b: (2 × 5)² = 100. 20 comes from squaring a instead of b: a² × b = 4 × 5 = 20. 10 comes from ignoring the square altogether and working out a × b = 2 × 5 = 10.
- (a) a = 20 − b — Method: get a on its own by adding a to both sides, then subtracting b from both sides. Working: b = 20 − a, so b + a = 20, so a = 20 − b. Answer: a = 20 − b. a = b − 20 comes from treating the formula as if it read b = 20 + a and then subtracting 20 from both sides. a = 20 + b comes from moving b across from b + a = 20 without changing its sign. a = −20 − b comes from a double sign error, changing the sign of the 20 as well as of b when rearranging.
- (d) 36 — (2n)² means the whole of 2n is squared, so with n = 3: (2n)² = (2 × 3)² = 6² = 36. Answering 18 instead works out 2n² — squaring only the n and then multiplying by 2 — which is a different expression because the brackets around 2n are missing. Answering 12 squares only the coefficient, treating (2n)² as 2² × n = 4 × 3 = 12, and forgets to square the n as well. Answering 9 ignores the coefficient of 2 altogether and works out n² on its own. The value of (2n)² when n = 3 is 36.
- (d) 6 — Method: a value satisfies x ≥ 7 when it is greater than 7 or exactly equal to 7, so test each value against the boundary. Working: 8 is greater than 7 and 100 is greater than 7, so both satisfy the inequality; 7 is equal to the boundary and ≥ includes equality, so 7 satisfies it as well; 6 is less than 7, so 6 is the one value that fails. Answer: 6. The distractors: 7 is chosen by candidates who read ≥ as a strict 'greater than' and so shut the boundary value out of the solution set; 8 is chosen by reading the question as asking which value DOES satisfy the inequality and taking the smallest such value; 100 is chosen by the same misreading, taking instead the value furthest above the boundary.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (b) 5 — Method: add 9 to both sides, then divide by 5. Working: 5x = 16 + 9 = 25, so x = 25 ÷ 5 = 5. Answer: 5. 3.2 comes from dividing 16 by 5 directly, without adding 9 first. 1.4 comes from a sign error, subtracting 9 from 16 instead of adding it, then dividing by 5. 25 comes from correctly working out 5x = 25 but stopping there, without dividing by 5 to find x.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
- (a) 11 — Substitute n=6 into 2n−1: 2×6−1=11. A candidate who adds 2 and 6 and then subtracts 1, instead of multiplying 2 by 6 first, would compute 2+6−1=7. A candidate who substitutes the wrong term number, n=5, would reach 2×5−1=9. A candidate who forgets to subtract 1 would compute just 2×6=12.
- (d) −4 — The coefficient of a letter is the number multiplying it, taken with the sign written in front of that number. The term containing p is being subtracted, so the term is −4p and the number multiplying p is −4. Quoting 4 drops the sign; 7 is a constant term with no letter attached to it; 9 is the number multiplying q, which is a different letter.
- (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.
- (c) 3 — Method: a point that lies on a graph makes its equation true, so substituting the coordinates into y = mx + 6 leaves an equation in m alone. Working: substituting x = −2 and y = 0 gives 0 = m × (−2) + 6, which rearranges to −2m = −6, so m = (−6) ÷ (−2) = 3. Answer: 3. The distractors: −3 comes from dividing −6 by 2 and keeping the minus sign, overlooking that the divisor is negative too, so the two signs cancel; −2 comes from writing down the x-coordinate of the given point in place of the gradient; 6 comes from reading the constant in y = mx + 6 as the gradient, confusing m with c.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (d) 4x + 5 — A minus sign directly before a bracket changes the sign of both terms inside it: 6x − (2x − 5) = 6x − 2x + 5 = 4x + 5. The option 4x − 5 comes from only changing the sign of the 2x term and not the −5, giving 6x − 2x − 5. The option 8x − 5 comes from adding 2x instead of subtracting it, as if the minus sign did not apply to the bracket, giving 6x + 2x − 5. The option 8x + 5 repeats that same addition mistake and also changes the sign of the −5 term.
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