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) 2x + 24 — Method: multiply each term inside the bracket by the 2 in front of it. Working: 2 × x = 2x and 2 × 12 = 24, and the products are added because the bracket contains an addition. Answer: 2x + 24. The distractors: 2x + 12 comes from multiplying only the x by 2 and copying the 12 across unchanged; 24x comes from multiplying 2 by 12 and then attaching the letter to that product, as though the two terms inside the bracket could be joined into one; 2x + 14 comes from adding 2 and 12 instead of multiplying them.
- (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.
- (a) 2 m/s² — Method: acceleration = change in speed ÷ time = (17 − 5) ÷ 6 = 12 ÷ 6 = 2 m/s². Distractor origins: 12 m/s² stops after finding the change in speed and forgets to divide by the time; 22 m/s² adds the two speeds instead of subtracting them, and also forgets to divide by time (5 + 17 = 22); 72 m/s² multiplies the change in speed by the time instead of dividing (12 × 6 = 72).
- (d) 8 — Method: compare the equation with y = mx + c, where c is the y-intercept. Working: in y = 3x + 8, the constant term is 8. Answer: the y-intercept is 8. 3 comes from confusing the y-intercept with the gradient. −8 comes from a sign error, treating the constant term as negative. 11 comes from wrongly adding the gradient and the constant term together.
- (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (d) The fixed call-out fee, charged before the hourly rate. — Method: compare the formula with C = (fixed charge) + (rate) × h, where the fixed charge is the part that does not depend on h. Working: in C = 40 + 25h, the term 25h depends on the number of hours, h, but 40 does not change however many hours the job takes. Answer: 40 is the fixed call-out fee. 'The hourly rate' confuses the fixed term with the coefficient of h, which is actually 25. 'The total cost for a job lasting 1 hour' comes from substituting h = 1 into the formula (40 + 25 = 65) rather than reading off the constant term. 'The number of hours worked before charging starts' wrongly treats the constant, which is in pounds, as if it were measured in hours.
- (a) 14 — Substituting x = 3: y = 5(3) − 1 = 15 − 1 = 14. A candidate who stops after the multiplication and forgets to subtract 1 gets 15. A candidate who adds 1 instead of subtracting gets 16. A candidate who incorrectly treats the expression as 5 × (3 − 1) gets 10.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (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.
- (a) 11 — Method: work out each term separately using the rule n² − 3, then subtract. Working: 6th term = 6² − 3 = 36 − 3 = 33. 5th term = 5² − 3 = 25 − 3 = 22. Difference: 33 − 22 = 11. Answer: 11. 8 comes from subtracting the constant −3 once at the end instead of it already being included in both terms, (36 − 25) − 3. 1 comes from working out (6 − 5)² instead of finding 6² and 5² separately and then subtracting. −11 comes from subtracting in the wrong order, the 5th term minus the 6th term instead of the 6th minus the 5th.
- (c) 70 — Method: the numbers of tiles form a sequence in which the same amount is added for each extra row, so the total for a number of rows is that amount multiplied by the number of rows. Working: 20 − 10 = 10 and 30 − 20 = 10, so each row adds 10 tiles; 7 rows therefore need 7 lots of 10, that is 7 × 10. Answer: 70. The distractors: 80 comes from counting one row too many and giving the total for 8 rows; 17 comes from adding the 10 tiles to the 7 rows instead of multiplying; 10 comes from giving the number of tiles in a single row rather than the total for all the rows.
- (a) (6, 0) — Method: a pair of coordinates records movement from the origin, the across movement written first and the up or down movement second. Working: C is 6 units to the right of the origin, so the across number is 6; it is 0 units up, so the up number is 0; written in order that gives (6, 0), a point on the x-axis. Answer: (6, 0). The distractors: (0, 6) comes from writing the two movements the wrong way round; (0, 0) comes from reading '0 units up' as meaning the point never left the origin at all, which ignores the movement across; (7, 0) comes from counting the origin itself as the first unit while counting 6 units to the right.
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