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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Algebra worksheet — GCSE Foundation
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- 1.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 2.A sequence has the position-to-term rule n² + 1, where n is the position number. Work out how much bigger the 5th term is than the 4th term.
- 3.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
- 4.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 5.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 6.The diagram shows part of the graph of a reciprocal function of the form , passing through the labelled point. Work out the value of k.
- 7.The formula for the circumference of a circle is C = 2πr, where r is the radius. Make r the subject of the formula.
- 8.A number machine multiplies its input by 3. The output is 15. Work out the input.
- 9.A box holds n pencils. A shop has 6 full boxes and 4 loose pencils. All of these pencils are shared equally between 2 classes. Write down the expression for the number of pencils each class receives.
- 10.Simplify k + k + k + k + k.
- 11.There are 50 adults on a coach to the Lake District. Every adult is either a man or a woman. There are 10 more men than women. Work out the number of men.
- 12.Which expression means the same as a × a × a?
- 13.y = 5x − 1. Work out the value of y when x = 3.y = 5x − 1
- 14.A pattern is made from tiles. Pattern 1 uses 4 tiles, pattern 2 uses 7 tiles, pattern 3 uses 10 tiles and pattern 4 uses 13 tiles, with each pattern using 3 more tiles than the one before. Work out an expression, in terms of n, for the number of tiles used in pattern n.
- 15.Solve the simultaneous equations 3x + 2y = 16 and x + y = 7. Work out the value of y.
Answer key
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (b) 6 — Method: for any point that lies on y = k/x, the value of k is found by multiplying the x-coordinate and the y-coordinate together, since k = x × y. Working: k = 2 × 3 = 6. Answer: k = 6. Distractor refutation: 1.5 comes from dividing the y-coordinate by the x-coordinate instead of multiplying them. 5 comes from adding the two coordinates instead of multiplying them. 9 comes from misreading the point's x-coordinate as 3 instead of 2, then multiplying 3 × 3.
- (a) r = C / (2π) — Method: undo the multiplication by 2π by dividing both sides by 2π. Working: C = 2πr, so dividing both sides by 2π gives r = C / (2π). The value r = C / π comes from dividing by π only and forgetting the factor of 2 in 2π. The value r = 2πC comes from multiplying by 2π instead of dividing. The value r = C − 2π comes from subtracting 2π instead of dividing by it.
- (d) 5 — Method: running a machine backwards means undoing its operation, and the operation that undoes multiplying by 3 is dividing by 3. Working: the input multiplied by 3 gives 15, so the input is 15 ÷ 3. Answer: 5, which checks because 3 × 5 = 15. The distractors: 45 comes from running the machine forwards on the output, 15 × 3; 12 comes from subtracting 3 from 15 instead of dividing; 18 comes from adding 3 to 15, which would undo a machine that subtracts rather than one that multiplies.
- (c) (6n + 4)/2 — Six full boxes hold 6 lots of n pencils, which is 6n, and the 4 loose pencils are added on, so the shop has 6n + 4 pencils altogether. Sharing them equally between 2 classes divides that whole total by 2, and brackets are what show that the division applies to all of it: (6n + 4)/2. Without the brackets, 6n + 4/2 halves only the loose pencils; 6(n + 4)/2 adds the loose pencils to every box before the division; 2(6n + 4) doubles the total instead of halving it.
- (a) 5k — Method: adding a letter to itself repeatedly means counting how many of that letter you have, written as a coefficient in front of the letter. Working: k + k + k + k + k is five lots of k, written 5k. Answer: 5k. k⁵ comes from treating repeated addition as repeated multiplication (raising to a power) instead of counting copies. 5 + k comes from adding the count of terms (5) to a single k instead of multiplying. k/5 comes from dividing instead of counting how many k's there are.
- (d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
- (c) a³ — a × a × a means a multiplied by itself three times, which is written using powers as a³ — the small 3 shows how many times a is multiplied by itself. Writing 3a instead uses the 3 as a coefficient, as if the expression meant a + a + a (three lots of a added together) rather than three a's multiplied together. Writing a² only accounts for two of the three a's being multiplied together, missing one factor. Writing 3 + a treats the repeated multiplication as an addition of 3 and a, which has no connection to the original expression. The expression that means a × a × a is a³.
- (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.
- (b) 3n + 1 — Method: find how many more tiles each pattern uses, then find the constant by adjusting pattern 1's total. Working: each pattern uses 3 more tiles than the last, so the coefficient of n is 3. The constant is pattern 1's total minus the common difference: 4 − 3 = 1. Answer: the nth term is 3n + 1. 3n + 4 comes from using pattern 1's total, 4, as the constant without subtracting the common difference. 3n − 2 comes from a slip in working out the constant, subtracting the common difference twice (4 − 3 − 3 = −2) instead of once. n + 3 comes from swapping the common difference and the constant.
- (b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.
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