Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher 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 Higher
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- 1.Solve the simultaneous equations 4x + 3y = 25 and 4x − y = 1. Work out the value of y.
- 2.Which of these equations gives a graph with two separate curved branches — one where x and y are both positive, and one where x and y are both negative?
- 3.Factorise 6x² − 7x − 3.
- 4.A triangle has vertices at (1, 1), (1, 5) and (6, 1). Work out the area of the triangle.
- 5.A candidate solves the simultaneous equations y = x − 2 and y = x² − 4x + 2 by substitution. They write: "x − 2 = x² − 4x + 2, so x² − 3x + 4 = 0." Which of these is a correct comment on the candidate's working?y = x − 2y = x² − 4x + 2
- 6.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 7.By completing the square, show that the curve y = x² − 14x + 50 never crosses the x-axis. Which statement gives the correct reason?y = x² − 14x + 50
- 8.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 9.A ramp rises 3 m over a horizontal distance of 12 m. Work out the gradient of the ramp.
- 10.Which expression is equivalent to 3(2x − 5) + 4x?
- 11.A hall is set out in rows of chairs for an assembly. Row 1 has 50 chairs, row 2 has 45 chairs, row 3 has 40 chairs, and row 4 has 35 chairs, with each row after the first having 5 fewer chairs than the one before. Work out an expression, in terms of n, for the number of chairs in row n.
- 12.n = 3. Work out the value of (2n)².
- 13.A theatre has rows of seats arranged so that row 1 has 12 seats, row 2 has 17 seats, row 3 has 22 seats and row 4 has 27 seats, with each row having 5 more seats than the row before. Work out an expression, in terms of n, for the number of seats in row n.
- 14.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 15.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
Answer key
- (a) 6 — Subtracting the second equation from the first: the x-terms, 4x and 4x, cancel; the y-terms combine as 3y − (−y) = 4y; and the right-hand sides give 25 − 1 = 24. This gives 4y = 24, so y = 6. A candidate who subtracts in the wrong order would get 4y = 1 − 25 = −24, so y = −6. A candidate who forgets the sign on the −y term, treating 3y − y as 2y, would get 2y = 24, so y = 12. A candidate who divides 24 by 6 instead of 4 would get y = 4.
- (b) y = 1/x — A graph with two separate curved branches must have a value of x that is excluded from it, creating a break in the curve. For y = 1/x, x = 0 is excluded, so the graph splits into one branch where x and y are both positive, and one where x and y are both negative. y = x² is a single U-shaped curve with no break, even though it has two 'arms'. y = x³ is one continuous curve that passes through the origin without any break. y = 2x + 1 is a single straight line; even though it passes through regions where x and y share the same sign, it is one continuous line, not two separate branches.
- (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
- (d) 10 — The right angle is at (1, 1). The vertical side has length 5 − 1 = 4 and the horizontal side has length 6 − 1 = 5, so the area is (4 × 5) ÷ 2 = 20 ÷ 2 = 10. A candidate who forgets to halve the product of the two sides gets 4 × 5 = 20. A candidate who forgets to subtract the shared vertex's coordinate and uses the raw coordinates 6 and 5 as the side lengths gets (6 × 5) ÷ 2 = 30 ÷ 2 = 15. A candidate who uses only one side length as the area gets 5.
- (c) Wrong — the correct equation is x² − 5x + 4 = 0. — Rearranging x − 2 = x² − 4x + 2 by moving every term across: 0 = x² − 4x + 2 − x + 2 = x² − 5x + 4. The candidate's x-term is wrong: combining −4x with the moved −x gives −4x − x = −5x, not −3x. So the correct equation is x² − 5x + 4 = 0. Distractor routes: "x² − 3x + 4 = 0 is correct" simply agrees with the candidate's sign error without checking it. "It should be x² − 5x − 4 = 0" correctly spots that something is wrong but blames the wrong term — the constant, 2 + 2 = 4, is correct; the error is in the x-term. "Subtracting is the same as adding" states something false: subtracting a term and adding it give different signs.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (d) (x − 7)² + 1 is at least 1 for every x, so y is never 0 — x² − 14x + 50 = (x − 7)² − 7² + 50 = (x − 7)² + 1. Since a squared term can never be negative, (x − 7)² ≥ 0 for every real x, so y = (x − 7)² + 1 ≥ 1 for every x — y is always at least 1, so it can never equal 0, and the curve never crosses the x-axis. Claiming the discriminant is positive is wrong on the facts: b² − 4ac = (−14)² − 4(1)(50) = 196 − 200 = −4, which is NEGATIVE, not positive — a negative discriminant is in fact the standard reason a quadratic has no real roots, so this option misstates its own sign. Claiming the turning point is at (7, −1) is wrong because completing the square gives (x − 7)² + 1, so the turning point is (7, 1) — the constant is +1, not −1, and a point with a negative y-coordinate would sit BELOW the x-axis, not above it, contradicting the option's own claim. Claiming that 50 being positive is enough is wrong because it ignores the −14x term completely — a positive constant term alone does not guarantee y stays positive for every x (the expression still dips down towards its turning point before growing again, and it is the completed square that shows how far it dips, not the constant on its own).
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (a) 0.25 — Method: gradient = vertical rise ÷ horizontal distance. Working: gradient = 3 ÷ 12 = 0.25. Answer: the gradient of the ramp is 0.25. 4 comes from inverting the gradient, dividing the horizontal distance by the vertical rise instead of the other way round. 9 comes from subtracting the rise from the horizontal distance instead of dividing. 36 comes from multiplying the rise and the horizontal distance instead of dividing.
- (c) 10x − 15 — Expand the bracket first: 3(2x − 5) = 6x − 15. Then add the 4x: 6x − 15 + 4x = 10x − 15. The option 10x − 5 comes from forgetting to multiply the 5 inside the bracket by 3 (treating it as 6x − 5), then adding 4x. The option 10x + 15 comes from a sign error when expanding, treating 3 × (−5) as +15 instead of −15, then adding 4x. The option 6x − 15 comes from expanding the bracket correctly but forgetting to add the 4x term at all.
- (a) 55 − 5n — The number of chairs decreases by 5 in each row after the first, so the common difference is d=−5, and the first term is a=50. The nth term is a+(n−1)d = 50+(n−1)(−5) = 50−5n+5 = 55−5n. A candidate who uses the common difference as the constant term instead of correctly finding 55, giving the constant as −5 instead, would write −5n−5. A candidate who uses the first term, 50, as the coefficient of n instead of the common difference, would write 50n−5. A candidate who does not multiply the common difference by n at all, treating the nth term as n+d instead of dn+c, would write n−5.
- (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.
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (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.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
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