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.
Algebra worksheet — GCSE Higher
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- 1.Line L has equation y = 3x − 2. Work out the gradient of a line perpendicular to L.y = 3x − 2
- 2.n = 3. Work out the value of (2n)².
- 3.A rule turns each input x into an output y. The inputs are x = 1, 2, 3, 4 and the matching outputs are y = −5, −3, −1 and one missing value. Work out the missing value of y.
- 4.Point P has coordinates (2, −5). P is reflected in the x-axis to point Q. Write down the coordinates of Q.
- 5.Harry is 14 years old and Mia is 8 years old. Will Harry ever be exactly twice as old as Mia? Give a reason for your answer.
- 6.A geometric sequence has first term 4 and common ratio √2. Work out the sum of the first three terms of the sequence, giving your answer in the form a + b√2.
- 7.Work out the value of 2(x + 2) when x = −5
- 8.Factorise fully 56x − 24
- 9.The nth term of a sequence is n² + 2n − 4. Work out the 7th term.
- 10.The solution set of a quadratic inequality is {x : x ≤ −3} ∪ {x : x ≥ 5}. Which of these inequalities has this solution set?
- 11.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 12.The graph of y = f(x) crosses the x-axis at x = −3 and x = 7, and crosses the y-axis at (0, 21). A second graph crosses the x-axis at x = −7 and x = 3, and crosses the y-axis at the same point, (0, 21). The second graph is y = g(x). Which of these could be the equation of g(x)?
- 13.A company charges for hiring chairs for a day. Hiring 1 chair costs £12, hiring 2 chairs costs £19, hiring 3 chairs costs £26, and hiring 4 chairs costs £33. Work out an expression, in terms of n, for the cost in pounds of hiring n chairs for a day.
- 14.A sequence has nth term 3n + 4. Two terms are missing from the list of its first five terms: 7, 10, _, 16, _. Work out the missing 3rd and 5th terms added together.
- 15.Solve 3x + 14 = 8x − 6
Answer key
- (a) −1/3 — The gradient of L is 3, the coefficient of x in y = mx + c form. The gradient of a line perpendicular to a line of gradient m is the negative reciprocal, −1/m. So the perpendicular gradient is −1/3. Distractor routes: 3 gives the gradient of L itself, forgetting to change it at all — that is the gradient of a PARALLEL line. 1/3 takes the reciprocal of 3 but keeps the same sign, missing the negative sign a perpendicular gradient requires. −3 negates the gradient of L but does not take its reciprocal, giving the gradient of a line with the opposite slope rather than a perpendicular one.
- (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) 1 — Method: find the step in the outputs for each step of 1 in the input, write the rule from that step and from one pair of values, then apply the rule to the last input. Working: the outputs −5, −3, −1 rise by 2 while x rises in ones, so x is multiplied by 2; at x = 1, 2 × 1 = 2 while y = −5, so 7 is subtracted, giving y = 2x − 7; at x = 4 the rule gives 2 × 4 = 8 and 8 − 7 = 1. Answer: y = 1. The distractors: 3 comes from carrying the outputs on one step too far, to x = 5; 0 comes from assuming the outputs −5, −3, −1 carry on by adding 1 rather than by adding 2; 8 comes from doubling the input and forgetting to subtract the 7.
- (c) (2, 5) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: Q = (2, 5). (−2, −5) comes from reflecting in the y-axis instead, which changes the sign of the x-coordinate. (−2, 5) comes from reflecting in both axes. (2, −5) comes from not applying the reflection at all.
- (c) No — that moment has already passed — Method: call the number of years from now x, add x to both ages, form the equation from the comparison and then interpret the value of x that comes out. Working: in x years Harry will be 14 + x and Mia will be 8 + x, so 14 + x = 2(8 + x); expanding gives 14 + x = 16 + 2x, and subtracting x and 16 from both sides gives x = −2. A negative value of x places the moment two years in the past, when Harry was 12 and Mia was 6 and 12 = 2 × 6, so it is not something still to come. Answer: no — that moment has already passed. The distractors: the claim that it has never happened and never will comes from reaching x = −2 and reading a negative number of years as no solution at all, when x = −2 does not say that no such moment exists but says where it is — two years before now; the claim that it happens when Harry is 16 comes from doubling Mia's present age, 2 × 8 = 16, and reading that as the age Harry has to reach; the claim that it happens when Harry is 20 comes from expanding 2(8 + x) as 8 + 2x, doubling only the x, which gives x = 6.
- (a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
- (c) −6 — Method: substitute the value, work out the bracket first, then multiply what the bracket comes to by 2. Working: the bracket gives (−5) + 2 = −3, and multiplying by 2 gives 2 × (−3) = −6. Answer: −6. The distractors: −3 comes from working out the bracket correctly but stopping there and never multiplying by 2; −1 comes from multiplying the 2 inside the bracket by the 2 outside before adding, giving −5 + 4; −8 comes from multiplying only the x by 2 and leaving the 2 inside the bracket unchanged, giving −10 + 2.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (c) 59 — Substitute n = 7: 7² + 2 × 7 − 4 = 49 + 14 − 4 = 59. A sign error on the +2n term, treating it as −2n, gives 49 − 14 − 4 = 31. Working out 7² + 2 × 7 but forgetting to subtract the final 4 gives 49 + 14 = 63. Using n = 6 instead of n = 7 gives 36 + 12 − 4 = 44.
- (c) x² − 2x − 15 ≥ 0 — The critical values are x = −3 and x = 5, so the quadratic factorises as (x + 3)(x − 5). Expanding: (x + 3)(x − 5) = x² − 2x − 15. Since the solution set is OUTSIDE the roots (x ≤ −3 or x ≥ 5), the quadratic must be ≥ 0 there, since a positive U-shape sits above the axis outside its roots. So the inequality is x² − 2x − 15 ≥ 0. Distractor routes: x² + 2x − 15 ≥ 0 comes from factorising as (x − 3)(x + 5), swapping the sign of each root when forming the brackets. x² − 2x − 15 ≤ 0 keeps the correct expansion but uses ≤ 0, which gives the region between the roots instead of outside them. x² − 8x + 15 ≥ 0 comes from using roots x = 3 and x = 5, dropping the negative sign on −3 before expanding.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.
- (d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
- (c) 32 — Using the rule 3n + 4: the 3rd term is 3 × 3 + 4 = 13, and the 5th term is 3 × 5 + 4 = 19, so their sum is 13 + 19 = 32. Forgetting to add the 4 for the 3rd term, 3 × 3 = 9, and adding the correct 5th term, gives 9 + 19 = 28. Rounding the 19 up to 20 to make the addition easier and then forgetting to take the extra 1 back off, 13 + 20 = 33, gives 33. Using the rule 4n + 3 instead of 3n + 4 gives 4 × 3 + 3 = 15 and 4 × 5 + 3 = 23, summing to 38.
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
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