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.
Answer key: Algebra worksheet — GCSE Higher
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- (c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
- (b) (3, 3) and (−1, −5) — Set the two expressions for y equal: x² − 6 = 2x − 3, which rearranges to x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Substitute into the linear equation, y = 2x − 3: x = 3 gives y = 3; x = −1 gives y = −5. The points are (3, 3) and (−1, −5). Distractor routes: (3, 3) and (1, −1) comes from mis-factorising x² − 2x − 3 as (x − 3)(x − 1), giving a second root of 1 instead of −1. (3, 9) and (−1, 1) comes from finding the y-coordinate from y = x² instead of substituting back into the given line equation y = 2x − 3. (3, 3) alone stops after finding only the first root of the quadratic.
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (b) 3 and 4 — Method: the graph of f(x) is continuous, so where it crosses the x-axis the value of f(x) changes sign; substitute consecutive integers until one value is negative and the next is positive. Working: f(2) = 8 − 6 − 20 = −18, f(3) = 27 − 9 − 20 = −2 and f(4) = 64 − 12 − 20 = 32. The sign changes from negative to positive between x = 3 and x = 4, so the solution lies there. Answer: 3 and 4. The distractors: 2 and 3 comes from ignoring the −3x term and solving x³ = 20, whose root is 2.71, one interval to the left; 6 and 7 comes from reading x³ as x² and solving x² − 3x − 20 = 0, whose positive root is 6.22; 4 and 5 is the interval immediately after the change of sign, named by a candidate who finds f(4) positive and quotes the interval beginning there instead of the one across which the sign actually turned.
- (b) −2, −1, 0, 1, 2, 3, 4 — Factorise x² − 2x − 8 = (x − 4)(x + 2), giving roots x = 4 and x = −2. Since the coefficient of x² is positive and the inequality is ≤ 0, the solution is the closed interval between the roots, −2 ≤ x ≤ 4, with the roots included because the inequality is not strict. The integers in this interval are −2, −1, 0, 1, 2, 3, 4. Distractor routes: −1, 0, 1, 2, 3 drops both endpoints, treating ≤ as if it were the strict inequality <. −2, −1, 0, 1, 2, 3, 4, 5 comes from mis-factorising as (x − 5)(x + 2), giving an upper root of 5 instead of 4. −3, −2, −1, 0, 1, 2, 3 comes from mis-factorising as (x − 4)(x + 3), giving a lower root of −3 instead of −2.
- (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.
- (b) 12.7 m/s — Find the total distance from the area under the graph, then divide by the total time. Break the graph into its four straight sections: a triangle from (0, 0) to (5, 20), where 5 × 20 = 100 gives an area of 50 m when halved; a rectangle from (5, 20) to (12, 20), area 7 × 20 = 140 m; a trapezium from (12, 20) to (16, 4), where (20 + 4) × 4 = 96 gives an area of 48 m when halved; and a rectangle from (16, 4) to (20, 4), area 4 × 4 = 16 m — a total distance of 50 + 140 + 48 + 16 = 254 m. Dividing by the 20 seconds gives an average speed of 254 ÷ 20 = 12.7 m/s. Writing 15.1 m/s comes from forgetting to halve the trapezium area for the third section, using 96 instead of 48: a total of 302 m, and 302 ÷ 20 = 15.1 m/s. Writing 11.9 m/s comes from leaving out the final section, from (16, 4) to (20, 4), entirely: a total of 238 m, and 238 ÷ 20 = 11.9 m/s. Writing 12.1 m/s comes from dividing the correct total distance by 21 instead of 20 — a fencepost slip, counting the whole seconds from t = 0 to t = 20 inclusive as 21 seconds of travel rather than reading the journey as a duration of 20 seconds: 254 ÷ 21 = 12.1 m/s (1 d.p.).
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (b) Translate +5 in x, then reflect in the x-axis. — Translating y = x² by 5 units in the positive x-direction gives y = (x − 5)². Reflecting this in the x-axis, which replaces y with −y, gives y = −(x − 5)², matching the target. Using a translation of −5 in x instead gives y = (x + 5)², and reflecting that in the x-axis gives y = −(x + 5)² — the sign inside the bracket is wrong. Reflecting in the y-axis first does nothing to y = x², since (−x)² = x², so translating afterwards only reaches y = (x − 5)² with no negative sign at all. Translating by 5 units in y instead of x gives y = x² + 5, and reflecting that in the x-axis gives y = −x² − 5, a different curve altogether — a vertical shift does not create the (x − 5)² term the target equation needs.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (b) 10 — Check n = 9: a₉ = 3 × 2⁸ = 3 × 256 = 768, below 1,000. Check n = 10: a₁₀ = 3 × 2⁹ = 3 × 512 = 1,536, above 1,000. So the first term greater than 1,000 is at position n = 10. Answering 9 comes from forgetting the −1 shift and using the formula as 3 × 2ⁿ instead of 3 × 2ⁿ⁻¹: checking 3 × 2⁹ = 1,536 (which exceeds 1,000) but then reporting the position as n = 9, the exponent used, instead of n = 10 — wrong, because the exponent in the real formula is n − 1, not n. Answering 11 comes from going one term too far: correctly finding that a₁₀ already exceeds 1,000, but then checking one position further and reporting n = 11 instead of stopping at the first position that already works — wrong, because n = 11 is not the FIRST term greater than 1,000. Answering 1,536 gives the VALUE of the term (a₁₀ itself) rather than its position — wrong, because the question asks which term it is (the value of n), not what that term is worth.
- (c) Yes; x = (y + 5)/2 is right — Sam's method is correct throughout: adding 5 to both sides gives y + 5 = 2x, and dividing both sides by 2 gives x = (y + 5)/2, so Sam is right. The option giving x = (y − 5)/2 assumes 5 should be subtracted again, but 2x − 5 = y means 5 has already been subtracted, so it must be added back, not taken away a second time. The option giving x = y/2 + 5 divides only the y term by 2 and leaves the 5 unhalved, which is not a valid rearrangement. The option agreeing Sam is correct but changing step 2 to x = 2(y + 5) confuses '2x' with 'x divided by 2' — since x is multiplied by 2, the inverse is division, not multiplication. Sam's working, and his final formula x = (y + 5)/2, are both correct.
- (d) Wrong — the correct solution is x < −3 or x > 3. — Squaring or unsquaring an inequality is not a safe one-step move: the solution must be split into two branches, since a number less than −3 also squares to more than 9. So x > 3 finds only half the solution set. The full answer is x < −3 or x > 3. Distractor routes: "Right — square rooting both sides gives x > 3" treats the square root of an inequality the same as the square root of an equation and misses the negative branch entirely. "Wrong — the correct solution is −3 < x < 3" applies the between-the-roots pattern that belongs to the opposite inequality, x² < 9. "Right, but x = −3 and x = 3 should also work" treats the inequality as if it also allowed equality, when x² > 9 is strict and excludes both x = 3 and x = −3.
- (c) y = 2x + 1 — Method: the gradient of the line through two points is the change in y divided by the change in x, and the constant is then found by substituting one of the points into y = mx + c. Working: m = (5 − 3) ÷ (2 − 1) = 2 ÷ 1 = 2, so the line is y = 2x + c; substituting x = 1 and y = 3 gives 3 = 2 × 1 + c, so c = 3 − 2 = 1. Answer: y = 2x + 1. The distractors: y = x + 2 comes from taking the gradient as the change in x, 2 − 1 = 1, and then substituting (1, 3) to reach a constant of 2; y = 2x − 1 comes from working out the constant as mx − y, 2 × 1 − 3 = −1, instead of y − mx; y = 2x + 3 comes from using the y-coordinate of (1, 3) as the constant without substituting at all.
- (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.
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