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.Work out the coordinates of the point where the graph of y = 2x − 6 crosses the x-axis.y = 2x − 6
- 2.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.
- 3.A straight line passes through the points (2, 7) and (5, 16). Work out the value of y when x = 0.
- 4.A sequence has the position-to-term rule n² + 2, where n is the position number. Work out the 6th term.
- 5.A geometric sequence begins 162, 54, 18, 6, ... Work out the next term in the sequence.
- 6.A car passes the 15 km marker on a road and passes the 195 km marker 3 hours later. Work out the gradient of the distance-time graph, that is the rate of change of distance with time.
- 7.Work out the period of the graph of y = cos x, in degrees.y = cos(x)
- 8.The equation x² − 4x + k = 0 has two different real solutions. Work out the range of values of k.
- 9.The tangent to a curve at the point where x = 4 passes through the points (2, 5) and (6, 21). Use these two points to estimate the gradient of the curve at x = 4.
- 10.Work out the value of n² − n when n = −3
- 11.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 12.A cyclist's speed-time graph is described as follows: speed increases steadily from 0 m/s to 8 m/s over the first 4 seconds, stays constant at 8 m/s for the next 6 seconds, then decreases steadily to 0 m/s over the final 2 seconds. Work out the total distance travelled.
- 13.Work out the coordinates of the reflection of (6, −3) in the y-axis.
- 14.Solve x² + 7x = 0.
- 15.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
Answer key
- (d) (3, 0) — Method: every point on the x-axis has y-coordinate 0, so substituting y = 0 into the equation and solving gives the x-coordinate of the crossing point. Working: 0 = 2x − 6 gives 2x = 6, so x = 6 ÷ 2 = 3 and the graph crosses the x-axis at (3, 0). Answer: (3, 0). The distractors: (0, −6) is where the graph crosses the y-axis, found by substituting x = 0 rather than y = 0; (−3, 0) comes from moving the 6 across the equals sign without changing its sign, giving 2x = −6; (6, 0) comes from reading the constant straight off as the crossing point and never dividing by the gradient 2.
- (a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (a) 38 — Method: substitute the position number into the rule and follow the order of operations, so the squaring is carried out before the 2 is added. Working: n = 6 gives 6² + 2; 6² means 6 × 6 = 36, and then 2 is added to 36. Answer: 38. The distractors: 14 comes from multiplying the position by 2 instead of squaring it, 6 × 2 + 2; 36 comes from squaring correctly and then forgetting to add the 2; 64 comes from adding the 2 first and squaring afterwards, (6 + 2)².
- (a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (c) 360° — The graph of y = cos x repeats itself every 360°, so its period is 360°. 180° is the period of y = tan x, not y = cos x — confusing the two graphs gives this. 90° is the horizontal distance from a maximum of y = cos x to the next point where the graph crosses the x-axis, a quarter of the period, mistaken for the whole period. 720° comes from counting the interval containing two full repeats of the graph, from one maximum to the second repeat of that maximum, instead of the first.
- (d) k < 4 — Method: the number of real solutions of ax² + bx + c = 0 is decided by the discriminant b² − 4ac, and two different real solutions need it to be greater than zero. Working: here a = 1, b = −4 and c = k, so b² − 4ac = 16 − 4k; the condition is 16 − 4k > 0, which gives 16 > 4k and then k < 4. Answer: k < 4; for example k = 3 gives x² − 4x + 3 = 0, whose solutions are 1 and 3. The distractors: k ≤ 4 comes from using b² − 4ac ≥ 0, which also allows the single repeated solution at k = 4; k > 4 comes from dividing −4k > −16 by −4 without reversing the inequality sign; k < 16 comes from leaving the factor 4 out of 4ac and solving 16 − k > 0.
- (a) 4 — The gradient of a straight line through two points is the change in y divided by the change in x. Change in y = 21 − 5 = 16. Change in x = 6 − 2 = 4. Gradient = 16 ÷ 4 = 4. Dividing x by y instead of y by x gives 0.25; forgetting to divide by the change in x at all leaves 16; dividing by only one of the two x-coordinates, 16 ÷ 2 = 8, uses the wrong denominator.
- (d) 12 — Method: substitute the value into both terms, remembering that subtracting a negative number has the same effect as adding the matching positive number. Working: n² = (−3) × (−3) = 9, and subtracting n means subtracting −3, which adds 3, so the calculation is 9 + 3 = 12. Answer: 12. The distractors: 6 comes from subtracting 3 rather than subtracting −3, giving 9 − 3; −6 comes from squaring −3 as −9 while still adding the 3, giving −9 + 3; −3 comes from reading n² as 2n, giving 2 × (−3) = −6 and then −6 + 3.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
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