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 midpoint of the line segment joining (−3, 5) and (7, 9).
- 2.Write down the coordinates of the point where the line y = 2x + 4 crosses the y-axis.y = 2x + 4
- 3.y = 5 − 2x. Work out the value of x when y = 11.
- 4.A tram sets off from a stop. Its velocity-time graph rises in a straight line from 0 m/s to 12 m/s over the first 8 seconds, then stays constant at 12 m/s for a further 10 seconds. Work out the total distance the tram travels in these 18 seconds, using the area under the graph.
- 5.A water butt holds 200 litres and is being drained at a steady 8 litres per minute. Write down the function for the amount of water y, in litres, left after x minutes.
- 6.Points A(−1, 2) and B(5, 8) are the endpoints of a line segment. Work out the equation of the perpendicular bisector of AB.
- 7.A garden path runs along the line 4x + y = 12. A drainage pipe must be laid perpendicular to the path, passing through the point (3, 1) where a sprinkler sits. Work out the equation of the pipe's line, giving your answer in the form y = mx + c.
- 8.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
- 9.A phone plan costs a fixed £15 per month plus 8p per minute of calls. In one month, Jamal's bill was £23.80. Form an equation using m for the number of minutes, and solve it to find how many minutes Jamal used that month.
- 10.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
- 11.A candle is 30 cm tall and burns down at a steady rate of 1 cm per hour. Write down the function for the height of the candle y, in centimetres, after x hours.
- 12.Work out the equation of the straight line through the points (−3, 4) and (1, −8).
- 13.Solve the inequality 3x − 1 ≤ 11.
- 14.The first five terms of a quadratic sequence are −1, 4, 13, 26, 43. Work out an expression, in terms of n, for the nth term.
- 15.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
Answer key
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
- (d) (0, 4) — Method: in the form y = mx + c the constant c is the y-coordinate of the point where the line crosses the y-axis, and every point on that axis has x-coordinate 0. Working: comparing y = 2x + 4 with y = mx + c gives c = 4, and substituting x = 0 confirms y = 2 × 0 + 4 = 4, so the crossing point is (0, 4). Answer: (0, 4). The distractors: (4, 0) is the same pair of numbers written in the wrong order and is a point on the x-axis; (0, 2) comes from reading the gradient 2 as the intercept, confusing m with c; (−2, 0) is where the line crosses the x-axis, found by solving 0 = 2x + 4 instead.
- (d) −3 — Substitute y = 11 into y = 5 − 2x, giving 11 = 5 − 2x. Subtracting 5 from both sides gives 6 = −2x, so x = 6 ÷ (−2) = −3. A candidate who mishandles the negative sign when rearranging, treating the equation as 6 = 2x, gets x = 3. A candidate who correctly finds −2x = 6 but forgets to divide by 2 at all gets x = 6. A candidate who adds 5 and 11 instead of subtracting, getting 2x = 16, gets x = 8.
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (d) x + y = 7 — The midpoint of AB is (−1 + 5)/2, (2 + 8)/2, which is (2, 5). The gradient of AB is (8 − 2)/(5 − (−1)) = 6/6 = 1. The perpendicular bisector has gradient −1 and passes through (2, 5): y − 5 = −(x − 2), which rearranges to x + y = 7. Distractor routes: x − y = −3 uses the gradient of AB itself, 1, rather than its negative reciprocal, giving a line PARALLEL to AB through the midpoint instead of perpendicular to it. x + y = 12 comes from adding the y-coordinates, 2 + 8 = 10, but forgetting to divide by 2, using the midpoint (2, 10) instead of (2, 5). x + y = 4 comes from using A's x-coordinate, −1, directly instead of averaging it with B's, giving the point (−1, 5) instead of the true midpoint (2, 5).
- (a) y = (1/4)x + 1/4 — Rearrange 4x + y = 12 to y = −4x + 12, so the path has gradient −4. The perpendicular gradient is 1/4. Substituting (3, 1) into y − 1 = (1/4)(x − 3): y = (1/4)x − 3/4 + 1 = (1/4)x + 1/4. Distractor routes: y = −4x + 13 uses the path's own gradient, −4, instead of the perpendicular gradient, giving a line PARALLEL to the path through (3, 1) rather than perpendicular to it. y = (1/4)x − 1/4 makes an arithmetic slip combining −3/4 and 1, landing on −1/4 instead of the correct 1/4. y = −(1/4)x + 1/4 takes the reciprocal of −4, which is −1/4, but forgets to change its sign, so it is not the true negative reciprocal.
- (b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (a) y = −3x − 5 — Gradient = (−8 − 4) ÷ (1 − (−3)) = −12 ÷ 4 = −3. Using the point (1, −8): −8 = −3(1) + c, so c = −5, giving y = −3x − 5. A candidate who drops the negative sign on the gradient, using m = 3 instead, would then solve −8 = 3(1) + c to get c = −11, writing y = 3x − 11. A candidate who makes a sign error isolating c, writing c = 5 instead of −5, would write y = −3x + 5. A candidate who mixes up both mistakes — keeping the correct gradient but the wrong, positive value of c from the flipped-gradient calculation — would write y = −3x + 11.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
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