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
MathsUKwww.geekhero.co.uk
- (c) Wrong — the product is −4/25, not −1. — The student's rearrangement is correct: 5y = −2x + 15 gives y = −(2/5)x + 3, gradient −2/5. But the test for perpendicularity is that the product of the two gradients equals exactly −1, not merely that it is negative. Here 2/5 × (−2/5) = −4/25, which is not −1, so the lines are NOT perpendicular. Distractor routes: "a negative product always means perpendicular" states a rule that does not exist — many pairs of lines have a negative gradient product without being perpendicular, as this pair shows. "The rearrangement is incorrect" wrongly blames a correct step; 5y = −2x + 15 does rearrange to y = −(2/5)x + 3. "2/5 and −2/5 are negatives of each other" notices a true but irrelevant fact — being negatives of each other is not the perpendicularity condition; an exact product of −1 is.
- (c) Line D — Method: the steeper of two lines is the one whose gradient is greater, and the gradient of a line through two points is the change in y divided by the change in x. Working: line C rises 2 for a run of 4, so its gradient is 2 ÷ 4 = 1/2; the other line rises 2 for a run of 2, so its gradient is 2 ÷ 2 = 1; since 1 is greater than 1/2, it is line D that is the steeper. Answer: Line D. The distractors: Line C is chosen by comparing the x-coordinates and calling the line that reaches further along the x-axis the steeper one; They are equally steep comes from comparing only the y-coordinates, which are both 2, without dividing by the two different runs; There is not enough information comes from believing that a gradient can only be measured off a drawn graph, when two points on a line are enough to work it out.
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
- (c) 55 — Continuing the pattern: 8+13=21 (5th term), 13+21=34 (6th term), 21+34=55 (7th term). A candidate who miscounts the position and stops one term early would give 34, the 6th term. A candidate who doubles the most recent term instead of adding the two before it would compute 34×2=68. A candidate who adds a non-adjacent pair — the 4th and 6th terms, skipping the 5th — would compute 13+34=47.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (c) −0.5 — fg(x) = f(g(x)) = f(x²) = x² + 2. gf(x) = g(f(x)) = g(x + 2) = (x + 2)² = x² + 4x + 4. Setting fg(x) = gf(x): x² + 2 = x² + 4x + 4. Subtract x² from both sides: 2 = 4x + 4. Subtract 4 from both sides: −2 = 4x, so x = −0.5. Writing 1.5 comes from adding the 4 instead of subtracting it: 4x = 2 + 4 = 6, giving x = 1.5. Writing 'no solution' comes from expanding (x + 2)² as x² + 4 using (a + b)² = a² + b², losing the middle term — the equation then reads x² + 2 = x² + 4, which has no solution, but the expansion itself is wrong. Writing 0 comes from treating gf(x) as g(x) + f(x) instead of g(f(x)): x² + (x + 2) = x² + 2 gives x = 0, but that adds the two functions rather than composing them.
- (a) A translation by vector (0, 3) — y = x² + 3 adds a constant outside the squaring, so it is a vertical translation of y = x² — every point moves the same distance parallel to the y-axis, with no change in x. The vector is (0, 3), matching the +3. A vector of (3, 0) confuses this with a horizontal shift; (0, −3) has the right axis but the wrong sign, since the graph moves up, not down; a stretch changes the shape of the curve, which adding a constant term never does.
- (c) 22 — p² means p × p. Substitute p = 2.5: p² = 6.25, so 4p² = 4 × 6.25 = 25, and 25 − 3 = 22. 97 comes from squaring 4p together instead of just p, (4 × 2.5)² − 3 = 10² − 3 = 97. 7 comes from using p instead of p², 4 × 2.5 − 3. 1 comes from squaring the 3 instead of the p, 4 × 2.5 − 3².
- (b) h = 2A/b — The height has been multiplied by the base and the result then divided by 2, so undo the division first. Multiplying both sides by 2 gives 2A = bh. Undoing the multiplication by the base comes next: dividing both sides by b gives 2A/b = h, so h = 2A/b. Dividing by 2 instead of multiplying gives A/(2b), a quarter of the correct height; multiplying by the base instead of dividing gives 2Ab; writing b/(2A) turns the final fraction upside down.
- (d) An equation, true for one value of x — 2x + 5 = 17 is only true when x = 6, so it is an equation. A candidate who ignores the equals sign and focuses only on the x terms and numbers present picks the expression option. A candidate who sees an equals sign and assumes it must relate two different quantities, as a formula does, picks that option, missing that there is only one quantity, x, involved here. A candidate who wrongly checks whether both sides look similar in structure rather than testing specific values picks the identity option, missing that the two sides are not equal for every value of x.
- (d) 128 — In 2p³ the index belongs to p only, so cube p first and multiply by the coefficient afterwards. Cubing gives 4 × 4 × 4 = 64, and then 2 × 64 = 128. Cubing the coefficient as well would mean working out (2 × 4)³, which is 512. Reading the index as an instruction to multiply by 3 gives 2 × 4 × 3 = 24, and ignoring the coefficient altogether leaves 64.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
Build your own mix at the worksheet builder.