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) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
- (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) An inequality, because ≤ compares the two sides — The symbol ≤ means 'is less than or equal to', so the statement compares the sizes of the two sides instead of saying they are equal: that makes it an inequality. Solving it gives n ≤ 5, a whole range of values rather than the single value an equation would give. An identity has to be true for every value of the letter, and this fails at n = 6, so it is not one. A formula works one quantity out from another, and there is only one letter here.
- (d) 16 boys and 14 girls — Method: write the number of boys in terms of the number of girls, then use the total for the class. Working: if there are g girls then there are g + 2 boys, so g + (g + 2) = 30, that is 2g + 2 = 30, so 2g = 28 and g = 14; the number of boys is 14 + 2 = 16. Answer: 16 boys and 14 girls, which total 30 and differ by 2. The distractors: 14 boys and 16 girls comes from substituting for the wrong group, writing b + (b + 2) = 30 and then calling b the number of boys; 15 boys and 15 girls comes from halving 30 and never using the difference; 17 boys and 13 girls comes from adding the whole 2 to one half of the class and taking the whole 2 off the other half, which leaves a difference of 4.
- (a) y = (4/3)x + 25/3 — The radius from (0, 0) to (−4, 3) has gradient 3 ÷ (−4) = −3/4. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal, 4/3. Using y − y₁ = m(x − x₁) with the point (−4, 3): y − 3 = (4/3)(x + 4), so y = (4/3)x + 16/3 + 3 = (4/3)x + 25/3. Using the radius's own gradient, −3/4, instead of taking the perpendicular gradient, gives y − 3 = (−3/4)(x + 4), which simplifies to y = −(3/4)x once the −3 and +3 in the constant cancel out. Taking the reciprocal of the radius's gradient but keeping the wrong sign, using −4/3 instead of 4/3, gives y = −(4/3)x − 7/3. Correctly finding the gradient 4/3 and expanding the bracket, but forgetting to add the y-coordinate 3 at the end, gives y = (4/3)x + 16/3.
- (b) 25 — The tangent at (9, 12) is 9x + 12y = 225. Setting y = 0 (the x-axis): 9x = 225, so x = 25. Choosing 18.75 comes from swapping the coefficients in the tangent equation (using 12x + 9y = 225) before setting y = 0. Choosing 15 is where the circle itself meets the x-axis (from x² = 225), not where the tangent does. Choosing 9 is just the x-coordinate of the original point (9, 12), not the point P.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (b) y = 2x and y = −(1/2)x + 1 — Method: two lines are perpendicular when the product of their gradients is −1, so the gradient of each line is read from the number multiplying x and the two are multiplied together. Working: for y = 2x and y = −(1/2)x + 1 the gradients are 2 and −1/2, and 2 × (−1/2) = −1, so that pair is perpendicular. Answer: y = 2x and y = −(1/2)x + 1. The distractors: the pair y = 2x and y = 2x + 1 has equal gradients, whose product is 4, so those two lines are parallel and never meet, and they are chosen by a candidate applying the parallel condition; the pair y = 2x and y = −2x + 1 comes from changing the sign of the gradient without turning it upside down, and its product is −4; the pair y = 2x and y = (1/2)x + 1 comes from turning the gradient upside down without changing its sign, and its product is 1.
- (a) 5.8 — The horizontal distance is 3 and the vertical distance is 5, so using Pythagoras' theorem the distance is √(3² + 5²) = √34 = 5.8 (1 d.p.). A candidate who adds the two differences instead of using Pythagoras gets 3 + 5 = 8.0. A candidate who works out 3² + 5² = 34 but forgets to take the square root gets 34.0. A candidate who subtracts the squares instead of adding them gets √(5² − 3²) = √16 = 4.0.
- (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
- (c) −4 — f(2) = 7, and y = f(x) + a passing through (2, 3) means f(2) + a = 3, so 7 + a = 3, giving a = 3 − 7 = −4. Writing a = 4 comes from subtracting the wrong way round, 7 − 3 instead of 3 − 7. Writing a = 10 comes from adding the two y-values instead of subtracting one from the other. Writing a = 3 comes from taking the new y-coordinate as the value of a directly, without accounting for the original y-value of 7 at all.
- (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) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (b) (x + 2)/3 — Start with y = 3x − 2 and swap x and y: x = 3y − 2. Add 2 to both sides: x + 2 = 3y. Divide both sides by 3: y = (x + 2)/3, so f⁻¹(x) = (x + 2)/3. Writing x/3 + 2 comes from dividing only the 3y term by 3 and leaving the +2 outside the division — the 2 must be added before you divide, not after. Writing (x − 2)/3 comes from keeping the subtraction sign instead of flipping it to addition when the −2 is moved across the equals sign. Writing 3x + 2 comes from swapping x and y but never actually solving for y — just changing the sign of the constant term.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
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