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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Answer key: Algebra worksheet — GCSE Higher
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- (a) y = −x³ + 5x — Reflecting a graph in the y-axis replaces every x in the equation with −x: y = (−x)³ − 5(−x) = −x³ + 5x. Writing y = −x³ − 5x comes from substituting −x into the x³ term only and leaving the −5x term as it was. Writing y = x³ + 5x comes from substituting −x into the −5x term only and leaving the x³ term as it was. Writing y = x³ − 5x is the original equation with no reflection applied at all — every term needs the substitution, not just one of them.
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
- (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) 3x = 12 — Adding the two equations: the y-terms, +y and −y, have opposite signs, so they cancel; the x-terms combine to 2x + x = 3x; and the right-hand sides add to 11 + 1 = 12. This gives 3x = 12. A candidate who forgets that the y-terms cancel, and instead adds them as if they had the same sign, would write 3x + 2y = 12. A candidate who subtracts the right-hand sides instead of adding them would get 3x = 10. A candidate who correctly reaches 3x = 12 but then treats 12 itself as the value of x, skipping the final division, would write x = 12.
- (a) −2 m/s² — The gradient of a line through two points is the change in the y-value divided by the change in the x-value: (6 − 22)/(16 − 8) = −16/8 = −2 m/s². The negative sign confirms the deceleration mentioned in the question. Writing 2 m/s² comes from subtracting the velocities in the wrong order, using (22 − 6) instead of (6 − 22), which loses the sign that shows the car is slowing down. Writing −0.5 m/s² comes from inverting the fraction, dividing the change in time by the change in velocity instead of the other way round. Writing −16 m/s² comes from finding the change in velocity, −16, but forgetting to divide by the change in time, 8 seconds.
- (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.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (d) x = (y + 3)/4 — Two operations have been applied to x: it has been multiplied by 4 and then 3 has been subtracted. Undo them in the opposite order. Adding 3 to both sides gives y + 3 = 4x. Dividing both sides by 4 then gives (y + 3)/4 = x, so x = (y + 3)/4. The bracket matters: writing y/4 + 3 divides only the y by 4 and leaves the 3 untouched. Writing (y − 3)/4 subtracts the 3 instead of adding it, and writing 4(y + 3) multiplies by 4 rather than dividing.
- (c) x = 3 or x = 1/2 — Method: factorise into two brackets whose x terms multiply to 2x² and whose numbers multiply to 3, checking that they produce the middle term −7x, then set each bracket equal to zero. Working: (2x − 1)(x − 3) expands to 2x² − 6x − x + 3 = 2x² − 7x + 3, so (2x − 1)(x − 3) = 0; then 2x − 1 = 0 gives x = 1/2 and x − 3 = 0 gives x = 3. Answer: x = 3 or x = 1/2. The distractors: x = 3/2 or x = 1 comes from factorising as (2x − 3)(x − 1), whose middle term is −5x and not −7x; x = −3 or x = −1/2 comes from reading the roots straight out of (2x − 1)(x − 3) without changing the signs; x = 6 or x = 1 comes from using the quadratic formula with the denominator written as a instead of 2a, dividing 7 ± 5 by 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.
- (d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (d) Yes: the 8th term is 50 − 8 × 7 = −6, which is negative. — Method: find the 8th term by subtracting 8 a total of 7 times from the first term, since the 1st term itself needs 0 subtractions. Working: 8th term = 50 − 8 × 7 = 50 − 56 = −6, which is negative, so Priya is correct. Answer: Yes, the 8th term is 50 − 8 × 7 = −6, which is negative. The "50 − 8 × 6 = 2" option subtracts 8 only six times instead of seven, an off-by-one error in counting the steps. The "50 − 8 × 8 = −14" option subtracts 8 eight times instead of seven, the opposite off-by-one error. The claim that repeated subtraction "can never go negative" ignores that subtracting enough times from any starting value eventually gives a negative result.
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