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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- (d) 2 — The gradient between two points on a line is the change in y divided by the change in x. Using (5, 2) and (0, −8): (2 − (−8)) ÷ (5 − 0) = 10 ÷ 5 = 2. Inverting the fraction, dividing the change in x by the change in y instead, gives 5 ÷ 10 = 0.5. Reversing the order of the x-values in the denominator, giving (2 − (−8)) ÷ (0 − 5) = 10 ÷ (−5), gives −2. Misreading the y-intercept as 8 instead of −8, giving (2 − 8) ÷ 5 = −6 ÷ 5, gives −1.2.
- (a) s/f — Sharing s sweets equally between f friends means dividing the total by the number of friends, written as a fraction: s/f. Writing f/s divides the wrong way round, sharing the number of friends between the sweets instead of the sweets between the friends. Writing s − f mistakes sharing for taking away, subtracting the number of friends from the number of sweets. Writing sf multiplies the two quantities together, which would make the total larger rather than splitting it into smaller equal parts. The number of sweets each friend receives is s/f.
- (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.
- (a) 18 cm² — Method: substitute the width into the formula, applying the index to the letter before multiplying by the 2 in front of it. Working: x² = 3 × 3 = 9, and then A = 2 × 9 = 18, so the area is 18 cm². Answer: 18 cm². The distractors: 36 cm² comes from multiplying 2 by 3 first and squaring afterwards, giving (2 × 3)²; 12 cm² comes from doubling instead of squaring, so that x² is replaced by 2x and the calculation becomes 2 × 2 × 3; 6 cm² comes from working out 2 × 3 and never applying the index at all.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (c) x = 0, y = 3 and x = 3, y = 0 — Substitute y = 3 − x into x² + y² = 9: x² + (3 − x)² = 9. Expanding (3 − x)² = 9 − 6x + x² gives x² + 9 − 6x + x² = 9, which simplifies to 2x² − 6x = 0, or 2x(x − 3) = 0, so x = 0 or x = 3. Using y = 3 − x: x = 0 gives y = 3; x = 3 gives y = 0. Distractor routes: x = 0, y = 3 alone stops after the factor 2x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = 3 and x = −3, y = 6 comes from factorising 2x² − 6x as 2x(x + 3), a sign error that gives a second root of −3 instead of 3. x = 0, y = 3 and x = 6, y = −3 comes from expanding (3 − x)² as 9 − 6x, dropping the x² term, which changes the quadratic to x² − 6x = 0 and its second root to 6.
- (b) 10 — Check n = 9: a₉ = 3 × 2⁸ = 3 × 256 = 768, below 1,000. Check n = 10: a₁₀ = 3 × 2⁹ = 3 × 512 = 1,536, above 1,000. So the first term greater than 1,000 is at position n = 10. Answering 9 comes from forgetting the −1 shift and using the formula as 3 × 2ⁿ instead of 3 × 2ⁿ⁻¹: checking 3 × 2⁹ = 1,536 (which exceeds 1,000) but then reporting the position as n = 9, the exponent used, instead of n = 10 — wrong, because the exponent in the real formula is n − 1, not n. Answering 11 comes from going one term too far: correctly finding that a₁₀ already exceeds 1,000, but then checking one position further and reporting n = 11 instead of stopping at the first position that already works — wrong, because n = 11 is not the FIRST term greater than 1,000. Answering 1,536 gives the VALUE of the term (a₁₀ itself) rather than its position — wrong, because the question asks which term it is (the value of n), not what that term is worth.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (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) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
- (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
- (c) {x : x ≤ −4} ∪ {x : x ≥ 4} — Rearrange so one side is zero: x² − 16 ≥ 0, then factorise: (x − 4)(x + 4) ≥ 0. The critical values are x = −4 and x = 4. Since the coefficient of x² is positive, the graph is a U-shape that is on or above the x-axis outside its roots, so the solution is x ≤ −4 or x ≥ 4, written as {x : x ≤ −4} ∪ {x : x ≥ 4}. Distractor routes: {x : −4 ≤ x ≤ 4} takes the region BETWEEN the roots, which is where x² − 16 is negative, the opposite region. {x : x ≥ 4} keeps only the positive square root and drops the negative branch entirely. {x : x ≤ 4} comes from a sign error, treating the inequality as if it were x² ≤ 16.
- (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
- (c) Second and fourth — If x + y = 0 then y = −x, so x and y always have opposite signs, one positive and one negative. A point with a negative x and a positive y lies in the second quadrant, and a point with a positive x and a negative y lies in the fourth quadrant, so the point lies in the second or the fourth. A candidate who reads x + y = 0 as meaning x and y have the same sign picks First and third, which is where x × y is positive, not where x + y = 0. A candidate who decides that y must be the positive coordinate picks the two quadrants above the x-axis, First and second. A candidate who decides that y must be the negative coordinate picks the two quadrants below the x-axis, Third and fourth.
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