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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- (c) 3 — Gradient = (change in y) ÷ (change in x) = (11 − 5) ÷ (4 − 2) = 6 ÷ 2 = 3. A candidate who puts the change in x over the change in y instead would get 2 ÷ 6 = 1/3. A candidate who subtracts the y-coordinates in the reverse order, but not the x-coordinates, would get (5 − 11) ÷ (4 − 2) = −3. A candidate who adds the coordinates instead of subtracting them would get (11 + 5) ÷ (4 + 2) = 16/6 = 8/3.
- (b) 4x − 4 = 20 — Method: subtract 2x from both sides of the equation, and simplify each side separately. Working: left side: 6x − 4 − 2x = 4x − 4. Right side: 2x + 20 − 2x = 20. Answer: 4x − 4 = 20. 4x = 20 drops the −4 from the left side, as though subtracting 2x also removes the constant term. 8x − 4 = 20 comes from moving the 2x across to the left without changing its sign: it is taken off the right side correctly, leaving 20, but added to the left side instead of subtracted, giving 6x + 2x = 8x. 4x − 4 = 2x + 20 comes from subtracting 2x from the left-hand side only and leaving the right-hand side unchanged; whatever is done to one side must be done to the other.
- (c) 5 — Method: substitute the value into the formula, work out the index first and the addition afterwards. Working: x² = (−2) × (−2) = 4, so y = 4 + 1 = 5. Answer: 5. The distractors: −3 comes from squaring only the 2 and keeping the minus sign outside the index, giving −4 + 1; 1 comes from adding the 1 to x before squaring, giving (−2 + 1)²; 4 comes from squaring correctly but stopping there and never adding the 1.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (a) x = 6 — Method: a minus sign in front of a bracket changes the sign of every term inside it, so expand the bracket first and then simplify. Working: expanding gives 4x − 2x + 6 = 18, which simplifies to 2x + 6 = 18; subtracting 6 from both sides gives 2x = 12, and dividing both sides by 2 gives x = 6. Answer: x = 6. The distractors: x = 12 comes from leaving the −6 unchanged when the bracket is removed, giving 4x − 2x − 6 = 18 and so 2x = 24; x = 24 comes from reaching 2x = 12 correctly and then multiplying by 2 instead of dividing by 2; x = 3 comes from dividing by 2 too early, at 2x + 6 = 18, and dividing only the 2x and the 18 while leaving the 6 untouched, which gives x + 6 = 9.
- (c) 360° — The graph of y = cos x repeats itself every 360°, so its period is 360°. 180° is the period of y = tan x, not y = cos x — confusing the two graphs gives this. 90° is the horizontal distance from a maximum of y = cos x to the next point where the graph crosses the x-axis, a quarter of the period, mistaken for the whole period. 720° comes from counting the interval containing two full repeats of the graph, from one maximum to the second repeat of that maximum, instead of the first.
- (d) $y = x^2 - 2x - 3$ — Method: read the two x-intercepts (roots) from the graph, write the quadratic as the product of the corresponding factors, then expand. Working: the curve crosses the x-axis at x = −1 and x = 3, so the equation factorises as (x + 1)(x − 3), which expands to x² − 2x − 3. Answer: y = x² − 2x − 3. Distractor refutation: y = x² − x − 6 comes from misreading the left-hand crossing point as x = −2 instead of x = −1, giving factors (x + 2)(x − 3). y = x² − x − 2 comes from misreading the right-hand crossing point as x = 2 instead of x = 3, giving factors (x + 1)(x − 2). y = x² + 2x − 3 comes from writing the factors as (x − 1)(x + 3), swapping which root gets the plus sign and which gets the minus sign, giving the wrong sign on the x term.
- (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.
- (a) 12 — Method: write both ages as they were six years ago, call Leo's present age x, and form an equation from the comparison at that time. Working: six years ago Harry was 24 − 6 = 18 and Leo was x − 6, so 18 = 3(x − 6); expanding gives 18 = 3x − 18, adding 18 to both sides gives 36 = 3x, and dividing by 3 gives x = 12. Checking: six years ago Harry was 18 and Leo was 6, and 18 = 3 × 6. Answer: 12. The distractors: 6 comes from solving 18 = 3(x − 6) as far as Leo's age six years ago, 18 ÷ 3 = 6, and giving that as his age now; 8 comes from dividing Harry's present age by 3, 24 ÷ 3 = 8, using the multiple at the wrong moment in time; 36 comes from stopping at 3x = 36 and giving 36 as Leo's age.
- (d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (c) x = 2, y = 3 — Method: no letter cancels straight away, so make one letter the subject of one equation and substitute that expression into the other. Working: 2x + y = 7 gives y = 7 − 2x, so x + 2y = 8 becomes x + 2(7 − 2x) = 8, that is x + 14 − 4x = 8, so −3x = −6 and x = 2; substituting back gives y = 7 − 4 = 3. Answer: x = 2, y = 3, which checks in 2 + 6 = 8. The distractors: x = 3, y = 2 comes from adding the equations to x + y = 5 and then subtracting them the wrong way round, writing x − y = 1 instead of x − y = −1; x = 2, y = 5 comes from substituting x = 2 into 2x + y = 7 as 2 + y = 7, dropping the coefficient; x = −2, y = 11 comes from dividing −3x = −6 as though two minus signs gave a negative answer.
- (b) 2n + 3 — Method: multiply the cost per mile by the number of miles to get an expression, then add the fixed fee as a separate term. Working: n miles at £2 each is 2n; add the £3 fixed fee: 2n + 3. Answer: 2n + 3. 3n + 2 comes from swapping the fee and the rate round, treating £3 as the rate per mile and £2 as the fixed fee. 5n comes from adding the fee and the rate together first (3 + 2 = 5) and multiplying the result by n, instead of keeping the fixed fee as its own term. 2n − 3 comes from subtracting the fixed fee instead of adding it.
- (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
- (d) (6, 6) — y = f(x − 4) + 1 combines two translations: the −4 inside the brackets moves the graph 4 units to the RIGHT, and the +1 outside the brackets moves it 1 unit UP. Applying both to (2, 5): add 4 to the x-coordinate 2, and add 1 to the y-coordinate 5, giving (6, 6). Moving left instead of right, as the minus sign inside the bracket might suggest, gives (−2, 6) or (−2, 4); forgetting the horizontal shift altogether and only applying the vertical one gives (2, 6).
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