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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- (a) 3y² — 'y squared, multiplied by 3' means the square is applied to y only, and the result is then multiplied by 3, written as 3y². Writing y³ mistakes the multiplication by 3 for an extra factor of y, adding to the power instead of using a coefficient. Writing (3y)² squares the whole of 3y, including the 3, which gives 9y² rather than 3y² — the square should apply to y alone. Writing 3 + y² adds the 3 instead of multiplying by it. The expression for 'y squared, multiplied by 3' is 3y².
- (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
- (a) n² + 4n + 4 — The whole patio-plus-border square has side length (n + 2), so the total is (n + 2)². Expanding this bracket correctly gives n² + 4n + 4, which matches 9, 16, 25, 36 for n = 1, 2, 3, 4. Expanding (n + 2)² by squaring each term separately, as if (a + b)² = a² + b², gives n² + 4, which is already wrong at n = 1 (it gives 5, not 9). Multiplying out (n + 2)(n + 2) as n² + 2n + 2n but forgetting the final 2 × 2 gives n² + 4n, which is 5 short at every value of n. Counting the patio twice — once as the n² inner square and again inside the (n + 2)² total — gives n² + (n² + 4n + 4) = 2n² + 4n + 4.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
- (c) 4 years — The yearly multiplier is 1 − 0.12 = 0.88. After 1 year: 2400 × 0.88 = 2112. After 2 years: 2112 × 0.88 = 1858.56. After 3 years: 1858.56 × 0.88 = 1635.5328. After 4 years: 1635.5328 × 0.88 = 1439.27 (2 d.p.), which is the first value below £1,500 — so it takes 4 complete years. Stopping after 3 years, where the value is still £1,635.53, above £1,500, gives 3 years — wrong, because the threshold has not been crossed yet at that point. Continuing for one year longer than needed gives 5 years — wrong, because the value has already fallen below £1,500 by year 4, so year 5 overcounts. Multiplying by the percentage itself instead of by the multiplier — using 0.12 in place of 0.88 — gives 2400 × 0.12 = £288 after a single year, apparently below £1,500 straight away, so 1 year — wrong, because 12% is the part lost each year, and the part left is 88%, so the multiplier is 0.88.
- (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (b) 1 at x = 90° — The graph of y = sin x rises from (0°, 0) to its maximum value of 1 at x = 90°, a quarter of the way through the period. Thinking the maximum occurs where the graph crosses the y-axis, at the start of the curve, gives 1 at x = 0°, but sin 0° = 0, not the maximum. Thinking the maximum occurs halfway to x = 360°, rather than a quarter of the way, gives 1 at x = 180°, but sin 180° = 0, not the maximum. Swapping the roles of the maximum value and the x co-ordinate at which it occurs gives 90 at x = 1, but the maximum value of sin x is 1, not 90.
- (c) 90° — The graph of y = tan x has a vertical asymptote at x = 90°, the smallest positive value for which it is undefined. tan 0° = 0, a defined value, not an undefined one — confusing where the graph equals zero with where it is undefined gives 0°. x = 270° is also a value where y = tan x is undefined, but it is not the smallest positive one, since 90° comes first. tan 45° = 1, a defined value — confusing the asymptote with the angle at which tan x = 1 gives 45°.
- (c) x = 3 and x = −5 — To factorise x² + 2x − 15, find two numbers that multiply to −15 and add to 2: these are 5 and −3, since 5 × (−3) = −15 and 5 + (−3) = 2. So x² + 2x − 15 = (x + 5)(x − 3). Setting each factor to zero gives x = −5 and x = 3. Choosing x = −3 and x = 5 comes from swapping the signs of the correct roots. Choosing x = 5 and x = 3 uses the right pair of numbers, 5 and 3, but forgets that one of them must be negative for the product to equal −15. Choosing x = −15 and x = 1 mistakes the constant term, −15, for one of the roots, and pairs it oddly with x = 1.
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (b) 3/2 — Rearrange 2x + 3y = 6 into y = mx + c: 3y = −2x + 6, so y = −(2/3)x + 2. The gradient of this line is −2/3. The perpendicular gradient is the negative reciprocal: 3/2. Distractor routes: −1/2 comes from reading the gradient straight off the x-coefficient, 2, without dividing by the y-coefficient, 3, first, then taking its negative reciprocal. −3/2 correctly finds the gradient −2/3 but only takes its reciprocal without also changing the sign, giving −3/2 instead of 3/2. 2/3 comes from negating the gradient −2/3 to 2/3, but forgetting to also take the reciprocal.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
- (d) 36 — (2n)² means the whole of 2n is squared, so with n = 3: (2n)² = (2 × 3)² = 6² = 36. Answering 18 instead works out 2n² — squaring only the n and then multiplying by 2 — which is a different expression because the brackets around 2n are missing. Answering 12 squares only the coefficient, treating (2n)² as 2² × n = 4 × 3 = 12, and forgets to square the n as well. Answering 9 ignores the coefficient of 2 altogether and works out n² on its own. The value of (2n)² when n = 3 is 36.
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