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) 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.
- (a) 4 — Method: set up the equation 35 + 20h = 115, then subtract the fixed fee and divide by the hourly rate. Working: 20h = 115 − 35 = 80; h = 80 ÷ 20 = 4. Answer: 4 hours. 5.75 comes from dividing the whole £115 by £20 without first subtracting the fixed fee: 115 ÷ 20 = 5.75. 2.71 comes from swapping the fee and the rate round, subtracting £20 and dividing by £35: (115 − 20) ÷ 35 ≈ 2.71. 7.5 comes from adding the fixed fee instead of subtracting it: (115 + 35) ÷ 20 = 7.5.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (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.
- (c) A reflection in the y-axis — Replacing x with −x reflects the graph in the y-axis: each point (x, y) maps to (−x, y). Reflecting the OUTPUT instead, y = −f(x), gives a reflection in the x-axis — that is a different function. Combining both reflections gives a rotation of 180° about the origin, and reflecting in the line y = x swaps the x- and y-values, which is what produces the inverse function, not f(−x). Check first which side of f the minus sign sits on.
- (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (d) It is n-shaped, since the x² coefficient is negative. — The coefficient of x² is −2, which is negative, so the quadratic curve opens downward — shaped like an n, with a maximum turning point. Saying it is U-shaped focuses only on x² being non-negative and ignores that the −2 in front of it flips the whole curve to open downward. Saying it is a straight line confuses having a constant term with being linear — any equation with an x² term is a curve, not a line. Saying it repeatedly rises and falls like a wave describes a trigonometric graph such as y = sin x, not a quadratic.
- (c) 3c + 2d — Method: 'triple c' is 3c, 'double d' is 2d, and 'add' joins the two separate terms with a plus sign. Working: 3c + 2d. Answer: 3c + 2d. 2c + 3d comes from swapping which letter gets tripled and which gets doubled. 6cd comes from multiplying the two terms together instead of adding them, and also multiplying the coefficients (3 × 2 = 6). 5(c + d) comes from adding the coefficients (3 + 2 = 5) and applying that single number to both letters together, as if c and d always came as a pair.
- (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
- (b) 2x + 3 — fg(x) means f(g(x)): apply g first, then apply f to the result. g(x) = 2x, so f(g(x)) = f(2x) = 2x + 3. Writing 2x + 6 comes from working out gf(x) instead — g(f(x)) = g(x + 3) = 2(x + 3) = 2x + 6 — which applies the functions in the wrong order. Writing 3x + 3 comes from adding f(x) and g(x) together, (x + 3) + 2x = 3x + 3, instead of composing them. Writing 2x² + 6x comes from multiplying f(x) and g(x) together, (x + 3)(2x) = 2x² + 6x, instead of substituting one into the other.
- (c) The sequence diverges, moving away from x = 3 — Starting from x₀ = 2.9: x₁ = 2.9² − 6 = 2.41, x₂ = 2.41² − 6 = −0.19, x₃ = (−0.19)² − 6 = −5.96, x₄ = (−5.96)² − 6 = 29.56 — the values swing away from 3 and grow rapidly, so the sequence diverges rather than settling anywhere. Choosing 'settles towards x = 3' assumes that starting close to a root is enough for a rearrangement to converge to it, which is not always true — this rearrangement changes values too steeply near x = 3 to stay there. Choosing 'settles towards x = −2' assumes a diverging sequence must eventually land on the other root; instead it runs away to increasingly large values. Choosing 'stays constant at 2.9' ignores that applying the formula changes the value at every step.
- (b) One turning point. — Every quadratic graph, one with an x² term and no higher power of x, has exactly one turning point, since it is a single U-shaped or n-shaped curve. Saying two turning points describes a cubic graph, which can rise, turn, then turn again. Saying no turning points describes a straight line, which has none. Saying four turning points greatly overestimates how many times a simple quadratic curve changes direction — that would need a much higher power of x.
- (d) −2 — Method: a gradient can only be read off an equation that is written in the form y = mx + c, so y has to be made the subject first. Working: subtracting 2x from both sides of 2x + y = 8 gives y = −2x + 8, and comparing that with y = mx + c gives m = −2. Answer: −2. The distractors: 2 comes from reading the coefficient of x straight off the equation as it is printed, without rearranging, so the change of sign is missed; 8 comes from reading the constant as the gradient, confusing m with c; −1/2 comes from rearranging correctly and then writing the gradient upside down, as the change in x over the change in y.
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