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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Algebra worksheet — GCSE Higher
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- 1.The diagram shows part of the graph of a reciprocal function of the form , passing through the labelled point. Work out the value of k.
- 2.A fundraising chain letter starts with 3 people taking part in round 1. Each following round has 4 times as many people taking part as the round before. Work out the number of people taking part in round 4.
- 3.Line L has equation y = 3x − 2. Work out the gradient of a line perpendicular to L.y = 3x − 2
- 4.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
- 5.A proof sets out to show that the sum of the squares of two consecutive odd numbers, written as 2n + 1 and 2n + 3, is always 2 more than a multiple of 8. Four attempts to expand (2n + 1)² + (2n + 3)² and reach a conclusion are shown below. Which attempt correctly proves this claim?
- 6.Factorise 3x² + 10x − 8.
- 7.The table shows some values of f(x): when x = 0, f(x) = 5; when x = 1, f(x) = 8; when x = 2, f(x) = 4; when x = 3, f(x) = 1. Work out the value of f(x − 1) when x = 2.
- 8.A pattern is made from tiles. Pattern 1 uses 4 tiles, pattern 2 uses 7 tiles, pattern 3 uses 10 tiles and pattern 4 uses 13 tiles, with each pattern using 3 more tiles than the one before. Work out an expression, in terms of n, for the number of tiles used in pattern n.
- 9.p = 4. Work out the value of 2p³.
- 10.The graph of y = sin x is reflected in the x-axis to give the graph of y = g(x). Work out the value of g(90), where x is measured in degrees.y = sin(x)
- 11.Simplify (2x² + 7x + 3) ÷ (x² − 9), giving your answer as a single fraction.
- 12.A proof that the product of two consecutive even numbers is always a multiple of 8 begins: Let the two consecutive even numbers be 2n and 2n + 2, so their product is 2n(2n + 2) = 4n(n + 1). Which line correctly completes the proof?
- 13.A student attempts to prove that the product of two consecutive integers is always even: (i) Let the two consecutive integers be n and n + 1. (ii) Since n(n + 1) is even, one of n and n + 1 must be an even number. (iii) Therefore, n(n + 1) is even. At which statement does the proof first assume the very fact it is trying to prove?
- 14.A circle has centre (0, 0) and equation x² + y² = 50. Work out the radius of the circle, giving your answer as a surd in its simplest form.
- 15.How many turning points does the graph of y = x² − 7x + 10 have?y = x² − 7x + 10
Answer key
- (b) 6 — Method: for any point that lies on y = k/x, the value of k is found by multiplying the x-coordinate and the y-coordinate together, since k = x × y. Working: k = 2 × 3 = 6. Answer: k = 6. Distractor refutation: 1.5 comes from dividing the y-coordinate by the x-coordinate instead of multiplying them. 5 comes from adding the two coordinates instead of multiplying them. 9 comes from misreading the point's x-coordinate as 3 instead of 2, then multiplying 3 × 3.
- (b) 192 — This is a geometric sequence with first term 3 and common ratio 4: round 2 has 3 × 4 = 12, round 3 has 12 × 4 = 48, round 4 has 48 × 4 = 192. A candidate who applies the ×4 multiplier four times instead of three gets 3 × 4⁴ = 768. A candidate who wrongly treats the growth as arithmetic, taking the round 2 figure of 12 as a fixed amount added each round, gets 3, 15, 27, 39. A candidate who forgets the starting 3 people and just works out 4⁴ gets 256.
- (a) −1/3 — The gradient of L is 3, the coefficient of x in y = mx + c form. The gradient of a line perpendicular to a line of gradient m is the negative reciprocal, −1/m. So the perpendicular gradient is −1/3. Distractor routes: 3 gives the gradient of L itself, forgetting to change it at all — that is the gradient of a PARALLEL line. 1/3 takes the reciprocal of 3 but keeps the same sign, missing the negative sign a perpendicular gradient requires. −3 negates the gradient of L but does not take its reciprocal, giving the gradient of a line with the opposite slope rather than a perpendicular one.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
- (c) (2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n + 1) + 2, and n² + 2n + 1 is an integer, so the sum is always 2 more than a multiple of 8. — Expand each square carefully: (2n + 1)² = 4n² + 4n + 1 and (2n + 3)² = 4n² + 12n + 9, since the cross term is 2 × 2n × 3 = 12n. Adding gives 8n² + 16n + 10, and factorising out 8 from every term that can hold one gives 8(n² + 2n + 1) + 2; since n² + 2n + 1 is always an integer, the sum is always 2 more than a multiple of 8. The attempt reaching 8(n² + 2n) + 10 has the correct expansion but stops the factorisation one step early — it never pulls a further 8 out of the 10 (10 = 8 + 2), so 'always 10 more than a multiple of 8' should be reduced to 'always 2 more than a multiple of 8'. The attempt reaching 2(4n² + 8n + 5) also has the correct expansion, and the factorisation is true, but 'always even' only shows the sum is a multiple of 2 — being even is necessary but nowhere near sufficient to be a multiple of 8, and the argument never finds the extra factor of 4. The fourth attempt makes an expansion slip, using (2n + 3)² = 4n² + 9 instead of 4n² + 12n + 9 — dropping the 12n cross term entirely — so it works from the wrong expression 8n² + 4n + 10 throughout, and no amount of correct working afterwards can recover the right conclusion.
- (d) (3x − 2)(x + 4) — For 3x² + 10x − 8, find two numbers multiplying to 3 × (−8) = −24 and adding to 10: these are 12 and −2. Rewrite: 3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4). Choosing (3x + 2)(x − 4) expands to 3x² − 10x − 8 — the correct factor pair but the wrong signs, giving the middle term the wrong sign. Choosing (x − 2)(3x + 4) expands to 3x² − 2x − 8 — the 3 is attached to the wrong bracket, changing which terms combine for the x-coefficient. Choosing (3x − 4)(x + 2) expands to 3x² + 2x − 8 — this uses 4 and 2 instead of the correct pair 12 and 2, so the middle term does not come to 10x.
- (a) 8 — f(x − 1) means every input to f is reduced by 1 before it is looked up. At x = 2, the input to f becomes 2 − 1 = 1, so f(x − 1) at x = 2 is f(1) = 8. Reading f(2) = 4 directly from the table, without applying the shift, gives 4. Shifting in the wrong direction, using x + 1 = 3 instead of x − 1 = 1, gives f(3) = 1. Confusing f(x − 1) with f(x) − 1 — taking f(2) = 4 and then subtracting 1 — gives 3.
- (b) 3n + 1 — Method: find how many more tiles each pattern uses, then find the constant by adjusting pattern 1's total. Working: each pattern uses 3 more tiles than the last, so the coefficient of n is 3. The constant is pattern 1's total minus the common difference: 4 − 3 = 1. Answer: the nth term is 3n + 1. 3n + 4 comes from using pattern 1's total, 4, as the constant without subtracting the common difference. 3n − 2 comes from a slip in working out the constant, subtracting the common difference twice (4 − 3 − 3 = −2) instead of once. n + 3 comes from swapping the common difference and the constant.
- (d) 128 — In 2p³ the index belongs to p only, so cube p first and multiply by the coefficient afterwards. Cubing gives 4 × 4 × 4 = 64, and then 2 × 64 = 128. Cubing the coefficient as well would mean working out (2 × 4)³, which is 512. Reading the index as an instruction to multiply by 3 gives 2 × 4 × 3 = 24, and ignoring the coefficient altogether leaves 64.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (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) n and n + 1 are consecutive integers, so one of them must be even; this makes n(n + 1) even, so 4n(n + 1) is 4 × an even number, which is a multiple of 8. — The proof needs a reason why n(n + 1) is even, not just an assertion. n and n + 1 are consecutive integers, so exactly one of them is even; multiplying by that even number keeps n(n + 1) even, so 4n(n + 1) = 4 × (an even number), and 4 × an even number always has a further factor of 2 hidden inside it, making the whole product a multiple of 8. The option 'is a multiple of 4, and because n and n + 1 are consecutive integers, it must be a multiple of 8' asserts the multiple-of-8 conclusion directly from 'multiple of 4' and 'consecutive integers' without ever showing that n(n + 1) itself is even — the missing step is exactly what earns the mark. The option '4n is always a multiple of 4 ... which means it is a multiple of 8' mistakes 4n being a multiple of 4 for the whole product 4n(n + 1) being a multiple of 8; that extra factor of 2 only comes from n(n + 1) being even, not from 4n alone. The option that expands to 4n² + 4n and calls it 'clearly a multiple of 8' never checks for a factor of 2 beyond the 4 already there — the word 'clearly' is standing in for a missing argument.
- (c) Statement (ii) — Statement (ii) opens with 'Since n(n + 1) is even', treating the very fact the proof is meant to establish as if it were already known — that is circular reasoning, assuming the conclusion to help derive itself. Statement (i) only names the two consecutive integers as n and n + 1; it makes no claim about whether their product is even, so it introduces nothing circular. Statement (iii) states the conclusion, and would be a valid final step if statement (ii) had reached 'one of n and n + 1 is even' by a genuine argument, such as considering the cases where n is even or odd separately. Saying the proof assumes nothing circular is wrong, because statement (ii)'s opening clause is exactly that assumption.
- (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.
- (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.
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