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.
Algebra worksheet — GCSE Higher
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- 1.The graph of y = x² − 4x is translated by the vector (3, 0). Work out the equation of the image, giving your answer in the form y = x² + bx + c.y = x² − 4xy = x²
- 2.Work out the value of 2(x + 2) when x = −5
- 3.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
- 4.Rosie is asked whether 4(x + 3) = 4x + 12 is an equation or an identity. Which statement is correct? Give a reason for your answer.
- 5.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
- 6.A Fibonacci-type sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₁ = 5 and a₃ = 17, work out a₂.
- 7.Explain why the equation 4x + 3 = 4x + 10 has no solution for x.
- 8.Solve the inequality x² − 5x + 6 < 0.
- 9.A person's body mass index is given by the formula B = m ÷ h², where m is the mass in kg and h is the height in metres. Work out B, to 1 decimal place, when m = 68 and h = 1.6.
- 10.Work out the value of 2x² when x = −3
- 11.Make x the subject of the formula y = 4x − 3.y = 4x − 3
- 12.A cyclist's velocity increases from 3 m/s to 11 m/s over 5 seconds, at a constant rate. Work out the cyclist's acceleration, in m/s².
- 13.A geometric sequence has first term 3 and common ratio √2. Work out the 6th term of the sequence, giving your answer in the form a√2.
- 14.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
- 15.A ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
Answer key
- (d) y = x² − 10x + 21 — A translation by the vector (3, 0) moves the graph 3 units in the positive x-direction, which means replacing every x in the equation with (x − 3). Substitute into x² − 4x: (x − 3)² − 4(x − 3). Expand (x − 3)² to x² − 6x + 9, and expand −4(x − 3) to −4x + 12. Collecting like terms, x² − 6x + 9 − 4x + 12 = x² − 10x + 21, so the image is y = x² − 10x + 21. Substituting (x + 3) instead of (x − 3) — translating in the wrong direction — gives y = x² + 2x − 3. Adding 3 straight onto the original equation, treating the translation as vertical instead of horizontal, gives y = x² − 4x + 3. Expanding (x − 3)² as x² − 3x + 9, using −3x instead of −6x for the middle term, and then combining with −4(x − 3) gives y = x² − 7x + 21.
- (c) −6 — Method: substitute the value, work out the bracket first, then multiply what the bracket comes to by 2. Working: the bracket gives (−5) + 2 = −3, and multiplying by 2 gives 2 × (−3) = −6. Answer: −6. The distractors: −3 comes from working out the bracket correctly but stopping there and never multiplying by 2; −1 comes from multiplying the 2 inside the bracket by the 2 outside before adding, giving −5 + 4; −8 comes from multiplying only the x by 2 and leaving the 2 inside the bracket unchanged, giving −10 + 2.
- (a) 6x + 12 — A regular hexagon has 6 equal sides, so the perimeter is 6(x + 2) = 6x + 12. A candidate who multiplies only the x-term by 6 and forgets to multiply the 2 gets 6x + 2. A candidate who multiplies only the number term by 6 and forgets to multiply the x gets x + 12. A candidate who adds 6 and 2 to make a single coefficient of x instead of expanding the brackets gets 8x.
- (a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (d) 12 — Since a₃ = a₂ + a₁, rearranging gives a₂ = a₃ − a₁ = 17 − 5 = 12. 22 comes from adding instead of subtracting: a₃ + a₁ = 17 + 5 = 22. 11 comes from treating the sequence as if it were arithmetic and averaging the two given terms: (5 + 17) ÷ 2 = 11 — but a Fibonacci-type sequence isn't arithmetic, so this doesn't apply. −12 comes from subtracting in the wrong order: a₁ − a₃ = 5 − 17 = −12.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (c) 2 < x < 3 — Factorise x² − 5x + 6 = (x − 2)(x − 3), giving roots x = 2 and x = 3. Since the coefficient of x² is positive, the graph is a U-shape that dips below the x-axis between its roots. So x² − 5x + 6 < 0 for 2 < x < 3. Distractor routes: x < 2 or x > 3 takes the region OUTSIDE the roots, where the graph is above the x-axis, the opposite of what is wanted. 2 ≤ x ≤ 3 uses ≤ instead of the strict < the question asks for, wrongly including the roots themselves, where the expression equals zero, not less than zero. −3 < x < −2 comes from factorising as (x + 2)(x + 3), reversing the sign of both roots.
- (c) 26.6 — h² = 1.6² = 2.56, so B = 68 ÷ 2.56 = 26.5625, which rounds to 26.6 (1 d.p.). A candidate who forgets to square the height gets 68 ÷ 1.6 = 42.5. A candidate who squares the mass instead of the height gets 68² ÷ 1.6 = 2890.0. A candidate who truncates 26.5625 instead of rounding it gets 26.5.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (d) x = (y + 3)/4 — Two operations have been applied to x: it has been multiplied by 4 and then 3 has been subtracted. Undo them in the opposite order. Adding 3 to both sides gives y + 3 = 4x. Dividing both sides by 4 then gives (y + 3)/4 = x, so x = (y + 3)/4. The bracket matters: writing y/4 + 3 divides only the y by 4 and leaves the 3 untouched. Writing (y − 3)/4 subtracts the 3 instead of adding it, and writing 4(y + 3) multiplies by 4 rather than dividing.
- (b) 1.6 m/s² — Acceleration is the change in velocity divided by the time taken: (11 − 3) ÷ 5 = 8 ÷ 5 = 1.6 m/s². Forgetting to subtract the initial velocity and dividing the final velocity by the time instead gives 11 ÷ 5 = 2.2 m/s². Inverting the fraction, dividing the time by the change in velocity, gives 5 ÷ 8 = 0.625 m/s². Finding the change in velocity, 8 m/s, but stopping without dividing by the time gives 8 m/s².
- (d) 12√2 — The nth term of a geometric sequence is a × rⁿ⁻¹. Here a = 3, r = √2, n = 6, so the 6th term is 3 × (√2)⁵. Since (√2)² = 2, (√2)⁴ = (2)² = 4, so (√2)⁵ = (√2)⁴ × √2 = 4√2. The 6th term is 3 × 4√2 = 12√2. 15√2 comes from treating (√2)⁵ as 5√2 — multiplying the index by the surd instead of raising √2 to that power — then multiplying by 3 gives 3 × 5√2 = 15√2, which is wrong because powers of a surd do not scale linearly with the index. 24√2 comes from miscalculating (√2)⁴ as 8 instead of 4 (a squaring slip, since (√2)² = 2 but (√2)⁴ should be 2² = 4, not 2 × 4), giving 3 × 8√2 = 24√2. 4√2 comes from forgetting to multiply by the first term a = 3, leaving just (√2)⁵ = 4√2.
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
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