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.Work out the gradient of the straight line with equation 3y = 12 − 6x.
- 2.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.
- 3.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.
- 4.The simultaneous equations 2x + 3y = 12 and x − y = 1 are given. Work out the value of y.
- 5.Solve 6x − 5 = 3x + 13.
- 6.The numbers x and y satisfy x + y = 10 and xy = 21. Work out the pair of values.
- 7.Write down the expression that means 5 more than half of n.
- 8.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?
- 9.f(x) = 2x² + 1 and g(x) = x − 1. Work out fg(x), giving your answer in expanded form.y = 2x² + 1
- 10.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
- 11.For the equation 2x + 3 = 11, and the inequality 2x + 3 > 11, which statement correctly compares their solutions?
- 12.Simplify (2x² + 7x + 3) ÷ (x² − 9), giving your answer as a single fraction.
- 13.Harry will spend at most £150 on a party. The cake costs £60 and each helium balloon costs £3. Solve an inequality to find all the possible numbers of balloons, x, that he can buy.
- 14.Point R has coordinates (3, −5). Point R is reflected in the line y = 2 to point S. Write down the coordinates of S.
- 15.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
Answer key
- (c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
- (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.
- (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.
- (a) y = 2 — Method: make x the subject of the simpler equation, substitute it into the other equation and then read off the letter the question asks for. Working: x − y = 1 gives x = y + 1, so 2x + 3y = 12 becomes 2(y + 1) + 3y = 12, that is 2y + 2 + 3y = 12, so 5y = 10 and y = 2. Answer: y = 2, and the matching value x = 3 checks in 2 × 3 + 3 × 2 = 12. The distractors: y = 3 comes from solving the pair correctly and then writing down the value of x; y = 2.2 comes from expanding 2(y + 1) as 2y + 1, which leaves 5y = 11; y = 10 comes from rearranging x − y = 1 as x = 1 − y, which turns the first equation into 2 + y = 12.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
- (a) 3 and 7 — Method: one equation is linear and the other is not, so rearrange the linear equation and substitute it into the other to leave a single quadratic in one letter. Working: x + y = 10 gives y = 10 − x, so xy = 21 becomes x(10 − x) = 21, which rearranges to x² − 10x + 21 = 0; factorising gives (x − 3)(x − 7) = 0, so the two values are 3 and 7, and 3 + 7 = 10 with 3 × 7 = 21. Answer: 3 and 7. The distractors: 1 and 21 comes from using only the product and taking the first factor pair of 21; −3 and −7 comes from factorising as (x + 3)(x + 7), which reverses the sign of both values; 5 and 5 comes from using only the sum and splitting 10 into two equal parts.
- (d) n/2 + 5 — Half of n is n ÷ 2, which is written as the fraction n/2. 'More than' means add, and the addition happens after the halving, so the expression is n/2 + 5. Writing (n + 5)/2 halves the 5 as well, because everything inside a bracket is divided; writing 2n + 5 doubles n instead of halving it; writing 5n/2 multiplies half of n by 5 instead of adding 5 to it.
- (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) 2x² − 4x + 3 — fg(x) means f(g(x)): substitute g(x) into f in place of x. g(x) = x − 1, so fg(x) = f(x − 1) = 2(x − 1)² + 1. Expanding (x − 1)² = x² − 2x + 1, so fg(x) = 2(x² − 2x + 1) + 1 = 2x² − 4x + 2 + 1 = 2x² − 4x + 3. Writing 2x² comes from working out gf(x) instead — g(f(x)) = f(x) − 1 = (2x² + 1) − 1 = 2x², which applies the functions in the wrong order. Writing 2x² − 1 comes from expanding (x − 1)² as x² − 1, dropping the middle term, so f(x − 1) becomes 2(x² − 1) + 1 = 2x² − 2 + 1 = 2x² − 1. Writing 2x² − 4x + 2 comes from expanding correctly but forgetting the final + 1 from f, stopping at 2(x² − 2x + 1) = 2x² − 4x + 2.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
- (a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
- (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) x ≤ 30 — Method: add the fixed cost to the cost of x balloons, set that total against the £150 limit with the sign that 'at most' calls for, then solve. Working: the total spend is 60 + 3x pounds, so 60 + 3x ≤ 150; subtracting 60 from both sides gives 3x ≤ 90; dividing both sides by 3, a positive number, gives x ≤ 30. Answer: x ≤ 30. The distractors: x ≤ 90 comes from taking the cake off the budget and stopping at 3x ≤ 90, reading the 90 as a number of balloons when it is the money left for them; x ≤ 50 comes from dividing the whole £150 by 3 and leaving the cake out of the calculation altogether; x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
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