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) (6, 8) — Method: a point lies on the circle x² + y² = 100 exactly when the squares of its two coordinates add to 100, so square both coordinates of each point and add them. Working: for (6, 8), 6² + 8² = 36 + 64 = 100, which matches the right-hand side of the equation. Answer: (6, 8) lies on the circle. The distractors: (3, 4) is the 3, 4, 5 right-angled triangle recalled but never scaled up to a radius of 10, and 3² + 4² = 25, so it lies on the far smaller circle x² + y² = 25; (5, 5) has coordinates adding to 10, which compares the sum of the coordinates with the radius instead of the sum of their squares with r², and 5² + 5² = 50; (10, 10) takes each coordinate separately to equal the radius, and 10² + 10² = 200, which is twice too big.
- (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.
- (d) 40 m — Each trapezium has width 2. Its area is width × the average of its two heights. Strip 1: average of 0 and 5 is 2.5, so area = 2 × 2.5 = 5. Strip 2: average of 5 and 9 is 7, so area = 2 × 7 = 14. Strip 3: average of 9 and 12 is 10.5, so area = 2 × 10.5 = 21. Total distance = 5 + 14 + 21 = 40 m. Leaving out the division by 2 in the averaging step doubles every strip, giving 80 m instead of 40 m. Using only the LEFT height of each strip as a rectangle, 0 × 2 + 5 × 2 + 9 × 2 = 28, or only the RIGHT height, 5 × 2 + 9 × 2 + 12 × 2 = 52, both ignore that the graph curves between the two ends of each strip — always average the two heights of a trapezium, never use just one of them.
- (c) 1 — At x = 0, f(0) = 4. Applying the transformations in order — reflect in the x-axis first, then translate up by 5 — gives −f(0) + 5 = −4 + 5 = 1. Applying the translation but forgetting the reflection gives f(0) + 5 = 9. Applying the reflection to the whole expression, including the +5, gives −f(0) − 5 = −9. Applying the reflection but forgetting the translation gives −f(0) = −4.
- (d) 1 — Method: find the step in the outputs for each step of 1 in the input, write the rule from that step and from one pair of values, then apply the rule to the last input. Working: the outputs −5, −3, −1 rise by 2 while x rises in ones, so x is multiplied by 2; at x = 1, 2 × 1 = 2 while y = −5, so 7 is subtracted, giving y = 2x − 7; at x = 4 the rule gives 2 × 4 = 8 and 8 − 7 = 1. Answer: y = 1. The distractors: 3 comes from carrying the outputs on one step too far, to x = 5; 0 comes from assuming the outputs −5, −3, −1 carry on by adding 1 rather than by adding 2; 8 comes from doubling the input and forgetting to subtract the 7.
- (b) An identity, since both sides are equal for every x. — Expanding the left-hand side: 5(2x−3)=10x−15, which is exactly the same as the right-hand side for every value of x, so it is an identity, not an equation that is only true for one particular x. A candidate who treats every equals-sign statement as an equation, without checking whether it holds for all values of x, would choose the equation option. A candidate who mistakes it for a formula is assuming it relates two different letters or quantities, but only x appears — there is no second variable such as area or cost — so it is not a formula. A candidate who mistakes it for an inequality is assuming the two sides are only equal for particular values of x, but expanding shows they are identical for every value, not just some.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (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.
- (b) 20 — a²b means a² multiplied by b: a² = 2² = 4, and 4 × 5 = 20. Reading the expression as (ab)² instead of a²b gives (2 × 5)² = 100. Squaring b instead of a, 2 × 5² = 50, squares the wrong letter. Adding a² and b instead of multiplying them, 2² + 5 = 9, uses the wrong operation.
- (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.
- (a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
- (d) £2000 — This is an arithmetic sequence with first term £500 and common difference £300. The 6th term is 500 + 5 × 300 = 2000. A candidate who uses 6 lots of the increase instead of 5 gets 500 + 6 × 300 = 2300. A candidate who forgets to add the first year's profit at all gets 5 × 300 = 1500. A candidate who miscounts the number of increases as 4 instead of 5 gets 500 + 4 × 300 = 1700.
- (b) 6 — Reflecting y = f(x) in the x-axis turns the minimum point (2, −3) into a maximum point at (2, 3), since −f(x) negates every y-value: −(−3) = 3. Adding a then gives 3 + a = 9, so a = 9 − 3 = 6. Forgetting the reflection and using the original y-value of −3 gives −3 + a = 9, so a = 12 — this ignores that −f(x) changes the sign of the y-coordinate before a is added. Writing a = −12 comes from subtracting in the wrong order, working out 9 − (−3) as −3 − 9 instead. Writing a = −6 comes from taking the negative of the correct answer, as if the final value of a needed to be reflected too, on top of the turning point.
- (a) Yes — the t² coefficient is negative, giving an n-shape. — The coefficient of t² in h = −5t² + 20t is −5, which is negative, so the graph is n-shaped with a maximum point — this matches the physical story of the rocket rising then falling, but the shape itself is decided by the negative coefficient of t², not by the story alone. Saying the shape comes from the story rather than the coefficient gets the reasoning backwards — the algebra determines the shape, and the story happens to agree with it. Saying it is U-shaped because height starts by increasing confuses the early part of the curve with its overall shape; a U-shaped curve would mean the rocket's height eventually increases again forever, which does not happen here. Saying it is n-shaped only because the rocket lands treats a consequence of the shape as if it were the cause.
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