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.Which expression means 'triple c, then add double d'?
- 2.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
- 3.By completing the square, show that the curve y = x² − 14x + 50 never crosses the x-axis. Which statement gives the correct reason?y = x² − 14x + 50
- 4.A ball is dropped and bounces. The height of each bounce after the first is 8 cm less than the bounce before it. The first bounce reaches 60 cm. Work out the height of the 5th bounce.
- 5.The graph of y = f(x) has a root (an x-intercept) at x = 5. Work out the x-coordinate of the corresponding root on the graph of y = f(x + 2).
- 6.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
- 7.A straight line has gradient 3 and passes through the point (1, 4). Work out the equation of the line.
- 8.A rectangle has width x cm. Its length is 3 cm more than its width. The perimeter of the rectangle is 38 cm. Form an equation and solve it to find x.
- 9.A plumber charges a call-out fee plus an hourly rate for a job. The total cost, £C, for a job lasting h hours is given by the formula C = 40 + 25h. What does the number 40 represent in this formula?
- 10.The function f(x) = x² for all real values of x has no inverse function, but g(x) = x² for x ≥ 0 does have one. Which statement correctly explains this?y = x²
- 11.Solve the inequality 3x − 1 ≤ 11.
- 12.A rule turns each input x into an output y. An input of 1 gives an output of 1, an input of 2 gives an output of 4 and an input of 3 gives an output of 9. Work out the rule.
- 13.A market stall's cost of hiring n tables is modelled by two formulas: Formula A: C = 3(2n + 5); Formula B: C = 6n + 15, where C is in pounds. A stallholder says the two formulas always give the same cost. Work out the cost given by each formula when n = 4, and use your results to decide whether the stallholder is correct.
- 14.Line L has equation y = 3x + 1. Which of these lines is perpendicular to L?y = 3x + 1
- 15.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.
Answer key
- (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.
- (c) 64 m — The distance travelled is the area under the speed-time graph. From t = 0 to t = 4, the shape is a triangle with base 4 and height 8, area 1/2 × 4 × 8 = 16. From t = 4 to t = 10, the shape is a rectangle with base 6 and height 8, area 6 × 8 = 48. Total distance: 16 + 48 = 64 m. Treating the whole 10 seconds as a single trapezium with parallel sides 0 and 8 and width 10, instead of splitting it into the triangle and rectangle, gives 40 m. Ignoring the acceleration phase completely and assuming the car travels at a constant 8 m/s for all 10 seconds gives 8 × 10 = 80 m. Misreading the second interval as running from t = 4 to t = 8 instead of t = 4 to t = 10 gives a rectangle area of 4 × 8 = 32, plus the correct triangle of 16, totalling 48 m.
- (d) (x − 7)² + 1 is at least 1 for every x, so y is never 0 — x² − 14x + 50 = (x − 7)² − 7² + 50 = (x − 7)² + 1. Since a squared term can never be negative, (x − 7)² ≥ 0 for every real x, so y = (x − 7)² + 1 ≥ 1 for every x — y is always at least 1, so it can never equal 0, and the curve never crosses the x-axis. Claiming the discriminant is positive is wrong on the facts: b² − 4ac = (−14)² − 4(1)(50) = 196 − 200 = −4, which is NEGATIVE, not positive — a negative discriminant is in fact the standard reason a quadratic has no real roots, so this option misstates its own sign. Claiming the turning point is at (7, −1) is wrong because completing the square gives (x − 7)² + 1, so the turning point is (7, 1) — the constant is +1, not −1, and a point with a negative y-coordinate would sit BELOW the x-axis, not above it, contradicting the option's own claim. Claiming that 50 being positive is enough is wrong because it ignores the −14x term completely — a positive constant term alone does not guarantee y stays positive for every x (the expression still dips down towards its turning point before growing again, and it is the completed square that shows how far it dips, not the constant on its own).
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (c) y = 3x + 1 — Method: a line of known gradient m has equation y = mx + c, and c is found by substituting the coordinates of a point known to lie on it. Working: the gradient is 3, so the line is y = 3x + c; substituting x = 1 and y = 4 gives 4 = 3 × 1 + c, so c = 4 − 3 = 1. Answer: y = 3x + 1. The distractors: y = 3x − 1 comes from working out the constant as mx − y, 3 − 4 = −1, instead of y − mx; y = x + 3 comes from swapping the two numbers over, putting the gradient 3 in the constant position and the x-coordinate 1 in front of x; y = 3x + 4 comes from using the y-coordinate 4 as the constant without substituting.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (d) The fixed call-out fee, charged before the hourly rate. — Method: compare the formula with C = (fixed charge) + (rate) × h, where the fixed charge is the part that does not depend on h. Working: in C = 40 + 25h, the term 25h depends on the number of hours, h, but 40 does not change however many hours the job takes. Answer: 40 is the fixed call-out fee. 'The hourly rate' confuses the fixed term with the coefficient of h, which is actually 25. 'The total cost for a job lasting 1 hour' comes from substituting h = 1 into the formula (40 + 25 = 65) rather than reading off the constant term. 'The number of hours worked before charging starts' wrongly treats the constant, which is in pounds, as if it were measured in hours.
- (d) g is one-to-one: f(3) = f(−3), removed by x ≥ 0 — A function has an inverse only if it is one-to-one: every output must come from exactly one input. f(3) = 9 and f(−3) = 9, so two different inputs give the same output, and there is no way to send 9 back to a single input — f is not one-to-one over all real x. Restricting the domain to x ≥ 0 removes one of the two inputs behind every such pair, so g is one-to-one and does have an inverse. 'g's outputs are positive; f's could be negative' is wrong because f(x) = x² also only gives outputs of 0 or more — the outputs of f and g are identical sets; it is the INPUTS that differ, not the outputs. 'Restricting any domain always creates an inverse' is wrong because a restriction only helps if it actually removes the repeated outputs: restricting f(x) = x² to x ≥ −3 still leaves f(1) = f(−1) = 1, so that restricted function is still not one-to-one and still has no inverse. 'Squares can never be reversed, under any conditions' is wrong because √9 = 3 does reverse 3² = 9 once you know the input was non-negative — a square root just cannot tell you WHICH of two inputs you started from unless the domain has already ruled one of them out.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (b) y = −(1/3)x + 2 — Method: two lines are perpendicular when the product of their gradients is −1, so the gradient of a line perpendicular to one of gradient m is the negative reciprocal, −1 divided by m. Working: reading m from y = 3x + 1 gives a gradient of 3, so the perpendicular gradient is −1 ÷ 3 = −1/3, and the line carrying that gradient is y = −(1/3)x + 2; the check 3 × (−1/3) = −1 confirms it. Answer: y = −(1/3)x + 2. The distractors: y = 3x − 4 comes from using an equal gradient, which is the test for parallel lines rather than perpendicular ones; y = (1/3)x + 2 comes from turning the gradient upside down but leaving out the change of sign, giving a product of 1 instead of −1; y = −3x + 2 comes from changing the sign of the gradient without turning it upside down, giving a product of −9.
- (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.
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