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.A coach company charges a booking fee plus a price for each passenger. A booking for 1 passenger costs £9, for 2 passengers costs £15, for 3 passengers costs £21, and for 4 passengers costs £27. Work out an expression, in terms of n, for the cost in pounds of a booking for n passengers.
- 2.Solve the inequality 6.3x + 1.8 ≤ 40. Hence write down the greatest number with exactly one decimal place that satisfies the inequality.
- 3.Write y = x² + 12x + 40 in the form (x + a)² + b, and hence write down the minimum value of y.y = x² + 12x + 40
- 4.The curve y = x² − 1 and the line y = 3x − 3 meet at two points. Work out the x-coordinates of those two points.y = 3x − 3y = x² − 1
- 5.Describe the single transformation that maps the graph of y = f(x) onto the graph of y = f(−x).
- 6.A walker's distance-time graph is described as follows: she walks 3 km in the first 15 minutes at a steady speed, rests for 10 minutes, then walks a further 3 km in the next 20 minutes, also at a steady speed. Work out her speed, in km/h, during the first 15 minutes.
- 7.Write down the first four terms of the sequence with nth term 6n − 5.
- 8.A student solves the simultaneous equations 2x + y = 11 and x − y = 1 by elimination, adding the two equations together. Which of these is the correct result of that step?
- 9.Work out the smallest positive value of x, in degrees, for which y = tan x is undefined.y = tan(x)
- 10.Line P has equation y = −2x + 5 and line Q has equation 2y = x + 6. Are lines P and Q perpendicular to each other?y = -2x + 5y = x + 6
- 11.A graph passes through the point (0, 1). As x increases through positive values, the graph rises more and more steeply, and as x decreases through negative values, the graph gets closer and closer to the x-axis without ever reaching it. Which of these could be the equation of the graph?
- 12.A plumber charges a £40 call-out fee plus £25 for each hour worked. The total charge, £C, for a job lasting h hours is given by C = 40 + 25h. Work out the charge for a job lasting 3 hours, and name what kind of statement C = 40 + 25h is.
- 13.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
- 14.f(x) = 2x² + 1 and g(x) = x − 1. Work out fg(x), giving your answer in expanded form.y = 2x² + 1
- 15.To solve 5x − 4 = 2x + 8, Chloe's working is shown. Line 1: 5x − 4 = 2x + 8. Line 2: 3x − 4 = 8. Line 3: 3x = 4. Line 4: x = 4/3. Chloe has made a mistake in her working. Which line contains the mistake?
Answer key
- (d) 6n + 3 — Method: find the price per passenger, then find the booking fee that fits a booking for 1 passenger. Working: the cost rises by £6 for each extra passenger (15 − 9 = 6, 21 − 15 = 6, 27 − 21 = 6), so the cost has the form 6n + c. Substituting n = 1: 6(1) + c = 9, so c = 3. Answer: the cost in pounds is 6n + 3. 6n ignores the booking fee altogether. 6n + 9 uses the cost of one passenger, £9, as the fee without first subtracting the £6 per-passenger price. 9n + 6 swaps the two numbers, using the first cost as the price per passenger and the difference as the fee.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (c) 4 — x² + 12x + 40 = (x + 6)² − 6² + 40 = (x + 6)² + 4, so a = 6 and b = 4. Since (x + 6)² can never be negative, y = (x + 6)² + 4 is smallest when the bracket is zero, so the minimum value of y is 4. Giving 40 instead reads off the ORIGINAL constant term and ignores the completing-the-square step entirely — wrong, because 40 is the value of y when x = 0, not the minimum value. Giving −6 instead answers with the x-coordinate of the turning point (where the bracket is zero) rather than the minimum y-value itself — wrong, because the question asks for the minimum value of y, not the value of x that produces it. Giving 36 instead stops after squaring half the coefficient, 6² = 36, without combining it with the 40 already in the expression — wrong, because the minimum value is 40 minus 36, not 36 on its own.
- (d) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
- (c) A reflection in the y-axis — Replacing x with −x reflects the graph in the y-axis: each point (x, y) maps to (−x, y). Reflecting the OUTPUT instead, y = −f(x), gives a reflection in the x-axis — that is a different function. Combining both reflections gives a rotation of 180° about the origin, and reflecting in the line y = x swaps the x- and y-values, which is what produces the inverse function, not f(−x). Check first which side of f the minus sign sits on.
- (a) 12 km/h — Method: convert 15 minutes to hours: 15 ÷ 60 = 0.25 h. Speed = distance ÷ time = 3 ÷ 0.25 = 12 km/h. Distractor origins: 9 km/h is the speed for the second section (3 km in 20 minutes) instead of the first; 8 km/h is the average speed for the whole journey (6 km in 45 minutes) instead of just the first section; 0.2 km/h divides 3 km by 15 without converting the minutes into hours.
- (c) 1, 7, 13, 19 — Method: substitute n = 1, 2, 3, 4 into the rule 6n − 5 in turn. Working: n = 1: 6 − 5 = 1. n = 2: 12 − 5 = 7. n = 3: 18 − 5 = 13. n = 4: 24 − 5 = 19. Answer: 1, 7, 13, 19. 6, 12, 18, 24 comes from using 6n on its own, forgetting to subtract 5. 5, 11, 17, 23 comes from using the rule 6n − 1 instead of 6n − 5, a slip in the constant. 0, 6, 12, 18 comes from using 6(n − 1) instead of 6n − 5, effectively shifting every term one position along.
- (a) 3x = 12 — Adding the two equations: the y-terms, +y and −y, have opposite signs, so they cancel; the x-terms combine to 2x + x = 3x; and the right-hand sides add to 11 + 1 = 12. This gives 3x = 12. A candidate who forgets that the y-terms cancel, and instead adds them as if they had the same sign, would write 3x + 2y = 12. A candidate who subtracts the right-hand sides instead of adding them would get 3x = 10. A candidate who correctly reaches 3x = 12 but then treats 12 itself as the value of x, skipping the final division, would write x = 12.
- (c) 90° — The graph of y = tan x has a vertical asymptote at x = 90°, the smallest positive value for which it is undefined. tan 0° = 0, a defined value, not an undefined one — confusing where the graph equals zero with where it is undefined gives 0°. x = 270° is also a value where y = tan x is undefined, but it is not the smallest positive one, since 90° comes first. tan 45° = 1, a defined value — confusing the asymptote with the angle at which tan x = 1 gives 45°.
- (a) Yes — the gradients multiply to −2 × 1/2 = −1. — Rearrange Q into the form y = mx + c: 2y = x + 6 gives y = (1/2)x + 3, so Q has gradient 1/2. P has gradient −2. Two lines are perpendicular exactly when the product of their gradients is −1: −2 × 1/2 = −1. Since this holds, P and Q are perpendicular. Distractor routes: "the product is −1, but perpendicular needs 1" works out the product correctly but misremembers the condition — the perpendicular test is a product of exactly −1, and parallel lines are spotted by their gradients being equal, not by a product of 1. "Q's gradient is 2, and −2 × 2 = −4" comes from reading the 2 in front of y in 2y = x + 6 as the gradient, instead of dividing the whole equation by 2 first to reach y = (1/2)x + 3, where the gradient is 1/2. "Both equations have a negative x-term" is not a valid test at all — P's equation does have a negative x-term, but Q's, once rearranged, does not, and matching signs say nothing about the actual gradients.
- (a) y = 2ˣ — An exponential graph y = 2ˣ passes through (0, 1) since 2⁰ = 1, rises more and more steeply for positive x, and has the x-axis as an asymptote as x becomes very negative, since 2ˣ gets closer to 0 without ever reaching it. y = x² + 1 also passes through (0, 1) and also rises steeply for positive x, but as x becomes very negative it rises to infinity too, rather than settling towards the x-axis — mistaking any curve that gets steeper for an exponential misses this. y = x³ + 1 passes through (0, 1) and rises for positive x, but as x becomes very negative it falls towards negative infinity rather than approaching the x-axis from above. y = 1 − x² also passes through (0, 1), but it falls for large positive x rather than rising — a candidate who checks only the y-intercept, without reading the described shape of the curve, could pick this.
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (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) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
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