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.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.
- 2.The nth term of a geometric sequence is given by aₙ = 3 × 2ⁿ⁻¹. Work out the position of the first term that is greater than 1,000.
- 3.Solve the simultaneous equations y = 3 − x and x² + y² = 9, giving both pairs of solutions.
- 4.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 5.The graph of y = f(x) passes through the point (0, 4). Work out the y-coordinate of the point where the graph of y = f(x) − 6 crosses the y-axis.
- 6.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.y = x² − 5x + 6
- 7.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
- 8.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
- 9.A number machine multiplies its input by 3 and then adds 7. The output is 1. Work out the input.
- 10.Find the common ratio of the geometric sequence 5, 5√2, 10, 10√2, ... Give your answer in surd form.
- 11.The point (5, −12) lies on the circle x² + y² = 169, which has centre (0, 0). Work out the equation of the tangent to the circle at (5, −12), giving your answer in the form y = mx + c.
- 12.A cyclist's speed rises from rest to a maximum — quickly at first, then more and more slowly — so her speed-time graph is a curve that is concave down (its gradient decreases as time goes on). A trapezium estimate is made for the distance travelled during this phase, using two points on the curve. Is the trapezium estimate an overestimate or an underestimate of the true distance, and why?
- 13.The tangent to a curve at the point where x = 4 passes through the points (2, 5) and (6, 21). Use these two points to estimate the gradient of the curve at x = 4.
- 14.f(x) = 2x − 1. Work out ff(x).y = 2x − 1
- 15.Ethan is 10 years old and his father is 40 years old. Work out how many years ago Ethan's father was seven times as old as Ethan.
Answer key
- (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.
- (b) 10 — Check n = 9: a₉ = 3 × 2⁸ = 3 × 256 = 768, below 1,000. Check n = 10: a₁₀ = 3 × 2⁹ = 3 × 512 = 1,536, above 1,000. So the first term greater than 1,000 is at position n = 10. Answering 9 comes from forgetting the −1 shift and using the formula as 3 × 2ⁿ instead of 3 × 2ⁿ⁻¹: checking 3 × 2⁹ = 1,536 (which exceeds 1,000) but then reporting the position as n = 9, the exponent used, instead of n = 10 — wrong, because the exponent in the real formula is n − 1, not n. Answering 11 comes from going one term too far: correctly finding that a₁₀ already exceeds 1,000, but then checking one position further and reporting n = 11 instead of stopping at the first position that already works — wrong, because n = 11 is not the FIRST term greater than 1,000. Answering 1,536 gives the VALUE of the term (a₁₀ itself) rather than its position — wrong, because the question asks which term it is (the value of n), not what that term is worth.
- (c) x = 0, y = 3 and x = 3, y = 0 — Substitute y = 3 − x into x² + y² = 9: x² + (3 − x)² = 9. Expanding (3 − x)² = 9 − 6x + x² gives x² + 9 − 6x + x² = 9, which simplifies to 2x² − 6x = 0, or 2x(x − 3) = 0, so x = 0 or x = 3. Using y = 3 − x: x = 0 gives y = 3; x = 3 gives y = 0. Distractor routes: x = 0, y = 3 alone stops after the factor 2x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = 3 and x = −3, y = 6 comes from factorising 2x² − 6x as 2x(x + 3), a sign error that gives a second root of −3 instead of 3. x = 0, y = 3 and x = 6, y = −3 comes from expanding (3 − x)² as 9 − 6x, dropping the x² term, which changes the quadratic to x² − 6x = 0 and its second root to 6.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
- (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (b) √2 — 5√2 ÷ 5 = √2. Checking between the third and second terms: 10 ÷ 5√2 = √2 as well (since 10 ÷ 5√2 = 2 ÷ √2 = √2), so the common ratio is confirmed as √2 throughout. Squaring the ratio instead of leaving it as a surd gives 2, which is wrong because 2 is the SQUARE of the common ratio, not the ratio itself. Rounding the exact surd to a decimal gives 1.41, which is wrong because the sequence is defined using an exact surd ratio, and a rounded decimal is not the same value. Writing down the SECOND TERM of the sequence, 5√2, instead of the ratio BETWEEN terms, gives 5√2, which is wrong because a term of the sequence is not the same thing as the common ratio.
- (c) y = (5/12)x − 169/12 — The gradient of the radius to (5, −12) is (−12 − 0) ÷ (5 − 0) = −12/5. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal, 5/12. Using y − y₁ = m(x − x₁) with (5, −12): y + 12 = (5/12)(x − 5), which gives y = (5/12)x − 169/12. y = −(12/5)x comes from using the radius's own gradient, −12/5, instead of turning it into the perpendicular gradient, and building the line through the origin (as the radius itself does). y = −(5/12)x − 119/12 comes from taking the reciprocal of −12/5 correctly as a size but keeping the wrong sign, using −5/12 instead of 5/12. y = (5/12)x − 25/12 comes from using the correct gradient 5/12 but building the line through (5, 0) instead of (5, −12) — dropping the point's y-coordinate.
- (a) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
- (a) 4 — The gradient of a straight line through two points is the change in y divided by the change in x. Change in y = 21 − 5 = 16. Change in x = 6 − 2 = 4. Gradient = 16 ÷ 4 = 4. Dividing x by y instead of y by x gives 0.25; forgetting to divide by the change in x at all leaves 16; dividing by only one of the two x-coordinates, 16 ÷ 2 = 8, uses the wrong denominator.
- (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
- (c) 5 — Method: call the number of years ago t, take t off both ages, and form an equation from the comparison at that time. Working: t years ago the father was 40 − t and Ethan was 10 − t, so 40 − t = 7(10 − t); expanding gives 40 − t = 70 − 7t, adding 7t to both sides gives 40 + 6t = 70, subtracting 40 gives 6t = 30, and dividing by 6 gives t = 5. Checking: five years ago the father was 35 and Ethan was 5, and 35 = 7 × 5. Answer: 5. The distractors: 35 comes from finding the right moment but giving the father's age at that time instead of the number of years; 30 comes from using Ethan's present age on the right-hand side, 40 − t = 7 × 10, which gives t = −30 and is then written as 30; 24 comes from reaching 6t = 30 correctly and subtracting 6 instead of dividing by 6.
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