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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Answer key: Algebra worksheet — GCSE Higher
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- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (d) 1 hour 32 minutes — T = 40 × 1.8 + 20 = 72 + 20 = 92 minutes. Since 92 = 60 + 32, the cooking time is 1 hour 32 minutes. A candidate who rounds the mass to 2 kg before substituting gets 40 × 2 + 20 = 100 minutes = 1 hour 40 minutes. A candidate who forgets to add the 20 minutes gets 40 × 1.8 = 72 minutes = 1 hour 12 minutes. A candidate who multiplies the mass by (40 + 20) = 60 instead of substituting into the formula gets 1.8 × 60 = 108 minutes = 1 hour 48 minutes.
- (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) 842.0 — The radius to (20, 21) has gradient 21/20, so the tangent's gradient is −20/21. The tangent line is y − 21 = −20/21(x − 20), i.e. y = −20/21x + 841/21. Setting y = 0 gives the x-intercept x = 841/20 = 42.05; setting x = 0 gives the y-intercept y = 841/21 ≈ 40.048. The area of triangle OPQ is 1/2 × 42.05 × 40.048 ≈ 842.0. 1684.0 comes from multiplying the two intercepts without the 1/2 that a triangle's area needs — twice the correct area. 580.7 comes from using the circle's radius, 29, as a side of the triangle instead of the x-intercept, 42.05: 1/2 × 29 × 40.048 ≈ 580.7. 2.0 comes from a sign error in the tangent's gradient — using 20/21 instead of −20/21 — which gives a different line, with intercepts x ≈ −2.05 and y ≈ 1.952, and area 1/2 × 2.05 × 1.952 ≈ 2.0.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (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.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (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.
- (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) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
- (a) 6x + 12 — A regular hexagon has 6 equal sides, so the perimeter is 6(x + 2) = 6x + 12. A candidate who multiplies only the x-term by 6 and forgets to multiply the 2 gets 6x + 2. A candidate who multiplies only the number term by 6 and forgets to multiply the x gets x + 12. A candidate who adds 6 and 2 to make a single coefficient of x instead of expanding the brackets gets 8x.
- (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.
- (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.
- (d) 2.7 — The height after 3 years of 10% compound growth is 2 × 1.1³ = 2.662 m, which rounds to 2.7 m. A candidate who adds 10% of the original height (0.2 m) in each of the 3 years, instead of compounding on the new height each time, would reach 2 + 3×0.2 = 2.6 m. A candidate who compounds for only 2 years would reach 2 × 1.1² = 2.42 m, rounding to 2.4 m. A candidate who compounds for 4 years instead of 3 would reach 2 × 1.1⁴ = 2.928 m, rounding to 2.9 m.
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