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.The equation x³ + 4x − 9 = 0 is to be solved by iteration. Work out which one of these iterative formulas comes from a correct rearrangement of that equation.
- 2.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 3.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
- 4.The iterative formula xₙ₊₁ = 5 − 3/xₙ is used with starting value x₀ = 2.5, so that x₁ is the value after the formula has been used once. Work out x₄ correct to 3 significant figures.
- 5.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
- 6.On a distance-time graph, a horizontal line segment shows a period when the graph's gradient is 0. What does this tell you about the journey during that time?
- 7.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
- 8.A circle has centre (0, 0) and equation x² + y² = 50. Work out the length of the diameter of the circle, correct to 1 decimal place.
- 9.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 10.The equation x² − x − 6 = 0 has roots x = 3 and x = −2. It can be rearranged as xₙ₊₁ = xₙ² − 6. This formula is used with starting value x₀ = 2.9, close to the root x = 3. Work out what happens to the sequence of values as n increases.
- 11.A train's distance-time graph shows it travelling 100 km at a steady speed in the first 1.5 hours, waiting at a signal for 0.25 hours, then travelling a further 47 km in 0.75 hours. Work out the average speed for the whole journey, correct to 1 decimal place.
- 12.A square tile has an area of 144 cm². Work out the side length of the tile.
- 13.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
- 14.The equation 7x = x² + 3 can be solved using the iterative formula xₙ₊₁ = (xₙ² + 3) ÷ 7. Taking x₀ = 0.4, x₁ = 0.4514 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
- 15.The trapezium rule is used to estimate the area under a curve between x = 2 and x = 6. The curve is concave up (it curves upwards, like the inside of a bowl) throughout this interval. Which statement about the estimate is correct?
Answer key
- (c) xₙ₊₁ = ∛(9 − 4xₙ) — Method: a formula xₙ₊₁ = f(xₙ) is a correct rearrangement when the equation x = f(x) turns back into the equation you started with, so rearrange x³ + 4x − 9 = 0 by making the cube the subject. Working: x³ + 4x − 9 = 0 gives x³ = 9 − 4x, because the 4x and the 9 each change sign as they cross the equals sign; taking the cube root of both sides gives x = ∛(9 − 4x), which is the formula xₙ₊₁ = ∛(9 − 4xₙ). Answer: xₙ₊₁ = ∛(9 − 4xₙ). The distractors: ∛(9 + 4xₙ) comes from writing x³ = 9 + 4x, moving the 4x across the equals sign without changing its sign; (9 + xₙ³)/4 comes from making the linear term the subject but keeping the sign of the cube, writing 4x = 9 + x³ when the equation gives 4x = 9 − x³; ∛(9 − 4xₙ³) cubes the x in the linear term as well, changing a term the original equation never cubed.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (c) 4.30 — Method: substitute the starting value into the right-hand side to get x₁, then feed each value back in, keeping the whole display and respecting the order of operations, which divides before it subtracts. Working: x₁ = 5 − 3 ÷ 2.5 = 5 − 1.2 = 3.8; x₂ = 5 − 3 ÷ 3.8 = 5 − 0.78947… = 4.21052…; x₃ = 5 − 3 ÷ 4.21052… = 5 − 0.7125 = 4.2875; x₄ = 5 − 3 ÷ 4.2875 = 5 − 0.69970… = 4.30029…, which is 4.30 correct to 3 significant figures. Answer: 4.30. The distractors: 4.29 is x₃ = 4.2875 rounded, reached by counting the starting value itself as the first iterate and so stopping one use of the formula early; 3.80 is x₁, the value after a single use of the formula; 2.50 comes from working out (5 − 3) ÷ xₙ instead of 5 − (3 ÷ xₙ), subtracting before dividing, which produces the sequence 0.8, 2.5, 0.8, 2.5 and lands on 2.5 at the fourth step.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (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.
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (c) The sequence diverges, moving away from x = 3 — Starting from x₀ = 2.9: x₁ = 2.9² − 6 = 2.41, x₂ = 2.41² − 6 = −0.19, x₃ = (−0.19)² − 6 = −5.96, x₄ = (−5.96)² − 6 = 29.56 — the values swing away from 3 and grow rapidly, so the sequence diverges rather than settling anywhere. Choosing 'settles towards x = 3' assumes that starting close to a root is enough for a rearrangement to converge to it, which is not always true — this rearrangement changes values too steeply near x = 3 to stay there. Choosing 'settles towards x = −2' assumes a diverging sequence must eventually land on the other root; instead it runs away to increasingly large values. Choosing 'stays constant at 2.9' ignores that applying the formula changes the value at every step.
- (b) 58.8 km/h — Total distance = 100 + 47 = 147 km. Total time = 1.5 + 0.25 + 0.75 = 2.5 hours. Average speed = 147 ÷ 2.5 = 58.8 km/h. Leaving out the 0.25 hours of waiting from the total time gives 147 ÷ 2.25 = 65.3 km/h (1 d.p.) — wrong, because the train is stationary but time is still passing on the whole journey. Using only the first leg gives 100 ÷ 1.5 = 66.7 km/h (1 d.p.) — wrong, because it ignores the second leg of the journey entirely. Working out each leg's own speed (100 ÷ 1.5 = 66.7 km/h and 47 ÷ 0.75 = 62.7 km/h) and then averaging those two speeds gives 64.7 km/h (1 d.p.) — wrong, because the average of two speeds over DIFFERENT times is not the same as total distance divided by total time.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
- (b) 0.458 — x₁ = (0.4² + 3) ÷ 7 = 3.16 ÷ 7 = 0.4514 (unrounded, 0.451428...). x₂ = (x₁² + 3) ÷ 7 = (0.2038 + 3) ÷ 7 = 3.2038 ÷ 7 = 0.458 (3 d.p.). Choosing 0.632 divides only the 3 by 7 instead of dividing the whole sum x₁² + 3 by 7. Choosing 0.451 repeats the calculation for x₁ instead of moving on to x₂. Choosing 0.493 uses x₁ itself instead of x₁² inside the formula.
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
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