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 rectangular sheep pen has perimeter P, length l and width w, connected by the formula P = 2l + 2w. A farmer has 60 m of fencing for the perimeter and sets the width at 8 m. Work out the length of the pen.
- 2.Solve x² + 3x − 10 = 0.
- 3.A gym charges a reduced price for the first month of membership and a standard price for every month after that. The total cost is £5 for 1 month, £12 for 2 months, £19 for 3 months and £26 for 4 months. Work out an expression, in terms of n, for the total cost, in pounds, of n months of membership.
- 4.Aisha saves money each week. In week 1 she has saved £15 in total, in week 2 she has saved £23 in total, in week 3 she has saved £31 in total and in week 4 she has saved £39 in total, with the total increasing by the same amount each week. Work out an expression, in terms of n, for the total amount she has saved, in pounds, after week n.
- 5.A line has equation y = −4x + 1. Work out the equation of the line perpendicular to it that passes through the point (0, 3).y = -4x + 1
- 6.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?
- 7.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.
- 8.A ribbon of length L metres is cut into 5 equal pieces, and then 3 metres is removed from one piece. Write down an expression, in metres, for the length of that piece after 3 metres is removed.
- 9.The first four terms of a sequence are 6, 13, 20, 27. Work out an expression, in terms of n, for the nth term.
- 10.A straight line has equation 4y = 12x + 20. Work out the gradient of the line.y = 12x + 20
- 11.A theatre has rows of seats arranged so that row 1 has 12 seats, row 2 has 17 seats, row 3 has 22 seats and row 4 has 27 seats, with each row having 5 more seats than the row before. Work out an expression, in terms of n, for the number of seats in row n.
- 12.A student is proving that (n + 1)² − n² is always an odd number. Which of these correctly completes the first line of algebra?
- 13.The curve y = x² − 3x. Use the chord between x = 1 and x = 4 to estimate the gradient of the curve at x = 2.5.y = x² − 3x
- 14.How many turning points does the graph of y = x² − 7x + 10 have?y = x² − 7x + 10
- 15.Line L has equation y = 2x + 3. Work out the equation of the line perpendicular to L that passes through the point (4, 1), giving your answer in the form x + 2y = c.y = 2x + 3
Answer key
- (c) 22 m — Rearranging P = 2l + 2w for l gives l = (P − 2w)/2. Substituting P = 60 and w = 8: l = (60 − 16)/2 = 44/2 = 22 m. Answering 30 m comes from ignoring the width altogether and working out l = P/2, which only holds when w = 0. Answering 38 m comes from adding 2w instead of subtracting it: (60 + 16)/2 = 38. Answering 11 m comes from dividing by 4 instead of 2, as if the formula were P = 4l + 4w. The pen's length is 22 m.
- (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (d) y = (1/4)x + 3 — The given line has gradient −4, so the perpendicular gradient is the negative reciprocal, 1/4. Since (0, 3) is the y-intercept, the perpendicular line is y = (1/4)x + 3. Distractor routes: y = −4x + 3 uses the original line's own gradient instead of the perpendicular gradient, only changing the intercept for the new point. y = (−1/4)x + 3 takes the reciprocal of −4 but forgets to change its sign, so it is not the true negative reciprocal. y = 4x + 3 negates the gradient −4 but does not take its reciprocal, giving 4 instead of 1/4.
- (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.
- (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.
- (c) L/5 − 3 — Each of the 5 equal pieces is L/5 metres long, and removing 3 metres from one piece gives L/5 − 3. Subtracting the 3 metres before dividing by 5, (L − 3)/5, divides the removed length between all 5 pieces instead of taking it from just one. Dividing only the 3 by 5 instead of dividing L by 5, L − 3/5, divides the wrong number. Writing 5/L − 3 inverts the fraction, swapping which number is the numerator.
- (c) 7n − 1 — Method: find the common difference, then find the constant that fits the first term. Working: 13 − 6 = 7, 20 − 13 = 7, 27 − 20 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 6, so c = −1. Answer: the nth term is 7n − 1. The value 7n comes from leaving out the constant altogether. The value 7n + 6 comes from using the first term as the constant directly, without subtracting the common difference first. The value 6n + 1 comes from using the first term, 6, as the coefficient of n instead of the common difference, and then attaching +1 with the sign of the constant flipped.
- (b) 3 — Method: rearrange the equation into the form y = mx + c first, then read off the gradient. Working: dividing 4y = 12x + 20 by 4 gives y = 3x + 5, so the gradient is 3. The value 12 comes from reading the coefficient of x before dividing the equation by 4. The value 20 comes from using the constant term of the unsimplified equation instead of the gradient. The value 5 comes from finding the y-intercept, 20 / 4 = 5, and giving that instead of the gradient.
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (b) (n + 1)² − n² = 2n + 1 — (n + 1)² = n² + 2n + 1, so (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1, which is odd because it is one more than the even number 2n. Expanding (n + 1)² as n² + 1 uses the false rule (a + b)² = a² + b², and subtracting n² from that leaves just 1 — always expand (a + b)² as a² + 2ab + b². Writing n² + 2n + 1 expands correctly but never carries out the subtraction of n². Writing 2n forgets the constant term left after subtracting.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (b) One turning point. — Every quadratic graph, one with an x² term and no higher power of x, has exactly one turning point, since it is a single U-shaped or n-shaped curve. Saying two turning points describes a cubic graph, which can rise, turn, then turn again. Saying no turning points describes a straight line, which has none. Saying four turning points greatly overestimates how many times a simple quadratic curve changes direction — that would need a much higher power of x.
- (c) x + 2y = 6 — L has gradient 2, so the perpendicular gradient is −1/2. Substituting (4, 1) into y − 1 = −(1/2)(x − 4): y = −(1/2)x + 2 + 1 = −(1/2)x + 3, so 2y = −x + 6, giving x + 2y = 6. Distractor routes: x − 2y = 2 comes from using gradient +1/2 instead of −1/2, forgetting the negative sign the perpendicular rule needs. x + 2y = 9 comes from swapping the point's coordinates, substituting (1, 4) instead of (4, 1). x + 2y = −2 comes from a sign error distributing the negative gradient over the bracket, computing −(1/2)(x − 4) as −(1/2)x − 2 instead of −(1/2)x + 2.
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