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.Solve 2x² = 18.
- 2.A sequence has the position-to-term rule n² − 3, where n is the position number. Work out the difference between the 6th term and the 5th term.
- 3.A garden path runs along the line 4x + y = 12. A drainage pipe must be laid perpendicular to the path, passing through the point (3, 1) where a sprinkler sits. Work out the equation of the pipe's line, giving your answer in the form y = mx + c.
- 4.f(x) = 2x − 1. Work out ff(x).y = 2x − 1
- 5.A car's speed, in m/s, during a journey is: 0 at t = 0 s, 8 at t = 4 s, and 8 (constant) from t = 4 s to t = 10 s, increasing at a constant rate between t = 0 and t = 4. Estimate the total distance the car travels, using the areas of a triangle and a rectangle.
- 6.The graph of y = sin x is reflected in the x-axis to give the graph of y = g(x). Work out the value of g(90), where x is measured in degrees.y = sin(x)
- 7.The table shows the rate at which water flows from a tank, in litres per minute, over a 6-minute period: 12 at t = 0 min, 12 at t = 2 min, 3 at t = 5 min, and 0 at t = 6 min. Between t = 0 and t = 2 the rate is constant. Between t = 2 and t = 5, and between t = 5 and t = 6, the rate decreases at a constant rate in each section. Estimate the total volume of water that has flowed out, using the areas of a rectangle and two trapeziums.
- 8.The curve y = x² − 6 and the line y = 2x − 3 intersect at two points. Which pair of points is correct?y = 2x − 3y = x² − 6
- 9.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.
- 10.How many turning points does the graph of y = x² − 7x + 10 have?y = x² − 7x + 10
- 11.A machine bought for £2,400 loses 12% of its value every year. After how many complete years does its value first fall below £1,500?
- 12.Work out the smallest positive value of x, in degrees, for which y = tan x is undefined.y = tan(x)
- 13.A square patio of side length n slabs is surrounded by a single border of square paving slabs of the same size. For a patio with side length n, the total number of slabs used for the patio and its border together is 9 when n = 1, 16 when n = 2, 25 when n = 3, and 36 when n = 4. Work out an expression, in terms of n, for the total number of slabs.
- 14.Kai writes the expression 8 − 3x + 5x² and says it has three terms. Ryan says it only has two terms because a number on its own is not a term. Work out the correct number of terms in the expression.
- 15.A zip-line runs from a platform 25 m high to a landing point 5 m high, over a horizontal distance of 40 m. Work out the gradient of the zip-line.
Answer key
- (d) x = 3 or x = −3 — Method: get x² on its own with a coefficient of 1, then take the square root of both sides and keep both the positive and the negative root. Working: dividing 2x² = 18 by 2 gives x² = 9, and the square root of 9 is 3, so x = 3 or x = −3; both check, because 2 × 9 = 18 either way. Answer: x = 3 or x = −3. The distractors: x = 9 or x = −9 comes from dividing by 2 and then forgetting to take the square root; x = 4 or x = −4 comes from subtracting 2 from 18 instead of dividing by it, giving x² = 16; x = 6 or x = −6 comes from multiplying by 2 instead of dividing, giving x² = 36.
- (a) 11 — Method: work out each term separately using the rule n² − 3, then subtract. Working: 6th term = 6² − 3 = 36 − 3 = 33. 5th term = 5² − 3 = 25 − 3 = 22. Difference: 33 − 22 = 11. Answer: 11. 8 comes from subtracting the constant −3 once at the end instead of it already being included in both terms, (36 − 25) − 3. 1 comes from working out (6 − 5)² instead of finding 6² and 5² separately and then subtracting. −11 comes from subtracting in the wrong order, the 5th term minus the 6th term instead of the 6th minus the 5th.
- (a) y = (1/4)x + 1/4 — Rearrange 4x + y = 12 to y = −4x + 12, so the path has gradient −4. The perpendicular gradient is 1/4. Substituting (3, 1) into y − 1 = (1/4)(x − 3): y = (1/4)x − 3/4 + 1 = (1/4)x + 1/4. Distractor routes: y = −4x + 13 uses the path's own gradient, −4, instead of the perpendicular gradient, giving a line PARALLEL to the path through (3, 1) rather than perpendicular to it. y = (1/4)x − 1/4 makes an arithmetic slip combining −3/4 and 1, landing on −1/4 instead of the correct 1/4. y = −(1/4)x + 1/4 takes the reciprocal of −4, which is −1/4, but forgets to change its sign, so it is not the true negative reciprocal.
- (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) 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.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (a) 48 litres — Split the area into three sections. The rectangle (t = 0 to 2) has area 2 × 12 = 24. The first trapezium (t = 2 to 5, parallel sides 12 and 3, width 3) has area 1/2 × (12 + 3) × 3 = 22.5. The second trapezium (t = 5 to 6, parallel sides 3 and 0, width 1) has area 1/2 × (3 + 0) × 1 = 1.5. Total volume: 24 + 22.5 + 1.5 = 48 litres. Forgetting the rectangle and adding only the two trapeziums gives 22.5 + 1.5 = 24 litres. Using a single trapezium across the whole 6 minutes, with parallel sides 12 and 0, ignoring that the first 2 minutes are constant, gives 1/2 × (12 + 0) × 6 = 36 litres. Correctly finding the rectangle and the first trapezium but forgetting to halve the second trapezium, using (3 + 0) × 1 = 3 instead of 1.5, gives 24 + 22.5 + 3 = 49.5 litres.
- (b) (3, 3) and (−1, −5) — Set the two expressions for y equal: x² − 6 = 2x − 3, which rearranges to x² − 2x − 3 = 0. Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1. Substitute into the linear equation, y = 2x − 3: x = 3 gives y = 3; x = −1 gives y = −5. The points are (3, 3) and (−1, −5). Distractor routes: (3, 3) and (1, −1) comes from mis-factorising x² − 2x − 3 as (x − 3)(x − 1), giving a second root of 1 instead of −1. (3, 9) and (−1, 1) comes from finding the y-coordinate from y = x² instead of substituting back into the given line equation y = 2x − 3. (3, 3) alone stops after finding only the first root of the quadratic.
- (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.
- (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) 4 years — The yearly multiplier is 1 − 0.12 = 0.88. After 1 year: 2400 × 0.88 = 2112. After 2 years: 2112 × 0.88 = 1858.56. After 3 years: 1858.56 × 0.88 = 1635.5328. After 4 years: 1635.5328 × 0.88 = 1439.27 (2 d.p.), which is the first value below £1,500 — so it takes 4 complete years. Stopping after 3 years, where the value is still £1,635.53, above £1,500, gives 3 years — wrong, because the threshold has not been crossed yet at that point. Continuing for one year longer than needed gives 5 years — wrong, because the value has already fallen below £1,500 by year 4, so year 5 overcounts. Multiplying by the percentage itself instead of by the multiplier — using 0.12 in place of 0.88 — gives 2400 × 0.12 = £288 after a single year, apparently below £1,500 straight away, so 1 year — wrong, because 12% is the part lost each year, and the part left is 88%, so the multiplier is 0.88.
- (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) n² + 4n + 4 — The whole patio-plus-border square has side length (n + 2), so the total is (n + 2)². Expanding this bracket correctly gives n² + 4n + 4, which matches 9, 16, 25, 36 for n = 1, 2, 3, 4. Expanding (n + 2)² by squaring each term separately, as if (a + b)² = a² + b², gives n² + 4, which is already wrong at n = 1 (it gives 5, not 9). Multiplying out (n + 2)(n + 2) as n² + 2n + 2n but forgetting the final 2 × 2 gives n² + 4n, which is 5 short at every value of n. Counting the patio twice — once as the n² inner square and again inside the (n + 2)² total — gives n² + (n² + 4n + 4) = 2n² + 4n + 4.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (b) −0.5 — Method: gradient = change in height ÷ horizontal distance, and the height decreases so the change is negative. Working: change in height = 5 − 25 = −20, horizontal distance = 40, so gradient = −20 ÷ 40 = −0.5. Answer: the gradient is −0.5. 0.5 comes from dropping the negative sign, ignoring that the zip-line descends. 2 comes from inverting the gradient, dividing the horizontal distance by the drop in height instead of the other way round. −0.8 comes from dividing the drop by the platform height, 25, instead of by the horizontal distance, 40.
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