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.
Non-calculator
Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.By completing the square, show that the curve y = x² − 14x + 50 never crosses the x-axis. Which statement gives the correct reason?y = x² − 14x + 50
- 2.Line L has equation y = 3x + 1. Which of these lines is perpendicular to L?y = 3x + 1
- 3.A theatre's front row has 18 seats. Each row behind has 4 more seats than the row in front. Which row has exactly 62 seats?
- 4.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.
- 5.The curve y = x² − 1 and the line y = 3x − 3 meet at two points. Work out the x-coordinates of those two points.y = 3x − 3y = x² − 1
- 6.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
- 7.A cup of tea cools after being made. Its temperature-time graph is a curve. Three tangents are drawn on the graph: at t = 2 minutes the tangent has gradient −8; at t = 10 minutes the tangent has gradient −3; at t = 20 minutes the tangent has gradient −0.5. Which of these statements about the tea's cooling is correct?
- 8.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
- 9.A circular pond has equation x² + y² = 20, with lengths in metres from the centre of the garden. A straight path runs along the line y = 2x, entering the pond and leaving it again. Work out the coordinates of the two points where the path meets the edge of the pond.y = 2x
- 10.A student solves the simultaneous equations 2x + y = 11 and x − y = 1 by elimination, adding the two equations together. Which of these is the correct result of that step?
- 11.The first five terms of a quadratic sequence are −1, 4, 13, 26, 43. Work out an expression, in terms of n, for the nth term.
- 12.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 13.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
- 14.Write down the expression that means the same as 'subtract 5 from n, then divide the result by 2'.
- 15.Make x the subject of the formula y = 5x + 3.y = 5x + 3
Answer key
- (d) (x − 7)² + 1 is at least 1 for every x, so y is never 0 — x² − 14x + 50 = (x − 7)² − 7² + 50 = (x − 7)² + 1. Since a squared term can never be negative, (x − 7)² ≥ 0 for every real x, so y = (x − 7)² + 1 ≥ 1 for every x — y is always at least 1, so it can never equal 0, and the curve never crosses the x-axis. Claiming the discriminant is positive is wrong on the facts: b² − 4ac = (−14)² − 4(1)(50) = 196 − 200 = −4, which is NEGATIVE, not positive — a negative discriminant is in fact the standard reason a quadratic has no real roots, so this option misstates its own sign. Claiming the turning point is at (7, −1) is wrong because completing the square gives (x − 7)² + 1, so the turning point is (7, 1) — the constant is +1, not −1, and a point with a negative y-coordinate would sit BELOW the x-axis, not above it, contradicting the option's own claim. Claiming that 50 being positive is enough is wrong because it ignores the −14x term completely — a positive constant term alone does not guarantee y stays positive for every x (the expression still dips down towards its turning point before growing again, and it is the completed square that shows how far it dips, not the constant on its own).
- (b) y = −(1/3)x + 2 — Method: two lines are perpendicular when the product of their gradients is −1, so the gradient of a line perpendicular to one of gradient m is the negative reciprocal, −1 divided by m. Working: reading m from y = 3x + 1 gives a gradient of 3, so the perpendicular gradient is −1 ÷ 3 = −1/3, and the line carrying that gradient is y = −(1/3)x + 2; the check 3 × (−1/3) = −1 confirms it. Answer: y = −(1/3)x + 2. The distractors: y = 3x − 4 comes from using an equal gradient, which is the test for parallel lines rather than perpendicular ones; y = (1/3)x + 2 comes from turning the gradient upside down but leaving out the change of sign, giving a product of 1 instead of −1; y = −3x + 2 comes from changing the sign of the gradient without turning it upside down, giving a product of −9.
- (d) 12 — Method: write the nth term of the sequence, 18 + 4(n − 1), set it equal to 62, and solve for n. Working: 18 + 4(n − 1) = 62, so 4(n − 1) = 44, giving n − 1 = 11, so n = 12. Answer: row 12. 11 comes from using 18 + 4n = 62 instead of 18 + 4(n − 1) = 62, an off-by-one error, giving n = 11. 48 comes from correctly simplifying to 4n = 48 but stopping there, without dividing by 4 to find n. 15.5 comes from dividing 62 by 4 directly, ignoring the 18 seats already in the front row.
- (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.
- (d) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
- (c) Fastest at t = 2 min — steepest gradient. — The rate of cooling is given by the size (magnitude) of the gradient, ignoring its sign — the steeper the tangent, the faster the temperature is changing. Of −8, −3 and −0.5, the gradient −8 has the greatest magnitude, so the tea is cooling fastest at t = 2 minutes. 'Fastest at t = 20 min — largest gradient' confuses the signed value with the size of the rate: −0.5 is the largest NUMBER of the three, but it's the smallest in magnitude, meaning the tea is barely cooling at all by then. 'Cools at the same rate throughout' ignores that the three gradients are different sizes, not just all negative. 'Fastest at t = 10 min — the middle reading' isn't a mathematical reason at all — the gradients themselves have to be compared, not their position in the list.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
- (a) 3x = 12 — Adding the two equations: the y-terms, +y and −y, have opposite signs, so they cancel; the x-terms combine to 2x + x = 3x; and the right-hand sides add to 11 + 1 = 12. This gives 3x = 12. A candidate who forgets that the y-terms cancel, and instead adds them as if they had the same sign, would write 3x + 2y = 12. A candidate who subtracts the right-hand sides instead of adding them would get 3x = 10. A candidate who correctly reaches 3x = 12 but then treats 12 itself as the value of x, skipping the final division, would write x = 12.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (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.
- (a) (n − 5)/2 — Method: do the subtraction first, keep it together as a single bracket, then divide that whole bracket by 2. Working: 'subtract 5 from n' is (n − 5); 'divide the result by 2' means the whole bracket goes over 2, giving (n − 5)/2. Answer: (n − 5)/2. n/2 − 5 comes from dividing n by 2 first and only then subtracting 5, the wrong order. 2(n − 5) comes from multiplying by 2 instead of dividing. (5 − n)/2 comes from subtracting n from 5 instead of subtracting 5 from n, the wrong way round.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
Build your own mix at the worksheet builder.