Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation 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 Foundation
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- 1.A car's speed increases steadily while it accelerates. Its speed v m/s after t seconds is given by v = u + at, where u is the starting speed in m/s and a is the acceleration in m/s². A car starts at u = 4 m/s and accelerates at a = 2 m/s² until it reaches v = 20 m/s. Work out t, the time taken in seconds.
- 2.A taxi firm charges a fixed fee of £3.50 plus £2.20 per mile. Work out the total cost of a journey of 6 miles.
- 3.Point P has coordinates (2, −5). P is reflected in the x-axis to point Q. Write down the coordinates of Q.
- 4.A rectangular field has length (3a + 2) metres and width (a − 1) metres. Write a simplified expression for the perimeter of the field.
- 5.A hiker's distance-time graph shows the following: he walks 5 km in the first 1 hour at a steady speed, rests for 1 hour, then walks a further 9 km in the next 2 hours, also at a steady speed. Work out his average speed for the whole journey, in km/h.
- 6.a = 3 and b = 5. Work out the value of a²b.
- 7.Work out the value of 5a + 2 when a = 3
- 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.Point A has coordinates (4, −5). Point B has coordinates (4, 3). Work out the distance between A and B.
- 10.A point lies on the y-axis. What can be said about the x-coordinate of that point?
- 11.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
- 12.Solve the simultaneous equations 2x + y = 7 and x − y = 2.
- 13.Make x the subject of the formula y = (x − 4)/5.
- 14.Solve the inequality 3x + 6 ≤ 0.
- 15.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
Answer key
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (c) (2, 5) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate: Q = (2, 5). (−2, −5) comes from reflecting in the y-axis instead, which changes the sign of the x-coordinate. (−2, 5) comes from reflecting in both axes. (2, −5) comes from not applying the reflection at all.
- (c) 8a + 2 — The perimeter of a rectangle is twice the length plus twice the width: P = 2(3a + 2) + 2(a − 1) = (6a + 4) + (2a − 2) = 8a + 2. Answering 5a adds the length and width once each but doubles only one of them, missing that a rectangle has two of each side. Answering 8a + 6 distributes the 2 into (a − 1) correctly as far as 2a, but then adds 2 instead of subtracting it, as though the bracket had been (a + 1). Answering 12a + 8 uses the length for all four sides instead of using the length twice and the width twice, as if the field were a square with side (3a + 2). The perimeter of the field is (8a + 2) metres.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (a) 45 — a²b means a × a × b, so the index applies to a only and b is multiplied on afterwards. Substituting the values gives 3 × 3 = 9, then 9 × 5 = 45. Reading a²b as (ab)² gives (3 × 5)², which is 15² = 225 and squares b as well. Substituting the two values the wrong way round works out 5 × 5 × 3 = 75, and reading the letters written side by side as an addition gives 9 + 5 = 14.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (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.
- (a) 8 — Both points share the x-coordinate, so the distance between them is the difference between the y-coordinates: 3 − (−5) = 8. A candidate who mistakenly uses the equal x-coordinates instead of the y-coordinates gets 4 − 4 = 0. A candidate who adds the y-coordinates instead of subtracting them gets 3 + (−5) = −2. A candidate who reads off only point B's y-coordinate as the distance gets 3.
- (b) x = 0 — Method: the first coordinate of a point measures how far to the left or right of the origin it is, so a point that is neither left nor right of the origin has a first coordinate of zero. Working: the y-axis is the vertical line through the origin; every point on it lies directly above or directly below the origin with no sideways movement at all, so its first coordinate is zero while its second coordinate may take any value. Answer: x = 0. The distractors: y = 0 is the condition for lying on the x-axis, the other axis; x = 1 comes from confusing the y-axis with the vertical line one unit to the right of it; x = y is the condition for the diagonal through the origin, which meets the y-axis at the origin only.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
- (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
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