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.ABCD is a parallelogram. A has coordinates (−3, 1), B has coordinates (2, 1) and C has coordinates (4, 4). Work out the coordinates of D.
- 2.A sequence begins at 7. Each term after the first is found by adding 6 to the term before it. Work out the 6th term of the sequence.
- 3.Solve 3x − 1 = 8
- 4.A formula is F = 4s. A pupil says that when s = 5 the value of F is 9, because they added 4 and 5 instead of multiplying. Work out the correct value of F when s = 5.
- 5.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
- 6.Write down the coordinates of the point that is 4 units to the left of the origin and 7 units up.
- 7.To rearrange y = 2x − 5 to make x the subject, Sam writes: 'Add 5 to both sides to get y + 5 = 2x, then divide both sides by 2 to get x = (y + 5)/2.' Which statement about Sam's method is correct?y = 2x − 5
- 8.Yusuf has £40 saved and adds £6 each week. His sister Aisha has £10 saved and adds £11 each week. After how many weeks will they have saved the same amount?
- 9.A cafe sells coffees at £c each and pastries at £p each. Three coffees and two pastries cost £9.60. Two coffees and two pastries cost £7.60. Work out the price of one coffee.
- 10.The volume of a cuboid is given by the formula V = lwh. Work out V, in cm³, when l = 2, w = 3 and h = 6.
- 11.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.
- 12.The formula for the volume of a cuboid is V = lwh, where l is the length, w is the width and h is the height. Make h the subject of the formula.
- 13.Solve the inequality 6x ≥ 18.
- 14.A gardener charges a fixed fee of £15 plus £8 for each hour worked. The total charge, £T, for a job lasting h hours is given by T = 15 + 8h. Work out the charge for a job lasting 4 hours, and identify what kind of statement T = 15 + 8h is.
- 15.A statement contains an equals sign and one unknown letter, and it is only true for particular value(s) of that letter. What is this type of statement called?
Answer key
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (b) 37 — The terms are 7, 13, 19, 25, 31, 37 — each found by adding 6 to the term before, so the 6th term is 37. Adding 6 six times to the first term instead of five times, 7 + 6 × 6 = 43, treats the first term as if it were before the sequence starts. Stopping one term early gives the 5th term, 31. Stopping two terms early gives the 4th term, 25.
- (d) x = 3 — Method: undo the operations on the left in reverse order — deal with the subtraction of 1 first, then with the multiplication by 3. Working: adding 1 to both sides gives 3x = 9, and dividing both sides by 3 gives x = 3. Answer: x = 3. The distractors: x = 9 comes from stopping at 3x = 9 and writing 9 as the value of x; x = 6 comes from subtracting 3 from 9 instead of dividing by 3; x = 27 comes from multiplying 9 by 3 instead of dividing by 3.
- (b) 20 — F = 4s, so when s = 5, F = 4 × 5 = 20. 9 comes from adding 4 and 5 instead of multiplying, the mistake described in the question. 25 comes from squaring s (5²) instead of multiplying by 4. 45 comes from writing the digits 4 and 5 next to each other instead of carrying out the multiplication.
- (a) It is 2(w + (w + 3)) = 4w + 6, not 2w + 3. — The perimeter of a rectangle is twice the width plus twice the length: 2 × w + 2 × (w + 3) = 2w + 2w + 6 = 4w + 6, so the gardener's 2w + 3 is wrong. Writing w + (w + 3) = 2w + 3 forgets to double the sides at all, only adding one width and one length once. Writing 4(w + 3) = 4w + 12 wrongly treats all four sides as equal to the length, as if the garden were a square. Writing 3w + 6 comes from doubling the length correctly but adding the width only once instead of doubling it too.
- (c) (−4, 7) — Left of the origin means negative x, and up means positive y, so the point is (−4, 7). (4, 7) comes from forgetting that 'left' means the x-coordinate is negative. (−4, −7) comes from treating 'up' as a negative direction instead of positive. (7, −4) comes from swapping the x- and y-coordinates.
- (c) Yes; x = (y + 5)/2 is right — Sam's method is correct throughout: adding 5 to both sides gives y + 5 = 2x, and dividing both sides by 2 gives x = (y + 5)/2, so Sam is right. The option giving x = (y − 5)/2 assumes 5 should be subtracted again, but 2x − 5 = y means 5 has already been subtracted, so it must be added back, not taken away a second time. The option giving x = y/2 + 5 divides only the y term by 2 and leaves the 5 unhalved, which is not a valid rearrangement. The option agreeing Sam is correct but changing step 2 to x = 2(y + 5) confuses '2x' with 'x divided by 2' — since x is multiplied by 2, the inverse is division, not multiplication. Sam's working, and his final formula x = (y + 5)/2, are both correct.
- (d) 6 — Method: write an expression for each person's savings after w weeks, and set them equal. Working: 40 + 6w = 10 + 11w. Subtract 6w from both sides: 40 = 10 + 5w. Subtract 10: 30 = 5w, so w = 6. Answer: 6 weeks. 0 comes from setting only the weekly amounts equal, 6w = 11w, and ignoring the different starting amounts entirely. 10 comes from adding the two starting amounts and dividing by the difference in weekly amounts, (40 + 10) ÷ (11 − 6), instead of forming and solving the correct equation. 1.76 comes from adding the two weekly amounts instead of subtracting them when rearranging, (40 − 10) ÷ (11 + 6).
- (b) £2.00 — Subtracting the second equation from the first eliminates the pastries: (3c + 2p) − (2c + 2p) = 9.60 − 7.60, so c = 2.00. A candidate who divides the first total by the number of coffees alone, ignoring the pastries, would get 9.60 ÷ 3 = £3.20. A candidate who finds the price of a pastry instead of a coffee — using c = 2.00 in 2c + 2p = 7.60 to get p = 1.80 — would answer £1.80. A candidate who reaches the correct difference of £2.00 but then mistakenly divides again or misplaces the decimal point would get £0.20.
- (d) 36 cm³ — V = lwh = 2 × 3 × 6 = 36 cm³. A candidate who adds all three numbers instead of multiplying gets 2 + 3 + 6 = 11 cm³. A candidate who multiplies only two of the three numbers, forgetting the length, gets w × h = 3 × 6 = 18 cm³. A candidate who multiplies l × w and w × h separately and adds the two products gets (2 × 3) + (3 × 6) = 6 + 18 = 24 cm³.
- (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.
- (c) h = V / (lw) — Method: undo the multiplication by lw by dividing both sides by lw. Working: V = lwh, so dividing both sides by lw gives h = V / (lw). The value h = V − lw comes from subtracting lw instead of dividing by it. The value h = Vlw comes from multiplying by lw instead of dividing. The value h = lw / V comes from inverting the fraction, dividing lw by V instead of V by lw.
- (c) x ≥ 3 — Method: x is multiplied by 6, so divide both sides by 6; dividing by a positive number leaves the direction of the inequality unchanged. Working: dividing both sides of 6x ≥ 18 by 6 gives x on the left and 18 ÷ 6 on the right, and 18 ÷ 6 = 3, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from turning the sign round on dividing, a step that is needed only when the divisor is negative; x ≥ 12 comes from subtracting 6 from both sides instead of dividing, giving 18 take away 6; x ≤ 12 comes from making both of those mistakes together.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (d) an equation — Method: recall what each piece of algebraic vocabulary means and match it to the description given. Working: a statement with an equals sign, one unknown letter, and only particular value(s) making it true is, by definition, an equation. Answer: an equation. A formula also has an equals sign, but it links two or more different letters so that one quantity can be calculated from the others — here there is only one letter. An identity is a statement with an equals sign that is true for every value of the letter, not just particular ones. An expression has no equals sign at all, so it cannot be 'true' or 'false'.
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