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.
Algebra worksheet — GCSE Foundation
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- 1.The formula for converting a distance in miles, m, to kilometres, k, is k = 8m ÷ 5. Work out k when m = 10.
- 2.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.
- 3.A rectangle has width x cm. Its length is 3 cm more than its width. The perimeter of the rectangle is 38 cm. Form an equation and solve it to find x.
- 4.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
- 5.The first five terms of a quadratic sequence are 5, 8, 13, 20, 29. Work out the next term in the sequence.
- 6.Solve the inequality x + 5 < 12.
- 7.y = x + 10. Work out the value of y when x = 0.y = x + 10
- 8.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 9.Meera writes the statement 3(x + 4) = 3x + 12. Which of these correctly describes what she has written, with a reason?
- 10.Oliver has m stickers and gives 5 of them away. The number he has left is L = m − 5. Work out L when m = 8.
- 11.Work out the value of 5y + 9 when y = 0
- 12.x = 5. Work out the value of 3x² − 4.
- 13.The graph of W = 840 ÷ L shows how the width, W metres, of a rectangular field of area 840 m² depends on its length, L metres. Use the relationship to work out the width when the length is 35 m, and hence the perimeter of the field.
- 14.The perimeter, P, of a rectangle with length l and width w is calculated using P = 2l + 2w. What kind of mathematical statement is P = 2l + 2w?
- 15.The point D has coordinates (3, −8). Write down the distance of D from the y-axis.
Answer key
- (a) 16 — k = 8m ÷ 5 = (8 × 10) ÷ 5 = 80 ÷ 5 = 16. A candidate who multiplies 8 by 10 but forgets to divide by 5 gets 80. A candidate who divides 10 by 8 and multiplies by 5, using the fraction upside down, gets 10 × 5 ÷ 8 = 6.25. A candidate who adds instead of substituting into the formula gets 8 + 10 = 18.
- (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.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (b) 10 — Method: substitute the value of x into the rule and carry out the addition; only the term containing x becomes zero. Working: y = 0 + 10, and adding to zero leaves the 10 unchanged. Answer: y = 10. The distractors: 0 comes from assuming that an input of x = 0 makes the whole rule zero, which is only true when nothing is added on; 11 comes from substituting x = 1 instead of x = 0; −10 comes from reading the rule as y = x − 10.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (a) 3 — Method: substitute the given value into the formula and carry out the subtraction in the order the formula is written. Working: replacing m with 8 gives L = 8 − 5, and 8 − 5 = 3. Answer: 3. The distractors: 13 comes from adding 5 to 8 instead of subtracting it; 40 comes from reading m − 5 as a multiplication and working out 8 × 5; −3 comes from subtracting the wrong way round and working out 5 − 8.
- (c) 9 — Method: replace the letter with the value given for it, then work out the multiplication before the addition. Working: 5y means 5 × y, and 5 × 0 = 0, so the expression becomes 0 + 9, which is 9. Answer: 9. The distractors: 14 comes from treating 5 × 0 as 5, a common slip with zero, and then adding the 9; 0 comes from assuming that an expression containing a zero must itself be zero, so the + 9 is ignored; 45 comes from multiplying the whole expression by 5 instead of multiplying only y, giving 5 × (0 + 9).
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (b) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (c) a formula — A statement that shows how to calculate one quantity from others, using an equals sign, is a formula. P = 2l + 2w tells us how to work out the perimeter, P, from the length and width, so it is a formula. Distractor origins: 'an equation' confuses a formula linking several letters with an equation solved for one unknown value; 'an identity' confuses a formula with a statement that is true for every value of a single variable; 'an expression' forgets that an expression has no equals sign at all.
- (b) 3 — Distance from the y-axis is given by the size of the x-coordinate, which is 3. 8 comes from using the size of the y-coordinate instead of the x-coordinate. −8 comes from using the y-coordinate and keeping its negative sign, but a distance is never negative. 11 comes from adding the sizes of both coordinates, 3 + 8, instead of using the x-coordinate alone.
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