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.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.
- 2.The graph of a straight line has equation y = 4x − 5. Beth says the y-intercept is 4. Work out the correct y-intercept.y = 4x − 5
- 3.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.
- 4.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.
- 5.Solve x² − 6x = 0.
- 6.Work out the equation of the straight line through the points (−2, 1) and (0, 7).
- 7.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?
- 8.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 9.A number machine multiplies its input by 3. The output is 15. Work out the input.
- 10.A number machine multiplies its input by 5. Work out the output when the input is −1.
- 11.Solve x/4 + 1 = 3
- 12.The point A lies on the x-axis. Which statement about A must be true?
- 13.Work out the gradient of the straight line with equation y = (2x/5) − 3.
- 14.Solve 6x + 4 = 28
- 15.Simplify 2b − 1 + 2b − 4 + 2b
Answer key
- (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.
- (d) −5 — In y = mx + c, the y-intercept is c, the constant term, not the coefficient of x. Here c = −5, so the y-intercept is −5. 4 is Beth's value, the coefficient of x (the gradient), not the intercept. 5 comes from taking the constant term but dropping its negative sign. −4 comes from taking the gradient but giving it the sign of the intercept.
- (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.
- (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.
- (d) x = 0 or x = 6 — Method: factorise by taking out the common factor x: x(x − 6) = 0, so x = 0 or x − 6 = 0, giving x = 0 or x = 6. Distractor origins: x = 6 comes from dividing both sides by x, which loses the solution x = 0; x = 0 stops after finding only one factor; x = 3 comes from halving the coefficient of x instead of factorising.
- (c) y = 3x + 7 — Method: the gradient is the change in y divided by the change in x with both taken in the same order, and a point whose x-coordinate is 0 gives the constant straight away because it lies on the y-axis. Working: m = (7 − 1) ÷ (0 − (−2)) = 6 ÷ 2 = 3; the point (0, 7) lies on the y-axis, so c = 7 and the line is y = 3x + 7. Answer: y = 3x + 7. The distractors: y = −3x + 7 comes from taking the y-difference as 1 − 7 while taking the x-difference as 0 − (−2), so the two subtractions run in opposite orders; y = 3x + 1 comes from using the y-coordinate of (−2, 1) as the constant instead of the point that actually lies on the y-axis; y = (1/3)x + 7 comes from writing the gradient upside down as the change in x over the change in y, 2 ÷ 6.
- (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.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (d) 5 — Method: running a machine backwards means undoing its operation, and the operation that undoes multiplying by 3 is dividing by 3. Working: the input multiplied by 3 gives 15, so the input is 15 ÷ 3. Answer: 5, which checks because 3 × 5 = 15. The distractors: 45 comes from running the machine forwards on the output, 15 × 3; 12 comes from subtracting 3 from 15 instead of dividing; 18 comes from adding 3 to 15, which would undo a machine that subtracts rather than one that multiplies.
- (a) −5 — Method: apply the machine's operation to the input, and carry the sign of the input through the multiplication. Working: the machine gives 5 × (−1); a positive number multiplied by a negative number gives a negative result, and 5 × 1 = 5, so the output is 5 below zero. Answer: −5, which is one lot of −1 taken five times. The distractors: 5 comes from working out 5 × 1 and losing the minus sign of the input; 4 comes from adding 5 to −1 instead of multiplying; −6 comes from subtracting 5 from −1 instead of multiplying.
- (b) x = 8 — Method: undo the addition first, then undo the division by 4. Working: subtracting 1 from both sides gives x/4 = 2, and multiplying both sides by 4 gives x = 8. Answer: x = 8. The distractors: x = 2 comes from stopping at x/4 = 2 and writing 2 as the value of x; x = 16 comes from adding 1 to both sides instead of subtracting it, giving x/4 = 4; x = 0.5 comes from dividing by 4 instead of multiplying by 4 at the last step.
- (a) y = 0 — Method: in the pair (x, y) the first coordinate measures how far left or right of the origin a point is and the second how far above or below the x-axis it is, so a point on an axis has one of those measurements equal to zero. Working: the x-axis is the horizontal line through the origin, so a point sitting on it is neither above nor below that line and its second coordinate is zero, while its first coordinate may be positive, negative or zero. Answer: y = 0. The distractors: x = 0 is the condition for lying on the y-axis, the other axis; x > 0 comes from assuming a point on the x-axis must be to the right of the origin, which is true only of part of that axis; x = y holds only at the origin, which is one point of the x-axis rather than a property shared by all of them.
- (a) 2/5 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = (2x/5) − 3, the coefficient of x is 2/5. Answer: the gradient is 2/5. −3 comes from confusing the gradient with the y-intercept. 5/2 comes from inverting the fraction that multiplies x. −2/5 comes from wrongly carrying the negative sign from the −3 term onto the coefficient of x.
- (a) x = 4 — Method: undo the addition of 4 first, then undo the multiplication by 6. Working: subtracting 4 from both sides gives 6x = 24, and dividing both sides by 6 gives x = 4. Checking: 6 × 4 + 4 = 28. Answer: x = 4. The distractors: x = 24 comes from stopping at 6x = 24 and writing 24 as the value of x; x = 18 comes from subtracting 6 from 24 instead of dividing by 6; x = 144 comes from multiplying 24 by 6 instead of dividing by 6.
- (a) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
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