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

Maths · GCSE Foundation · Algebra · 15 questions

MathsUKwww.geekhero.co.uk

1.Factorise fully 56x − 24
(a)8(7x − 16)
(b)8(56x − 3)
(c)8(7x + 3)
(d)8(7x − 3)
2.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
(a)It is an identity, because both sides begin with the term 3x, so they must be equivalent.
(b)It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true.
(c)It is not an identity, because the student forgot to multiply the 4 by 3 on both sides of the equation.
(d)It is not an identity, because the student needs to substitute a specific value of x before comparing the two sides.
3.A rectangle has length k cm and width 5 cm less than its length. Write down an expression for the width of the rectangle, in cm.
(a)k + 5
(b)k − 5
(c)5k
(d)5 − k
4.Solve 4x + 3 = 19
(a)3.75
(b)5.5
(c)4
(d)4.75
5.Solve 2(x + 3) = 10
(a)x = 3.5
(b)x = 8
(c)x = 5
(d)x = 2
6.A water tank empties according to y = −5x + 200, where y is the number of litres left after x minutes. Work out the coordinates of the point where this graph crosses the x-axis.
y = -5x + 200
(a)(40, 0)
(b)(0, 40)
(c)(0, 200)
(d)(200, 0)
7.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
(a)12 km/h
(b)8 km/h
(c)6 km/h
(d)4 km/h
8.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.
y = x² − 2x − 8
(a)x = −2
(b)x = −1 or x = 3
(c)x = −2 or x = 4
(d)x = 0 or x = −8
9.There are 30 students in a Year 10 maths class. There are 2 more boys than girls. Work out the number of boys and the number of girls.
(a)15 boys and 15 girls
(b)17 boys and 13 girls
(c)14 boys and 16 girls
(d)16 boys and 14 girls
10.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
(a)720
(b)184
(c)176
(d)45
11.The diagram shows the graph of $y = x^{2} - x - 2$ , together with the horizontal line $y = 3$ . Use the graphs to estimate, to 1 decimal place, the two solutions of $x^{2} - x - 2 = 3$ .
(a)x = −1.8 and x = 2.8
(b)x = −3.0 and x = 4.0
(c)x = 1.8 and x = 2.8
(d)x = −1.0 and x = 2.0
12.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.
y = x² − 7x + 2
(a)x ≈ 6.70
(b)x ≈ 6.30
(c)x ≈ 7.30
(d)x ≈ 0.70
13.Expand 4(2x − 3).
(a)2x + 1
(b)8x + 12
(c)8x − 12
(d)8x − 3
14.Solve 3x − 1 = 8
(a)x = 9
(b)x = 6
(c)x = 27
(d)x = 3
15.Make x the subject of the formula y = 5x + 3.
y = 5x + 3
(a)x = (y + 3)/5
(b)x = (y − 3)/5
(c)x = y/5 − 3
(d)x = 5(y − 3)

Answer key

(d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
(b) It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true. — Expanding the left-hand side, 3(x − 4) = 3x − 12. Setting this equal to the right-hand side, 3x − 12 = 3x − 4, gives −12 = −4 once the 3x terms are removed from both sides — a statement that is never true, so no value of x satisfies the equation at all, and it is certainly not an identity. The option about substituting a specific value misunderstands algebraic expansion, which holds for every x, not one chosen value. The option matching the first term wrongly assumes that is enough to prove equivalence. The option about multiplying the 4 by 3 on both sides is nonsensical, since there is only one bracket to expand, on the left-hand side.
(b) k − 5 — Method: the width is 5 less than the length, so subtract 5 from the length. Working: length − 5 = k − 5. Answer: k − 5. 5 − k comes from subtracting in the wrong order, taking the length away from 5 instead of the other way round. k + 5 comes from adding instead of subtracting, missing that the width is smaller than the length. 5k comes from multiplying the length by 5 instead of subtracting 5 from it.
(c) 4 — Method: subtract 3 from both sides, then divide by 4. Working: 4x = 19 − 3 = 16, so x = 16 ÷ 4 = 4. Answer: 4. 4.75 comes from dividing 19 by 4 directly, without subtracting 3 first. 3.75 comes from subtracting 4 instead of 3 before dividing. 5.5 comes from a sign error, adding 3 to 19 instead of subtracting it.
(d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
(a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
(c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
(c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
(d) 16 boys and 14 girls — Method: write the number of boys in terms of the number of girls, then use the total for the class. Working: if there are g girls then there are g + 2 boys, so g + (g + 2) = 30, that is 2g + 2 = 30, so 2g = 28 and g = 14; the number of boys is 14 + 2 = 16. Answer: 16 boys and 14 girls, which total 30 and differ by 2. The distractors: 14 boys and 16 girls comes from substituting for the wrong group, writing b + (b + 2) = 30 and then calling b the number of boys; 15 boys and 15 girls comes from halving 30 and never using the difference; 17 boys and 13 girls comes from adding the whole 2 to one half of the class and taking the whole 2 off the other half, which leaves a difference of 4.
(d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
(a) x = −1.8 and x = 2.8 — Method: the solutions of x² − x − 2 = 3 are the x-coordinates of the points where the curve y = x² − x − 2 meets the line y = 3, read off the grid to 1 decimal place. Working: the curve meets the line y = 3 at approximately x = −1.8 and at x = 2.8. Answer: x ≈ −1.8 and x ≈ 2.8. Distractor refutation: x = −1.0 and x = 2.0 comes from reading where the curve crosses the x-axis (y = 0), solving x² − x − 2 = 0, instead of where it meets the line y = 3. x = −3.0 and x = 4.0 comes from taking the two ends of the drawn curve as the intersection points, instead of finding where it actually crosses the line y = 3. x = 1.8 and x = 2.8 comes from a sign slip on the left-hand intersection, reading it as positive instead of negative.
(a) x ≈ 6.70 — Since the two roots sum to 7, the other root is 7 − 0.30 = 6.70. The option 7.30 comes from adding the given root to 7 instead of subtracting it. The option 6.30 comes from subtracting 0.70 (one minus the given root) rather than the given root itself. The option 0.70 confuses the required root with the amount by which the given root falls short of 1.
(c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
(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) 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.

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