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.Expand 7(a + 8)
- 2.The solution to an inequality is n ≤ 5. Write down the largest integer value of n that satisfies this inequality.
- 3.The area of a circle is given by the formula A = πr², where r is the radius. Rearrange the formula to make r the subject.
- 4.Solve the inequality 2(x − 1) ≤ 8.
- 5.Given that (x + 2)(x − 7) = 0, write down the two solutions of x.
- 6.A rectangular field has length (3a + 2) metres and width (a − 1) metres. Write a simplified expression for the perimeter of the field.
- 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.
- 8.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
- 9.Which expression means the same as a × a × a?
- 10.Write down the coefficient of xy in the expression 4x²y − 7xy + 2.
- 11.Grace says that 2(x + 1) and 2x + 2 always have the same value. Work out the value of both expressions when x = 3.
- 12.A quadratic curve has a root at x = −2 and its turning point has x-coordinate 3. Work out the curve's other root, using the symmetry of the graph.
- 13.Write down the expression that means the same as 'subtract 5 from n, then divide the result by 2'.
- 14.Work out the value of 2x² when x = −3
- 15.Write down the expression that means the same as w ÷ 6.
Answer key
- (d) 7a + 56 — Method: multiplying a single term over a bracket means every term inside the bracket is multiplied by the term outside. Working: 7 × a = 7a and 7 × 8 = 56, and the two products are added because the bracket contains an addition. Answer: 7a + 56. The distractors: 7a + 15 comes from adding 7 and 8 instead of multiplying them; 7a + 8 comes from multiplying only the first term inside the bracket and copying the 8 across unchanged; a + 56 comes from multiplying only the 8 and copying the a across unchanged.
- (a) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
- (c) r = √(A/π) — A = πr² means r has been squared and then multiplied by π. To make r the subject, first divide both sides by π to get A/π = r², then take the square root of both sides: r = √(A/π). Writing r = A/π stops after dividing by π and forgets that r is still squared — it never undoes the square. Writing r = √A/π takes the square root before dividing by π, which square-roots only the A and not the whole of A/π. Writing r = (A/π)² squares A/π instead of taking its square root — the opposite of what is needed to undo r². The correct rearrangement is r = √(A/π).
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (c) x = −2 or x = 7 — Method: each factor equals zero in turn. From x + 2 = 0, x = −2. From x − 7 = 0, x = 7. So x = −2 or x = 7. Distractor origins: x = 2 or x = −7 flips both signs the wrong way; x = −2 or x = −7 wrongly makes both solutions negative; x = 2 or x = 7 ignores the signs in the brackets completely.
- (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.
- (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.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
- (c) a³ — a × a × a means a multiplied by itself three times, which is written using powers as a³ — the small 3 shows how many times a is multiplied by itself. Writing 3a instead uses the 3 as a coefficient, as if the expression meant a + a + a (three lots of a added together) rather than three a's multiplied together. Writing a² only accounts for two of the three a's being multiplied together, missing one factor. Writing 3 + a treats the repeated multiplication as an addition of 3 and a, which has no connection to the original expression. The expression that means a × a × a is a³.
- (a) −7 — The coefficient of xy is the number multiplying that exact term, including its sign: in 4x²y − 7xy + 2, the term in xy is −7xy, so the coefficient is −7. Answering 7 correctly picks out the number but drops the minus sign that belongs to it. Answering 4 picks the coefficient of the x²y term instead of the xy term — x²y and xy are different terms because their powers of x differ. Answering −5 combines −7 with the constant term +2, as though the coefficient of xy included the number with no letter attached to it. The coefficient of xy is −7.
- (c) 8 — Method: substitute x = 3 into each expression in turn, working out the bracket first in the expression that has one and the multiplication first in the expression that does not. Working: 2(x + 1) gives 2 × (3 + 1) = 2 × 4 = 8, and 2x + 2 gives 2 × 3 + 2 = 6 + 2 = 8, so both expressions take the value 8, as Grace claims. Answer: 8. The distractors: 7 comes from multiplying only the x inside the bracket and then adding the 1, giving 2 × 3 + 1; 10 comes from reading 2x + 2 as 2(x + 2) and working out 2 × 5; 6 comes from substituting into 2x and stopping before the 2 is added.
- (c) x = 8 — The turning point lies exactly halfway between the two roots. If the other root is r, the midpoint of −2 and r must be 3, so (−2 + r) ÷ 2 = 3, giving r = 8. Choosing x = 5 comes from adding 2 and 3 rather than using the midpoint relationship correctly. Choosing x = 1 comes from subtracting 2 from 3 instead of reflecting −2 across the turning point. Choosing x = −8 finds the right distance but then reflects in the y-axis instead of in the line of symmetry x = 3, so the sign of the answer is flipped.
- (a) (n − 5)/2 — Method: do the subtraction first, keep it together as a single bracket, then divide that whole bracket by 2. Working: 'subtract 5 from n' is (n − 5); 'divide the result by 2' means the whole bracket goes over 2, giving (n − 5)/2. Answer: (n − 5)/2. n/2 − 5 comes from dividing n by 2 first and only then subtracting 5, the wrong order. 2(n − 5) comes from multiplying by 2 instead of dividing. (5 − n)/2 comes from subtracting n from 5 instead of subtracting 5 from n, the wrong way round.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (c) w/6 — Method: a ÷ b is written as a fraction a/b, with the number being divided (w) on top. Working: w ÷ 6 = w/6. Answer: w/6. 6/w comes from writing the numbers the wrong way round, putting the 6 on top instead of w. 6w comes from reading the ÷ sign as ×, multiplying instead of dividing. w − 6 comes from reading ÷ as −, subtracting instead of dividing.
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