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.Work out the value of n² − n when n = −3
- 2.The point P has coordinates (0, 12). Write down the axis that P lies on and the y-coordinate of P.
- 3.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
- 4.A tank already contains some water and is then filled at a constant rate. After 2 minutes it contains 50 litres in total; after 7 minutes it contains 150 litres in total. Work out the rate at which water is added, in litres per minute.
- 5.Make b the subject of the formula a = 5b.
- 6.A parcel delivery company's cost graph shows the following: the cost is a flat £5 for parcels weighing up to 2 kg, and then rises by £2 for each additional kg above 2 kg. Use the graph to work out the weight of a parcel that costs £17.
- 7.Write down the gradient of the line with equation y = 7x − 2.y = 7x − 2
- 8.Solve 2(x + 3) = 10
- 9.A car passes the 15 km marker on a road and passes the 195 km marker 3 hours later. Work out the gradient of the distance-time graph, that is the rate of change of distance with time.
- 10.Which expression means 'add 3 to n, then multiply the result by 4'?
- 11.A geometric sequence begins 162, 54, 18, 6, ... Work out the next term in the sequence.
- 12.Solve 2x − 2 = 2
- 13.Which of these statements is an identity?
- 14.y = 5 − 2x. Work out the value of x when y = 11.
- 15.Solve the inequality 9 − 2x ≥ 1.
Answer key
- (d) 12 — Method: substitute the value into both terms, remembering that subtracting a negative number has the same effect as adding the matching positive number. Working: n² = (−3) × (−3) = 9, and subtracting n means subtracting −3, which adds 3, so the calculation is 9 + 3 = 12. Answer: 12. The distractors: 6 comes from subtracting 3 rather than subtracting −3, giving 9 − 3; −6 comes from squaring −3 as −9 while still adding the 3, giving −9 + 3; −3 comes from reading n² as 2n, giving 2 × (−3) = −6 and then −6 + 3.
- (d) the y-axis, and 12 — Method: coordinates are written (x, y), so the first number is the distance across and the second the distance up; a point whose first coordinate is 0 has not moved across from the origin and therefore lies on the vertical axis. Working: in (0, 12) the first number is 0, so P is on the y-axis, and the second number, 12, is the y-coordinate of P. Answer: the y-axis, and 12. The distractors: 'the x-axis, and 12' comes from mixing up which axis the condition 'the first coordinate is 0' describes; 'the y-axis, and 0' comes from placing P correctly but reading the pair the wrong way round, so that the first number is quoted as the y-coordinate; 'the x-axis, and 0' comes from making both of those mistakes at once.
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (b) b = a/5 — Writing 5b means 5 × b, so b has been multiplied by 5. The inverse of multiplying by 5 is dividing by 5, and dividing both sides by 5 gives a/5 = b, which is written b = a/5. Multiplying both sides by 5 instead gives 5a, which applies the operation a second time; treating the 5 as though it were added gives a − 5; writing 5/a turns the fraction upside down.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (b) 7 — Method: compare the equation with y = mx + c, where m is the gradient. Working: in y = 7x − 2, the coefficient of x is 7. Answer: the gradient is 7. −2 comes from confusing the gradient with the y-intercept. −7 comes from a sign error, treating the coefficient of x as negative. 2 comes from using the y-intercept but dropping its sign.
- (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.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (c) 4(n + 3) — 'Add 3 to n' must happen before 'multiply the result by 4', so the addition needs brackets to show it happens first: 4(n + 3). Writing 4n + 3 multiplies n by 4 immediately and only adds the 3 afterwards, which reverses the order the words describe. Writing n + 3 × 4 only multiplies the 3 by 4, and adds n as a separate, unmultiplied term — it treats 'the result' as just the 3, not the whole of n + 3. Writing 3(n + 4) keeps the correct structure but swaps which number is added and which is multiplied. The expression for 'add 3 to n, then multiply the result by 4' is 4(n + 3).
- (a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
- (a) x = 2 — Method: undo the subtraction first, then undo the multiplication, doing the same to both sides each time. Working: adding 2 to both sides gives 2x = 4, and dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 4 comes from stopping at 2x = 4 and writing 4 as the value of x; x = 0 comes from subtracting 2 from both sides instead of adding it, giving 2x = 0; x = 8 comes from multiplying 4 by 2 instead of dividing by 2.
- (c) 2(x + 4) ≡ 2x + 8 — 2(x + 4) ≡ 2x + 8 is an identity because expanding the bracket on the left gives exactly the right-hand side for every value of x — try any number and both sides match. 2(x + 4) = 20 is an equation: expanding gives 2x + 8 = 20, which is only true for the single value x = 6. 2(x + 4) = 2x + 4 is not true for any value of x at all: expanding the left side gives 2x + 8, which can never equal 2x + 4 since 8 ≠ 4. x + 4 = 2x is also an equation, true only for the single value x = 4. The identity is 2(x + 4) ≡ 2x + 8.
- (d) −3 — Substitute y = 11 into y = 5 − 2x, giving 11 = 5 − 2x. Subtracting 5 from both sides gives 6 = −2x, so x = 6 ÷ (−2) = −3. A candidate who mishandles the negative sign when rearranging, treating the equation as 6 = 2x, gets x = 3. A candidate who correctly finds −2x = 6 but forgets to divide by 2 at all gets x = 6. A candidate who adds 5 and 11 instead of subtracting, getting 2x = 16, gets x = 8.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
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