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 5a + 2 when a = 3
- 2.Work out the coordinates of the reflection of (6, −3) in the y-axis.
- 3.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 4.Work out the value of 5y + 9 when y = 0
- 5.There are 50 adults on a coach to the Lake District. Every adult is either a man or a woman. There are 10 more men than women. Work out the number of men.
- 6.a = 2 and b = 5. Work out the value of a²b.
- 7.The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4. Work out the x-coordinate of the turning point of the curve, using the symmetry of the graph.y = x² − 2x − 8
- 8.A taxi firm charges a fixed fee of £3.50 plus £2.20 per mile. Work out the total cost of a journey of 6 miles.
- 9.Work out the value of m² + 3m when m = −2
- 10.Work out the value of 2(x + 2) when x = −5
- 11.Make c the subject of the formula k = 6c.
- 12.Make x the subject of the formula y = 3(x + 2).
- 13.The graph of y = mx + 6 passes through the point (−2, 0). Work out the value of m.
- 14.Which expression is equivalent to 3(x + 4) − 2(x − 1)?
- 15.A straight line has equation 4y = 12x + 20. Work out the gradient of the line.y = 12x + 20
Answer key
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (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).
- (d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
- (b) 20 — a²b means a² multiplied by b: a² = 2² = 4, and 4 × 5 = 20. Reading the expression as (ab)² instead of a²b gives (2 × 5)² = 100. Squaring b instead of a, 2 × 5² = 50, squares the wrong letter. Adding a² and b instead of multiplying them, 2² + 5 = 9, uses the wrong operation.
- (d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (d) −2 — Method: substitute the value into both terms, working out the index and the multiplication before the addition. Working: m² = (−2) × (−2) = 4 and 3m = 3 × (−2) = −6, so the expression becomes 4 + (−6), which is −2. Answer: −2. The distractors: −10 comes from squaring −2 as −4, giving −4 + (−6); 10 comes from working out 3m as +6 and losing the minus sign, giving 4 + 6; −14 comes from working from left to right instead of multiplying first, giving (4 + 3) × (−2).
- (c) −6 — Method: substitute the value, work out the bracket first, then multiply what the bracket comes to by 2. Working: the bracket gives (−5) + 2 = −3, and multiplying by 2 gives 2 × (−3) = −6. Answer: −6. The distractors: −3 comes from working out the bracket correctly but stopping there and never multiplying by 2; −1 comes from multiplying the 2 inside the bracket by the 2 outside before adding, giving −5 + 4; −8 comes from multiplying only the x by 2 and leaving the 2 inside the bracket unchanged, giving −10 + 2.
- (b) c = k/6 — k = 6c means c has been multiplied by 6, so to make c the subject, divide both sides by 6: c = k/6. Writing c = 6k multiplies by 6 again instead of undoing the multiplication. Writing c = k − 6 mistakes multiplying for adding, and subtracts 6 rather than dividing. Writing c = 6 − k reverses the order as well as the operation. The correct rearrangement is c = k/6.
- (b) x = y/3 − 2 — The bracket containing x has been multiplied by 3, so divide both sides by 3 first, which gives y/3 = x + 2. Subtracting 2 from both sides then leaves y/3 − 2 = x, so x = y/3 − 2. Taking the 2 away before dividing gives (y − 2)/3, which divides the 2 by 3 as well, although the 2 was never divided in the original formula. Adding 2 rather than subtracting it gives y/3 + 2, and multiplying by 3 instead of dividing gives 3y − 2.
- (c) 3 — Method: a point that lies on a graph makes its equation true, so substituting the coordinates into y = mx + 6 leaves an equation in m alone. Working: substituting x = −2 and y = 0 gives 0 = m × (−2) + 6, which rearranges to −2m = −6, so m = (−6) ÷ (−2) = 3. Answer: 3. The distractors: −3 comes from dividing −6 by 2 and keeping the minus sign, overlooking that the divisor is negative too, so the two signs cancel; −2 comes from writing down the x-coordinate of the given point in place of the gradient; 6 comes from reading the constant in y = mx + 6 as the gradient, confusing m with c.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
- (b) 3 — Method: rearrange the equation into the form y = mx + c first, then read off the gradient. Working: dividing 4y = 12x + 20 by 4 gives y = 3x + 5, so the gradient is 3. The value 12 comes from reading the coefficient of x before dividing the equation by 4. The value 20 comes from using the constant term of the unsimplified equation instead of the gradient. The value 5 comes from finding the y-intercept, 20 / 4 = 5, and giving that instead of the gradient.
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