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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Answer key: Algebra worksheet — GCSE Foundation
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- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
- (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.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (c) 3x + 5 = 17 — An equation contains an equals sign and is true only for particular value(s) of the unknown — solving 3x + 5 = 17 gives the single value x = 4. 3x + 5 is an expression: it has no equals sign, so it cannot be solved, only simplified or evaluated. A = πr² is a formula: it shows the general relationship between different quantities (area and radius), rather than asking for one unknown value. 3(x + 5) ≡ 3x + 15 is an identity: the ≡ sign shows it is true for every value of x, not just one. The equation is 3x + 5 = 17.
- (c) No — that moment has already passed — Method: call the number of years from now x, add x to both ages, form the equation from the comparison and then interpret the value of x that comes out. Working: in x years Harry will be 14 + x and Mia will be 8 + x, so 14 + x = 2(8 + x); expanding gives 14 + x = 16 + 2x, and subtracting x and 16 from both sides gives x = −2. A negative value of x places the moment two years in the past, when Harry was 12 and Mia was 6 and 12 = 2 × 6, so it is not something still to come. Answer: no — that moment has already passed. The distractors: the claim that it has never happened and never will comes from reaching x = −2 and reading a negative number of years as no solution at all, when x = −2 does not say that no such moment exists but says where it is — two years before now; the claim that it happens when Harry is 16 comes from doubling Mia's present age, 2 × 8 = 16, and reading that as the age Harry has to reach; the claim that it happens when Harry is 20 comes from expanding 2(8 + x) as 8 + 2x, doubling only the x, which gives x = 6.
- (a) 12 — Method: write both ages as they were six years ago, call Leo's present age x, and form an equation from the comparison at that time. Working: six years ago Harry was 24 − 6 = 18 and Leo was x − 6, so 18 = 3(x − 6); expanding gives 18 = 3x − 18, adding 18 to both sides gives 36 = 3x, and dividing by 3 gives x = 12. Checking: six years ago Harry was 18 and Leo was 6, and 18 = 3 × 6. Answer: 12. The distractors: 6 comes from solving 18 = 3(x − 6) as far as Leo's age six years ago, 18 ÷ 3 = 6, and giving that as his age now; 8 comes from dividing Harry's present age by 3, 24 ÷ 3 = 8, using the multiple at the wrong moment in time; 36 comes from stopping at 3x = 36 and giving 36 as Leo's age.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
- (a) 1.8 — E = (1/2)kx² = 0.5 × 40 × 0.3² = 0.5 × 40 × 0.09 = 1.8 joules. 3.6 comes from forgetting the (1/2) at the front, 40 × 0.09. 72 comes from squaring kx together instead of squaring only x, 0.5 × (40 × 0.3)² = 0.5 × 144 = 72. 6 comes from using x instead of x², 0.5 × 40 × 0.3.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (a) 4 — Rectangle 1: P = 2(9 + 4) = 2 × 13 = 26 cm. Rectangle 2: P = 2(6 + 5) = 2 × 11 = 22 cm. Difference = 26 − 22 = 4 cm. 2 comes from finding the difference between (l + w) for each rectangle, 13 − 11 = 2, but forgetting to double it. 26 comes from giving rectangle 1's whole perimeter instead of the difference between the two. 6 comes from doubling only the change in length, 2 × (9 − 6), and ignoring that the width also changed.
- (b) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
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