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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- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (c) 22 — p² means p × p. Substitute p = 2.5: p² = 6.25, so 4p² = 4 × 6.25 = 25, and 25 − 3 = 22. 97 comes from squaring 4p together instead of just p, (4 × 2.5)² − 3 = 10² − 3 = 97. 7 comes from using p instead of p², 4 × 2.5 − 3. 1 comes from squaring the 3 instead of the p, 4 × 2.5 − 3².
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (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) 5 — Method: call the number of years ago t, take t off both ages, and form an equation from the comparison at that time. Working: t years ago the father was 40 − t and Ethan was 10 − t, so 40 − t = 7(10 − t); expanding gives 40 − t = 70 − 7t, adding 7t to both sides gives 40 + 6t = 70, subtracting 40 gives 6t = 30, and dividing by 6 gives t = 5. Checking: five years ago the father was 35 and Ethan was 5, and 35 = 7 × 5. Answer: 5. The distractors: 35 comes from finding the right moment but giving the father's age at that time instead of the number of years; 30 comes from using Ethan's present age on the right-hand side, 40 − t = 7 × 10, which gives t = −30 and is then written as 30; 24 comes from reaching 6t = 30 correctly and subtracting 6 instead of dividing by 6.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (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.
- (d) 5 — Method: rearrange the formula to make m the subject, then substitute F = 15.50. Working: F = 3 + 2.5m, so subtracting 3 from both sides gives F − 3 = 2.5m, then dividing by 2.5 gives m = (F − 3) / 2.5. Substituting F = 15.50: m = (15.50 − 3) / 2.5 = 12.50 / 2.5 = 5. The value 6.2 comes from dividing 15.50 by 2.5 without subtracting the fixed £3 first. The value 3.2 comes from dividing first and subtracting afterwards, in the wrong order: (15.50 / 2.5) − 3 = 3.2. The value 7.4 comes from adding £3 instead of subtracting it before dividing: (15.50 + 3) / 2.5 = 7.4.
- (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.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
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