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.
Answer key: Algebra worksheet — GCSE Foundation
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- (d) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
- (d) m³n — m × m × m × n means three m's multiplied together, then multiplied by n, which is written as m³n. Putting the power of 3 on n instead of m, mn³, puts the power on the wrong letter. Writing 3mn treats the repeated multiplication as if it were repeated addition, as if it meant 3 × m × n. Writing m³ + n adds the n instead of multiplying it in.
- (d) 15 — Substituting x = 3 into y = 5x gives y = 5 × 3 = 15. A candidate who adds instead of multiplying would get 5 + 3 = 8. A candidate who ignores the coefficient 5 and just copies the value of x would write y = 3. A candidate who multiplies correctly but treats x as −3 by mistake would get y = 5 × (−3) = −15.
- (c) x = 5 — Method: every point on a vertical line has the same x-coordinate however far up or down the line it lies, so the equation of a vertical line fixes x at a number and does not involve y at all. Working: of the four equations only x = 5 fixes x; it is satisfied by (5, 0), (5, 1), (5, 7) and by every other point whose x-coordinate is 5, and those points form a vertical line. Answer: x = 5. The distractors: y = 5 fixes the y-coordinate instead of the x-coordinate, which gives a horizontal line; y = 5x is a line through the origin with gradient 5, steep but not vertical, and it has a different y-value for every x; x = y fixes neither coordinate and is the line through the origin with gradient 1.
- (a) 6c — Each box has 6 chocolates, so c boxes have 6 × c = 6c chocolates. A candidate who adds the number of boxes to the number per box instead of multiplying gets 6 + c. A candidate who subtracts instead of multiplying gets c − 6. A candidate who divides instead of multiplying gets c ÷ 6.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
- (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) 3 + 2m — Method: add the fixed fee to the cost of the miles travelled. Working: the miles cost 2 × m = 2m pounds, and the fixed fee is £3, so the total is 3 + 2m. Answer: 3 + 2m. 2 + 3m comes from swapping which number is the fixed fee and which is the per-mile rate. 5 + m comes from adding the fixed fee and the rate together into one number, 3 + 2 = 5, before adding the miles, instead of multiplying the rate by the number of miles. 5m comes from multiplying that combined total by the miles, treating the whole £5 as if it were a per-mile rate.
- (b) h = 2A/b — The height has been multiplied by the base and the result then divided by 2, so undo the division first. Multiplying both sides by 2 gives 2A = bh. Undoing the multiplication by the base comes next: dividing both sides by b gives 2A/b = h, so h = 2A/b. Dividing by 2 instead of multiplying gives A/(2b), a quarter of the correct height; multiplying by the base instead of dividing gives 2Ab; writing b/(2A) turns the final fraction upside down.
- (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.
- (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (d) x = 6 — Method: with an unknown on both sides, first collect the x terms on one side by subtracting the smaller x term from both sides, then deal with the numbers. Working: subtracting x from both sides gives x + 7 = 13, and subtracting 7 from both sides gives x = 6. Answer: x = 6. The distractors: x = 20 comes from adding 7 to 13 instead of subtracting it once the x terms have been collected; x = 2 comes from collecting the x terms by adding them, giving 3x + 7 = 13 and then 3x = 6; x = −6 comes from subtracting 2x from both sides to get 7 = −x + 13, reaching −6 = −x and then copying the sign straight across instead of dividing by −1.
- (d) $y = 2x - 1$ — Method: pick two points the line passes through, work out the gradient as vertical change ÷ horizontal change, and read the y-intercept from where the line crosses the y-axis. Working: the line passes through (0, −1) and (1, 1), so the gradient is (1 − (−1)) ÷ (1 − 0) = 2, and it crosses the y-axis at −1. Answer: y = 2x − 1. Distractor refutation: y = 2x + 1 comes from reading the y-intercept as +1 instead of −1, misreading which side of the origin the line crosses. y = x − 1 comes from taking the gradient as 1, counting the same number of squares across and up instead of checking the rise is twice the run. y = −2x − 1 comes from a sign error on the gradient, treating the line as sloping downward from left to right when it actually rises.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
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