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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- (c) 3 — Gradient = (change in y) ÷ (change in x) = (11 − 5) ÷ (4 − 2) = 6 ÷ 2 = 3. A candidate who puts the change in x over the change in y instead would get 2 ÷ 6 = 1/3. A candidate who subtracts the y-coordinates in the reverse order, but not the x-coordinates, would get (5 − 11) ÷ (4 − 2) = −3. A candidate who adds the coordinates instead of subtracting them would get (11 + 5) ÷ (4 + 2) = 16/6 = 8/3.
- (a) a = 20 − b — Method: get a on its own by adding a to both sides, then subtracting b from both sides. Working: b = 20 − a, so b + a = 20, so a = 20 − b. Answer: a = 20 − b. a = b − 20 comes from treating the formula as if it read b = 20 + a and then subtracting 20 from both sides. a = 20 + b comes from moving b across from b + a = 20 without changing its sign. a = −20 − b comes from a double sign error, changing the sign of the 20 as well as of b when rearranging.
- (d) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (c) x = 2y + 10 — Method: undo the operations done to x in reverse order — add 5, then multiply by 2. Working: y = x/2 − 5, so y + 5 = x/2, so x = 2(y + 5) = 2y + 10. Answer: x = 2y + 10. x = 2y + 5 comes from multiplying only the x/2 term by 2 and forgetting to multiply the 5 as well. x = 2y − 10 comes from a sign error, subtracting 5 instead of adding it before multiplying by 2. x = (y + 5)/2 comes from dividing by 2 instead of multiplying, the wrong operation to undo a division.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (d) 1 hour 32 minutes — T = 40 × 1.8 + 20 = 72 + 20 = 92 minutes. Since 92 = 60 + 32, the cooking time is 1 hour 32 minutes. A candidate who rounds the mass to 2 kg before substituting gets 40 × 2 + 20 = 100 minutes = 1 hour 40 minutes. A candidate who forgets to add the 20 minutes gets 40 × 1.8 = 72 minutes = 1 hour 12 minutes. A candidate who multiplies the mass by (40 + 20) = 60 instead of substituting into the formula gets 1.8 × 60 = 108 minutes = 1 hour 48 minutes.
- (b) x = 0 — Method: the first coordinate of a point measures how far to the left or right of the origin it is, so a point that is neither left nor right of the origin has a first coordinate of zero. Working: the y-axis is the vertical line through the origin; every point on it lies directly above or directly below the origin with no sideways movement at all, so its first coordinate is zero while its second coordinate may take any value. Answer: x = 0. The distractors: y = 0 is the condition for lying on the x-axis, the other axis; x = 1 comes from confusing the y-axis with the vertical line one unit to the right of it; x = y is the condition for the diagonal through the origin, which meets the y-axis at the origin only.
- (b) 4, 12, 36, 108 — Method: multiply the previous term by 3 each time, starting from the first term. Working: 4 × 3 = 12, 12 × 3 = 36, 36 × 3 = 108. Answer: 4, 12, 36, 108. 4, 7, 10, 13 comes from adding 3 each time instead of multiplying by 3, confusing this with an arithmetic sequence. 4, 12, 15, 18 comes from multiplying correctly to get the second term, then switching to adding 3 for the rest. 12, 36, 108, 324 comes from listing the terms after the first term, missing off the starting value of 4.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (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.
- (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.
- (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
- (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.
- (d) It is n-shaped, since the x² coefficient is negative. — The coefficient of x² is −2, which is negative, so the quadratic curve opens downward — shaped like an n, with a maximum turning point. Saying it is U-shaped focuses only on x² being non-negative and ignores that the −2 in front of it flips the whole curve to open downward. Saying it is a straight line confuses having a constant term with being linear — any equation with an x² term is a curve, not a line. Saying it repeatedly rises and falls like a wave describes a trigonometric graph such as y = sin x, not a quadratic.
- (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.
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