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) s/f — Sharing s sweets equally between f friends means dividing the total by the number of friends, written as a fraction: s/f. Writing f/s divides the wrong way round, sharing the number of friends between the sweets instead of the sweets between the friends. Writing s − f mistakes sharing for taking away, subtracting the number of friends from the number of sweets. Writing sf multiplies the two quantities together, which would make the total larger rather than splitting it into smaller equal parts. The number of sweets each friend receives is s/f.
- (c) (6n + 4)/2 — Six full boxes hold 6 lots of n pencils, which is 6n, and the 4 loose pencils are added on, so the shop has 6n + 4 pencils altogether. Sharing them equally between 2 classes divides that whole total by 2, and brackets are what show that the division applies to all of it: (6n + 4)/2. Without the brackets, 6n + 4/2 halves only the loose pencils; 6(n + 4)/2 adds the loose pencils to every box before the division; 2(6n + 4) doubles the total instead of halving it.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (c) x = 4 and x = −1, because the graph crosses the x-axis where x − 4 = 0 or x + 1 = 0. — The graph crosses the x-axis where y = 0, which happens when either bracket equals zero. Solving x − 4 = 0 gives x = 4, and solving x + 1 = 0 gives x = −1. The option x = −4 and x = 1 incorrectly reverses both signs. The option x = 4 and x = 1 misreads the second bracket, ignoring that x + 1 = 0 requires x to be negative. The option x = −4 and x = −1 wrongly assumes both roots must be the negative of the constants shown, which only matches the second bracket, not the first.
- (c) (2, 6) — Method: substitute the x-coordinate of each point into the rule for the line (y = 3 × x) and compare it with the point's y-coordinate. Working: the line passes through the origin and rises 3 squares for every 1 square across, so at x = 2 the line's y-value is 3 × 2 = 6, giving the point (2, 6). Answer: (2, 6). Distractor refutation: (2, 3) comes from counting only 3 squares up in total between the origin and x = 2, instead of 3 squares up for every 1 square across, halving the true rise. (3, 2) comes from swapping the x-coordinate and the y-coordinate round. (2, 5) comes from a miscounted gridline, landing one square below the line.
- (c) y = 9x + 3 — Method: in a linear cost model the charge that is paid once is the constant term and the charge made for each item is the coefficient of x. Working: each pizza costs £9, so x pizzas cost 9x pounds; the £3 delivery fee is paid once whatever the value of x, so it is added on as a constant and the total cost is y = 9x + 3. Answer: y = 9x + 3. The distractors: y = 3x + 9 comes from swapping the two charges over, treating the £3 delivery as a per-pizza cost and the £9 as the one-off charge; y = 9x leaves out the delivery fee altogether, which undercharges every order by £3; y = 12x comes from adding the delivery fee to the price of each pizza, £9 + £3 = £12 each, so the fee is charged once per pizza instead of once per order.
- (b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.
- (c) Second and fourth — If x + y = 0 then y = −x, so x and y always have opposite signs, one positive and one negative. A point with a negative x and a positive y lies in the second quadrant, and a point with a positive x and a negative y lies in the fourth quadrant, so the point lies in the second or the fourth. A candidate who reads x + y = 0 as meaning x and y have the same sign picks First and third, which is where x × y is positive, not where x + y = 0. A candidate who decides that y must be the positive coordinate picks the two quadrants above the x-axis, First and second. A candidate who decides that y must be the negative coordinate picks the two quadrants below the x-axis, Third and fourth.
- (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.
- (c) 100 — Method: apply the position-to-term rule straight to the position asked for, since it does not depend on the term before it. Working: the rule squares the position number and the position is 10, so the term is 10², which means 10 × 10. Answer: 100. The distractors: 81 comes from squaring 9 instead of 10, one position short; 20 comes from multiplying the position by 2 instead of squaring it; 1000 comes from cubing the position, 10 × 10 × 10, instead of squaring it.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (d) 68, and it is a formula — Substituting C = 20 gives F = 1.8 × 20 + 32 = 36 + 32 = 68. The statement F = 1.8C + 32 is a formula because it expresses a general relationship between F and C that holds for any value of C, not just C = 20. A candidate who gets the correct value but calls it an equation is confusing "has an equals sign" with "is an equation" — an equation is only true for particular value(s) of the unknown, whereas this relationship is used to calculate rather than to be solved. A candidate who forgets to multiply C by 1.8 and just adds 32 to 20 gets 52. A candidate who multiplies correctly but forgets to add 32 gets 36.
- (c) x = 4 — Method: undo the addition of 1 first, then undo the multiplication by 2. Working: subtracting 1 from both sides gives 2x = 8, and dividing both sides by 2 gives x = 4. Answer: x = 4. The distractors: x = 8 comes from stopping at 2x = 8 and writing 8 as the value of x; x = 5 comes from adding 1 to both sides instead of subtracting it, giving 2x = 10; x = 16 comes from multiplying 8 by 2 instead of dividing by 2.
- (c) 2(x + 4) ≡ 2x + 8 — 2(x + 4) ≡ 2x + 8 is an identity because expanding the bracket on the left gives exactly the right-hand side for every value of x — try any number and both sides match. 2(x + 4) = 20 is an equation: expanding gives 2x + 8 = 20, which is only true for the single value x = 6. 2(x + 4) = 2x + 4 is not true for any value of x at all: expanding the left side gives 2x + 8, which can never equal 2x + 4 since 8 ≠ 4. x + 4 = 2x is also an equation, true only for the single value x = 4. The identity is 2(x + 4) ≡ 2x + 8.
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