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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Algebra worksheet — GCSE Foundation
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- 1.Write down the expression that means 5 more than half of n.
- 2.A rule turns each input into an output. An input of 0 gives an output of −1, an input of 1 gives an output of 1, and an input of 2 gives an output of 3. Work out the rule, writing the input as x and the output as y.
- 3.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 4.Isla says that the lines y = 2x + 1 and y = −2x + 3 are parallel. Write down the statement that gives the correct answer for the correct reason.y = 2x + 1y = -2x + 3
- 5.y = 5x − 1. Work out the value of y when x = 3.y = 5x − 1
- 6.Solve x² − 2x − 24 = 0.
- 7.Solve 3x − 1 = 8
- 8.Make m the subject of the formula w = m − 8.
- 9.Point R has coordinates (3, −5). Point R is reflected in the line y = 2 to point S. Write down the coordinates of S.
- 10.p = 4. Work out the value of 2p³.
- 11.A number, w, satisfies the statement "three times w, minus 5, is at least 16". Which of these is the correct inequality for w?
- 12.Which of these statements is an identity?
- 13.The point A is at (5, −3). In which quadrant does A lie?
- 14.A conversion graph shows that 5 miles is equivalent to 8 kilometres. Use this to work out how many kilometres are equivalent to 20 miles.
- 15.Work out the coordinates of the midpoint of the line segment joining (−3, 5) and (7, 9).
Answer key
- (d) n/2 + 5 — Half of n is n ÷ 2, which is written as the fraction n/2. 'More than' means add, and the addition happens after the halving, so the expression is n/2 + 5. Writing (n + 5)/2 halves the 5 as well, because everything inside a bracket is divided; writing 2n + 5 doubles n instead of halving it; writing 5n/2 multiplies half of n by 5 instead of adding 5 to it.
- (b) y = 2x − 1 — Method: in a rule that multiplies and then adds, the multiplier is the step in the outputs for each step of 1 in the input, and the number added on is the output when the input is 0. Working: the inputs 0, 1, 2 rise in ones while the outputs −1, 1, 3 rise by 2 each time, so the input is multiplied by 2; an input of 0 gives 2 × 0 = 0 and the output must be −1, so 1 is subtracted. Answer: y = 2x − 1, checked against the last pair by 2 × 2 − 1 = 3. The distractors: y = 2x + 1 comes from finding the multiplier 2 correctly and then reading the output at an input of 0 as +1 instead of −1; y = x − 1 comes from taking the multiplier as 1 because the inputs go up in ones, instead of using the step in the outputs; y = 3x − 1 comes from reading the largest output, 3, as the multiplier.
- (b) 28 — Method: find the common difference, then use the position-to-term rule (or extend the sequence) to reach the 9th term. Working: the common difference is 7 − 4 = 3, so the nth term is 4 + 3(n − 1). For n = 9: 4 + 3 × 8 = 4 + 24 = 28. Answer: 28. 27 comes from using 3n instead of 3n + 1, dropping the constant term from the rule, 3 × 9 = 27. 31 comes from extending from the 5th term by one step too many, adding 3 five times instead of four, 16 + 3 × 5. 25 comes from extending by one step too few, adding 3 three times instead of four, 16 + 3 × 3.
- (b) No, because their gradients are 2 and −2 — Method: two lines are parallel exactly when their gradients are equal as signed numbers, so m is read from each equation written in the form y = mx + c and the two are compared. Working: y = 2x + 1 has gradient 2 and y = −2x + 3 has gradient −2; those are not equal, so the lines are not parallel, and indeed one slopes upwards while the other slopes downwards. Answer: No, because their gradients are 2 and −2. The distractors: saying yes because both gradients have size 2 comes from comparing the sizes of the gradients and ignoring their signs; saying yes because the gradients add to 0 comes from using a sum of zero as the test for parallel lines instead of equality of gradients; saying no because the y-intercepts are 1 and 3 reaches the right verdict by the wrong route, since the intercepts decide where the lines sit rather than whether they are parallel.
- (a) 14 — Substituting x = 3: y = 5(3) − 1 = 15 − 1 = 14. A candidate who stops after the multiplication and forgets to subtract 1 gets 15. A candidate who adds 1 instead of subtracting gets 16. A candidate who incorrectly treats the expression as 5 × (3 − 1) gets 10.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (d) x = 3 — Method: undo the operations on the left in reverse order — deal with the subtraction of 1 first, then with the multiplication by 3. Working: adding 1 to both sides gives 3x = 9, and dividing both sides by 3 gives x = 3. Answer: x = 3. The distractors: x = 9 comes from stopping at 3x = 9 and writing 9 as the value of x; x = 6 comes from subtracting 3 from 9 instead of dividing by 3; x = 27 comes from multiplying 9 by 3 instead of dividing by 3.
- (b) m = w + 8 — To make m the subject of w = m − 8, add 8 to both sides so the subtraction is undone: m = w + 8. Writing m = w − 8 forgets to change the operation at all — it just relabels the equation without moving the 8 across. Writing m = 8 − w swaps the order and puts the wrong sign on w, as if the equation had been w = 8 − m instead. Writing m = w/8 mistakes subtraction for a scaling relationship and divides by 8 instead of adding it. The correct rearrangement is m = w + 8.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (d) 128 — In 2p³ the index belongs to p only, so cube p first and multiply by the coefficient afterwards. Cubing gives 4 × 4 × 4 = 64, and then 2 × 64 = 128. Cubing the coefficient as well would mean working out (2 × 4)³, which is 512. Reading the index as an instruction to multiply by 3 gives 2 × 4 × 3 = 24, and ignoring the coefficient altogether leaves 64.
- (b) 3w − 5 ≥ 16 — "Three times w, minus 5" translates to 3w − 5, and "is at least 16" means it must be 16 or more, giving 3w − 5 ≥ 16. A candidate who reads "at least" as a strict inequality writes 3w − 5 > 16. A candidate who misreads the wording and applies the subtraction before the multiplication writes 3(w − 5) ≥ 16. A candidate who reverses the direction of the inequality writes 3w − 5 ≤ 16.
- (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.
- (c) the fourth quadrant — Method: read the sign of each coordinate in turn, then count round the regions, which are numbered anticlockwise from the one where both coordinates are positive. Working: the x-coordinate 5 is positive, so A is to the right of the y-axis; the y-coordinate −3 is negative, so A is below the x-axis; right of the y-axis and below the x-axis is the fourth of the four regions. Answer: the fourth quadrant. The distractors: 'the first quadrant' comes from ignoring the minus sign on the y-coordinate and treating the point as (5, 3); 'the second quadrant' comes from writing the pair the wrong way round and locating (−3, 5) instead; 'the third quadrant' comes from assuming that any point with a negative coordinate belongs to the region where the minus signs are, without checking the other coordinate.
- (b) 32 km — Method: work out the scale factor connecting the two mile values: 20 miles is 4 times 5 miles (20 ÷ 5 = 4), so multiply the km value by 4 too: 8 × 4 = 32 km. Distractor origins: 24 km comes from subtracting the mile values (20 − 5 = 15), dividing by 5 to get a wrong scale factor of 3, then multiplying 8 × 3; 23 km comes from adding the difference in miles, 15, directly to the km value, 8; 16 km comes from using a scale factor of 2, doubling the km value once instead of scaling by the correct factor of 4.
- (a) (2, 7) — Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((−3+7)/2, (5+9)/2) = (4/2, 14/2) = (2, 7). (4, 14) comes from adding the coordinates correctly but forgetting to divide by 2. (2, 9) comes from correctly averaging the x-coordinates but simply copying the y-coordinate of the second point instead of averaging the y-coordinates. (5, 2) comes from subtracting the coordinates instead of adding them before halving: ((7−(−3))/2, (9−5)/2) = (5, 2).
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