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.
Algebra worksheet — GCSE Foundation
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- 1.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.
- 2.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 3.Solve 3x − 1 = 8
- 4.Three vertices of a rectangle are (−4, −1), (2, −1) and (2, 3). The sides of the rectangle are parallel to the axes. Write down the coordinates of the fourth vertex.
- 5.Write down the gradient of the line with equation y = −x + 3.y = −x + 3
- 6.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 7.The volume of a cuboid is given by the formula V = lwh. Work out V, in cm³, when l = 2, w = 3 and h = 6.
- 8.The point A lies on the x-axis. Which statement about A must be true?
- 9.Solve 6x − 5 = 3x + 13.
- 10.The first four terms of a sequence are 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 11.Work out the gradient of the straight line with equation 2x + y = 8.
- 12.Simplify 2b − 1 + 2b − 4 + 2b
- 13.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 14.Make x the subject of the formula y = 3(x + 2).
- 15.A number machine adds 1 to its input. Isla says that an input of 5 gives an output of 6. Which statement is correct?
Answer key
- (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) 11 — Substitute n=6 into 2n−1: 2×6−1=11. A candidate who adds 2 and 6 and then subtracts 1, instead of multiplying 2 by 6 first, would compute 2+6−1=7. A candidate who substitutes the wrong term number, n=5, would reach 2×5−1=9. A candidate who forgets to subtract 1 would compute just 2×6=12.
- (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) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (b) −1 — Method: in the form y = mx + c the gradient m is the number multiplying x, and an x term written with no number in front of it has a coefficient of 1. Working: y = −x + 3 is the same equation as y = (−1)x + 3, so comparing it with y = mx + c gives m = −1 and c = 3. Answer: −1. The distractors: 1 comes from taking the coefficient as 1 and leaving the minus sign out of the answer; 3 comes from reading the constant as the gradient, confusing m with c; −3 comes from moving the minus sign across to the constant term and then reading the gradient off that term instead of the x term.
- (d) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
- (d) 36 cm³ — V = lwh = 2 × 3 × 6 = 36 cm³. A candidate who adds all three numbers instead of multiplying gets 2 + 3 + 6 = 11 cm³. A candidate who multiplies only two of the three numbers, forgetting the length, gets w × h = 3 × 6 = 18 cm³. A candidate who multiplies l × w and w × h separately and adds the two products gets (2 × 3) + (3 × 6) = 6 + 18 = 24 cm³.
- (a) y = 0 — Method: in the pair (x, y) the first coordinate measures how far left or right of the origin a point is and the second how far above or below the x-axis it is, so a point on an axis has one of those measurements equal to zero. Working: the x-axis is the horizontal line through the origin, so a point sitting on it is neither above nor below that line and its second coordinate is zero, while its first coordinate may be positive, negative or zero. Answer: y = 0. The distractors: x = 0 is the condition for lying on the y-axis, the other axis; x > 0 comes from assuming a point on the x-axis must be to the right of the origin, which is true only of part of that axis; x = y holds only at the origin, which is one point of the x-axis rather than a property shared by all of them.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
- (d) −2 — Method: a gradient can only be read off an equation that is written in the form y = mx + c, so y has to be made the subject first. Working: subtracting 2x from both sides of 2x + y = 8 gives y = −2x + 8, and comparing that with y = mx + c gives m = −2. Answer: −2. The distractors: 2 comes from reading the coefficient of x straight off the equation as it is printed, without rearranging, so the change of sign is missed; 8 comes from reading the constant as the gradient, confusing m with c; −1/2 comes from rearranging correctly and then writing the gradient upside down, as the change in x over the change in y.
- (a) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (b) x = y/3 − 2 — The bracket containing x has been multiplied by 3, so divide both sides by 3 first, which gives y/3 = x + 2. Subtracting 2 from both sides then leaves y/3 − 2 = x, so x = y/3 − 2. Taking the 2 away before dividing gives (y − 2)/3, which divides the 2 by 3 as well, although the 2 was never divided in the original formula. Adding 2 rather than subtracting it gives y/3 + 2, and multiplying by 3 instead of dividing gives 3y − 2.
- (d) Isla is right, because 5 + 1 = 6 — Method: put the input through the machine yourself, then compare what comes out with the output claimed, and check that the reason given with the verdict is itself true. Working: the machine adds 1 to whatever is put in, so an input of 5 gives 5 + 1 = 6, which is exactly the output claimed. Answer: Isla is right, and the reason is that 5 + 1 = 6. The distractors: the statement that the machine adds 1 to its output describes a different machine — this one adds 1 to what goes in — so the verdict is right but the reason is false; 5 − 1 = 4 comes from running the machine backwards and subtracting instead of adding; 5 × 1 = 5 comes from reading ‘adds 1’ as ‘multiplies by 1’, which leaves the input unchanged.
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