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.Yuki is y years old. Her brother is 4 years younger than Yuki. Write down an expression, in terms of y, for the brother's age in 5 years' time.
- 2.Make x the subject of the formula y = 5x + 3.y = 5x + 3
- 3.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 4.A curve has equation y = x² − 9. Which statement about its graph is correct?y = x² − 9
- 5.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 6.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 7.a = 2 and b = 5. Work out the value of ab².
- 8.Write down the expression that means the same as m × m × m × n.
- 9.Simplify p + p + p.
- 10.A straight line passes through the points (2, 7) and (5, 16). Work out the value of y when x = 0.
- 11.Oliver has m stickers and gives 5 of them away. The number he has left is L = m − 5. Work out L when m = 8.
- 12.A delivery van's distance-time graph shows the following: it travels 9 km in the first 15 minutes at a steady speed, stops for 45 minutes to unload, then travels a further 21 km in the next 30 minutes, also at a steady speed. Work out the average speed, in km/h, for the whole journey.
- 13.The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4. Work out the x-coordinate of the turning point of the curve, using the symmetry of the graph.y = x² − 2x − 8
- 14.Write down the gradient of the line with equation y = 6 + 4x.
- 15.A company's profit was £2000 in its first year. Each following year, the profit increases by £800. Work out the first year in which the profit is more than £7000.
Answer key
- (a) y + 1 — The brother's age now is y − 4. In 5 years' time this becomes y − 4 + 5 = y + 1. A candidate who adds 4 instead of subtracting it, then adds 5, gets y + 4 + 5 = y + 9. A candidate who subtracts 5 instead of adding it gets y − 4 − 5 = y − 9. A candidate who works out the brother's current age but forgets to add on the 5 years gets y − 4.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
- (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
- (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.
- (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.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (a) 50 — Method: in ab², only the b is squared, so square b first, then multiply by a. Working: b² = 5² = 25, then a × b² = 2 × 25 = 50. Answer: 50. 100 comes from squaring the product ab instead of just b: (2 × 5)² = 100. 20 comes from squaring a instead of b: a² × b = 4 × 5 = 20. 10 comes from ignoring the square altogether and working out a × b = 2 × 5 = 10.
- (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.
- (b) 3p — p + p + p means three lots of p added together, and repeated addition of the same term is written as a coefficient: 3p. Writing p³ mistakes the repeated addition for repeated multiplication, as if the expression had been p × p × p. Writing 3 + p adds the number of terms (3) onto p as a separate constant, instead of writing 3 as a coefficient of p. Writing just p forgets to count the terms at all, as though repeating the same letter makes no difference. The simplified expression is 3p.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (a) 3 — Method: substitute the given value into the formula and carry out the subtraction in the order the formula is written. Working: replacing m with 8 gives L = 8 − 5, and 8 − 5 = 3. Answer: 3. The distractors: 13 comes from adding 5 to 8 instead of subtracting it; 40 comes from reading m − 5 as a multiplication and working out 8 × 5; −3 comes from subtracting the wrong way round and working out 5 − 8.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
- (d) 4 — Method: compare the equation with the general form y = mx + c after writing the x term first, then read off the coefficient of x. Working: y = 6 + 4x can be written as y = 4x + 6, so comparing with y = mx + c gives m = 4. Answer: the gradient is 4. The value 6 comes from reading off the constant, written first in the equation, and treating it as the gradient because it comes before the x term. The value 10 comes from adding the two numbers in the equation, 6 + 4, instead of reading off the coefficient of x. The value 24 comes from multiplying the two numbers in the equation, 6 × 4, instead of reading off the coefficient of x.
- (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.
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