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 sequence has the position-to-term rule: the term is the square of the position number. Its first four terms are 1, 4, 9, 16. Work out the 10th term.
- 2.Work out the gradient of the straight line with equation 3y = 12 − 6x.
- 3.Solve the simultaneous equations x + y = 1 and 2x + y = 5.
- 4.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
- 5.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.
- 6.Which of these quadratic graphs does NOT cross the x-axis at all?
- 7.Which expression is equivalent to 4(x − 2) + 10?
- 8.x = 5. Work out the value of 3x² − 4.
- 9.A number machine multiplies its input by 2 and then subtracts 5. Work out the output when the input is 6.
- 10.A gardener charges a fixed fee of £15 plus £8 for each hour worked. The total charge, £T, for a job lasting h hours is given by T = 15 + 8h. Work out the charge for a job lasting 4 hours, and identify what kind of statement T = 15 + 8h is.
- 11.A coach company charges a booking fee plus a price for each passenger. A booking for 1 passenger costs £9, for 2 passengers costs £15, for 3 passengers costs £21, and for 4 passengers costs £27. Work out an expression, in terms of n, for the cost in pounds of a booking for n passengers.
- 12.Which of these equations gives a graph shaped like a U, symmetrical about a vertical line?
- 13.The diagram shows the graph of a function. Which equation could it represent?
- 14.Two expressions are 6(x − 1) and 6x − 6. A student says these are equivalent for every value of x. Is the student correct?
- 15.Solve 5x − 9 = 16
Answer key
- (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.
- (c) −2 — Method: rearrange the equation into the form y = mx + c, then read off the gradient. Working: 3y = 12 − 6x, so dividing every term by 3 gives y = 4 − 2x, so the gradient is −2. Answer: the gradient is −2. 2 comes from dropping the negative sign after dividing by 3. 4 comes from using the y-intercept, 4, instead of the gradient. −6 comes from reading off the coefficient of x before dividing the whole equation by 3.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (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.
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (a) 4x + 2 — Expand the bracket: 4(x − 2) = 4x − 8. Then add the 10 that follows the bracket: 4x − 8 + 10 = 4x + 2. The option 4x − 8 comes from expanding the bracket correctly but then forgetting to add the 10. The option 4x + 8 comes from forgetting to multiply the 2 inside the bracket by 4 (treating it as 4x − 2 instead of 4x − 8), then adding 10. The option 4x + 18 comes from a sign error when expanding — treating 4 × (−2) as +8 instead of −8, giving 4x + 8, then adding 10.
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (d) 6n + 3 — Method: find the price per passenger, then find the booking fee that fits a booking for 1 passenger. Working: the cost rises by £6 for each extra passenger (15 − 9 = 6, 21 − 15 = 6, 27 − 21 = 6), so the cost has the form 6n + c. Substituting n = 1: 6(1) + c = 9, so c = 3. Answer: the cost in pounds is 6n + 3. 6n ignores the booking fee altogether. 6n + 9 uses the cost of one passenger, £9, as the fee without first subtracting the £6 per-passenger price. 9n + 6 swaps the two numbers, using the first cost as the price per passenger and the difference as the fee.
- (a) y = x² + 2 — A quadratic equation, with x², gives a graph shaped like a symmetrical U (or an upside-down U). Of the four equations, y = x² + 2 is quadratic, so it gives this U-shape. y = x³ + 2 is cubic, giving an S-shaped curve rather than a symmetrical U. y = 2x + 2 is linear, giving a straight line. y = 2/x is reciprocal, giving two separate curved branches rather than a single U-shape.
- (b) $y = x^3 - 4x$ — Method: count how many times the curve crosses the x-axis and check whether it is a cubic (an S-shape with up to three crossing points) rather than a lower power, and note which way it runs overall from bottom-left to top-right or the reverse. Working: the curve crosses the x-axis at three points, x = −2, 0 and 2, and runs from bottom-left to top-right, which matches y = x³ − 4x = x(x − 2)(x + 2). Answer: y = x³ − 4x. Distractor refutation: y = −x³ + 4x comes from a sign error on every term, which would flip the curve so it ran from top-left to bottom-right instead. y = x³ + 4x comes from a sign error on the x term only, which removes two of the three crossing points, since x(x² + 4) has only x = 0 as a real root. y = x² − 4x comes from dropping the cubic term altogether, giving a parabola with only two crossing points instead of an S-shaped curve with three.
- (d) Yes, since expanding 6(x − 1) gives 6x − 6. — Expand the bracket: 6(x − 1) = 6 × x − 6 × 1 = 6x − 6, which matches the second expression exactly, so the student is correct for every value of x. Saying 6(x − 1) means 6x − 1 comes from multiplying only the x and forgetting to multiply the 1 by 6. Saying brackets always change an expression's value is not true — expanding here gives back an equivalent expression, not a different one. The identity holds for every value of x, not just whole numbers, since both sides are expanded algebraically, not tested by substitution.
- (b) 5 — Method: add 9 to both sides, then divide by 5. Working: 5x = 16 + 9 = 25, so x = 25 ÷ 5 = 5. Answer: 5. 3.2 comes from dividing 16 by 5 directly, without adding 9 first. 1.4 comes from a sign error, subtracting 9 from 16 instead of adding it, then dividing by 5. 25 comes from correctly working out 5x = 25 but stopping there, without dividing by 5 to find x.
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