Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher 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 Higher
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- 1.Work out the value of a² − 2b when a = −3 and b = 4.
- 2.Make x the subject of the formula y = x/2 − 5.y = x
- 3.A triangle has vertices at (1, 1), (1, 5) and (6, 1). Work out the area of the triangle.
- 4.Solve (2x + 1)/3 = 5
- 5.Factorise fully 12x² + 18x.
- 6.Four sequences are shown below. Sequence P: 3, 6, 12, 24, … Sequence Q: 1, 4, 9, 16, … Sequence R: 2, 4, 8, 14, … Sequence S: 5, 10, 15, 20, … Work out which of the sequences P, Q, R and S is geometric.
- 7.p = 4. Work out the value of 2p³.
- 8.Make x the subject of the formula y = 5x + 3.y = 5x + 3
- 9.The simple interest, £I, earned on a savings account is given by the formula I = PRT/100, where P is the amount invested in pounds, R is the interest rate per year and T is the time in years. Rearrange the formula to make P the subject.
- 10.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².y = x²
- 11.A graph has equation y = x² − 6x + 5. A student says its turning point has x-coordinate 6, because that's the coefficient of x. Which statement corrects the student's mistake?y = x² − 6x + 5
- 12.A plumber charges a call-out fee of £30 plus £25 for each hour worked. The total bill for a job was £130. Work out how many hours the plumber worked.
- 13.The graph of y = f(x) has a minimum point at (4, −1). Write down the coordinates of the corresponding turning point on the graph of y = −f(x).
- 14.f(x) = x³ − 5x − 6. Given that f(2.6) = −1.424 and f(2.7) = 0.183, work out what this shows about the equation x³ − 5x − 6 = 0.y = x
- 15.Which of these quadratic graphs does NOT cross the x-axis at all?
Answer key
- (a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
- (c) x = 2y + 10 — Method: undo the operations done to x in reverse order — add 5, then multiply by 2. Working: y = x/2 − 5, so y + 5 = x/2, so x = 2(y + 5) = 2y + 10. Answer: x = 2y + 10. x = 2y + 5 comes from multiplying only the x/2 term by 2 and forgetting to multiply the 5 as well. x = 2y − 10 comes from a sign error, subtracting 5 instead of adding it before multiplying by 2. x = (y + 5)/2 comes from dividing by 2 instead of multiplying, the wrong operation to undo a division.
- (d) 10 — The right angle is at (1, 1). The vertical side has length 5 − 1 = 4 and the horizontal side has length 6 − 1 = 5, so the area is (4 × 5) ÷ 2 = 20 ÷ 2 = 10. A candidate who forgets to halve the product of the two sides gets 4 × 5 = 20. A candidate who forgets to subtract the shared vertex's coordinate and uses the raw coordinates 6 and 5 as the side lengths gets (6 × 5) ÷ 2 = 30 ÷ 2 = 15. A candidate who uses only one side length as the area gets 5.
- (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.
- (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
- (a) P — A sequence is geometric when consecutive terms share a constant ratio. For P: 6 ÷ 3 = 2, 12 ÷ 6 = 2, 24 ÷ 12 = 2 — the ratio is constant at 2, so P is geometric. Q is the square numbers (1², 2², 3², 4²), a quadratic sequence: its ratios are 4, 2.25, 1.78, … — not constant. R looks geometric at first (2, 4, 8 doubles each time), but the pattern breaks: 8 to 14 is a ratio of 1.75, not 2. Its first differences are 2, 4, 6 — a constant second difference of 2 — so R is a quadratic sequence, not geometric. S has a constant DIFFERENCE of 5 (it is arithmetic), but its ratios (2, 1.5, 1.33, …) are not constant, so it is not geometric.
- (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) 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.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (b) Translate +5 in x, then reflect in the x-axis. — Translating y = x² by 5 units in the positive x-direction gives y = (x − 5)². Reflecting this in the x-axis, which replaces y with −y, gives y = −(x − 5)², matching the target. Using a translation of −5 in x instead gives y = (x + 5)², and reflecting that in the x-axis gives y = −(x + 5)² — the sign inside the bracket is wrong. Reflecting in the y-axis first does nothing to y = x², since (−x)² = x², so translating afterwards only reaches y = (x − 5)² with no negative sign at all. Translating by 5 units in y instead of x gives y = x² + 5, and reflecting that in the x-axis gives y = −x² − 5, a different curve altogether — a vertical shift does not create the (x − 5)² term the target equation needs.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (a) (4, 1) — y = −f(x) reflects the graph of y = f(x) in the x-axis: every point (x, y) maps to (x, −y). Applying this to (4, −1): the x-coordinate stays 4, and the y-coordinate −1 becomes its negative, 1. Leaving the y-coordinate unchanged skips the reflection entirely, giving (4, −1); reflecting the x-coordinate instead, or reflecting both, mixes this up with a reflection in the y-axis or a rotation, giving (−4, −1) or (−4, 1).
- (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
- (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.
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