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.The formula connecting speed, distance and time is v = s ÷ t. Work out v, in km/h, when s = 180 and t = 4.
- 2.The formula for the perimeter of a rectangle is P = 2(l + w). One rectangle has l = 9 cm and w = 4 cm. A second rectangle has l = 6 cm and w = 5 cm. Work out how many cm greater the perimeter of the first rectangle is than the perimeter of the second.
- 3.A car park charges by the hour. Parking for 1 hour costs £5, 2 hours costs £9, 3 hours costs £13 and 4 hours costs £17, with the cost increasing by the same amount for each extra hour. Work out an expression, in terms of n, for the cost, in pounds, of parking for n hours.
- 4.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
- 5.Kai writes the expression 8 − 3x + 5x² and says it has three terms. Ryan says it only has two terms because a number on its own is not a term. Work out the correct number of terms in the expression.
- 6.A train's distance-time graph shows it travelling 100 km at a steady speed in the first 1.5 hours, waiting at a signal for 0.25 hours, then travelling a further 47 km in 0.75 hours. Work out the average speed for the whole journey, correct to 1 decimal place.
- 7.A rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
- 8.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
- 9.A coastguard radar at the origin covers a circular region modelled by x² + y² = 400, where each unit represents 1 kilometre. A boat travels along the straight line that touches the boundary of the region at the point (12, 16). Work out the equation of the line the boat travels along.
- 10.The graph of y = 60 × (0.5)ˣ is sketched for x ≥ 0. Which statement correctly describes what happens to the curve as x increases?
- 11.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.y = x
- 12.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
- 13.The iterative formula xₙ₊₁ = xₙ³ − 2 is used repeatedly, starting from x₀ = 2. Which of these describes what happens to the sequence of values as n increases?
- 14.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.y = x² − 2x − 8
- 15.The equation x² − 7 = 0 has a positive root. Let f(x) = x² − 7. Given that x₁ = 2.6 and x₂ = 2.65, work out which of these is correct.y = x² − 7
Answer key
- (d) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (a) 4 — Rectangle 1: P = 2(9 + 4) = 2 × 13 = 26 cm. Rectangle 2: P = 2(6 + 5) = 2 × 11 = 22 cm. Difference = 26 − 22 = 4 cm. 2 comes from finding the difference between (l + w) for each rectangle, 13 − 11 = 2, but forgetting to double it. 26 comes from giving rectangle 1's whole perimeter instead of the difference between the two. 6 comes from doubling only the change in length, 2 × (9 − 6), and ignoring that the width also changed.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (b) 58.8 km/h — Total distance = 100 + 47 = 147 km. Total time = 1.5 + 0.25 + 0.75 = 2.5 hours. Average speed = 147 ÷ 2.5 = 58.8 km/h. Leaving out the 0.25 hours of waiting from the total time gives 147 ÷ 2.25 = 65.3 km/h (1 d.p.) — wrong, because the train is stationary but time is still passing on the whole journey. Using only the first leg gives 100 ÷ 1.5 = 66.7 km/h (1 d.p.) — wrong, because it ignores the second leg of the journey entirely. Working out each leg's own speed (100 ÷ 1.5 = 66.7 km/h and 47 ÷ 0.75 = 62.7 km/h) and then averaging those two speeds gives 64.7 km/h (1 d.p.) — wrong, because the average of two speeds over DIFFERENT times is not the same as total distance divided by total time.
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (b) y = −3x/4 + 25 — Method: a straight line that touches a circle at one point is a tangent there, so it is perpendicular to the radius drawn to that point; find the gradient of the radius, take its negative reciprocal, then substitute the point of contact into y − y₁ = m(x − x₁). Working: the radius from (0, 0) to (12, 16) has gradient 16 ÷ 12, which cancels to 4/3, so the tangent has gradient −3/4. Substituting gives y − 16 = −3/4(x − 12), so y − 16 = −3x/4 + 9 and y = −3x/4 + 25. Answer: y = −3x/4 + 25. The distractors: y = 3x/4 + 7 turns the gradient of the radius upside down but leaves it positive, so the perpendicular step is only half done; y = −4x/3 + 32 changes the sign of the radius gradient without turning it upside down, which is the other half left undone; y = −3x/4 − 25 uses the correct gradient but substitutes the point of contact with both signs reversed, writing y + 16 = −3/4(x + 12).
- (a) It gets closer to zero but never actually reaches it. — y = 60 × (0.5)ˣ is always positive, however large x is, because a positive number raised to any power stays positive. As x increases, (0.5)ˣ gets smaller and smaller but never equals zero, so y approaches zero without ever reaching it. Believing the curve reaches zero when x = 60, because 60 is the starting value, mixes up the y-intercept with a stopping point — an exponential decay curve like this one never actually reaches the x-axis, so this is wrong. Believing the curve goes negative for large x is wrong because multiplying a positive number by (0.5) any number of times can never produce a negative result. Believing the curve levels off at y = 0.5, confusing the base of the exponential with its eventual level, is wrong: the base only controls how fast the curve falls, not where it settles — this curve settles at y = 0, not y = 0.5. Whenever you sketch an exponential decay curve, draw it approaching the x-axis but never touching it.
- (d) It has a solution between x = 2.2 and x = 2.3 — Since f(2.2) is negative and f(2.3) is positive, the graph of f crosses the x-axis somewhere between x = 2.2 and x = 2.3, so the equation has a solution in that interval. Choosing 'between x = −2.2 and x = −2.3' confuses the negative f-VALUE at 2.2 with a negative x-value. Choosing 'no solution' misreads a change of sign as meaning the opposite of what it shows. Choosing 'exactly two solutions' assumes a single change of sign must give two roots, which is not what the rule guarantees.
- (b) Student B's — narrower strips fit the curve more closely. — The trapezium rule replaces the curve with straight-line segments; the narrower each strip, the more closely its straight edge follows the curve, so Student B's estimate with 8 narrower strips is more likely to be closer to the true area. Reasoning from the number of arithmetic steps rather than from how well the straight lines fit the curve gives the wrong claim that Student A makes fewer rounding errors. Believing the trapezium rule is exact, rather than an estimate that improves with narrower strips, gives the wrong claim that the two are always the same. Believing wider strips smooth out the curve better, rather than following it less closely, gives the wrong claim in favour of Student A's wider strips.
- (d) It diverges, growing rapidly without limit. — x₁ = 2³ − 2 = 8 − 2 = 6. x₂ = 6³ − 2 = 216 − 2 = 214. x₃ = 214³ − 2 = 9800344 − 2 = 9800342. The values 6, 214, 9800342, … grow far larger at every step, so the sequence diverges rather than settling anywhere. Checking whether the sequence converges to a fixed value near 2 fails, since the terms grow enormously instead of levelling off. Checking for a repeating pair of values also fails, since 6, 214 and 9800342 are all different, with no sign of a return to 6. x₀ = 2 is a fixed point only if 2³ − 2 = 2, but 2³ − 2 = 6, not 2, so the sequence does not stay constant.
- (c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
- (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
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