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

Maths · GCSE Higher · Algebra · 15 questions

MathsUKwww.geekhero.co.uk

1.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
(a)4.67
(b)−4
(c)2
(d)4
2.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
(a)40
(b)11
(c)20
(d)80
3.By completing the square, find the turning point of the curve y = x² − 10x + 30.
y = x² − 10x + 30
(a)x = 10, y = −70
(b)x = 5, y = −5
(c)x = 5, y = 5
(d)x = −5, y = 5
4.The graph of y = f(x) has a maximum turning point at (−1, 6). Write down the coordinates of the maximum turning point of the graph of y = f(x − 3).
(a)(−1, 9)
(b)(2, 6)
(c)(−1, 3)
(d)(−4, 6)
5.A stack of firewood has 3 logs in the top layer. Each layer below has 4 more logs than the layer above it. Work out the number of logs in the 6th layer from the top.
(a)23
(b)27
(c)24
(d)20
6.Solve the inequality 9 − 2x ≥ 1.
(a)x ≤ 5
(b)x ≥ 4
(c)x ≤ 4
(d)x ≤ −4
7.A ribbon of length L metres is cut into 5 equal pieces, and then 3 metres is removed from one piece. Write down an expression, in metres, for the length of that piece after 3 metres is removed.
(a)5/L − 3
(b)(L − 3)/5
(c)L/5 − 3
(d)L − 3/5
8.A machine bought for £2,400 loses 12% of its value every year. After how many complete years does its value first fall below £1,500?
(a)1 year
(b)3 years
(c)4 years
(d)5 years
9.A geometric sequence has first term 4 and common ratio √2. Work out the sum of the first three terms of the sequence, giving your answer in the form a + b√2.
(a)12 + 4√2
(b)36 + 4√2
(c)4 + 12√2
(d)8 + 4√2
10.Make x the subject of the formula y = (x − 4)/5.
(a)x = 5(y + 4)
(b)x = 5y − 4
(c)x = y/5 + 4
(d)x = 5y + 4
11.The numbers x and y satisfy x + y = 10 and xy = 21. Work out the pair of values.
(a)3 and 7
(b)−3 and −7
(c)5 and 5
(d)1 and 21
12.For the equation 2x + 3 = 11, and the inequality 2x + 3 > 11, which statement correctly compares their solutions?
(a)The equation has one solution; the inequality has many.
(b)Both have exactly one solution.
(c)Both have many solutions.
(d)The equation has many solutions; the inequality has one.
13.A number, w, satisfies the statement "three times w, minus 5, is at least 16". Which of these is the correct inequality for w?
(a)3(w − 5) ≥ 16
(b)3w − 5 ≥ 16
(c)3w − 5 > 16
(d)3w − 5 ≤ 16
14.The table shows the speed, in m/s, of a car at times, in seconds, all 2 seconds apart: 0 at t = 0, 3 at t = 2, 8 at t = 4, 15 at t = 6, 24 at t = 8. A student uses the trapezium rule with these five readings to estimate the distance the car travels. By finding the second differences of the speed values, decide whether this trapezium estimate is an overestimate or an underestimate of the true distance.
(a)Underestimate — second differences are constant.
(b)Overestimate — the speed values are increasing.
(c)Underestimate — the curve bends upward (convex).
(d)Overestimate — the curve bends upward (convex).
15.Rosie is asked whether 4(x + 3) = 4x + 12 is an equation or an identity. Which statement is correct? Give a reason for your answer.
(a)An identity, because it is true for every value of x
(b)An identity, because the brackets can be expanded
(c)An equation, because solving it gives x = 3
(d)A formula, because it works out 4x + 12 from x

Answer key

(d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
(a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
(c) x = 5, y = 5 — x² − 10x + 30 = (x − 5)² − 5² + 30 = (x − 5)² + 5. Substituting x = 5: 5² = 25, 10 × 5 = 50, so 25 − 50 + 30 = 5, confirming the minimum value 5 at x = 5: turning point (5, 5). Using −10 instead of half of it inside the bracket gives (x − 10)² − 70, turning point (10, −70) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−5, 5) — wrong, because (x − 5)² is zero at x = 5, not x = −5. Computing 25 − 30 = −5 instead of 30 − 25 = 5 flips the sign of the constant, giving (5, −5) — wrong, since the completed square's constant must be evaluated as 30 minus 25, not 25 minus 30. Always check a turning point by substituting its x-value back into the original equation.
(b) (2, 6) — y = f(x − 3) translates y = f(x) horizontally by 3 units to the RIGHT — inside the brackets, subtracting moves the graph in the positive x-direction. Turning point (−1, 6) → (−1 + 3, 6) = (2, 6). The common slip is to move LEFT instead, since the sign inside the bracket is negative — that gives (−4, 6). Changing the y-coordinate instead of the x-coordinate, as in (−1, 3) or (−1, 9), treats this as a vertical shift, which y = f(x − 3) is not.
(a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
(c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
(c) L/5 − 3 — Each of the 5 equal pieces is L/5 metres long, and removing 3 metres from one piece gives L/5 − 3. Subtracting the 3 metres before dividing by 5, (L − 3)/5, divides the removed length between all 5 pieces instead of taking it from just one. Dividing only the 3 by 5 instead of dividing L by 5, L − 3/5, divides the wrong number. Writing 5/L − 3 inverts the fraction, swapping which number is the numerator.
(c) 4 years — The yearly multiplier is 1 − 0.12 = 0.88. After 1 year: 2400 × 0.88 = 2112. After 2 years: 2112 × 0.88 = 1858.56. After 3 years: 1858.56 × 0.88 = 1635.5328. After 4 years: 1635.5328 × 0.88 = 1439.27 (2 d.p.), which is the first value below £1,500 — so it takes 4 complete years. Stopping after 3 years, where the value is still £1,635.53, above £1,500, gives 3 years — wrong, because the threshold has not been crossed yet at that point. Continuing for one year longer than needed gives 5 years — wrong, because the value has already fallen below £1,500 by year 4, so year 5 overcounts. Multiplying by the percentage itself instead of by the multiplier — using 0.12 in place of 0.88 — gives 2400 × 0.12 = £288 after a single year, apparently below £1,500 straight away, so 1 year — wrong, because 12% is the part lost each year, and the part left is 88%, so the multiplier is 0.88.
(a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
(d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
(a) 3 and 7 — Method: one equation is linear and the other is not, so rearrange the linear equation and substitute it into the other to leave a single quadratic in one letter. Working: x + y = 10 gives y = 10 − x, so xy = 21 becomes x(10 − x) = 21, which rearranges to x² − 10x + 21 = 0; factorising gives (x − 3)(x − 7) = 0, so the two values are 3 and 7, and 3 + 7 = 10 with 3 × 7 = 21. Answer: 3 and 7. The distractors: 1 and 21 comes from using only the product and taking the first factor pair of 21; −3 and −7 comes from factorising as (x + 3)(x + 7), which reverses the sign of both values; 5 and 5 comes from using only the sum and splitting 10 into two equal parts.
(a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
(b) 3w − 5 ≥ 16 — "Three times w, minus 5" translates to 3w − 5, and "is at least 16" means it must be 16 or more, giving 3w − 5 ≥ 16. A candidate who reads "at least" as a strict inequality writes 3w − 5 > 16. A candidate who misreads the wording and applies the subtraction before the multiplication writes 3(w − 5) ≥ 16. A candidate who reverses the direction of the inequality writes 3w − 5 ≤ 16.
(d) Overestimate — the curve bends upward (convex). — The first differences of the speeds are 3, 5, 7 and 9, so the second differences are 2, 2 and 2 — constant and positive, which means the speed-time graph curves upwards (is convex). On a convex curve, each straight chord used by the trapezium rule lies above the curve, so the trapezium rule overestimates the true distance. 'Underestimate — the curve bends upward' states the same correct geometry but gets the conclusion backwards — a chord above the curve means too much area is counted, not too little. 'Overestimate — the speed values are increasing' uses the wrong evidence: increasing speed alone doesn't tell you whether the curve bends up or down, only the second differences do. 'Underestimate — second differences are constant' confuses a constant second difference with a steady rate of change in speed, which isn't what the second difference of a speed-time table measures.
(a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.

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