Algebraic fractions – GCSE Maths exam questions with mark scheme

1Grade 4–5HigherNon-calculator4 marks

(a) Simplify fully $\dfrac{12a^3b}{4ab^2}$ [2]

(b) Write $\dfrac{x}{3} + \dfrac{2x}{5}$ as a single fraction in its simplest form. [2]

(a) for two of the three parts correct in a single term, e.g. $\frac{3a^2}{b^k}$, $\frac{3a^k}{b}$ or $\frac{ka^2}{b}$B1
(a) (total B2) for $\frac{3a^2}{b}$B1
(b) for writing both fractions over a common denominator, e.g. $\frac{5x}{15} + \frac{6x}{15}$M1
$\frac{11x}{15}$A1
Answer: (a) $\dfrac{3a^2}{b}$ (b) $\dfrac{11x}{15}$
Examiner tip: In (b) add the numerators only – $\frac{3x}{8}$ (adding tops and bottoms) scores nothing.

2Grade 4–5HigherNon-calculator2 marks

Simplify

(a) $\dfrac{6x - 9}{3}$ [1]

(b) $\dfrac{x^2 + 3x}{x}$ [1]

(a) $2x - 3$B1
(b) $x + 3$B1
Answer: (a) $2x - 3$ (b) $x + 3$
Examiner tip: Divide EVERY term on the top by the bottom – $2x - 9$ (dividing only the first term) is wrong.

3Grade 6–7HigherNon-calculator3 marks

Simplify fully $\dfrac{x^2 - 16}{x^2 + 2x - 8}$ [3]

for factorising the numerator: $(x - 4)(x + 4)$M1
for factorising the denominator: $(x + 4)(x - 2)$M1
$\frac{x - 4}{x - 2}$A1
Answer: $\dfrac{x - 4}{x - 2}$
Examiner tip: You can only cancel whole FACTORS – cancelling the $x^2$ terms or the $x$'s gets no marks.

4Grade 6–7HigherNon-calculator3 marks

Write $\dfrac{3}{x + 2} - \dfrac{2}{x - 1}$ as a single fraction in its simplest form. [3]

for using the common denominator $(x + 2)(x - 1)$, e.g. $\frac{3(x - 1)}{(x + 2)(x - 1)} - \frac{2(x + 2)}{(x + 2)(x - 1)}$M1
for a correct numerator with brackets expanded: $3x - 3 - 2x - 4$M1
$\frac{x - 7}{(x + 2)(x - 1)}$ (accept denominator $x^2 + x - 2$)A1
Answer: $\dfrac{x - 7}{(x + 2)(x - 1)}$
Examiner tip: The minus sign applies to ALL of $2(x + 2)$, giving $-2x - 4$; writing $-2x + 4$ gives $x + 1$ on top and loses the A mark.

5Grade 6–7HigherNon-calculator3 marks

Show that $\dfrac{x + 5}{x^2 - 9} - \dfrac{1}{x - 3}$ can be written as $\dfrac{2}{(x + 3)(x - 3)}$ [3]

for factorising $x^2 - 9 = (x + 3)(x - 3)$M1
for writing as a single fraction over $(x + 3)(x - 3)$: $\frac{x + 5 - (x + 3)}{(x + 3)(x - 3)}$M1
for simplifying the numerator to 2 and completing the proof with no errorsA1
Answer: Shown: $\dfrac{x + 5 - x - 3}{(x + 3)(x - 3)} = \dfrac{2}{(x + 3)(x - 3)}$
Examiner tip: Spot that $x^2 - 9$ already contains $(x - 3)$ – using $(x^2 - 9)(x - 3)$ as the denominator makes the algebra much harder.

6Grade 8–9HigherCalculator4 marks

Solve $\dfrac{4}{x + 1} + \dfrac{3}{x - 2} = 2$
Give your solutions correct to 2 decimal places. [4]

for a correct method to remove the fractions, e.g. multiplying through by $(x + 1)(x - 2)$: $4(x - 2) + 3(x + 1) = 2(x + 1)(x - 2)$ (oe, e.g. $\dfrac{4(x - 2) + 3(x + 1)}{(x + 1)(x - 2)} = 2$ or $\dfrac{7x - 5}{(x + 1)(x - 2)} = 2$)M1
for a correct 3-term quadratic $= 0$: $2x^2 - 9x + 1 = 0$ oe (e.g. $-2x^2 + 9x - 1 = 0$ or $2x^2 = 9x - 1$ rearranged to $=0$)A1
for correct substitution into the quadratic formula for their 3-term quadratic (ft): e.g. $x = \dfrac{9 \pm \sqrt{(-9)^2 - 4 \times 2 \times 1}}{2 \times 2}$, or a correct completing-the-square method, e.g. $\left(x - \frac{9}{4}\right)^2 = \frac{73}{16}$M1
cao: awrt $4.39$ and awrt $0.11$ (both required; exact form $\frac{9 \pm \sqrt{73}}{4}$ alone does not earn this mark); no need to mention the excluded values $x=-1$, $x=2$A1
Answer: $x = 4.39$ or $x = 0.11$
Examiner tip: Multiply EVERY term by $(x + 1)(x - 2)$ – including the 2 on the right-hand side, which is often left as just 2.

7Grade 8–9HigherNon-calculator4 marks

Simplify fully

$$\frac{2x^2 + 5x - 3}{4x^2 - 1} \div \frac{x + 3}{2x^2 + x}$$ [4]

for correctly factorising one of the quadratic expressions, e.g. $(2x - 1)(x + 3)$, $(2x - 1)(2x + 1)$ or $x(2x + 1)$M1
for inverting the second fraction and multiplying: $\frac{2x^2 + 5x - 3}{4x^2 - 1} \times \frac{2x^2 + x}{x + 3}$M1
for all three quadratic expressions correctly factorisedM1
$x$A1
Answer: $x$
Examiner tip: Flip only the SECOND fraction, then factorise everything before cancelling – $4x^2 - 1$ is a difference of two squares.

8Grade 4–5HigherNon-calculator3 marks

(a) Simplify $\dfrac{18y^2}{6y}$ [1]

(b) Write $\dfrac{3n}{4} - \dfrac{n}{6}$ as a single fraction in its simplest form. [2]

(a) $3y$ caoB1
(b) for writing both fractions over a common denominator, e.g. $\dfrac{9n}{12} - \dfrac{2n}{12}$ (allow any common multiple of 4 and 6, e.g. $\dfrac{18n}{24} - \dfrac{4n}{24}$)M1
(b) $\dfrac{7n}{12}$ cao (accept $\dfrac{7}{12}n$; an unsimplified answer such as $\dfrac{14n}{24}$ scores M1 A0)A1
Answer: (a) $3y$ (b) $\dfrac{7n}{12}$
Examiner tip: Using 24 as the denominator is fine for the method mark, but you must cancel $\frac{14n}{24}$ down to $\frac{7n}{12}$ to get the answer mark.

9Grade 4–5Non-calculator3 marks

Solve $\dfrac{x}{2} + \dfrac{x}{5} = 14$ [3]

for a correct first step to deal with the fractions, e.g. multiplying every term by 10 (or any common multiple of 2 and 5), or writing $\dfrac{5x}{10} + \dfrac{2x}{10}$M1
for a correct equation without fractions or with a single fraction, e.g. $5x + 2x = 140$ or $\dfrac{7x}{10} = 14$ or $7x = 140$M1
$x = 20$ caoA1
Answer: $x = 20$
Examiner tip: When you multiply through by 10, the 14 on the right must become 140 too – leaving it as 14 gives $x = 2$, which scores no marks.

10Grade 6–7HigherNon-calculator3 marks

Simplify fully $\dfrac{2x^2 + 7x - 15}{x^2 - 25}$ [3]

for factorising the numerator: $(2x - 3)(x + 5)$M1
for factorising the denominator: $(x + 5)(x - 5)$M1
$\dfrac{2x - 3}{x - 5}$ caoA1
Answer: $\dfrac{2x - 3}{x - 5}$
Examiner tip: You may only cancel whole brackets (factors) – cancelling the $x^2$ terms or the 5s before factorising scores nothing.

11Grade 6–7HigherNon-calculator3 marks

Simplify fully

$$\frac{x^2 - 9}{6x} \times \frac{4x^2}{x - 3}$$ [3]

for factorising $x^2 - 9$ as $(x + 3)(x - 3)$M1
for a correct multiplication and cancelling of at least one common factor, e.g. $\dfrac{(x + 3)(x - 3) \times 4x^2}{6x(x - 3)}$ reduced to $\dfrac{4x^2(x + 3)}{6x}$ or $\dfrac{4x(x+3)}{6}$M1
$\dfrac{2x(x + 3)}{3}$ oe (accept $\dfrac{2x^2 + 6x}{3}$ or $\frac{2}{3}x(x+3)$)A1
Answer: $\dfrac{2x(x + 3)}{3}$
Examiner tip: Simplify fully means cancelling the number factors too: $\frac{4}{6}$ must become $\frac{2}{3}$.

12Grade 6–7HigherNon-calculator3 marks

The diagram shows a rectangle.

The length of the rectangle is $\dfrac{3}{x - 2}$ cm and the width is $\dfrac{1}{x + 4}$ cm.

Show that the perimeter, in cm, of the rectangle can be written as

$$\frac{4(2x + 5)}{(x - 2)(x + 4)}$$ [3]

for a correct expression for the perimeter, e.g. $2\left(\dfrac{3}{x - 2} + \dfrac{1}{x + 4}\right)$ or $\dfrac{6}{x - 2} + \dfrac{2}{x + 4}$M1
for using the common denominator $(x - 2)(x + 4)$ with correct numerators, e.g. $\dfrac{6(x + 4) + 2(x - 2)}{(x - 2)(x + 4)}$ or $2 \times \dfrac{3(x + 4) + (x - 2)}{(x - 2)(x + 4)}$M1
for complete, correct working leading to the given result, including a correct expanded numerator such as $8x + 20$ or $2(4x + 10)$ then factorised to $4(2x + 5)$A1
Answer: Perimeter $= \dfrac{6(x + 4) + 2(x - 2)}{(x - 2)(x + 4)} = \dfrac{8x + 20}{(x - 2)(x + 4)} = \dfrac{4(2x + 5)}{(x - 2)(x + 4)}$
Examiner tip: In a 'show that', every step must appear – jumping from the common denominator straight to the given answer loses the final mark.

13Grade 8–9HigherCalculator5 marks

A boat travels 6 km up a river and then 6 km back down the river.

The speed of the boat in still water is $x$ km/h. The river flows at 2 km/h, so the boat travels up the river at $(x - 2)$ km/h and down the river at $(x + 2)$ km/h.

The whole journey takes 2 hours.

(a) Show that $x^2 - 6x - 4 = 0$ [3]

(b) Find the speed of the boat in still water. Give your answer correct to 2 decimal places. [2]

(a) for an equation for the total time, $\dfrac{6}{x - 2} + \dfrac{6}{x + 2} = 2$M1
(a) for removing the fractions correctly, e.g. $6(x + 2) + 6(x - 2) = 2(x - 2)(x + 2)$ or $12x = 2(x^2 - 4)$M1
(a) for complete working to reach $x^2 - 6x - 4 = 0$, e.g. $12x = 2x^2 - 8$ so $2x^2 - 12x - 8 = 0$ then dividing by 2A1
(b) for correct substitution into the quadratic formula, e.g. $x = \dfrac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times (-4)}}{2}$, or completing the square $(x - 3)^2 = 13$M1
(b) $6.61$ (km/h) cao, with the negative root ($-0.61$) rejected or not given as a final answer; awrt 6.61 from $3 + \sqrt{13}$A1
Answer: (b) $6.61$ km/h
Examiner tip: Time = distance ÷ speed, so each part of the journey is a fraction; then reject $x = -0.61$ because a speed (and $x - 2$) must be positive.

14Grade 6–7HigherNon-calculator4 marks

Ben is asked to write $\dfrac{4}{x^2 - 4} + \dfrac{1}{x + 2}$ as a single fraction in its simplest form.

Here is his working.

$$\frac{4(x + 2) + 1(x^2 - 4)}{(x^2 - 4)(x + 2)} = \frac{x^2 + 4x + 4}{(x^2 - 4)(x + 2)}$$

Ben says, 'This fraction cannot be simplified.'

(a) Show that Ben is wrong by writing the expression as a single fraction in its simplest form. [3]

(b) Give a simpler common denominator that Ben could have used. [1]

(a) for factorising $x^2 - 4$ as $(x + 2)(x - 2)$ (in Ben's denominator or in the original expression)M1
(a) for a correct method leading to a fraction with a factor that cancels, e.g. $\dfrac{(x + 2)^2}{(x - 2)(x + 2)^2}$, or $\dfrac{4 + (x - 2)}{(x + 2)(x - 2)} = \dfrac{x + 2}{(x + 2)(x - 2)}$M1
(a) $\dfrac{1}{x - 2}$ caoA1
(b) $(x + 2)(x - 2)$ or $x^2 - 4$B1
Answer: (a) $\dfrac{1}{x - 2}$ (b) $(x + 2)(x - 2)$, i.e. $x^2 - 4$
Examiner tip: Always factorise every denominator first – here $x^2 - 4 = (x + 2)(x - 2)$ already contains $x + 2$, so the lowest common denominator is just $x^2 - 4$.

15Grade 4–5HigherNon-calculator2 marks

Kiran simplifies $\dfrac{x + 8}{2}$ and writes

$$\frac{x + 8}{2} = x + 4$$

(a) Explain what Kiran has done wrong. [1]

(b) Write $\dfrac{x + 8}{2}$ correctly as the sum of two terms. [1]

(a) explanation that Kiran has only divided the 8 by 2 and not the $x$ (oe, e.g. 'both terms on top must be divided by 2', or 'he should have got $\frac{x}{2} + 4$'); a counter-example on its own (e.g. $x = 2$: $\frac{10}{2} = 5$ but $2 + 4 = 6$) shows he is wrong but does not say what he did wrong, so scores C0; do not accept just 'he has done it wrong'C1
(b) $\dfrac{x}{2} + 4$ (accept $\frac{1}{2}x + 4$ or $0.5x + 4$)B1
Answer: (a) He divided only the 8 by 2; the $x$ must be divided by 2 as well. (b) $\dfrac{x}{2} + 4$
Examiner tip: An explanation must say WHAT is wrong (the $x$ was not divided by 2); 'he should have divided' with no detail does not earn the mark.

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