Fractions GCSE Questions, Worked Examples and Answers

Recognise it: simplify without changing value. Divide both numbers by their highest common factor, $15$.

$\frac{45}{60}=\frac{3}{4}$

$\boxed{\frac34}$

Recognise it: each part asks for the same value in another form.

(a) Divide the numerator by the denominator, then multiply the decimal by $100$.

$\frac38=3\div8=0.375$

$0.375\times100=37.5\%$

(b) Six tenths simplifies by dividing top and bottom by $2$.

$0.6=\frac6{10}=\frac35$

$\boxed{\frac38=0.375=37.5\%,\qquad 0.6=\frac35}$

Recognise it: count whole groups of $6$ in $29$.

$29=4\times6+5$

$\boxed{4\frac56}$

Recognise it: ascending means from the most negative value to the greatest value. Use denominator $24$.

$-\frac34=-\frac{18}{24},\quad -\frac23=-\frac{16}{24},\quad -\frac58=-\frac{15}{24}$

On a number line, $-\frac{18}{24}$ is furthest left.

$\boxed{-\frac34,\ -\frac23,\ -\frac58}$

Recognise it: addition needs a common denominator. The lowest common multiple of $8$ and $12$ is $24$.

$\frac9{24}+\frac{10}{24}=\boxed{\frac{19}{24}}$

Recognise it: subtraction needs equal-sized parts. Use denominator $18$.

$\frac{14}{18}-\frac3{18}=\boxed{\frac{11}{18}}$

Recognise it: convert mixed numbers so one addition handles every part.

$\frac74+\frac83=\frac{21}{12}+\frac{32}{12}=\frac{53}{12}$

$\boxed{4\frac5{12}}$

Recognise it: multiply, so cancel before multiplying rather than finding a common denominator.

$\frac8{21}\times\frac{14}{15}=\frac8{3}\times\frac2{15}=\boxed{\frac{16}{45}}$

Recognise it: divide by $\frac{15}{16}$, so multiply by its reciprocal.

$\frac58\times\frac{16}{15}=\frac{10}{15}=\boxed{\frac23}$

Recognise it: $260$ is the whole. Divide by $20$, then multiply by $11$.

$260\div20\times11=13\times11=\boxed{143}$

Recognise it: $45$ is five parts, not the whole. Undo the fraction.

$\text{one part}=45\div5=9$

$\text{seven parts}=9\times7=\boxed{63}$

Recognise it: the second fraction uses a new whole: the money remaining after the ticket.

$\text{ticket}=\frac38\times120=45$

$\text{remaining}=120-45=75$

$\text{food}=\frac25\times75=30$

$\boxed{\text{£}45\text{ left}}$

Recognise it: this is still fraction addition, but the numerators contain $x$. Use a common denominator of $12$.

$\frac{x}{3}=\frac{4x}{12},\qquad \frac{x}{4}=\frac{3x}{12}$

$\frac{4x}{12}+\frac{3x}{12}=\boxed{\frac{7x}{12}}$

Recognise it: the $21$ litres equals the change between two fractions of the same capacity.

$\frac34-\frac25=\frac{15}{20}-\frac8{20}=\frac7{20}$

So $\frac7{20}$ of the capacity is $21$ litres.

$\text{capacity}=21\div7\times20=\boxed{60\text{ litres}}$

Recognise it: the block $27$ repeats, so multiplying by $100$ lines up the recurring digits.

Let $x=0.\overline{27}$.

$100x=27.\overline{27}$

Subtract the original equation so the recurring parts cancel.

$100x-x=27.\overline{27}-0.\overline{27}$

$99x=27$

$x=\frac{27}{99}=\boxed{\frac3{11}}$