Reduce the following fractions to simplest form:

(i) $\dfrac{12}{27}$

(ii) $\dfrac{150}{350}$

(iii) $\dfrac{18}{81}$

(iv) $\dfrac{276}{115}$

(i) $\dfrac{12}{27}$

By prime factorisation,

$\Rightarrow \dfrac{12}{27} = \dfrac{2 \times 2 \times 3}{3 \times 3 \times 3} \\[1em] = \dfrac{2 \times 2}{3 \times 3} \\[1em] = \dfrac{4}{9}.$

Hence, $\dfrac{12}{27}$ in simplest form is $\dfrac{4}{9}$.

(ii) $\dfrac{150}{350}$

By prime factorisation,

$\Rightarrow \dfrac{150}{350} = \dfrac{2 \times 3 \times 5 \times 5}{2 \times 5 \times 5 \times 7} \\[1em] = \dfrac{3}{7}.$

Hence, $\dfrac{150}{350}$ in simplest form is $\dfrac{3}{7}$.

(iii) $\dfrac{18}{81}$

By prime factorisation,

$\Rightarrow \dfrac{18}{81} = \dfrac{2 \times 3 \times 3}{3 \times 3 \times 3 \times 3} \\[1em] = \dfrac{2}{3 \times 3} \\[1em] = \dfrac{2}{9}.$

Hence, $\dfrac{18}{81}$ in simplest form is $\dfrac{2}{9}$.

(iv) $\dfrac{276}{115}$

By prime factorisation,

$\Rightarrow \dfrac{276}{115} = \dfrac{2 \times 2 \times 3 \times 23}{5 \times 23} \\[1em] = \dfrac{2 \times 2 \times 3}{5} \\[1em] = \dfrac{12}{5} \\[1em] = 2\dfrac{2}{5}.$

Hence, $\dfrac{276}{115}$ in simplest form is $\dfrac{12}{5}$ or $2\dfrac{2}{5}$.