Find the simplest form of:
(i) $\frac{69}{92}$
(ii) $\frac{473}{645}$
(iii) $\frac{1095}{1168}$
(iv) $\frac{368}{496}$
(i) Prime factorisation of 69 and 92 is:
69 = 3 × 23
$92=2^{2} \times 23$
Therefore, $\frac{69}{92}=\frac{3 \times 23}{2^{2} \times 23}=\frac{3}{2^{2}}=\frac{3}{4}$
Thus, simplest form of $\frac{69}{92}$ is $\frac{3}{4}$.
(ii) Prime factorisation of 473 and 645 is:
473 = 11 × 43
645 = 3 × 5 × 43
Therefore, $\frac{473}{645}=\frac{11 \times 43}{3 \times 5 \times 43}=\frac{11}{15}$
Thus, simplest form of $\frac{473}{645}$ is $\frac{11}{15}$.
(iii) Prime factorisation of 1095 and 1168 is:
1095 = 3 × 5 × 73
$1168=2^{4} \times 73$
Therefore, $\frac{1095}{1168}=\frac{3 \times 5 \times 73}{2^{4} \times 73}=\frac{15}{16}$
Thus, simplest form of $\frac{1095}{1168}$ is $\frac{15}{16}$.
(iv) Prime factorisation of 368 and 496 is:
$368=2^{4} \times 23$
$496=2^{4} \times 31$
Therefore, $\frac{368}{496}=\frac{2^{4} \times 23}{2^{4} \times 31}=\frac{23}{31}$
Thus, simplest form of $\frac{368}{496}$ is $\frac{23}{31}$.