# Find the simplest form of:

Question:

Find the simplest form of:

(i) $\frac{69}{92}$

(ii) $\frac{473}{645}$

(iii) $\frac{1095}{1168}$

(iv) $\frac{368}{496}$

Solution:

(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 $\frac{23}{31}$.