Question

Find the simplest form of:
(i) $\frac{69}{92}$
(ii) $\frac{473}{645}$
(iii) $\frac{1095}{1168}$
(iv) $\frac{368}{496}$

Answer 1

(i) Prime factorisation of 69 and 92 is:

69 = 3 × 23
92 = 2² × 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 = 2⁴ × 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 = 2⁴ × 23
496 = 2⁴ × 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 $\frac{23}{31}$.