Question

Find the simplest form of

(i) $\frac{69}{92}$
(ii) $\frac{561}{748}$
(iii) $\frac{1095}{1168}$

Answer 1

(i) $\frac{69}{92}=\frac{69÷23}{92÷23}=\frac{3}{4}$; ∴​ HCF (69, 92) = 23

(ii) $\frac{561}{748}=\frac{561÷187}{748÷187}=\frac{3}{4}$; ∴​ HCF (561, 748) = 187

(iii) $\frac{1095}{1168}=\frac{1095÷73}{1168÷73}=\frac{15}{16}$; ∴​ HCF (1095, 1168) = 73