Question

Reduce each of the following fractions to the lowest terms:

(i) $\frac{161}{207}$

(ii) $\frac{296}{481}$

Answer 1

(i) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.
Now, we have to find the HCF of 161 and 207.
Prime factorisation of 161 = 7 × 23
Prime factorisation of 207 = 3 × 3 × 23
∴ HCF of 161 and 207 = 23

Now, $\frac{161÷23}{207÷23}=\frac{7}{9}$
Hence, $\frac{7}{9}$ is the required fraction.

(ii) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.
Now, we have to find the HCF of 296 and 481.
Prime factorisation of 296 = 2 × 2 × 2 × 37
Prime factorisation of 481 = 13 × 37
∴ HCF of 296 and 481 = 37

Now, $\frac{296÷37}{481÷37}=\frac{8}{13}$
Hence, $\frac{8}{13}$ is the required fraction.