Question

Find the least fraction which must be multiplied to $(1\frac{2}{4}-\frac{1}{2}+3\frac{6}{16}-\frac{1}{16}+\frac{2}{8}$) to make it an integer.

• A. $\frac{1}{73}$
• B. $\frac{73}{1}$
• C. $\frac{16}{1}$
• D. $\frac{24}{1}$

Answer 1

The correct option is C

$\frac{16}{1}$

$(1\frac{2}{4}$ -$\frac{1}{2}$ +3$\frac{6}{16}$ -$\frac{1}{16}$ +$\frac{2}{8}$)

=$\frac{6}{4}-\frac{1}{2}+\frac{54}{16}-\frac{1}{16}+\frac{2}{8}$

Taking LCM of denominators and solving,

$\frac{24-8+54-1+4}{16}$=$\frac{73}{16}$

Now, If we multiply $\frac{73}{16}$ by $\frac{16}{1}$, it becomes 73, which is an integer

Hence, $\frac{16}{1}$ is the required fraction.