Question

Find the sum of $\frac{41}{4}$ and $13\frac{5}{3}$.

• A. $24\frac{9}{12}$
• B. $24\frac{7}{12}$
• C. $24\frac{11}{12}$
• D. $24\frac{21}{12}$

Answer 1

The correct option is C

$24\frac{11}{12}$

We shall first convert $13\frac{5}{3}$ into an improper fraction.

$13\frac{5}{3}$ = $\frac{13\times 3+5}{3}$ = $\frac{44}{3}$

Now, $\frac{41}{4}+\frac{44}{3}$

Making the denominator same , we have

$\frac{44}{3}$ = $\frac{44\times 4}{3\times 4}$

= $\frac{176}{12}$

$\frac{41}{4}$ = $\frac{41\times 3}{4\times 3}$

= $\frac{123}{12}$

Adding we have

=$\frac{123+176}{12}$

=24$\frac{11}{12}$