$\frac{2}{9} + \frac{5}{11}$ =

6799

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{7}{20}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1099

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 6799
Consider the fraction $\frac{2}{9}$ . We multiply both the numerator and denominator by the denominator of the other fraction
i.e., $\frac{2}{9} = \frac{2 \times 11}{9 \times 11} = \frac{22}{99}$ .
Similarly, $\frac{5}{11} = \frac{5 \times 9}{11 \times 9} = \frac{45}{99}$ .

Thus, $\frac{2}{9} + \frac{5}{11}$

$= \frac{22}{99} + \frac{45}{99}$

$= \frac{22 + 45}{99}$

$= \frac{67}{99}$

Suggest Corrections

Similar questions

\frac{4}{5} \times \frac{11}{2} - \frac{9}{10} \times \frac{11}{2} = \frac{11}{2} \times

\frac{- 2}{11}, \frac{- 5}{11}, \frac{- 9}{11}

\frac{- 2}{11}, \frac{- 5}{11}, \frac{- 9}{11}

x

{(\frac{11}{5})}^{- 5} \times {(\frac{11}{5})}^{- 2} = {(\frac{11}{5})}^{x - 9}

{(\frac{2}{9})}^{2 x + 1} \div {(\frac{2}{9})}^{3} = {(\frac{2}{9})}^{x + 1}

\div

\times

\times

\div

Join BYJU'S Learning Program