Question

Jenny has two cows on her farm. The first cow drinks $\frac{9}{2}$ liters of water and the second cow drinks $\frac{11}{5}$ liters of water every day. How many liters of water will both cows drink in $30$ days?

• A. $\frac{67}{10}L$
• B. $201L$
• C. $2010L$
• D. $\frac{600}{7}L$

Answer 1

The correct option is B $201L$

The first cow drinks $\frac{9}{2}L$ of water in one day.
The second cow drinks $\frac{11}{5}L$ of water in one day.
Total quantity of water both cows drink in one day $=\frac{9}{2}+\frac{11}{5}$
To add both fractions, we have to make them like fractions by making their denominators equal. To make the denominators equal, we will multiply both the numerator and denominator of $\frac{9}{2}$ by $5$, and the numerator and denominator of $\frac{11}{5}$ by $2$ to make both denominators $10$.
$\frac{9}{2}=\frac{9\times 5}{2\times 5}=\frac{45}{10}$
$\frac{11}{5}=\frac{11\times 2}{5\times 2}=\frac{22}{10}$
Total quantity of water both cows drink in one day $=\frac{9}{2}+\frac{11}{5}=\frac{45}{10}+\frac{22}{10}=\frac{67}{10}L$
Total quantity of water both cows will drink in $30$ days $=\frac{67}{10}\times 30=201L$