Question

A man buys oranges at $9$ for $Rs.16$ and sells $11$ for $Rs.20$. What does he gain or lose per cent?

• A. $16$%
• B. $2\frac{1}{11}$%
• C. $2\frac{3}{11}$%
• D. $1\frac{3}{11}$%

Answer 1

Answer: (C)

$\left(2\frac{3}{11}\right)$%

Given, a man buys $9$ oranges for $Rs.16$
So, the cost price(C.P) of each orange is $Rs.\frac{16}{9}$
And also, he sells $11$ oranges for $Rs.20$.
So, the selling price(S.P) of each orange is $\frac{20}{11}$
To find gain or loss, we need to compare both the fractions $\frac{16}{9}$ and $\frac{20}{11}$.
LCM of $11$ and $9$ is $99$
Consider$\frac{16}{9}\times \frac{11}{11}=\frac{176}{99}$ and $\frac{20}{11}\times \frac{9}{9}=\frac{180}{99}$
Since, $180>176$
$\Rightarrow \frac{180}{99}>\frac{176}{99}$
$\therefore \frac{20}{11}>\frac{16}{9}$
Thus, S.P is greaten than C.P.
So, the man gets profit and the gain(profit) amount $=$S.P$-$C.P
$=\frac{20}{11}-\frac{16}{9}$
$=\frac{180}{99}-\frac{176}{99}$
$=\frac{4}{99}$
Therefore, profit percentage $=\frac{\text{profit}}{\text{C.P}}\times 100$
$=\frac{\left(\frac{4}{99}\right)}{\left(\frac{16}{9}\right)}\times 100$
$=\frac{4}{99}\times \frac{9}{16}\times 100$
$=\frac{25}{11}$
$=2\frac{3}{11}\%$