Question

A fruit seller has 24 kg of apples. He sells a part of these at a gain of 20% and the balance at a loss of 5%. If on the whole, he earns a profit of 10%, the quantity of apples sold at loss is ....?

• A. 4.6 kg
• B. 6 kg
• C. 9.6 kg
• D. 11.4 kg

Answer 1

Answer: (C)

9.6 kg

Given: Total weight of apples $=24$ kg

Let the cost price of each $kg$ of apple be $Rs.y$

The total cost price of $24$ kgs of apples is $Rs.24y$

Let us consider the weight of apples sold at $5\%$ loss as $x$ kg

Then the cost price of $x$ kg of apples will be $Rs.xy$ and selling price be $S{P}_1$

We know $Loss\%=\frac{CP-SP}{CP}\times 100$

$\Rightarrow 5=\frac{xy-S{P}_1}{xy}\times 100$

$\Rightarrow \frac{xy}{20}=xy-S{P}_1$

$\Rightarrow S{P}_1=xy-\frac{xy}{20}$

$\Rightarrow S{P}_1=\frac{19}{20}xy$ $...\left(1\right)$

Given: Rest of the apples apples i.e., $(24-x)$ kgs are sold at a gain of $20\%$

Then, the CP of $(24-x)$ kg of apples will be $(24-x)y$ and its SP be $S{P}_2$

We know $Profit\%=\frac{SP-CP}{CP}\times 100$

$\Rightarrow 20=\frac{S{P}_2-(24-x)y}{(24-x)y}\times 100$

$\Rightarrow \frac{(24-x)y}{5}=S{P}_2-(24-x)y$

$\Rightarrow S{P}_2=\frac{(24-x)y}{5}+(24-x)y$

$\Rightarrow S{P}_2=\frac{6}{5}(24-x)y$ $...\left(2\right)$

Hence, Total $SP$ $=S{P}_1+S{P}_2$

$=Rs.[\frac{19}{20}xy+\frac{6}{5}(24-x)y]$

$=Rs.[\frac{19}{20}xy+\frac{144}{5}y-\frac{6}{5}xy]$

$=Rs.[\frac{144}{5}y-\frac{1}{4}xy]$

Given, the seller earns a profit of $10\%$ on the whole
$\Rightarrow 10=\frac{[\frac{144}{5}y-\frac{1}{4}xy]-24y}{24y}\times 100$

$\Rightarrow \frac{12y}{5}=[\frac{144}{5}y-\frac{1}{4}xy]-24y$

$\Rightarrow \frac{1}{4}xy=\frac{144}{5}y-\frac{12}{5}y-24y$

$\Rightarrow \frac{1}{4}xy=\frac{132}{5}y-24y$

$\Rightarrow \frac{1}{4}xy=\frac{132-120}{5}y$

$\Rightarrow \frac{1}{4}xy=\frac{12}{5}y$

$\Rightarrow x=\frac{48}{5}$

$\Rightarrow x=9.6kg$

The amount of apples sold at a loss of $5\%$ is $9.6kg$