Question

How many kilograms of sugar costing $\mathrm{Rs}.9$ per $kg$ must be mixed with $27\mathrm{kg}$ of sugar costing $\mathrm{Rs}.7$ per $kg$ so that there may be gain of $10\%$ by selling the mixture at $\mathrm{Rs}.9.24$ per $kg$?

• A. $36$
• B. $42$
• C. $54$
• D. $63$

Answer 1

The correct option is D

$63$

Explanation for the correct option

Given, selling price of mixture ${\text{SP}}_{\mathrm{M}}=\mathrm{Rs}.9.24$
Cost of expensive sugar, ${\text{CP}}_{\mathrm{E}}=\mathrm{Rs}.9$
Cost of cheaper sugar, ${\text{CP}}_{\mathrm{C}}=\mathrm{Rs}.7$
Weight of cheaper sugar in the mixture $m_{\mathrm{C}}=27\mathrm{kg}$
Let $x$ be the quantity of the expensive sugar needed

We know that,
Gain$\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$

Thus, the cost price of $1\mathrm{kg}$ of mixture is
$\begin{array}{cc}& 10=\frac{9.24-\text{CP}}{\text{CP}}\times 100\\ \Rightarrow & \frac{10}{100}\text{CP}=9.24-\text{CP}\\ \Rightarrow & \text{CP}=\mathrm{Rs}.8.4\end{array}$

By the rule of allegation,
$\frac{\text{Quantity of cheaper variety}}{\text{Quantity of costlier variety}}=\frac{\text{Price of costlier variety}-\text{Mean price}}{\text{Mean price}-\text{Price of cheaper variety}}$

Thus,
$\begin{array}{cc}& \frac{m_C}{x}=\frac{{\text{CP}}_{\mathrm{E}}-{\text{CP}}_{\mathrm{M}}}{{\text{CP}}_{\mathrm{M}}-{\text{CP}}_{\mathrm{C}}}\\ \Rightarrow & \frac{27}{x}=\frac{9-8.4}{8.4-7}\\ \Rightarrow & x=63\mathrm{kg}\end{array}$

Hence, option (D) i.e. $63$ is correct.