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A trader sells $10$ litres of a mixture of paints $\text{A}$ and $\text{B}$, where the amount of $\text{B}$ in the mixture does not exceed that of $\text{A}$. The cost of paint $\text{A}$ per litre is Rs. $8$ more than that of paint $\text{B}$. If the trader sells the entire mixture for Rs. $264$ and makes a profit of $10\%$, then the highest possible cost of paint $\text{B}$, in Rs. per litre, is

- $26$
- $16$
- $20$
- $22$

1 vote

Given that,

A trader sells $10$ litres of a mixture of paints $ \text{A} $ and $\text {B},$ where the amount of $\text{B}$ in the mixture does not exceed that of $\text{A}$. That means mixture of $\text{A}$ and $\text{B}$ should be same to get the highest cost of paint $\text{B}$.

- $ \text{A} \rightarrow 5 \; \text{litres} $
- $ \text {B} \rightarrow 5 \; \text{litres} $

Let the cost price of paint $\text {B}$ be $x.$

So, the cost price of $ 5 \; \text {litres}$ mixture of paint $\text{B} = 5x \quad \longrightarrow (1)$

And the cost price of paint $ \text {A}$ per litre is $ \text {Rs.}\;8$ more than that of paint $ \text{B} = x + 8$

So, the cost price of $ 5 \; \text {litres}$ mixture of paint $\text{A} = (x + 8) \times 5 = 5x+40 \quad \longrightarrow (2)$

If the trader sells the entire mixture for $ \text {Rs.} \;264$ and makes a profit of $ 10 \%.$

We can get the cost price of $10 \; \text{litres}$ of a mixture.

Using the formula of cost price$: \boxed {\text {cost price} = \frac{ \text{selling price}} { 1 + \frac{\text{profit}}{100}}}$

Now, the cost price of $10$ litres mixture $= \frac {264} { 1 + \frac {10}{100}} $

$ \qquad = \frac {264} { \frac {11}{10}}$

$ \qquad = \frac { 264 \times 10}{11}$

$\qquad = 240 $

From the equation $(1),$ and $(2),$ we can write

$ 5x + 5x + 40 = 240 $

$ \Rightarrow 10x = 240 – 40 $

$ \Rightarrow 10x = 200 $

$ \Rightarrow x = 20 $

$\therefore$ The cost price of paint $ \text{B} = x = 20.$

Correct Answer $: \text {C}$