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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

1. $26$
2. $16$
3. $20$
4. $22$

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 $: \text {C}$

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