Question

The liquids $X$ and $Y$ are mixed in the ratio of $3:2$ and the mixture is sold at $Rs11$ per liter at a profit of $10\%$. If the liquid $X$ costs $Rs2$ more per liter than $Y$, the cost of $X$ per liter is (in Rs.):

• A. $9.50$
• B. $10.80$
• C. $11.75$
• D. $11$

Answer 1

Answer: (B)

$10.80$

Let Cost of $X$ be $x$

Cost of $Y$ be $y$

and their volumes be $3K$ and $2K$

Now, $x=y+2$

Let original price be $P$

So, $P+10$% of $P=11$

$P+0.1P=11$

$1.1P=11$

$P=10$

So, $\frac{3Kx+2Ky}{3K+2K}=\frac{3x+2y}{5}=10$

$3x+2y=50$....(1)

and $x=y+2$....(2)

Using (2) in (1), we get

$\Rightarrow 3y+6+2y=50\Rightarrow 5y=44\Rightarrow y=\frac{44}{5}=8.8$ Rs.

$x=10.8$Rs.