Question

The liquids X and Y are mixed in the ratio $3:2$ and the mixture is sold at Rs. 11 per litre at a profit of $10\% $. If the liquid X costs Rs. 2 more per litre than Y, the cost of X per litre is (in Rs.):
A) 9.50
B) 10.80
C) 11.75
$) 11

Answer 1

Hint: We can assume the cost of each liquid to each variable. We can find the original price by equating 110% of the original price to the selling price. From the ratio, we can find how much part of 1 litre contains each liquid. Then we can equate the sum quantity of each liquid multiplied with its cost to the cost of 1 litre of the mixture. We can get another equation from the difference in the price of X and Y. Now we have 2 equations and solving for the variable gives us the required answer.

Complete step by step solution: Let us assume that the cost of liquid X per litre is x and the cost of Y per litre be y.
It is given that cost X is 2 more than the cost of Y.
$ \Rightarrow x = y + 2$
$ \Rightarrow x - y = 2$… (1)
We are given that the mixture is sold at a profit of 10% at Rs. 11. We need to find the original price.
Let the original price be A, then,
$A + 10\% A = 11$
$ \Rightarrow A + 0.1A = 11$
$ \Rightarrow 1.1A = 11$
$ \Rightarrow A = \dfrac{{11}}{{1.1}} = 10$
Therefore, the original price is Rs.10.
Now we have the liquids X and Y in the ratio $3:2$. We can interpret this in the following way,
If the mixture has 5 parts, then 3 parts of the mixture will be X and 2 parts will be Y.
For 1 litre of the mixture, $\dfrac{3}{5}$litre will be liquid X and $\dfrac{2}{5}$litre will be liquid Y.
So, we can write the cost of one litre of the mixture as,
$\dfrac{3}{5}x + \dfrac{2}{5}y = 10$
Multiplying throughout with 5, we get,
$3x + 2y = 50$ … (2)
Subtracting two times equation (1) from (2), we get,
\[
  {\text{ }}3x + 2y = 50 \\
  \underline {( + )2x - 2y = 4} \\
  {\text{ }}5x + 0y = 54 \\
 \]
$ \Rightarrow x = \dfrac{{54}}{5} = 10.8$

Therefore, the cost of liquid X per litre is Rs. 10.8

Note: In this problem, we are using the concepts of percentages, ratios, mathematical modelling and solving of linear equations. Mathematical modelling is the process of converting a word problem into mathematical equations. A ratio is used to compare the quantity of something to another thing. We can multiply or divide a ratio with the same value. While handling ratios, their order is important. Here the given ratio is a part to part ratio. In the given ratio we have 3 parts of X for every 2 parts of Y. Here the total parts will the sum of the ratio, i.e.,5. So we divide the ratio with the sum of the ratio. Then the new sum of the ratio will be equal to 1 and can apply to the equation.