The liquids $X$ and $Y$ are mixed in the ratio $3:2$ and the mixture is sold at Rs. 11 per liter at a profit of 10%. If the liquid $X$ costs Rs. 2 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
607.8k+ views
Hint: Assume the volume of liquids to be some variable according to the ratio given. Find the total selling price of liquid using Rs. 11 per liter rate. Then determine their cost using the profit percent given. Finally use the cost per liter rate of liquids X and Y and compare it with total cost.
Complete step-by-step answer:
According to the question, the liquids $X$ and $Y$ are mixed in the ratio $3:2$.
Let $3x$ and $2x$ are the respective volumes in which they are mixed. So the total volume of the mixture is $5x$.
Now, this mixture is sold at Rs. 11 per liter. So the total selling price i.e. S.P. is given as:
$ \Rightarrow $S.P. $ = 11 \times 5x = 55x$
And this is sold at 10% profit. So we have:
$ \Rightarrow $C.P. + 10% of C.P. = S.P.
where C.P. is the total cost price.
Putting the value of S.P. and simplifying further, we’ll get:
$ \Rightarrow $C.P. + $\dfrac{{10}}{{100}} \times $ C.P. $ = 55x$
$ \Rightarrow $C.P. + $\dfrac{1}{{10}} \times $ C.P. $ = 55x$
$ \Rightarrow $$\dfrac{{11}}{{10}} \times $ C.P. $ = 55x$
$ \Rightarrow $C.P. $ = 50x$
So we have a total cost price as $50x$.
Further it is given that, liquid $X$ costs Rs. 2 more per liter than $Y$. So the cost price of liquid $X$ is Rs. $a + 2$ per liter and that of liquid $Y$ is Rs. $a$ per liter.
Thus the total cost of $3x$ liter of $X$ and $2x$ liter of $Y$ is given as:
$ \Rightarrow $C.P. $ = 3x\left( {a + 2} \right) + 2xa = 5xa + 6x$
Putting the value of total cost price calculated above, we’ll get:
$
\Rightarrow 5xa + 6x = 50x \\
\Rightarrow 5xa = 44x \\
\Rightarrow a = \dfrac{{44}}{5} = 8.8 \\
$
Thus the cost price of liquid $X$ is $a + 2 = 10.8$.
(B) is the correct option.
Note: Profit or loss is always calculated over cost price and not over selling price. In this problem, the profit percentage was given but cost price was not given. So we were required to find the cost price first to compare the quantities and solve the problem.
If the cost price is more than the selling price, loss is seen. If the selling price is more than the cost price, profit is seen.
Complete step-by-step answer:
According to the question, the liquids $X$ and $Y$ are mixed in the ratio $3:2$.
Let $3x$ and $2x$ are the respective volumes in which they are mixed. So the total volume of the mixture is $5x$.
Now, this mixture is sold at Rs. 11 per liter. So the total selling price i.e. S.P. is given as:
$ \Rightarrow $S.P. $ = 11 \times 5x = 55x$
And this is sold at 10% profit. So we have:
$ \Rightarrow $C.P. + 10% of C.P. = S.P.
where C.P. is the total cost price.
Putting the value of S.P. and simplifying further, we’ll get:
$ \Rightarrow $C.P. + $\dfrac{{10}}{{100}} \times $ C.P. $ = 55x$
$ \Rightarrow $C.P. + $\dfrac{1}{{10}} \times $ C.P. $ = 55x$
$ \Rightarrow $$\dfrac{{11}}{{10}} \times $ C.P. $ = 55x$
$ \Rightarrow $C.P. $ = 50x$
So we have a total cost price as $50x$.
Further it is given that, liquid $X$ costs Rs. 2 more per liter than $Y$. So the cost price of liquid $X$ is Rs. $a + 2$ per liter and that of liquid $Y$ is Rs. $a$ per liter.
Thus the total cost of $3x$ liter of $X$ and $2x$ liter of $Y$ is given as:
$ \Rightarrow $C.P. $ = 3x\left( {a + 2} \right) + 2xa = 5xa + 6x$
Putting the value of total cost price calculated above, we’ll get:
$
\Rightarrow 5xa + 6x = 50x \\
\Rightarrow 5xa = 44x \\
\Rightarrow a = \dfrac{{44}}{5} = 8.8 \\
$
Thus the cost price of liquid $X$ is $a + 2 = 10.8$.
(B) is the correct option.
Note: Profit or loss is always calculated over cost price and not over selling price. In this problem, the profit percentage was given but cost price was not given. So we were required to find the cost price first to compare the quantities and solve the problem.
If the cost price is more than the selling price, loss is seen. If the selling price is more than the cost price, profit is seen.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Give me the opposite gender of Duck class 8 english CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Advantages and disadvantages of science

Full form of STD, ISD and PCO

Right to vote is a AFundamental Right BFundamental class 8 social science CBSE

