Question

A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains $20\%$. Calculate the percentage of water in the mixture.

Answer 1

Hint: Let us assume that the milk of “m L” he professes to sell so the cost price of this pure milk of “m L” is Rs x. Now, let us assume he mixes n L of water into the milk then the volume of pure milk reduces to “(m – n) L” which he claims to be pure m L milk so he is selling this milk at the same cost price of “m L” pure milk but actually the price that milkman is bearing becomes less than x. Now, we know that m L milk costs Rs x so by unitary method 1 L milk cost is equal to the division of x by m then multiply this result by (m – n) L we get the actual cost of the milk that milkman is bearing which is the cost price of the milk that he is selling. It is given that the milkman is having a gain of $20\%$ so using the formula of gain which is given as $Gain=\dfrac{S.P.-C.P.}{C.P.}\times 100$ substitute the cost price, selling price and Gain. Now, we are asked to find the percentage of milk in the mixture which is equal to the division of n by m followed by multiplication with 100 so solve the equation of gain and find this percentage.

Complete step by step answer:
Let us assume that the cost price of “m L” pure milk which he professes to sell is Rs. x. Now, let us assume that he mixes “n L” of water in such a way that total litre of the solution remains “m L” then the volume of milk reduces to “m – n L” which he sells at the price of “m L” pure milk so his selling price becomes Rs x but his cost price for the milk has reduced because the quantity of milk has reduced.
Cost price of m L milk $=x$
Then by unitary method, the cost price of 1 L milk is equal to x divided by m.
Cost price of 1 L milk $=\dfrac{x}{m}$
Then cost price of (m – n) L milk $=\dfrac{x}{m}\left( m-n \right)$
Hence, the cost price of the impure milk that the milkman is selling is equal to;
$Rs\dfrac{x}{m}\left( m-n \right)$
Selling price of this “m L” impure milk which contains water is equal to Rs. x.
It is given that the gain percent that a milkman is having is equal to $20\%$.
We know the formula for gain percent as:
$Gain=\dfrac{S.P.-C.P.}{C.P.}\times 100$
In the above formula, S.P. represents selling price and C.P. represents the cost price. Substituting gain as 20, S.P. as x and C.P. as $\dfrac{x}{m}\left( m-n \right)$ in the above equation we get,
$20=\dfrac{x-\dfrac{x}{m}\left( m-n \right)}{\dfrac{x}{m}\left( m-n \right)}\times 100$
In the above equation, x will be cancelled out from the numerator and denominator.
$\begin{align}
  & 20=\dfrac{1-\dfrac{1}{m}\left( m-n \right)}{\dfrac{1}{m}\left( m-n \right)}\times 100 \\
 & \Rightarrow 20=\dfrac{\dfrac{m-\left( m-n \right)}{m}}{\dfrac{1}{m}\left( m-n \right)}\times 100 \\
 & \Rightarrow 20=\dfrac{n}{m-n}\times 100 \\
\end{align}$
Dividing 100 on both the sides we get,
$\begin{align}
  & \dfrac{20}{100}=\dfrac{n}{m-n} \\
 & \Rightarrow \dfrac{1}{5}=\dfrac{n}{m-n} \\
 & \Rightarrow m-n=5n \\
 & \Rightarrow m=6n \\
 & \Rightarrow \dfrac{n}{m}=\dfrac{1}{6} \\
\end{align}$
Now, we are asked to find the percentage of water in the mixture which is equal to n divided by m followed by multiplication with 100.
$\dfrac{n}{m}\times 100$
We have already found the ratio of n and m so putting that in the above equation we get,
$\begin{align}
   \dfrac{1}{6}\times 100 \\
  =\dfrac{50}{3}=16.66\% \\
\end{align}$

Hence, the percentage of water in the mixture is 16.66%.

Note: This question demands the knowledge of gain, how to calculate percentage. If you know these things then the question is pretty simple. In this problem they just add a little twist by not giving the exact selling price and cost price so here we are assuming the quantity of selling price and cost price and then solve the problem.