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A mild vendor has $2$ cans of milk. $\dfrac{3}{4}l \times x$ The first one contains $25\%$ water and rest as milk while the second can contain $50\%$ water. How much milk should he mix from each of the containers such that he gets $12$ litres milk such that the ratio of water to milk is $3:5$?

Last updated date: 13th Jun 2024
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Hint: Assume the cost of milk in every mixture as a variable. Then find the cost price of milk in both the mixtures. At last use allegation to get the ratio of milk we want from the mixture.

Let the cost of milk be Rs$x$.
Now the first one contains $25\%$ water and $75\%$as milk.
Hence milk in $1$ litre mixture in can $1$ $= 75\%$ of $1$Litre
$= \dfrac{3}{4}l$
Cost price of $1l$ mixture of can $1$$= \dfrac{3}{4}l \times x$$ =$ Rs $\dfrac{3}{4}x$
Similarly for can $2$ it contains $50\%$ water and $50\%$ milk
Milk in $1$Litre mixture in can $2$$= 50\% \times 1l =$$\dfrac{1}{2}l$.
Cost price of $1l$ mixture of can $2$$=$ Rs $\dfrac{x}{2}$.
Ratio of water to milk is $3:5$ in the final mixture.
Hence for every $1l$ of mixture, we will have $\dfrac{5}{{5 + 3}} = \dfrac{5}{8}l$ of milk.
Cost price of $1l$ milk in the final mixture$=$ Rs$\dfrac{5}{8}x$
Hence mean price$=$ Rs$\dfrac{5}{8}x$
By using the concept of allegation,
$\left( {\dfrac{{{\text{Quantity from can 1}}}}{{{\text{Quantity from can 2}}}}} \right) = \left( {\dfrac{{{\text{Cost price of can 1}} - {\text{mean price}}}}{{{\text{mean price }} - {\text{cost price of can 2}}}}} \right)$
$\dfrac{{\dfrac{3}{4}x - \dfrac{5}{8}x}}{{\dfrac{5}{8}x - \dfrac{x}{2}}} = \dfrac{{\dfrac{x}{8}}}{{\dfrac{x}{8}}} = 1:1$
Hence quantity from both the cans should be equal’
Since the total quantity is $12l$, therefore the quantity taken from each can should be $6l$.

Note: We can also do it by using the basic method where we can use the data interpretation which is that as in can $1$, $1l$ of water comes with the $3l$ of milk and from can $2$,$1l$ of water comes with the $1l$ of milk. Then we can find the amount of water in the final mixture by using the ratio given. Finally we can use hit and trial to get the capacity of each drum.