Question

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 liters of milk such that the ratio of water to milk is 3:5?

Answer 1

Hint: Suppose the volumes of milk (Mixture) in both the cans as 100L and the volumes taken from both the cans to make another mixture as two different variables. Calculate the amount of pure milk and water in the volumes taken from the cans with the help of the percentages of water and pure milk given in both the cans. From two equations in terms of both the variables and solve them to get the answer.

Complete Step-by-Step solution:
Let us suppose the amount of milk(pure milk + water) taken from both the cans are ${{V}_{1}}$ liters and ${{V}_{2}}$ liters respectively.
Now it is given that the first can contains 25% of water and rest as milk i.e. amount of milk should be 100 – 25 = 75%. And the amount of water in the second can is 50%, it means the amount of pure milk is 100 – 50 = 50%.

So, let us suppose the total amount of milk (pure milk + water) in both the cans be 100 L. It means 25% of total amount i.e. $\dfrac{25}{100}\times 100=25L$ is an amount of water, and hence amount of pure milk in first can would be 100 – 25 = 75L.
Similarly, amount of water and pure milk in the second can would be 50L and 50L both as percentage of them in the mixture is 50% and total amount of milk in the second tank is 100L
Now, coming to the question, it is given that the amount taken from both the cans is collectively 12L i.e. the sum of amounts of milk from both the cans is 12L. Hence, we can get equation in ${{V}_{1}}$ and ${{V}_{2}}$ as
${{V}_{1}}+{{V}_{2}}=12.............(i)$
where ${{V}_{1}}$ and ${{V}_{2}}$ are the amounts of milk (mixture) taken from both the cans as supposed initially in the solution.
Now, it is also given in the question that the mixture formed by taking ${{V}_{1}}L$ from first can and ${{V}_{2}}L$ from second can, have the ratio of water to water as 3 : 5
Now, Let us calculate the exact amounts of pure milk and water in the ${{V}_{1}}$ volume taken from first can and similarly, we need to calculate the exact amounts of pure milk and water in volume ${{V}_{2}}$ taken from second can
Now, as we know,
100L of volume in first can is containing 25L of water
So, 1L of volume will contain $\dfrac{25}{100}L$ of water
So, ${{V}_{1}}$ volume taken from first can will contain
$\dfrac{25}{100}\times {{V}_{1}}=\dfrac{{{V}_{1}}}{4}\text{ of water}$
Similarly, we can calculate volume of pure milk in volume ${{V}_{1}}$ as:
The 100L volume in the first can is containing 75L of pure milk.
So, 1L volume will contain $\dfrac{75}{100}=\dfrac{3}{4}L$ of pure milk
Hence, ${{V}_{1}}L$ will contain ${{V}_{1}}\times \dfrac{3}{4}=\dfrac{3{{V}_{1}}}{4}L$ of pure milk
Now, with the similar approach we can calculate amounts of pure milk and water in volume of ${{V}_{2}}$ , taken from second can as:
100L of volume in the second can contains 50L of water.
So, ${{V}_{2}}L$ of volume will contain $\dfrac{50}{100}\times {{V}_{2}}=\dfrac{{{V}_{2}}}{2}L$of water
Similarly,
100L of volume in the second can contains 50L of pure milk.
So, ${{V}_{2}}L$ of volume will contain $\dfrac{50}{100}\times {{V}_{2}}=\dfrac{{{V}_{2}}}{2}L$of pure milk
Hence, amount of water and pure milk in the mixture can be calculate by using the above information as
Total amount of water in 12L mixture = $\dfrac{{{V}_{1}}}{4}+\dfrac{{{V}_{2}}}{2}=\dfrac{{{V}_{1}}+2{{V}_{2}}}{4}$
And total amount of pure milk in 12L mixture $=\dfrac{3{{V}_{1}}}{4}+\dfrac{{{V}_{2}}}{2}=\dfrac{3{{V}_{1}}+2{{V}_{2}}}{4}$
Now, it is given that ratio of the amount of water to milk in a 12L mixture in 3:5. So, can get equation as
\[\begin{align}
  & \dfrac{\text{Total amount of water in 12L mixture}}{\text{Total amount of pure milk in 12L mixture}}=\dfrac{3}{5} \\
 & \Rightarrow \dfrac{\left( \dfrac{{{V}_{1}}+2{{V}_{2}}}{4} \right)}{\left( \dfrac{3{{V}_{1}}+2{{V}_{2}}}{4} \right)}=\dfrac{3}{5} \\
 & \Rightarrow \dfrac{{{V}_{1}}+2{{V}_{2}}}{3{{V}_{1}}+2{{V}_{2}}}=\dfrac{3}{5} \\
\end{align}\]
On cross multiplying the above equation, we get expression as
\[\begin{align}
  & 5{{V}_{1}}+10{{V}_{2}}=9{{V}_{1}}+6{{V}_{2}} \\
 & 10{{V}_{2}}-6{{V}_{2}}=9{{V}_{1}}-5{{V}_{1}} \\
 & 4{{V}_{2}}=4{{V}_{1}} \\
 & \Rightarrow {{V}_{1}}={{V}_{2}}..........(ii) \\
\end{align}\]
Now, we know sum of volume ${{V}_{1}}$ and ${{V}_{2}}$ is 12L from the expression of equation (i).So, Put \[{{V}_{1}}={{V}_{2}}\] from the equation (ii) to the equation (i).Hence, we get
\[\begin{align}
  & {{V}_{1}}+{{V}_{2}}=12 \\
 & {{V}_{1}}+{{V}_{1}}=2{{V}_{1}}=12 \\
 & {{V}_{1}}=6 \\
\end{align}\]
Hence, ${{V}_{2}}=6$ from equation (ii)
So, the volume of mixture (pure milk + water) from both the cans are 6L from each.
Hence answer is 6L and 6L

Note: One may suppose the total volume of the cans as variables or any other value except 100L as well. As we don’t need the exact volume of the cans, we need to use the ratios of milk and water only. So, the answer will be wrong if someone supposes the volume of cans to any other volume except 100L. So, don’t confuse the volumes. 100L is used for the simplicity of the problem. As 25% of 100 is 25 and 50% of 100 is 50. So, 100 is the number which plays an important role with these kind of questions
Please make sure that amount of water in 12L mixture calculates in terms of ${{V}_{1}}$ and ${{V}_{2}}$ should be directly proportional to the value of it in the given ratio and same with the amount of pure milk as well. As one may go wrong if he/she write the equation as \[\dfrac{\dfrac{{{V}_{1}}+2{{V}_{2}}}{4}}{\dfrac{3{{V}_{1}}+2{{V}_{2}}}{4}}=\dfrac{5}{3}\] ,because left hand side is the ratio of amount from water to pure milk and right hand side is denoting ratio of amounts from pure milk to water. So, don’t confuse this step as well.