Question

A solution containing water and a liquid concentrated has $60\%$ concentrate. By adding $20$ liters of water to the solution the concentrates reduced to $40\%$ of the solution. How many liters of the original solution is concentrated?
A. $36$
B. $30$
C. $24$
D. $16$

Answer 1

Hint: First we will assume the volume of water and concentrated liquid and find the volume of the solution. From the given concentration we will find the reduced volume. Now the $20$ liters of water are added then find the volume of the solution. After adding water to the solution, the concentration of the solution is reduced and the volume of the solution. Now equate both volumes to find the volume of the solution.

Complete step-by-step solution
Let us assume the
Volume of water is ${{V}_{W}}=Y$ liters
Volume of liquid is ${{V}_{l}}=Z$ liters
Now the volume of the solution is
$\begin{align}
  & {{V}_{s}}={{V}_{w}}+{{V}_{l}} \\
 & =Y+Z\text{ }
\end{align}$
Given that the Solution is concentrated at $60\%$ then volume \[\]
$\begin{align}
  & {{V}_{con}}=60\%\times {{V}_{s}}\text{ liters} \\
 & {{V}_{con}}=0.6{{V}_{s}}\text{ liters}
\end{align}$
If we add $20$ liters of water to the solution then the volume of water is changed to
${{V}_{w}}=Y+20\text{ liters}$
And the volume of the solution is changed to
${{V}_{s+20}}=\left( Y+20 \right)+Z\text{ liters}$
After adding the $20$ liters of water the concentration of the solution is reduced to $40\%$ then the volume is
$\begin{align}
  & {{V}_{con}}=40\%\times {{V}_{s+20}} \\
 & =0.4{{V}_{s+20}}
\end{align}$
Equating the both the volumes then we get
$\begin{align}
  & 0.6{{V}_{s}}=0.4{{V}_{s+20}} \\
 & 0.6{{V}_{s}}=0.4\left( Y+Z+20 \right) \\
 & 0.6{{V}_{s}}=0.4{{V}_{s}}+0.4\left( 20 \right) \\
 & 0.6{{V}_{s}}-0.4{{V}_{s}}=8 \\
 & 0.2{{V}_{s}}=8 \\
 & {{V}_{s}}=40
\end{align}$
Hence $40$ liters of the solution are present.

Note: Find the volumes of the solutions using the concentrations carefully. You can also directly equate
$\begin{align}
  & 0.6{{V}_{s}}=0.4\left( {{V}_{s}}+20 \right) \\
 & 0.6{{V}_{s}}-0.4{{V}_{s}}=0.4\left( 20 \right) \\
 & 0.2{{V}_{s}}=8 \\
 & {{V}_{s}}=40
\end{align}$
From the above calculation we can say $40$ liters of solution are present.