Question

A litre of water is evaporated from 6 litres of sugar solution containing 4% of sugar. Find the percentage of sugar in the remaining solution?
\[\begin{align}
  & (\text{A) 4 }\!\!\%\!\!\text{ } \\
 & \text{(B) 5 }\!\!\%\!\!\text{ } \\
 & \text{(C) 4}\text{.2 }\!\!\%\!\!\text{ } \\
 & (\text{D) 4}\text{.8 }\!\!\%\!\!\text{ } \\
\end{align}\]

Answer 1

Hint: In the question, we were given that there is 4% of sugar in 6 litres of sugar solution. We should calculate the number of litres of sugar in sugar solution. We were given that 1 litre of water evaporated. So, we will calculate the number of litres of water after evaporation. We will find the percentage of sugar in the volume of water after evaporation.

Complete step by step solution:
In the question, we were given that there are 6 litres of sugar solution.
We were given that there is 4% of sugar in 6 litres of sugar solution.
Number of litres of sugar in sugar solution \[=\dfrac{4}{100}(6)=\dfrac{24}{100}\]
So, now we know that there are 0.24 litres of sugar in sugar solution.
In the question, we were given that one litre of water evaporated.
So,
Number of litres of water at present \[=6-1=5\]
So,
Now we know that after evaporation of one litre, we are having 5 litres of water remaining.
So, we are having 0.24 litres of sugar in 5 litres of water.
Now we have to find the percent of sugar in 5 litres of water.
Let the percentage of sugar in 5 litres of water is equal to x.
Now we get
\[\Rightarrow \dfrac{24}{100}=\left( \dfrac{x}{100} \right)\left( 5 \right)\]
Now by cross multiplication, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{24}{5}\% \\
 & \Rightarrow x=4.8\% \\
\end{align}\]
So, it is clear that there is \[4.8\%\] of sugar in 5 litres of water.
Hence, option (D) is correct.

Note: Students may have a misconception that if there are 0.24 litres of sugar in the 6 litres of solution, then there will be x litres of volume in 5 litres of volume.
\[\begin{align}
  & 0.24\to 6 \\
 & x\to 5 \\
\end{align}\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow 6x=(0.24)(5) \\
 & \Rightarrow x=0.2 \\
\end{align}\]
By using this x=0.2, they will calculate the percentage of sugar solution in 5 litres. Students should have a clear view that there will be no change in sugar content in the solution even after evaporation.