Question

How many kilograms of pure water to be added to 100 kilograms of a 30% saline solution to make it a 10% saline solution.

Answer 1

Hint: Assume that x kilograms of pure water is added to 100 kilograms of a 30% saline solution to make it a 10% saline solution. The initial weight of the saline solution is 100 kilograms. There is 30% salt in the saline solution. The initial weight of salt is 30% of 100 kilograms. After adding x kilograms of pure water, the weight of the new saline solution is
\[\left( 100+x \right)\] . Since we have only added pure water to the initial solution, the weight of the salt in the new saline solution is also the same as that of initial saline solution. The weight of the salt in the new solution is 30 kg. Now, calculate the percentage of the salt in the \[\left( 100+x \right)\] kilograms of saline solution. It is given that the percentage of salt in the new solution is 10%. Now, solve it further and get the value of x.

Complete step-by-step answer:
According to the question, we have 100 kilograms of a 30% solution and we have to find the amount of pure water that can be added to make it a 10% saline solution.
First of all, let us assume that x kilograms of pure water are added to 100 kilograms of a 30% saline solution to make it a 10% saline solution.
We know that a saline solution is composed of salt and pure water.
In \[{{1}^{st}}\] case, it is given that there is 30% salt in 100 kilograms of saline solution.
The weight of salt present in 100 kilograms of saline solution = 30% of 100 kilograms =
\[\dfrac{30}{100}\times 100=30\] kilograms …………………………………(1)
The weight of the saline solution = 100 kilograms ………………………………..(2 )
In \[{{2}^{nd}}\] case, we have
After adding x kilograms of pure water to 100 kilograms of a 30% saline solution, we get a 10% saline solution.
As x kilograms of pure water is being added to 100 kilograms of saline solution. So,
The weight of the new solution after adding x kilograms of pure water = \[\left( 100+x \right)\] kilograms ………………………..(3)
The percentage of salt in the new saline solution = 10% ………………………(4)
Since we have only added pure water to the initial solution, the weight of the salt in the new saline solution is also the same as that of initial saline solution.
From equation (1), we weigh salt in the initial saline solution. So,
The weight of salt present in the new saline solution = 30 kg.
The percentage of salt in the new saline solution = \[\left( \dfrac{30}{100+x}\times 100 \right)\] ………………………………..(5)
From equation (4), we have the percentage of the salt in the new saline solution.
Now, comparing equation (4) and equation (5), we get
\[\begin{align}
  & \Rightarrow 10=\dfrac{30}{100+x}\times 100 \\
 & \Rightarrow 10=\dfrac{3000}{100+x} \\
 & \Rightarrow 10\left( 100+x \right)=3000 \\
 & \Rightarrow 1000+10x=3000 \\
 & \Rightarrow 10x=3000-1000 \\
 & \Rightarrow 10x=2000 \\
 & \Rightarrow x=\dfrac{2000}{10} \\
 & \Rightarrow x=200 \\
\end{align}\]
Therefore, 200 kilograms of pure water must be added to 100 kilograms of a 30% saline solution to make it a 10% saline solution.

Note: In this question, one might make a silly mistake while comparing 10% and \[\left( \dfrac{30}{100+x}\times 100 \right)\] . While comparing these two, one might take the equation as \[10%=\left( \dfrac{30}{100+x}\times 100 \right)\] and then solve it as \[\dfrac{10}{100}=\left( \dfrac{30}{100+x}\times 100 \right)\] . This is wrong because the term \[\left( \dfrac{30}{100+x}\times 100 \right)\] is already the percentage of the weight of salt in the new saline solution. So, there is no need to add ‘%’ after 10. Hence, the correct equation that we should have after comparing 10% and \[\left( \dfrac{30}{100+x}\times 100 \right)\] is \[10=\left( \dfrac{30}{100+x}\times 100 \right)\] .