Question

How many Kilograms of Pure water is to be added to $100$ Kilograms of a $30\%$ saline solution to make it a $10\%$ saline solution.

• A. $200$ kg
• B. $300$ kg
• C. $400$ kg
• D. $500$ kg

Answer 1

The correct option is A $200$ kg

Let $x$ be the weights, in kilograms, of pure water to be added.

Let $y$ be the weight in kilograms of the $10\%$ solution.

$\therefore$ $x+100=y$ ---- ( 1 )

Let us now express the fact that the amount of salt in the pure water pluse the amount of salt in the $30\%$ solution is equal to the amount of salt in the final saline solution at $10\%$

$\therefore$ $0+30\%100=10\%y$ ---- ( 2 )

Substituting ( 1 ) in ( 2 ) we get,

$\Rightarrow$ $30\%100=10\%(x+100)$

$\Rightarrow$ $\frac{30}{100}\times 100=\frac{10}{100}(x+100)$

$\Rightarrow$ $3000=10x+1000$

$\Rightarrow$ $10x=2000$

$\Rightarrow$ $x=\frac{2000}{10}$

$\therefore$ $x=200$

$\therefore$ The required answer is $200$ kilograms.