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

Let $x$ be the weights, in Kilograms, of pure water to be added. Let $y$ be the weight, in Kilograms, of the 10% solution.

Hence $x+100=y$
Let us now express the fact that the amount of salt in the pure water (which is 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%.
0 + 30% 100 = 10%$y$

Now by substituting the value of $y$ in terms of $x$, we get the equation as,
30% 100 = 10%$(x+100)$

$30\times 100=10(x+100)$

$3000=10x+1000$

$10x=2000$

$x=200$

Answer $x=200$ Kilograms.