# How many Kilos of Pure water to be added to 100 Kilos of 30% saline solution to make it 10% saline solution.

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Let x be the weights of pure water to be added.

Let y be the weight of the 10% solution.

Then:

x + 100 = y

Now Let us auume that the amount of salt in the pure water is 0.

Given the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%.

Then we could write:

0 + 30% 100 = 10% y

Substitute y by x + 100 in the last equation and solve.

30% 100 = 10% (x + 100)

==> 30 = (x+100)/10

==> 300 = x+ 100

==> x= 200 kg.

Given:

Weight of original Saline solution = w1 = 100 kg

Concentration of the original solution = 30%

Since the quantity of solution and additional water is given in terms of weight, we assume that the concentration is in terms of weight.

Therefore:

Weight of salt in the original solution = 100*(30/100) = 30 kg

When we add water to the original solution the weight of salt in continues to be 30 kg.

Let:

w2 = weight of solution with when diluted to 10 percent concentration.

Then:

(Weight of salt)*100/w2 = Concentration percentage of solution

Substituting the actual weight of salt and required concentration percentage:

30*100/w2 = 10

==> w2 = 30*100/10 = 300 kg

The water to be added = (Weight of 10% solution) - (Weight of 10% solution)

= w2 - w1 = 300 - 100 = 200kg

Answer:

200 kg of pure water is to be added