In a 60 liter solution of acid and water, the ratio of acid and water is 3 : 7. How much(in liter) acid is to be mixed in the solution so that the ratio of acid and water in the resulting solution is 1 : 2?
In a 60 liter solution of acid and water, the ratio of acid and water is 3 : 7.
quantity of acid initially = $$60\times\ \frac{3}{10}$$ = 18
quantity of water initially =Â $$60\times\ \frac{7}{10}$$ = 42
Let's assume the quantity of acid mixed in the solution is 'y'.
$$\frac{18+y}{42}\ =\ \frac{1}{2}$$
36+2y = 42
2y = 42-36 = 6
quantity of acid mixed in the solution =Â y = 3
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