Question

A container has a mixture of kerosene and water in a ratio of 7 : 5. when 9 litres of mixture are taken off and the container is filled with 9 litres of water, the ratio between kerosene and water becomes 7 : 9. how many litres of kerosene were initially in the container?

Answer 1

21 liters of kerosene

Answer 2

Answer: 21 L

Explanation: Given : Ratio of kerosene to water = 7: 5

let x =the multiplier to determine the total amount of solution

Then quantity of kerosene = 7 x

quantity of water= 5 x

Total quantity = 7x+5x = 12x

Mixture removed = 9 liters

Thus kerosene removed =$\frac{7x}{12x}\times 9=\frac{21}{4}$

Water removed =$\frac{5x}{12x}\times 9=\frac{15}{4}$

Now water added = 9 L

New ratio = 7: 9

Thus $\frac{7x-\frac{21}{4}}{5x-\frac{15}{4}+9}=\frac{7}{9}$

Solving for x , x= 3

Thus kerosene present will be =$7\times x=7\times 3=21L$