DRDO Multi tasking staff Model paper 2

Let the initial quantity of alcohol in the mixture = x liter
So, the initial quantity of water in the mixture = (20 – x) liters
According to the question ∶
(x + 1) = (20 – x) × 50/100
(x + 1) = (20 – x) × 1/2
2x + 2 = 20 – x
3x = 18
x = 6
The initial quantity of alcohol in the mixture = 6 liters
The initial quantity of water in the mixture = 20 – 6 = 14 liters
Required ratio = 6 ∶ 14 = 3 ∶ 7