Question : 80 litres of a mixture of alcohol and water contains 60% alcohol. If 'X' litres of water is added to this mixture, then the alcohol percentage in the new mixture becomes 40%. What is the value of 'X'?
Option 1: 20
Option 2: 80
Option 3: 40
Option 4: 60
Correct Answer: 40
Solution :
Given,
The mixture of alcohol and water = 80 litres
Initial alcohol concentration = 60%
Amount of alcohol in 80 litres of mixture = 60% of 80
= $\frac{60}{100}\times80$ = 48 litres
Also, 40 percent alcohol in the new mixture after the addition of X litres of water.
So, according to the question,
48 = $\frac{40}{100}$ × (80 + X)
⇒ 48 = $\frac{2}{5}$ (80 + X)
⇒ 240 = 2(80 + X)
⇒ 240 = 160 + 2X
⇒ 2X = 80
⇒ X = 40
Hence, the correct answer is 40.
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