Question 69

80 liters of a mixture of alcohol and water contains 60% alcohol. If 'X' liters of water is added to this mixture, then the percentage of alcohol in the new mixture becomes 40%. What is the value of X?

Solution

80 liters of a mixture of alcohol and water contains 60% alcohol.

Initial quantity of alcohol = 80 of 60% = 48 liters

Initial quantity of water = 80-48 = 32 liters

If 'X' liters of water is added to this mixture, then the percentage of alcohol in the new mixture becomes 40%.

$$\frac{48}{32+X}\ =\frac{40}{100-40}$$

$$\frac{48}{32+X}\ =\frac{40}{60}$$
$$\frac{24}{32+X}\ =\frac{1}{3}$$
32+X = 72

X = 72-32

Value of X = 40 liters

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