Quantitative Aptitude Ques 1578
Question: 20 L of a mixture contains 20% alcohol and the rest water. If 4 L of water be mixed in it, the percentage of alcohol in the new mixture, will be
Options:
A) $ 33\frac{1}{3} $ %
B) $ 16\frac{2}{3} $ %
C) $ 25 $ %
D) $ 12\frac{1}{2} $ %
Show Answer
Answer:
Correct Answer: B
Solution:
- Amount of alcohol in 20 L of mixture $ =20%\text{of 20}\text{L=}\frac{20\times 20}{100}=4L $
$ \therefore $ Water in the mixture $ =20-4=16L $ Now, 4 L of water is further added
$ \therefore $ Amount of water $ =16+4=20L $
$ \therefore $ Percentage of alcohol in new mixture $ =\frac{Amountofalcohol}{Totalmixture}\times 100% $ $ =\frac{4}{24}\times 100=\frac{100}{6} $ % $ =\frac{50}{3} $ % $ =16\frac{2}{3} $ %
BETA
