Question: In a 120 L of mixture of milk and water, water is only 25%. The milkman sold 20 L of this mixture and then he added 16.2 L of pure milk and 3.8 L of pure water in the remaining mixture. What is the percentage of water in the final mixture? [LIC (AAO) 2014]
Options:
A) 22%
B) 21%
C) 24%
D) 25%
E) 20%
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Quantity of water in original mixture
$ =120\times \frac{25}{100}=30L $
and quantity of milk in original mixture
$ =120-30=90L $
Now, milkman sold 20 L of mixture,
So, remaining mixture $ =120-20=100L $
$ \therefore $ Quantity of water in 100 L mixture
$ =100\times \frac{25}{100}=25L $
and quantity of milk in 100 L mixture
$ =100-25=75L $
Now, milkman made new mixture.
$ \therefore $ Quantity of milk in new mixture $ 75+16.2=91.2,L $
and quantity of water in new mixture
$ =25+3.8=28.8L $
Now, percentage of water in new mixture
$ =\frac{28.8}{(91.2+28.8)}\times 100 $ %
$ =\frac{2880}{120} $ % $ =\frac{288}{12} $ %=24%
BETA
