Options:
A) 7 : 2
B) 3 : 5
C) 3 : 4
D) 4 : 9
Show Answer
Answer:
Correct Answer: A
Solution:
- In vessel A,
Quantity of milk $ =\frac{5}{7} $ of the weight of mixture.
In vessel B,
Quantity of milk $ =\frac{8}{13} $ of the weight of mixture.
Average weight of milk in the mixture $ =\frac{9}{13} $
$ \therefore $ Required ratio $ =\frac{1}{13}:\frac{2}{91}=7:2 $