Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 4)
4.
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
Answer: Option
Explanation:
Let the cost of 1 litre milk be Re. 1
Milk in 1 litre mix. in 1st can = | 3 | litre, C.P. of 1 litre mix. in 1st can Re. | 3 |
4 | 4 |
Milk in 1 litre mix. in 2nd can = | 1 | litre, C.P. of 1 litre mix. in 2nd can Re. | 1 |
2 | 2 |
Milk in 1 litre of final mix. = | 5 | litre, Mean price = Re. | 5 |
8 | 8 |
By the rule of alligation, we have:
C.P. of 1 litre mixture in 1st can C.P. of 1 litre mixture in 2nd can | ||||||||
|
Mean Price
|
|
||||||
|
|
![]() |
1 | : | 1 | = 1 : 1. |
8 | 8 |
So, quantity of mixture taken from each can = | ![]() |
1 | x 12 | ![]() |
= 6 litres. |
2 |
Discussion:
74 comments Page 8 of 8.
Parvy Govil said:
10 years ago
Let x litres of A and y litres of B is extracted x+y = 12.
Milk quantity = 1/5x+1/2y = 3/5(4/5x+1/2 y).
x = 5 litres.
y = 7 litres.
Milk quantity = 1/5x+1/2y = 3/5(4/5x+1/2 y).
x = 5 litres.
y = 7 litres.
Shashi bhushan said:
10 years ago
Let x and y liters from the can to be mixed to make 12 liters of milk.
Can 1 has x/4 (water) and 3x/4 (milk).
Can 2 has y/2 (water) and y/2 (milk).
So new ratio of water to milk is.
(x/3+y/2)/(3x/4+y/2) = 3/5.....(1).
By the question.
x+y = 12.....(2).
To solve these equation.
x = 6, y = 6.
Can 1 has x/4 (water) and 3x/4 (milk).
Can 2 has y/2 (water) and y/2 (milk).
So new ratio of water to milk is.
(x/3+y/2)/(3x/4+y/2) = 3/5.....(1).
By the question.
x+y = 12.....(2).
To solve these equation.
x = 6, y = 6.
Megh said:
10 years ago
Can any one tell me how did you get 5/8?
Anu said:
10 years ago
Please somebody clarify the logic behind that question in short way.
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