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?
Question: 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?
[A].
4 litres, 8 litres
6 litres, 6 litres
5 litres, 7 litres
7 litres, 5 litres
[A].
[B].
[C].
[D].
Answer: Option B
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
|
|
||||||
|
|
|||||||
| Ratio of two mixtures = | 1 | : | 1 | = 1 : 1. |
| 8 | 8 |
| So, quantity of mixture taken from each can = | 1 | x 12 | = 6 litres. | ||
| 2 |