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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 12litres of milk such that the ratio of water to milk is 3:5?


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Solution

Step 1: Determine the mean value.

It is given that the ratio of water to milk in the first can is 25%:75%i.e. 1:3 and in the second can is 50%:50% i.e. 1:1.

Let the cost of 1litre milk be Re.1.

The amount of milk contained in 1litre mixture from the first can is 34litres.

Therefore, the cost price of 1litre mixture from the first can will be 34×Re.1=Rs.34.

Similarly, the amount of milk contained in 1litre mixture from the second can is 12litres.

Therefore, the cost pricr of 1litre mixture from the second can will be 12×Re.1=Rs.12.

The required ratio of water to milk in 12litres of mixture from both the cans is 3:5.

Hence, the amount of milk contained in 1litre of the required mixture will be 58litres.

Therefore, the mean price will be 58×Re.1=Rs.58.

Step 2: Apply the rule of alligation to find the required ratio.

According to the rule of alligation, when different quantities of different ingredients are mixed together to produce a mixture of a mean value, the ratio of their quantities is inversely proportional to the difference in their cost from the mean value.

xy=34-5858-12=1818=11

Therefore, the required ratio of the quantities of the mixtures from the two cans is 1:1.

Hence, the quantity of mixture taken from each can should be 12×12litres=6litres.

Therefore, to get 12litres of milk such that the ratio of water to milk is 3:5, the milk vendor should mix 6litres of milk from each can.


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