A milk vendor has 2 cans of milk. The 1st 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 12l of milk such that the ratio of water to milk is 3:5?
A milk vendor has 2 cans of milk. The 1st 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 12l of milk such that the ratio of water to milk is 3:5?
Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the the first container = .75x
Amount of water in x litres of the the first container = .25x
Amount of milk in (12-x) litres of the the second container = .5(12-x)
Amount of water in (12-x) litres of the the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3 : 5
⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5⇒(6−.25x)/(.25x+6)=3/5⇒30−1.25x=.75x+18⇒2x=12⇒x=6
Since x = 6, 12-x = 12-6 = 6
Hence 6 and 6 litres of milk should mixed from the first and second container respectively
Let x and (12-x) litres of milk be mixed from the first and second container respectively
Amount of milk in x litres of the the first container = .75x
Amount of water in x litres of the the first container = .25x
Amount of milk in (12-x) litres of the the second container = .5(12-x)
Amount of water in (12-x) litres of the the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3 : 5
⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5⇒(6−.25x)/(.25x+6)=3/5⇒30−1.25x=.75x+18⇒2x=12⇒x=6
Since x = 6, 12-x = 12-6 = 6
Hence 6 and 6 litres of milk should mixed from the first and second container respectively
