Carcass wrote:
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
B. 6 litres, 6 litres
C. 5 litres, 7 litres
D. 7 litres, 5 litres
E 8 litres, 7litres
GIVEN: The first can contains 25% water, and second can contains 50% water
We want 12 liters of milk such that the ratio of water to milk is 3 : 5
In other words, we want the resulting mixture to be
3/8 water
Let x = the volume of 25% water needed
So, 12 - x = the volume of 50% water needed (since we want a total volume of 12 liters)
So, the total volume of water in the resulting 12 liters of mixture =
0.25x + 0.5(12 - x)Since we want the resulting
12-liter mixture to be
3/8 water, we can write: [
0.25x + 0.5(12 - x)]/
12 =
3/8 Simplify numerator: (6 - 0.25x)/12 = 3/8
Cross multiply: 48 - 2x = 36
Solve: x = 6
So we want 6 liters of the first milk, and 6 liters of the second milk
Answer: B