In a vessel,milk is 60% less than water. When X liter milk is added to mixture then ratio of milk and water become 3:5. Now Y liter water is extracted from mixture then this ratio become 7:10. Now 30 liter of milk is added again then new ratio of milk and eater eater becomes 4:5. Find the value of x+y/3?
Answers
Given : In a vessel,milk is 60% less than water. When X liter milk is added to mixture then ratio of milk and water become 3:5. Now Y liter water is extracted from mixture then this ratio become 7:10. Now 30 liter of milk is added again then new ratio of milk and eater eater becomes 4:5
To find : the value of (x+y)/3
Solution:
Let Say water = W
ilk is 60% less than water.
=> Milk = W - (60/100)W = 0.4W
Water = W
Milk = 0.4W
Mixture = 1.4W
X liter milk is added ratio of milk and water become 3:5.
=> (0.4W + X) : W = 3 : 5
=> 2W + 5X = 3W
=> W = 5X
=> X = W/5 = 0.2W
Water = W
Milk = 0.4W + 0.2W = 0.6W
Mixture = 1.6W
Y liter water is extracted from mixture then this ratio become 7:10
0.6W : (W - Y) = 7 : 10
=> 6W = 7W - 7Y
=> W = 7Y
=> Y = W/7
Water = W - W/7 = 6W/7
Milk = 0.6W
Mixture = 6W/7 + 0.6W = 10.2W/7
Now 30 liter of milk is added again then new ratio of milk and water becomes 4:5
(0.6W + 30) : 6W/7 = 4 : 5
=> 21 W + 1050 = 24W
=> 3W = 1050
=> W = 350
X = 0.2W = 0.2 * 350 = 70 litre
Y = W/7 = 350/7 = 50 litre
( X + Y)/3 = (70 + 50)/3 = 40 Litre
Milk Water
Initial 140 350
70 L Milk added 210 350 3 : 5
50 L water extracted 210 300 7 : 10
70 L Milk added 240 300 4 : 5
value of (x+y)/3 = 40
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