Math, asked by sabbahkhan4654, 8 months ago

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

Answered by amitnrw
2

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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