Question

A 6 litre vessel contains milk and water in ratio 71:93 and another 12 litre vessel contains the ratio of milk and water 21:29.x litres of liquid is withdrawn from each vessel and then put it into another vessel simultaneously. if the ratio of milk and water became equal in the two vessels find the value of x. quora

Answer 1

6 Liter vessel:
Liquid removed contains 71x/164 liters of milk and 93x/164 liters of water.
Liquid remaining contains 71(6-x)/164 liters of milk and 93(6-x)/164 liters of water
Liquid added contains 21x/50 liters of milk and 29x/50 liters of water
Resulting contents are 71(6-x)/164 + 21x/50 liters of milk and 93(6-x)/164 + 29x/50 liters of water
Milk: (426 - 71x)/164 + 21x/50 = (21300 - 106x)/8200
Water: (558 - 93x)/164 + 29x/50 = (27900 + 106x)/8200

12 Liter vessel:
Liquid removed contains 21x/50 liters of milk and 29x/50 liters of water.
Liquid remaining contains 21(12-x)/50 liters of milk and 29(12-x)/50 liters of water
Liquid added contains 71x/164 liters of milk and 93x/164 liters of water.
Resulting contents are 21(12-x)/50 + 71x/164 liters of milk and 29(12-x)/50 + 93x/164 liters of water
Milk: (252 - 21x)/50 + 71x/164 = (41328 + 106x)/8200
Water: (348 - 29x)/50 + 93x/164 = (57072 - 106x)/8200

The ratio of milk to water is the same in both vessels:
(41328 + 106x)/(57072 - 106x) = (21300 - 106x)/(27900 + 106x)
(41328 + 106x)(27900 + 106x) = (21300 - 106x)(57072 - 106x)
(41328)(27900) + (27900 + 41328)(106x) + (106x)² = (21300)(57072) - (21300 + 57072)(106x) + (106x)²
(27900 + 41328 + 21300 + 57072)(106x) = (21300)(57072) - (41328)(27900)
15645600x = 62582400
x = 4

Answer 2

Answer:

If both drawn quantity is same from two beaker use this formula, Here both quantity is x ltr, so

Step-by-step explanation:

xy/x+y =6x12/6+12 =72/18====>>4