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##### This topic has expert replies
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### milk

by goyalsau » Wed Dec 08, 2010 2:22 am

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If k litres of a solution of x% milk, 2k litres of a solution of y% milk, and 3k litres of a solution of (x + y)% milk are all mixed in a vessel, the resultant solution has 50% milk. Find the resultant concentration of milk when 40 litres of the first solution and 50 litres of the second solution are mixed.
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by Rahul@gurome » Wed Dec 08, 2010 2:51 am
When k, 2k and 3k liters are mixed:
• Total amount of the resultant solution = (k + 2k + 3k) liters = 6k liters
Total amount of milk in the solution = (kx/100 + 2ky/100 + 3k(x + y)/100) liters = k*(4x + 5y)/100 liters

Percentage of milk in the solution = 100*[k*(4x + 5y)/100]/(6k) = (4x + 5y)/6
According to question, (4x + 5y)/6 = 50 => (4x + 5y) = 300
Now when 40 and 50 liters are mixed:
• Total amount of the resultant solution = (40 + 50) liters = 90 liters
Total amount of milk in the solution = (40x/100 + 50y/100) liters = (40x + 50y)/100 liters

Percentage of milk in the solution = 100*[(40x + 50y)/100]/(90) = (4x + 5y)/9 = 300/9 = 100/3 = 33.33
The resultant concentration of milk is 33.33 percent.
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by shovan85 » Wed Dec 08, 2010 2:55 am
goyalsau wrote:If k litres of a solution of x% milk, 2k litres of a solution of y% milk, and 3k litres of a solution of (x + y)% milk are all mixed in a vessel, the resultant solution has 50% milk. Find the resultant concentration of milk when 40 litres of the first solution and 50 litres of the second solution are mixed.
k(x/100) + 2k(y/100) + 3k[(x+y)/100] = 6k(50/100)

=> 4kx/100 + 5ky/100 = 300k/100

=> x(0.04) + y(0.05) = 3

40 lts type 1 contains .4x milk
50 lts type 2 contains .5y milk

Thus 0.4x+0.5y = 10 [x(0.04) + y(0.05)] = 10*3 = 30 lts

Thus concentration = 30/90 = 1/3 = 33.33%
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