## milk

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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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### GMAT/MBA Expert

- Rahul@gurome
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**Posts:**1179**Joined:**11 Apr 2010**Location:**Milpitas, CA**Thanked**: 447 times**Followed by:**88 members

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

- 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

Rahul Lakhani

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Quant Expert

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- shovan85
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**Posts:**991**Joined:**23 Sep 2010**Location:**Bangalore, India**Thanked**: 146 times**Followed by:**24 members

k(x/100) + 2k(y/100) + 3k[(x+y)/100] = 6k(50/100)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.

=> 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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