milk

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

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

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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Source: Beat The GMAT — Problem Solving | Wed Dec 08, 2010 2:22 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

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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