Solution:
Let the quantity of petrol, diesel and kerosene in the first vessel be x, 2x and 4x and let the same in the other vessel be 3y, 5y and 6y.
So x+2x+4x or 7x is the quantity of mixture in the first vessel and 3y+5y+6y or 14y is the quantity of mixture in the second vessel.
Now 7x parts of mixture in the first vessel has quantity of petrol, diesel and kerosene as x, 2x and 4x.
So 1 part of mixture has quantity of petrol, diesel and kerosene as x/7x, 2x/7x and 4x/7x which is 1/7, 2/7, 4/7.
Similarly 1 part of mixture in the second vessel has petrol, diesel and kerosene as 3/14, 5/14 and 6/14.
In the resulting mixture quantity of petrol, diesel and kerosene is 1/7+3/14, 2/7+5/14, 4/7+6/14 which is 5/14, 9/14 and 14/14.
Or the required ratio is 5:9:14.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
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