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by mk101 » Wed Feb 23, 2011 10:22 am
assume that initially he has 100gallons of milk that costs 100 USD @ USD 1 per gallon

To earn 20 % profit, he must earn USD 120, by selling the same qty of milk

But the selling price of milk = the cost price of the milk.


qty of milk and water mixture sold = 120/1 = 120 gallons

of the 120 lgallons, 100gallons are milk so the remaining must be water added.

Thus the amount of water added = 20 gallons

ratio of milk :water = 100:20 = 5:1
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by Anurag@Gurome » Mon Feb 28, 2011 7:13 pm
Solution:
Let the ratio of milk and water in every gallon be r : 1.
Let the milk vendor sell 'g' gallons of milk at price 'p'.
Now this 'g' gallons of milk has 'g*r/(r+1)' amount of milk.
So cost price of 'g*r/(r+1)' amount of milk is 'p*r/(r+1)'.
So, 'p*r/(r+1)' is the cost price of milk which he is selling at price 'p'.
Or profit % = {p - p*r/(r+1)}/{p*r/(r+1)}*100 = {1/r}*100 = 20.
Or 1/r = 1/5.
Or r = 5.
So, ratio of milk is to water is 5:1.
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