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Manhattan 700 + Mixing Problem

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by Rahul@gurome » Sat Oct 23, 2010 12:55 am
Gasohol is a mixture consisting of 5% Ethanol and 95% Gasoline.
5% 0f 20 gallon = 20*5/100 = 1 gallon

Therefore, 20 gallons of Gasohol contains 1 gallon of Ethanol and 19 gallons of Gasoline.

Say, x gallons of Ethanol is mixed to achieve a mixture consisting of 10% of Ethanol and 90% Gasoline.
After mixing x gallons of Ethanol, the mixture will contain (x + 1) gallons of ethanol and 19 gallons of Gasoline.

Now, percentage of Ethanol in the mixture = [(New amount of Ethanol)/(New amount of mixture)]*100 = [(x + 1)/(x + 20)]*100

So, [(x + 1)/(x + 20)]*100 = 10
=> (x + 1) = 0.1*(x + 20) = 0.1x + 2
=> (x - 0.1x) = 2 - 1 = 1
=> 0.9x = 1
=> x = 1/0.9 = 10/9

The correct answer is C.
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by sixpointer » Sat Oct 23, 2010 5:26 am
Ratio of ethanol and gasoline :- 1/19

needed ratio of ethanol and gasoline :-1/9

Let x amount of ethanol is added to get ratio of 1/9


So, 1+x/19 =1/9

X=10/9
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