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MGMAT CAT 600-700 ratio question

Expert replies
by ssy » Mon Oct 22, 2007 6:14 am
Taken from: MGMAT CAT, I didn't quite understand their explanation.

Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?

9/10
1
10/9
20/19
2

ANS:[spoiler]10/9[/spoiler]
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Source: — Problem Solving |

by naren_nayak » Mon Oct 22, 2007 6:36 am
Bob just filled his car's gas tank with 20 gallons of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If his car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many gallons of ethanol must he add into the gas tank for his car to achieve optimum performance?

20 gallons of gasohol => 1 gal ethanol (5%) & 19 gals gasoline (95%)
If x gals of ethanol are added to the gasohol, the amount of ethanol in the new mixture is supposed to be 10% concentration.
x + 1 = (10/100) * *20+x)
Solving for x gives us 10/9 gals
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by ssy » Mon Oct 22, 2007 6:56 am
Got it, thank you!
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by wongee » Mon Oct 22, 2007 6:26 pm
Just FYI...the other option is to use the portion that stays the same i.e. gasoline.

95%=19 this stays the same.

So, 19=90/100*x
x=190/9

Total new solution - Gasoline = Total new ethanol
190/9 - 19= 19/1

Total new ethanol - Old ethanol: Added portion of ethanol
19/1-1 = 10/9

This may be a bit longer, but wanted to throw this option out there!
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