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Mixture

by heshamelaziry » Tue Dec 22, 2009 12:39 pm
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?

a) 9/10
b) 1
c) 10/9
d) 20/19
e) 2
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by Testluv » Tue Dec 22, 2009 1:13 pm
Hi Hesham,

We start with 20 gallons of solution that is 5% ethanol and 95% gasoline.

Initial:

eth 1 gallon
gas 19 gallons
Total 20 gallons

Then we are adding only ethanol, and we want ethanol to be 10% of the final solution. But we are not adding any gas, and so the 19 gallons of gas will be 90% of the final solution.

19 is 90% of what number?--> 19 = (9/10) * x (let x be total gallons in the final solution)

x = 190/9

Thus, the final solution will be 190/9 gallons. We started with 20 or 180/9 gallons of solution. Because we have added only ethanol, we have added 190/9 - 180/9 = 10/9 gallons of ethanol. Choice C.

You could have also backsolved this question.

I should add that I don't think I've ever seen an official GMAT mixture problem.
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by pranali » Sat Dec 26, 2009 6:58 am
Hi Testluv,

Your approach is good. There is another way in which I have solved this problem. Correct me if I am wrong, I feel this is less time consuming.

Initially, the ratio of ethanol to Gasoline is;

eth:gas = 1 :19 --------------- (Since, the mixture contains 5% of Eth and 95% of Gas)

Now, lets consider 'x' amount of ethanol is added to make the mixture

eth:gas = 1:9 ---------------- (Since,the second mixture has 10% Eth and 90% Gas)

Therefore, (1+x)/19 = 1/9,
x= 10/9

Answer option C
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by valleeny » Sat Dec 26, 2009 11:09 pm
Hi

I always ended up using algebra for most problems.

In the initial solution, there is 1 gallon of ethanol and 19 gallons of gasoline.

Let the amount of ethanol added = x

So the solution becomes
(0.1)(20+x) + 19 = 20 + x

Solve for x = 10/9
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by pjr87 » Thu Feb 11, 2010 6:10 pm
You can also use Kaplans balancing method

quantity weak * (% wanted- % weak) = quantity strong * (% strong - %wanted)

20 (10-5) = x(100-10) (since it asks how much ethanol should be added, we assume 100% ethanol)

100=90x

100/90 = x

10/9 = x
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by komal » Tue Feb 16, 2010 11:19 am
heshamelaziry wrote: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?

a) 9/10
b) 1
c) 10/9
d) 20/19
e) 2
First find the current ratio:

E (ethanol) = 5% x 20 = 1
G (gasoline) = 20 - 1 = 19

Now, set up the new ratio that we are trying to obtain using the current one:

1/9 = (1 + x)/19
19 = 9 + 9x
10 = 9x
x = 10/9

1/9 represents the new ratio we are trying to get of 10% ethanol and 90% gasoline. x is for the amount of gallons we need to add to the current 1 gal. of ethanol to achieve the new 10%.

(C) is correct
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by amittilak » Tue Feb 16, 2010 1:16 pm
Hi,
This is the method I use for most of the mixture problems:

20 + x = 20 + x
(5% E/95%G) (100% E) (10% E/90%G)

(5/100)*20 + x = (10/100) * (20 + x)
100 + 100x = 200 + 10x
90x = 100
x = 10/9 gal.
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