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Sample of 20 Gallons

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Sample of 20 Gallons

by vinay1983 » Wed Aug 21, 2013 3:38 am

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A sample of 20 Gallons of gasoline is found to be adulterated to the extent of 15%. How much pure gasoline (in gallons) should be added to bring up the level of purity to 95%?


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by ganeshrkamath » Wed Aug 21, 2013 6:55 am
vinay1983 wrote:A sample of 20 Gallons of gasoline is found to be adulterated to the extent of 15%. How much pure gasoline (in gallons) should be added to bring up the level of purity to 95%?


40
20 gallons has 85% pure gasoline = 85/100*20 = 17 gallons of pure gasoline

Now to bring the purity level to 95%, let's say we have to add x gallons of pure gasoline.

So 95% of (20+x) = 17+x
95/100*(20+x) = 17+x
1900 + 95x = 1700 + 100x
5x = 200
[spoiler]x = 40 gallons[/spoiler]

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by [email protected] » Wed Aug 21, 2013 12:09 pm
Hi vinay1983,

These types of questions will have answers that are numbers, so you can either solve with algebra (using the "mixture" formula as ganeshrkamath has done) or you can TEST THE ANSWERS by plugging them into the prompt and finding the value that would give you 95% purity (as the question asks for).

The weighted average formula also tells you that the number of pure gallons of gas that would have to be added would be > 20 gallons, since...

20 gallons of 85% pure + 20 gallons of 100% pure = average of 92.5% pure which is TOO LOW, so we have to add more than 20 gallons.

You could also use that info against the answers to narrow down the possibilities.

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by Java_85 » Wed Aug 21, 2013 4:17 pm
I would say 15% of 20 gallons is 3 gallons.
If we want to have the purity of 95% it means that the ratio of pure gallons to whole should be 95% i.e. ((x-3)/x)=95/100 ==> x=60 therefore we should add 40 more gallons to 20 gallons.
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by GMATGuruNY » Wed Aug 21, 2013 4:54 pm
vinay1983 wrote:A sample of 20 Gallons of gasoline is found to be adulterated to the extent of 15%. How much pure gasoline (in gallons) should be added to bring up the level of purity to 95%?
Since 15% of the original 20 gallons is NON-gasoline, the amount of non-gasoline = .15(20) = 3 gallons.

After gasoline is added to increase the gasoline percentage to 95%, these 3 gallons of non-gasoline must constitute 5% of the new total:
3 = .05t
t = 3/.05 = 300/5 = 60.

Since the new total = 60 gallons, the amount of gasoline added to the original 20 gallons = 40 gallons.
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Re: Sample of 20 Gallons

by Scott@TargetTestPrep » Sun Mar 29, 2020 4:01 am
vinay1983 wrote: ↑
Wed Aug 21, 2013 3:38 am
 A sample of 20 Gallons of gasoline is found to be adulterated to the extent of 15%. How much pure gasoline (in gallons) should be added to bring up the level of purity to 95%?


40
Solution:

We see that currently there are 0.85 x 20 = 17 gallons of pure gasoline in the mixture. Let g = the number of pure gasoline should be added to bring up the level of purity to 95%, we can create the proportion:

(17 + g) / (20 + g) = 95/100

(17 + g) / (20 + g) = 19/20

20(17 + g) = 19(20 + g)

340 + 20g = 380 + 19g

g = 40

Alternate Solution:

We have 20 gallons of 85% solution, and we will add x gallons of 100% solution, obtaining (20 + x) gallons of 95% solution. The equation representing this is:

0.85 * 20 + 1 * x = 0.95 * (20 + x)

17 + x = 19 + 0.95x

0.05x = 2

x = 40

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