Sample of 20 Gallons

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.

GMAT assassins aren't born, they're made,
Rich

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