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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please show me the right way to do this.

A container contained 80Kg of milk. from this container 8Kg of milk was taken and replaced by water. the process of taking out 8kg of liquid was repeated 2 times. How much milk is now there in the vessel.

A. 54
B. 56
C. 58.32
D. 53.18
E. I m not that smart.....

A container contained 80Kg of milk. from this container 8Kg of milk was taken and replaced by water. the process of taking out 8kg of liquid was repeated 2 times. How much milk is now there in the vessel.

A. 54 B. 56 C. 58.32 D. 53.18 E. I m not that smart.....

The way to do this is
1) realize that you remove a "solution" and add "water" which creates a new solution of X% milk.
2) everytime you remove 8kg of the solution, keep track of how much of that solution is milk by multiplying 8kg x X%.
repeat.
3) The amount of milk left is 80kg less that amount you removed.

P.S. strickly speaking, we must assume milk and water have same density.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

8*3 = 24kg milk, so if we pretend that we remove 8kg of milk only three times there will be 56kg milk left. But now we know that some of what we remove will be water so there has to be more than 56kg of milk left.

First drain:
8kg of milk taken out. You are left with 72 kg of milk and 8kg of water

Second drain:
The liquid left is with a concentration of 90% (72kg) of milk
Then when you take 8kg out, you are taking 90% milk and 10% water
Therefore, the quantity of milk left will be 72kg * 0.9 = 64.8kg milk after the second drain

Third drain:
Same process. 64.8kg of milk * 0.9 = 58.32kg of milk left
_________________

A container contained 80Kg of milk. from this container 8Kg of milk was taken and replaced by water. the process of taking out 8kg of liquid was repeated 2 times. How much milk is now there in the vessel.

A. 54 B. 56 C. 58.32 D. 53.18 E. I m not that smart.....

Good Question. But, the Question should be reworded to
".... This process of taking out 8Kg of liquid was repeated 2 *MORE*
times....".

I interpreted the question as taking 8kg liquid only two times.
Did anyone else interpret in the same way?

Interesting question. Keep these kind of questions flooding the forum !

I agree with you kpadma. It got me confused too as to the number of times the process was repeated. I guess the answer choices gave me the hint.
_________________

kpadma: I agree that the phrasing is suspect, but your correction isn't any better. To be "repeated 2 more times" it must have already been repeated. "Performed two more times" would be the clearest way to ask this.

I know this is PS, not SC, but these little differences are important.

kpadma: I agree that the phrasing is suspect, but your correction isn't any better. To be "repeated 2 more times" it must have already been repeated. "Performed two more times" would be the clearest way to ask this.

I know this is PS, not SC, but these little differences are important.

Stoolfi: You are most welcome to correct the mistakes!
How else are we going to learn?

I agree with stoolfi about this question. It is not clear whether the diluation occured 2 times or 3 times.
if it occured 2 times then remaining milk is 0.9*0.9 * 80 = 64.8
if it occured 3 times then remaining milk is 0.9*0.9*0.9 * 80 = 58.32

( 8Kg is 10 percent of 80kg so 90% milk remains at each dilution )