What percent of an antifreeze solution was originally ethylene glycol?

Hi rekhabishop,

Shouldn't the percentage of ethylene glycol in the solution be 6/(6+20/3) ?

Yes the percentage should be 6/(6+20/3)= 9/19

My math is working out the same as others who have posted, so I wont reiterate the same information. I just have one question.

Statement 1 says that - if 4 liters were to be added to the original amount, then the resulting amount would be 16L. Thus, the original amount (per statement 1) is = 12L (16L-4L added = 12L). However, using numbers provided in statement 2, we can calculate a total original volume of 6+(20/3) = 12.66L, not 12L as we calculated from statement 1. We cannot have contradictory information in the two statements, correct? Shouldn't statement 1 say, "there would be 16.66L if 4L were added"? This doesn't change the answer, but I cannot figure out what I am missing. I've read the question repeatedly, and I know I must be missing something. Can someone help me with this?

from statement 1 we can tell that initially there is 12L of the solution and after adding 4L we have 16L.
from statement 2 , the initial volume of the solution is 6+(20/3) = 38/3

Hence both the statements are not giving the same volume of initial solution. This is a contradiction. Initial volume should be same.

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