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What percent of an antifreeze solution was originally ethylene glycol?

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What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 18 Jan 2017, 06:32
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What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.

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Re: What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 18 Jan 2017, 07:50
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Bunuel wrote:
What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.



(1) insuff
%age of ethylene glycol is not known in the original solution

(2) 10 lts have 60 % ethylene glycol means 6 lts of ethylene glycol
as 40%water evaporated thus when we add this 40 % of water to 10 lts of solution we get say Xlts
so %age of ethylene glycol is 6lts in X lts
Suff

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What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 01 Jul 2017, 09:55
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Bunuel wrote:
What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.


(1) Insuff. Nothing about the %age of ethylene glycol in the solution before adding the 4 litres.
(2) (60/100)*10 =6L of ethylene glycol. and hence 4L of water is left in the solution after 6 hours.

Now the statement says 40% of the water has evaporated

which means that 60% of the initial water content is still left.

Let T was the initial total water content.

(60/100)*T=4
T=20/3

Thus we can easily find out the initial % of ethylene glycol.

4/(4+(20/3))
Please give KUDOS, if you like my solution. :-D
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What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 22 Aug 2017, 03:23
rekhabishop wrote:
Bunuel wrote:
What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.


(1) Insuff. Nothing about the %age of ethylene glycol in the solution before adding the 4 litres.
(2) (60/100)*10 =6L of ethylene glycol. and hence 4L of water is left in the solution after 6 hours.

Now the statement says 40% of the water has evaporated

which means that 60% of the initial water content is still left.

Let T was the initial total water content.

(60/100)*T=4
T=20/3

Thus we can easily find out the initial % of ethylene glycol.

4/(4+(20/3))
Please give KUDOS, if you like my solution. :-D


Hi rekhabishop,

Shouldn't the percentage of ethylene glycol in the solution be 6/(6+20/3) ?
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Re: What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 29 Aug 2017, 11:11
Dkingdom wrote:
rekhabishop wrote:
Bunuel wrote:
What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.


(1) Insuff. Nothing about the %age of ethylene glycol in the solution before adding the 4 litres.
(2) (60/100)*10 =6L of ethylene glycol. and hence 4L of water is left in the solution after 6 hours.

Now the statement says 40% of the water has evaporated

which means that 60% of the initial water content is still left.

Let T was the initial total water content.

(60/100)*T=4
T=20/3

Thus we can easily find out the initial % of ethylene glycol.

4/(4+(20/3))
Please give KUDOS, if you like my solution. :-D


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
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Re: What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 18 Feb 2018, 08:06
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?
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Re: What percent of an antifreeze solution was originally ethylene glycol?  [#permalink]

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New post 06 May 2019, 03:26
Bunuel wrote:
What percent of an antifreeze solution was originally ethylene glycol?

(1) There would be 16 liters of solution if 4 liters of 50% ethylene glycol solution were added.
(2) After 6 hours in the sun 40% of the water has evaporated, leaving 10 liters of a 60% ethylene glycol solution.


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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Re: What percent of an antifreeze solution was originally ethylene glycol?   [#permalink] 06 May 2019, 03:26
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