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DS: Solution

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mirhaque
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DS: Solution [#permalink] New post   08 Jul 2005, 07:23
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A
B
C
D
E

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Didn't agree with OA. Can anybody explain?



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Re: DS: Solution [#permalink] New post   08 Jul 2005, 07:37
mirhaque wrote:
Didn't agree with OA. Can anybody explain?


E.

If x grams of water is added to x grams solution of any concentration , the new concentration of the resultant solution is reduced to half of it's original concentration.

HMTG.
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Re: DS: Solution [#permalink] New post   08 Jul 2005, 07:45
HowManyToGo wrote:
mirhaque wrote:
Didn't agree with OA. Can anybody explain?


E.

If x grams of water is added to x grams solution of any concentration , the new concentration of the resultant solution is reduced to half of it's original concentration.

HMTG.


I see. Thanks :oops:
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rthothad
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Re: DS: Solution [#permalink] New post   08 Jul 2005, 10:55
HMTG, Isn't the question saying that the 80 gms of solution contains 100% acid?

What clued you in to make you understand that it contains x% acid
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Re: DS: Solution [#permalink] New post   08 Jul 2005, 11:30
rthothad wrote:
HMTG, Isn't the question saying that the 80 gms of solution contains 100% acid?

What clued you in to make you understand that it contains x% acid


My chemistry is a bit rusty, But i think that a strong soln. doesn't always mean a 100% solution.

HMTG.
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Re: DS: Solution [#permalink] New post   08 Jul 2005, 14:41
HMTG,
I don't understand why the answer is (E)... Per my analysis it should be (C). Here is what I am doing...

Assuming the solution originally has 'a' units of acid & 'w' of water

a+w=80

Original Concentration of Acid in solution = a/80*100 = 5a/4
New Concentration of Acid in solution = 100a/(80+x) [x units of water added]

Now, since the addition of water dropped the concentration of Acid by Y% => 5a/4 - 100a/(80+x) = y

(1) X=80

We still have 2 unknowns (a & y).

(2) Y=50

We still have 2 unknowns (a & x).

(1) & (2) together lead us to the answer that the original concentration was 100%.

Can you please explain how you got (E)?

thanks!
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Re: DS: Solution [#permalink] New post   08 Jul 2005, 22:35
himjhamb wrote:
Now, since the addition of water dropped the concentration of Acid by Y% => 5a/4 - 100a/(80+x) = y


the equation should be [5a/4 - 100a/(80+x)]/[5a/4]= y%

i.e. the change is not an absolute 50%, but concentration reduced by 50% (of it's original concentration)

in this equation if x is 80 y is 50 ,and vice-versa, without still having to know a.

HMTG.

PS : eg. if concentration changes for 30% to 15% ,it is still a drop of 50% in concentration.



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Re: DS: Solution [#permalink]
08 Jul 2005, 22:35
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