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IrinaOK
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 06:01
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please help



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KillerSquirrel
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 06:46
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Let s be the acid and w the water in the solution.

s+w = 80

after adding amount of x water to the solution:

s+(w+x) = 80+x

so the acid that was s/80 of the solution is now s/(80+x)

so y is equal to:

[s/80 - s/(80+x)]/s/80 = y/100

statement 1

even after we plug in x as 80 in all the equations above we will not find a solution.

insufficient

statement 2

even after we plug in y as 50 in all the equations above we will not find a solution.

insufficient

statements 1&2

solving:

[s/80 - s/(80+80)]/s/80 = 50/100

[s/80 - s/160]/s/80 = 50%

s/160/s/80 = 50%

s can get any value - so still insufficient

the answer is (E)

:)
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IrinaOK
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 06:58
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KillerSquirrel
Let s be the acid and w the water in the solution.

s+w = 80

after adding amount of x water to the solution:

s+(w+x) = 80+x

so the acid that was s/80 of the solution is now s/(80+x)

so y is equal to:

[s/80 - s/(80+x)]/s/80 = y/100

statement 1

even after we plug in x as 80 in all the equations above we will not find a solution.

insufficient

statement 2

even after we plug in y as 50 in all the equations above we will not find a solution.

insufficient

statements 1&2

solving:

[s/80 - s/(80+80)]/s/80 = 50/100

[s/80 - s/160]/s/80 = 50%

s/160/s/80 = 50%

s can get any value - so still insufficient

the answer is (E)

:)
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Thank you!

I got it, so clearly explained.
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 12:18
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E..
both d options gives d same equation
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IrinaOK
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 12:46
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bkk145
IrinaOK
please help

Is the OA = E?
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Yes, it is.

The OA is E.
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zzyyxx
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 12:54
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Hi Killer Squirrel:

Why would you further divide [s/80 - s/(80+x)] by s/80?

% of acid initially is (s/80)*100
% of acid after x qty of water is added is (s/(80+x))*100

%difference between these two is given as Y

so wouldn't the equation be
[s/80 - s/(80+x)]*100 = Y

In this case, statements taken together will be sufficient. (C).
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KillerSquirrel
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SOLUTIONS, acid and water, help to understand the OE please 01 Sep 2007, 13:10
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zzyyxx
Hi Killer Squirrel:

Why would you further divide [s/80 - s/(80+x)] by s/80?

% of acid initially is (s/80)*100
% of acid after x qty of water is added is (s/(80+x))*100

%difference between these two is given as Y

so wouldn't the equation be
[s/80 - s/(80+x)]*100 = Y

In this case, statements taken together will be sufficient. (C).
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Here is a simplified example to illustrate my logic:

lets say the acid (s) = 20

20/80 = 1/4 ---> acid in old solution

20/160 = 1/8 ---> acid in new solution

1/4 - 1/8 = 2/8 - 1/8 = 1/8 ---> note !! this is NOT y this is only the difference between the acid percentages not the change ! of acid percentages in the solutions.

In order to find the change, you need to divide the original amout (1/4) by the diffrance (1/8) meaning 1/8/1/4 = 1/8/2/8 = 1/2 ---> since y is 50 the change is 50/100 = 1/2

The problem is, that for every amount (s) you will get the same resualt !

You are welcome to try for yourself :wink:

:)



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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Thank you for understanding, and happy exploring!