There will be problems that will require you really know how

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There will be problems that will require you really know how [#permalink]

05 Aug 2003, 14:01

There will be problems that will require you really know how to manipulate algebra. Don't make the mistake I did and rely on picking numbers like I did for the SAT.

:!:

When 2/9 of the certain votes have been counted, 3/4 of those counted are in favor of the resolution. What fraction of the remaining votes must be against the resolution so that the total count will result in a vote of 2 to 1 against the resolution?

This is extremely challenging not to pick in numbers.

Victor
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[#permalink]

05 Aug 2003, 14:13

11/14

Let the total # of voters be 36x

Already counted = 36x * 2/9 = 8x
Out of which,
For the resolution = 8x*3/4 = 6x
So, already counted and against = 8x-6x=2x

Counting left for 36x - 8x = 28x votes
Let, the "for" candidate gets y votes out of the 28x
So, "against" gets 28x-y

"For" gets in total = 6x+y
"Against" gets = 2x+28x-y=30x-y

Given, (30x-y)/(6x+y)=2
=> y=6x

So, the "against" party has to get 28x-y=28x-6x=22x out of the remaining 28x votes. So, ratio = 22/28 = 11/14

Eternal Intern

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Hi Prakunda [#permalink]

05 Aug 2003, 15:51

I find it easier to do it the other way;
3 + x
------- = 1/2
1 + 14 -x

I don't know what "X" stands for your in your problem. Mine is votes.

VT
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[#permalink]

05 Aug 2003, 22:25

If all votes are either against or for the resolution, then the answer should be 33/42.

SVP

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[#permalink]

21 Jan 2004, 22:35

Let 36 be total votes

Counted = 36 * 2/9 = 8 votes
F = 8*3/4 = 6
A = 8-6 = 2

Remaining votes = 28
left f be for votes of this remaining 28 votes
total F = f+6
total against = 2+28-f = 30-f

now (30-f)/(f+6) = 2/1
30-f = 2f+12
3f = 18 or f = 6
so 22 out of 28 must be against
22/28 = 11/14

[#permalink] 21 Jan 2004, 22:35

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There will be problems that will require you really know how

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