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Of the 2500 votes participating in a local election, three

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Director
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Joined: 01 May 2007
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Of the 2500 votes participating in a local election, three [#permalink] New post 16 Jun 2007, 17:06
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A
B
C
D
E

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0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Of the 2500 votes participating in a local election, three times as many voted for candidate A as for candidate B, and twice as many people voted for candidate A as for candidate C. If each voter chose either candidate A, B, or C, or abstained, and 300 voters abstained, how many people voted for candidate C?

400
600
1200
1800
2000

Know how to solve using brute force. Looking for a equation and explanation of how to solve.

Thanks. OA to come.
Manager
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Joined: 02 May 2007
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Re: PS: Word Problem [#permalink] New post 16 Jun 2007, 17:26
jimmyjamesdonkey wrote:
Of the 2500 votes participating in a local election, three times as many voted for candidate A as for candidate B, and twice as many people voted for candidate A as for candidate C. If each voter chose either candidate A, B, or C, or abstained, and 300 voters abstained, how many people voted for candidate C?

400
600
1200
1800
2000

Know how to solve using brute force. Looking for a equation and explanation of how to solve.

Thanks. OA to come.


(B) it is

A + B +C + O (abstain) = 2,500
=> A + B + C = 2,200
A = 3B
A = 2C

=> A + A/2 + A/3 = 2,200

=> C = 600
Manager
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 [#permalink] New post 16 Jun 2007, 17:32
B

a = ppl who voted for candidate A
b = ppl who voted for candidate B
c = ppl who voted for candidate C

from the stem we have a=3b and a=2c. it gives us the amount that abstained, and asks us for the value of c.

2200 = a + b + c

we have 3 variables and one equation, so we need to manipulate this to include only 1 variable, which should obviously be c so that we can solve for it directly.

a = 2c, and by substitution, we have 2c=3b, which is c = 3b/2

2200 = 2c + 3b/2 + c

solve for c and we get 600.
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Re: PS: Word Problem [#permalink] New post 16 Jun 2007, 19:10
jimmyjamesdonkey wrote:
Of the 2500 votes participating in a local election, three times as many voted for candidate A as for candidate B, and twice as many people voted for candidate A as for candidate C. If each voter chose either candidate A, B, or C, or abstained, and 300 voters abstained, how many people voted for candidate C?

400
600
1200
1800
2000

Know how to solve using brute force. Looking for a equation and explanation of how to solve.

Thanks. OA to come.


Easy. :P

kirakira solved in terms of A and plaguerabbit solved in terms of C.

This is how to do it in terms of B.

Given
A=3B
A=2C

Equation:
A+B+C=2500-300
A+B+C=2200
3B + B + C = 2200

2C=3B
C=3B/2

therefore...
3B + B + 3B/2 = 2200
multiply entire equation by 2 to get rid of denominator

6B + 2B + 3B = 2200
11B = 2200
B=200

C= 3B = 3*200 = 600
Re: PS: Word Problem   [#permalink] 16 Jun 2007, 19:10
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Of the 2500 votes participating in a local election, three

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