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Mr. Kramer, the losing candidate in a two-candidate

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19 Feb 2006, 13:33

We look for x such as: x*0.6*N+0.4*N>0.5*N
x*0.6>(0.5-0.4)
x>1/6=16.66%

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20 Feb 2006, 03:15
Or...

940.000 votes he received (which we now is 40% of total) translates into 2.350.000 of total votes....

The question asks what percent of the REMAINING votes he would need, so the REMAINING votes equates to 2.350.000-940.000=1.410.000

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18 Nov 2008, 13:15
bindrakaran001 wrote:
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all the votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?
(A) 10%
(B) 12%
(C) 15%
(D) 17%
(E) 20%

It is D

Assuming he got 40 of the 100 votes. He needs 10 more==> 10/60X100=16.66
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18 Nov 2008, 16:00
can this really be done by assuming 100 ? Can you pls confirm the QA bindrakaran ?
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11 May 2011, 18:50
you really dont have to use a lot of algebra.

given (40/100)(V) = 942568 where V is the total number of votes

=> (60V/100) is the remaining and we were asked to find what % of remaining votes does he need to win

( he needs 10% more votes to win)

=> $$(p/100)(60V/100) = 10V/100$$

=> p = 17%

if you look carefully we dont even have to use the 942568 any where in our calculation. Hope it helps.

tonebeeze wrote:
I got this problem correct using brute force algebra, but the process took to long. What is the most efficient method to solve problems like this one?

Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?

a. 10%
b. 12%
c. 15%
d. 17%
e. 20%
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Re: PS: What's a Quick Way to Get This? [#permalink]

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11 May 2011, 21:22
Good method to solve these kinds of questions.

Equating to 10,1 and 5 respectively.
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Re: PS: What's a Quick Way to Get This? [#permalink]

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11 May 2011, 22:58
942568 = 0.4x

942568 + y = 0.5x

y = 0.1x

=> 0.1x/0.6x * 100 = 100/6 = 50/3 = 16.66 ~ 17%

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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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25 Apr 2014, 00:19
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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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17 Nov 2014, 08:03
Let x = total # of votes casted
so, 0.4x + y*0.6x = 0.5x

solve for x. Ans
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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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27 Dec 2015, 20:11
Hello from the GMAT Club BumpBot!

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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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07 Jul 2016, 23:14
Curly05 wrote:
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?

A. 10%
B. 12%
C. 15%
D. 17%
E. 20%

No need to use the given number of votes..

The total votes are = 100%

Kramer = 40%

Kramer needs 10% of the total votes to become 50%..we will take this out from Opponent's votes

Thus,

$$\frac{10}{60}$$

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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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08 Jul 2016, 06:19
Curly05 wrote:
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 40 percent of all votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast?

A. 10%
B. 12%
C. 15%
D. 17%
E. 20%

Given number is irrelevant here.

x= .10n (.50n-.40n)

if .60n is 100% then .10n = .10*100/.60= 17% (Approximately)

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Re: Mr. Kramer, the losing candidate in a two-candidate   [#permalink] 08 Jul 2016, 06:19

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