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

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New post 19 Feb 2006, 12:33
N=number of votes
Remaining votes=0.6*N

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%

Answer is D.

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New post 20 Feb 2006, 02:15
Or...

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

So 50% of the votes would equal to 1.175.000, thus he would need additional 1175-940=235.000 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

The answer 235/1410=0.166666667 or 17%

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Re: PERCENTAGE PROB.. [#permalink]

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New post 18 Nov 2008, 12:15
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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Re: PERCENTAGE PROB.. [#permalink]

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New post 18 Nov 2008, 15:00
can this really be done by assuming 100 ? Can you pls confirm the QA bindrakaran ?

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Re: Mr. Kramer's votes [#permalink]

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New post 11 May 2011, 17:50
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%

Answer is D.

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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New post 11 May 2011, 20:22
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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New post 11 May 2011, 21:58
942568 = 0.4x

942568 + y = 0.5x

y = 0.1x

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

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

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New post 24 Apr 2014, 23:19
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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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New post 17 Nov 2014, 07:03
Let x = total # of votes casted
y= % of additional votes needed
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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New post 27 Dec 2015, 19:11
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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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New post 07 Jul 2016, 22:14
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%

Opponent(Remaining Votes) = 60%

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

Thus,

\(\frac{10}{60}\)

Answer 16.67% or 17%(D)

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

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New post 08 Jul 2016, 05:19
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.

Let's say total number of votes = n and additional number of votes required = x

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

Remaining votes = .60n

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

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

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New post 24 Jul 2017, 07:33
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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New post 24 Jul 2017, 08:28
60x/100 = 10
x = 1000/60 = ~17.
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Re: Mr. Kramer, the losing candidate in a two-candidate [#permalink]

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New post 23 Aug 2017, 09:30
Curly05 wrote:
But how did you get six remaining parts, partner?


We have taken the total as 10. Since, he received 40% i.e., 4 this implies he didn't get 60% i.e., 6 votes.

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

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New post 08 Sep 2017, 20:11
No need to get bogged down with calculations. Since it is a % problem.
Let total votes be 100
40% of 100=40
Left=60(100-40)
Now we need a % of 60 so that the no obtained can be added to 40 to make it 50 or 50% of the total votes.
10% of 60=6
15% of 60=9
17% of 60=10.2-Answer(question says approx value)

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

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New post 09 Sep 2017, 04:23
It is a very tricky question the number of voters seem to suggest a lot of calculation but we do not have to use that number
Kramer got 40% of the votes
Remaining votes =60%

Now to get 50 percent he must get 10 percent more
So so 10/60*100=16.66
or 17% hence D is the answer
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Re: Mr. Kramer, the losing candidate in a two-candidate   [#permalink] 09 Sep 2017, 04:23

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