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

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 [#permalink] New post 14 Oct 2005, 13:20
Let me try a simpler one.

Lets assume that candidate got 40% votes and total votes is 100.
Candidate won = 40
Remaining = 60

To get 50%, candidate requires 10 votes from 100 which is 10% and 10 votes from 60.

10/60 = 1/6 = .166 = 16.67%

Which is approx 17%. Hence the answer is D.

Whatz OA?
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 [#permalink] New post 14 Oct 2005, 13:23
sudhagar wrote:
Let me try a simpler one.

Lets assume that candidate got 40% votes and total votes is 100.
Candidate won = 40
Remaining = 60

To get 50%, candidate requires 10 votes from 100 which is 10% and 10 votes from 60.

10/60 = 1/6 = .166 = 16.67%

Which is approx 17%. Hence the answer is D.

Whatz OA?


Right. Its D.
Good explanation. It doesnt matter what are the actual numbers as we are dealing with %ge only.
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 [#permalink] New post 14 Oct 2005, 13:32
sudhagar, that's a good one.

i also got D, but by the longer method! :(
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 [#permalink] New post 16 Oct 2005, 19:54
Kramer received 40% of the votes, so 60% = (60/40)(942568) = (3/2)(942568)
Replace 942568 with T.

He needs 50% so (5/4)(T)
Difference = 3/2T - 5/4T = 1/4T
% = (1/4T)/(3/2T) *100% = 16.66% ~ 17% (D)
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Need help with problem please, Thanks! [#permalink] New post 18 Feb 2006, 19:39
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%

Answer is D, but somehow i'm not using the write numbers or I'm rounding wrong or something cause I can't get it. Thanks for the help!

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Re: Need help with problem please, Thanks! [#permalink] New post 18 Feb 2006, 19:49
Kramer ( Cosmo ? ) has 40 %. He needs 50-40 = 10 % of the remaining ( 60%) votes to win at least 50 % of all the votes cast.

Thus, the percentage of the remaining votes = (10/60)*100 = 16.66 %.


Dumpling 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%

Answer is D, but somehow i'm not using the write numbers or I'm rounding wrong or something cause I can't get it. Thanks for the help!

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 [#permalink] New post 19 Feb 2006, 13:33
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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 [#permalink] New post 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....

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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PERCENTAGE PROB.. [#permalink] New post 18 Nov 2008, 07:19
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%
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Re: PERCENTAGE PROB.. [#permalink] New post 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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Re: PERCENTAGE PROB.. [#permalink] New post 18 Nov 2008, 16:00
can this really be done by assuming 100 ? Can you pls confirm the QA bindrakaran ?
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Re: PERCENTAGE PROB.. [#permalink] New post 19 Nov 2008, 00:32
Yep, the fastest way to solve such problems.


hibloom wrote:
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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Mr. Kramer's votes [#permalink] New post 11 May 2011, 11:40
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: Mr. Kramer's votes [#permalink] New post 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%

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] New post 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] New post 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%

Answer - D
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Re: PS: What's a Quick Way to Get This?   [#permalink] 11 May 2011, 22:58
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