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

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21 Jul 2003, 16:35

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Question Stats:

60% (02:47) correct
40% (02:05) wrong based on 472 sessions

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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?

Re: PS: What's a Quick Way to Get This? [#permalink]

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21 Jul 2003, 18:16

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Curly05 wrote:

14. 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?

Divide the vote into 10 parts. He got 4 parts, he needs 5 parts. So he need 1 part out of the remaining 6 parts or .166666666...% -- plus one vote.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

The other candidate received the six remaining "parts" (or six-tenths of the vote).

For the vote to have been 50/50, one of the other candidates "tenths" would have to convert to Mr. Kramer. So one tenth out of six tenths is 1/6 = .1666667. That is the fraction of the TOTAL vote that would have to be shifted in order for Mr. Kramer to have tied the vote 50/50.

Great point Stolyar!! Hopefully we would recognize this in the choices (and also remember to round up rather than down), but in other questions such as CR, if we omit something small like "EXCEPT", we will be cooked!

AkamaiBrah, reading what you wrote two posts above made me laugh out loud at work!!! Curly, can you repeat that question/statement please?

but the stem asks about approximate percentage! The correct answer is 17%

Strictily speaking, since 1/6 has infinitely repeating decimals, an approximate could be:

17%
16.7%
16.67%
etc.

Just have to look at the answers and pick appropriate choice.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

1 min
Assume there are 100 votes
Kramer has 40 and the winner has 60
Kramer needs atleast 50 which means atleast 10 out of 60 which is 16.67%
Hence I'll go with D

14. 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%

Assume the total votes cast 100- votes recd 40, remaining votes 60. K needs 10 more votes to get 50% . hence percent out of remaining = 100*10/60 = 17%

Note to self. Big numbers mean its universally applicable. so pick easy numbers like 100 and go ahead with the problem and would end up with the same result.

Re: Need help with problem please, Thanks! [#permalink]

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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!

Dumpling

gmatclubot

Re: Need help with problem please, Thanks!
[#permalink]
18 Feb 2006, 19:49

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