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Old 2007 November 5th, 07:02 PM   #1 (permalink)
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quants doubts..tricky questions

If 72.42= k*(24+n/100) , where k and n are positive integers and
n>100, then what is the value of k+n?
a)17
b)16
c)15
d)14
e)13

- From the set of six letters, A,B,C,D,E,F there can be 20 sets of 3
letters formed

Col A
The number of 3 letter sets with the letter F in them

Col B
10
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Old 2007 November 5th, 07:19 PM   #2 (permalink)
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L.H.S = 72.42 = 72 + 42/100
R.H.S = 24k + k*n/100

24k = 72
k=3
n=14

k+n = 17
Answer : A
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Old 2007 November 5th, 07:25 PM   #3 (permalink)
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Quote:
Originally Posted by vensgal View Post
- From the set of six letters, A,B,C,D,E,F there can be 20 sets of 3
letters formed

Col A
The number of 3 letter sets with the letter F in them

Col B
10
Col A : 5C2 = 10
Col B : 10

Answer : C.
Am I right?
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Old 2007 November 5th, 07:37 PM   #4 (permalink)
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Nice #1 solution, wouldn't have thought of it.

#2 is what I got as well.
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Old 2007 November 6th, 05:20 AM   #5 (permalink)
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same approach for both the answer!! wats the OA?
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Old 2007 November 7th, 03:52 PM   #6 (permalink)
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ques no 1 is repost
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Old 2007 November 7th, 04:25 PM   #7 (permalink)
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Hi coyote..
Thank you for explaining i have no idea abt the answers but the logic seems to be right.
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Old 2007 November 11th, 07:50 AM   #8 (permalink)
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number 2 is easy if u think in terms of probabilities.

you have a set of three letters out of six. what is the probability that F is in set of three letters? it's 50%...
why means that out of the 20 combinations there are 10 of them containing one F.

why is it 50%? if u don't get it instinctively construct a smaller example...
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Old 2007 November 11th, 11:21 PM   #9 (permalink)
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#1 I got stumped.

#2 Was a much easier combination problem... 5 choose 3 = 5!/2!3! = 10 combinations that do not have F. Thus, the columns are equal.

EDIT: I wrote the problem wrong for #1, twice. No wonder I couldn't find a solution, heh. Coyote's trick was awesome though.
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