quants doubts..tricky questions

[vensgal]

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

[coyote]

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

[coyote]

Originally Posted by vensgal

- 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?

[DWarrior]

Nice #1 solution, wouldn't have thought of it.

#2 is what I got as well.

[peachy_snare]

same approach for both the answer!! wats the OA?

[tanmoy]

ques no 1 is repost

[vensgal]

Hi coyote..
Thank you for explaining i have no idea abt the answers but the logic seems to be right.

[doubtful]

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...

[Olm]

#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.