Data Sufficiency

There are 42 students in a group. If each student is either a freshmen or a senior, how many of the students are seniors?

(1) The group has more than four times as many seniors as it has freshmen.

(2) The group has more than 7 freshmen.

S + F = 42

stmt 1 says that S > 4F. We get many possibilities
stmt 2 says that F > 7. Again, we get many possibilities

1 & 2 Let pick numbers. If F = 8, S > 32. S can be 33 or 34. Since total need to be 42, S must be 34.


Hence C.

Edited.

Not clear.
If x=freshman,then from stmt 1 we get seniors=5x,6x,7x,8x...as more than four times is mentioned.

Stmt 2 insuff..

Combining,
6x=42, x=7 stmt 2 says more than 7 freshmen so not poss.
7x=42,x=6 which is again not greater than 7

I am messing it up..could u pls pt out my mistake?

crackgmat007 wrote:S + F = 42

stmt 1 says that S > 4F. We get many possibilities
stmt 2 says that F > 7. Again, we get many possibilities

1 & 2 Let pick numbers. If F = 8, S > 32. S can be any number greater than 32. If F = 9, S > 36.

Hence E.

For both statements together
You missed out on the fact that S+F =42.
If F=9 => S> 36 => S+F > 45... not possible,
so we are left with F= 8 and S =34.
So, choice C

I think answer is C

stat 1 and 2 clearly insuff.

taking together we get following possibilities

f s
8 34

9 36 not possible bcos 9+36=45

we have just 42 students.

m i missing something?

guess me and mohit bumped into each other..

aah..missed that. Silly mistake, but it costs.

Thanks for pointing out. C it is

uptowngirl92 wrote: 6x=42

This is creating problems.

Actually, question gives 3 equations

From question stem S + X = 42
Stmt 1 - S > 4X
Stmt 2 - X > 7

Continuing your working, X must be > 7 (stmt 2). Let X be 8. Stmt 1 says that S > 4X. Plugging in 8, we get S > 32. Now refer to the equation (hiding) in the question. S + X = 42

HTH