Tricky DS - How to solve Company, Divison x Division y

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 416
Joined: Wed Oct 03, 2007 9:08 am
Thanked: 10 times
Followed by:1 members
Can this be solved in two minutes.. spent a lot of time on this problem
Attachments
q28-t1.jpg

Master | Next Rank: 500 Posts
Posts: 468
Joined: Sat Mar 03, 2007 10:17 pm
Thanked: 5 times

by moneyman » Tue Nov 13, 2007 9:55 am
Well here is my approach,

The question says that all of the employees either belong to Div X or Div Y.

(1) gives the required answer.SUFFICIENT.

(2) says that more than half of full time employees of X belong to Div X and more than half of part time emplyees of X belong to Div Y. This gives us the exact information we are looking for.

To understand this better lets take an example

Lets assume company X has 100 full time and 100 part time exployees.So the ration of company X is 1:1. According to (2) lets say 60 FT emplyees belong to X and 60 PT employees belong to Y.The question says that all employees either belong to X or Y.So if there are 60 FT emps in X, there would be 40 FT emps in Y.Similarly, if there are 60 PT emps in Y then there would be 40 PT emps in X.This way we can find out the ratio for Division X.Hence SUFFICIENT.


Do you get me??
Maxx

Junior | Next Rank: 30 Posts
Posts: 29
Joined: Sat Oct 27, 2007 9:49 pm
Location: Vancouver, BC, Canada
Thanked: 1 times

Here is another approach....

by silivest60 » Mon Nov 19, 2007 1:37 pm
I found this problem solved on a website (I don't remember what website though) some time ago....I think it's a very straightforward approach. Here it goes:

Let denote:
f1 = number of FT employee for div X
p1 = number of PT employee for div X
f2 = number of FT employee for div Y
p2 = number of PT employee for div Y
Then: f=f1+f2 and p=p1+p2

We are asked to show that:
f1/p1 > (f1+ f2)/(p1 + p2)
develop this inequation and we get:
f1p2 > f2p1

Stat.1 says: f2/p2 less then (f1+f2) divided by (p1+p2), develop it and we get f1p2 > f2p1 => stat.1 sufficient

stat.2 says: f1>f2 & p2>p1, multiply these two inequalities and you'll get:
f1p2 > f2p1 => stat.2 is sufficient

Ans. is D

Is that simple, or what??

Senior | Next Rank: 100 Posts
Posts: 41
Joined: Thu Nov 08, 2007 3:22 pm

by aimhigh715 » Mon Nov 19, 2007 10:59 pm
thanks silivest60!
Work Hard, Play Harder!

Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

Re: Here is another approach....

by 4meonly » Thu Nov 20, 2008 9:25 am
silivest60 wrote: stat.2 says: f1>f2 & p2>p1, multiply these two inequalities and you'll get:
f1p2 > f2p1 => stat.2 is sufficient
Can anybody elaborate?

Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

by 4meonly » Tue Nov 25, 2008 9:01 am
ANYBODY?

Master | Next Rank: 500 Posts
Posts: 207
Joined: Sun Mar 11, 2007 6:16 pm
Location: Mumbai
Thanked: 11 times

by jimmiejaz » Tue Nov 25, 2008 9:14 am
4meonly wrote:ANYBODY?
In this equation,

since we know all the numbers are positive...
since employees are offcourse in positive.
let's say
3>2 and 5>4
offcourse 15>8 that's what is done here.

Hope it clears the confusion. Also, it can't be fractions.... only positive integers....
What if i have not yet beat the beast, I know i will beat it!!!!!!!!

Master | Next Rank: 500 Posts
Posts: 207
Joined: Sun Mar 11, 2007 6:16 pm
Location: Mumbai
Thanked: 11 times

by jimmiejaz » Tue Nov 25, 2008 9:16 am
one more thing to add.....
as long as the variables on both sides in inequalities are positive, we can treat them same as equations.
What if i have not yet beat the beast, I know i will beat it!!!!!!!!