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

[gmatrant]

Can this be solved in two minutes.. spent a lot of time on this problem

[moneyman]

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??

[silivest60]

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??

[aimhigh715]

thanks silivest60!

[4meonly]

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

Can anybody elaborate?

[4meonly]

ANYBODY?

[jimmiejaz]

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....

[jimmiejaz]

one more thing to add.....
as long as the variables on both sides in inequalities are positive, we can treat them same as equations.