Tricky DS - How to solve Company, Divison x Division y
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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??
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
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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??
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??
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- Master | Next Rank: 500 Posts
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In this equation,4meonly wrote:ANYBODY?
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!!!!!!!!
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- Master | Next Rank: 500 Posts
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one more thing to add.....
as long as the variables on both sides in inequalities are positive, we can treat them same as equations.
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!!!!!!!!