Each employee of Company Z is an employee of either Division X or Division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for Division X than for Company Z?

(1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Company Z.
(2) More than half the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.

I dont understand why you have to solve in 60 seconds and guess. You can take 90 seconds and be sure.

Yes %Full-Z will be a weighted average of %Full-X and %Full-Y. So either X>Z>Y or X<Z<Y, Z will always be in the middle.

So (1) is enough, as if Y<Z, then 'X must >Z'

For 2 we see that X has more full time and less part time than Y, therefore X>Y, and by extension also X>Z.

So both are enough,(D)

@bostonrb

For 2 we see that X has more full time and less part time than Y, therefore X>Y, and by extension also X>Z.

jsu to calrify, are you saying that For x -> (Higher numerator)/(lower denominator) > For Y -> (Lower numerator)/(higher denominator) ?
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@ subashghosh.....got it.....ur detailied explanation was helpful
regards

Lets say ratio was initially 1:1 for Z and for the divisions X 1:1 and for Y 1:1

Now I make the ratio for X as 1:4. Since X is 1:4 then Y will become 4:1 So that the weighted ratio is still 1:1. The same is true vice-versa.

So basically the DS is based on truism.

Pardon me. a) ratios are fractions and not certain numbers and b) there is no statement - saying one division is superior over the other. I dont know what you mean by weight.

High score requires common sense not math

Actually... I did not use weights, just that z will always be between x and y regardless of their weights.

Dont mean to be rude but you gotta go read up on weights too, because you also dont understand them. e.g. If you make the ratio of X as 1:4 as gmat1220 says, it is completely INCORRECT to say that Y's becomes 4:1. And this is because of weights.

z= weightX*AverageX+weightY*AverageY. NOT (AverageX+AverageY)/2

Belie' dat

Don't worry. I understand weighted average. What is intriguing is that we use these concepts often without even meaning to use them. When I say 'he is using weights', it means the general concept of 'weighted averaging'

Let me give you an example:

I have two solutions X and Y, 1 lt each, of wine and water. Solution X has wine:water in the ratio 2:3. I mix these two solutions to get wine:water ratio in the mixture as 1:1. What is wine:water in Y?
Guess what, the ratio of wine:water in Y is 3:2.

You can do it in various ways
Say working with concentration on wine:
1/2 = (2/5 * 1 + x*1)/2
x = 3/5

or we can simply say that wine in X is 40%, so wine in Y must be 60% to get 50% in mixture. This is averaging (or more generically weighted averaging... ) Here the weights are the same....

But the thing to note is that it doesn't matter whether we have the weights or now because due to averaging, the mixture will have concentration of wine in between X and Y.
If one understands this, one may not need to use any math and do it intuitively. Others may use the equations above to arrive at the same conclusion.

gmat1220, I believe, was using a generic example to explain why his intuition said what it did. If you want to get a ratio of 1:1 in the combined mixture and one mixture has a ratio of 1:4, the other will have 4:1 provided they both have equal weights (but we don't have to worry about the weights since eventually they do not matter in our question) So basically, it is the same general concept of weighted averages.

@gmat1220: I hope this answers your question too.
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