GMATPrep Ratio Problem

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by GMATGuruNY » Sat Sep 25, 2010 7:06 pm
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 of 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.


Z is the ratio when X and Y are combined. So we have 3 options:

X<Z<Y (The ratio in X is less than the ratio in Y, so that Z -- the ratio when X and Y are combined -- is somewhere in the middle.)
X=Y=Z (All 3 have the same ratio.)
Y<Z<X (The ratio in Y is less than the ratio in X, so that Z -- the ratio when X and Y are combined -- is somewhere in the middle.)

Statement 1:
If Y<Z, then Y<Z<X.
Thus, X>Z. Sufficient.

Statement 2:
To see the situation more clearly, let's plug in values that satisfy the statement.

Let's say that X has 51 full-time employees, and Y has 49.
Let's say that Y has 51 part-time employees, and X has 49.
This means Z has 100 full-time employees and 100 part-time employees.
In X, full-time:part-time = 51:49.
In Y, full-time:part-time = 49:51.
In Z, full-time:part-time = 100:100 = 1:1.
Then Y<Z<X, because 49:51 < 1:1 < 51:49.
Thus X>Z.

Let's plug in numbers that are further apart.
Let's say that X has 90 full-time employees, and Y has 10.
Let's say that Y has 90 part-time employees, and X has 10.
This means Z has 100 full-time employees and 100 part-time employees.
In X, full-time:part-time = 90:10 = 9:1. (X gets bigger.)
In Y, full-time:part-time = 10:90 = 1:9. (Y gets smaller.)
In Z, full-time:part-time = 100:100 = 1:1. (Z is unchanged.)
Then Y<Z<X, because 1:9 < 1:1 < 9:1.
Thus, X>Z.

So X>Z. Sufficient.

The correct answer is D.
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