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Very hard. Full time vs Part time in company z

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Very hard. Full time vs Part time in company z

by gmatruler » Sat Jul 17, 2010 1:27 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

Not sure how to setup equations to test statements. should I be plugging in??
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Source: — Data Sufficiency |
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by Patrick_GMATFix » Sat Jul 17, 2010 1:36 pm
(1) is easily sufficient because the ratio of full:part for the company will be somewhere between the ratios in the Divisions. Since the ratio for Y is less than for the ratio of the whole company, the ratio for X must be greater than the ratio for the whole company. SUFFICIENT.

(2) If we call p the number of part timers company wide and f the number of full timers. we know that division X has more than .5f but less than .5p. So ratio of full time to part time is something like .6f/.4p for division X, but .4f/.6p for division Y. We can be sure that the ratio is higher for division X. SUFFICIENT

The answer is D

A more elaborate algebraic solution as well as a step by step video solution is available at GMATPrep Question 1028. To practice similar questions, set topic='FDPs & Ratios' and difficulty='700+' in the Drill Engine.

Good luck,
-Patrick
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by gmatruler » Sun Jul 18, 2010 6:41 am
Nice reasoning. If I could have thought of that the question might have taken me 1 minute instead of 5!

The detailed explanation you linked to proves why the 2nd statement is sufficient algebraically, but I have one question about the reasoning you used here. In statement 2, how do you know to break up the groups into .6 and .4? all we know is where more than 1/2 of each group belongs; we are not given the specific ratio.

Please clarify
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by sumanr84 » Sun Jul 18, 2010 11:05 pm
gmatruler wrote:Nice reasoning. If I could have thought of that the question might have taken me 1 minute instead of 5!

The detailed explanation you linked to proves why the 2nd statement is sufficient algebraically, but I have one question about the reasoning you used here. In statement 2, how do you know to break up the groups into .6 and .4? all we know is where more than 1/2 of each group belongs; we are not given the specific ratio.

Please clarify
.6 and .4 are just reference that Patrick has used here to demonstrate one possibility, since the statement says Full time employees at div X >.5f

You can take anything i.e. ( 0.7 f , 0.3 p), (0.8 f, 0.2 p)
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by Patrick_GMATFix » Mon Jul 19, 2010 1:47 pm
Thanks Sunmar. You got what I was trying to explain.
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by Vipulvp » Tue Jul 20, 2010 2:44 am
gmatruler wrote: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

Not sure how to setup equations to test statements. should I be plugging in??
You can use equations in this question to proceed but that method won't be half as elegant and intuitive as the one already descibed. Still....Let us construct a matrix as follows:

X Y
______________________
P.T a b

F.T c d

we are asked if c/a > d/b.

From (1), we can set up the equation (d + c)/(b +a) > d/b.
now this increase in the original fraction d/b is only possible if the numerator increases more than does the denominator. i.e. c > a. If c were to be equal to a, the original fraction d/b would have remained the same. Thus c/a > d/b.

From (2), c > d i.e. c/d > 1 and b > a i.e. b/a > 1. We can multiply the two inequalities knowing that the LHS would always exceed unity. So we have bc/ad > 1. Rearranging the terms we have c/a > d/b. Hence sufficient.

As I said, this question was solved much more elegantly without resorting to any equations.
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by Patrick_GMATFix » Tue Jul 20, 2010 5:52 am
Hi Vipulvp,

Thank you for your solution. There is one problem with your math in statement 1 that I want to make you aware of so it doesn't burn you later. You write:
Vipulvp wrote:From (1), we can set up the equation (d + c)/(b +a) > d/b.
now this increase in the original fraction d/b is only possible if the numerator increases more than does the denominator. i.e. c > a. If c were to be equal to a, the original fraction d/b would have remained the same. Thus c/a > d/b.
In fact is is not true that given a fraction d/b, increasing the top and bottom by the same number would result in the same fraction. For instance 1/2 is not equal to (1+5)/(2+5). The fraction would remain the same if the top and bottom are changed by the same percentage, not the same number: 1/2 = (1+3)/(2+6) [top and bottom increased by 300%]

So to get back to your analysis of statement 1, since (d+c)/(b+a) > (d/b), the only thing we can safely conclude is that the numerator was increased by a greater % than the denominator was. In other words, c is a greater percentage of d than a is of b. Mathematically c/d > a/b. To see how this is sufficient mathematically, consider the question as you rephrased it.
Vipulvp wrote:we are asked if c/a > d/b.
Cross multiply to rephrase to "is cb > ad?". Now look at our mathematical deduction from above: c/d > a/b. Cross multiply that and you will get cb > ad. The deduction directly answers your rephrase. Thus statement 1 is SUFFICIENT

-Patrick

PS. As you noticed above, I cross multiplied your rephrase to transform "is c/a > d/b" into "is cb > ad?". Since we are dealing with positive values only (number of employees), it's safe to cross multiply (the inequality symbol cannot flip). As a general rule, in DS if you can get rid of fractions, you should. Fractions do a good job of hiding data, so they are more difficult to evaluate in DS.
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by Vipulvp » Tue Jul 20, 2010 6:18 am
Patrick_GMATFix wrote:Hi Vipulvp,

Thank you for your solution. There is one problem with your math in statement 1 that I want to make you aware of so it doesn't burn you later. You write:
Vipulvp wrote:From (1), we can set up the equation (d + c)/(b +a) > d/b.
now this increase in the original fraction d/b is only possible if the numerator increases more than does the denominator. i.e. c > a. If c were to be equal to a, the original fraction d/b would have remained the same. Thus c/a > d/b.
In fact is is not true that given a fraction d/b, increasing the top and bottom by the same number would result in the same fraction. For instance 1/2 is not equal to (1+5)/(2+5). The fraction would remain the same if the top and bottom are changed by the same percentage, not the same number: 1/2 = (1+3)/(2+6) [top and bottom increased by 300%]

So to get back to your analysis of statement 1, since (d+c)/(b+a) > (d/b), the only thing we can safely conclude is that the numerator was increased by a greater % than the denominator was. In other words, c is a greater percentage of d than a is of b. Mathematically c/d > a/b. To see how this is sufficient mathematically, consider the question as you rephrased it.
Vipulvp wrote:we are asked if c/a > d/b.
Cross multiply to rephrase to "is cb > ad?". Now look at our mathematical deduction from above: c/d > a/b. Cross multiply that and you will get cb > ad. The deduction directly answers your rephrase. Thus statement 1 is SUFFICIENT

-Patrick

PS. As you noticed above, I cross multiplied your rephrase to transform "is c/a > d/b" into "is cb > ad?". Since we are dealing with positive values only (number of employees), it's safe to cross multiply (the inequality symbol cannot flip). As a general rule, in DS if you can get rid of fractions, you should. Fractions do a good job of hiding data, so they are more difficult to evaluate in DS.
Thanks Patrick indeed for pointing this out!
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