please help me with these tricky DS

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saeed: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Tue Nov 27, 2007 9:40 pm

please help me with these tricky DS

by saeed » Sat Jul 26, 2008 12:53 am

Q1. If X & Y are positives integers, what is the value of X?
1. Y=3
2. 3^x5^y=1,125

Q2. Each of the 45 boxes on self J weighs less than each of the 44 boxes on self K. What is the median weight of the 89 boxes on these shelves?
1. The heaviest box on self J weighs 15 pounds.
2. The lightest box on self k weighs 20 pounds.

Q3. If XYZ>0, is XY^2Z^3>0?
1. Y>0
2. X>o
What is the best approach to solve for the last one?

Quote

Source: Beat The GMAT — Data Sufficiency | Sat Jul 26, 2008 12:53 am

Mani_mba: Master | Next Rank: 500 Posts; Posts: 267; Joined: Wed Jul 16, 2008 12:20 am; Thanked: 4 times; Followed by:1 members

by Mani_mba » Sat Jul 26, 2008 2:34 am

3) IMO A.

Here is my approach:

Given: xyz>0
a) Y>0. It implies that x and Z should be both positive or both negative.
If they are both positive, then x and z^3 would be both positive. Hence xy^2z^3 >0.
If they are both negative, then x and z^3 would be both negative. Hence xy^2z^3 > 0
b) x>0, It implies that y and z should be both positive or both negative.
If they are both positive, then y^2 and z^3 would be positive. Hence xy^2z^3 >0, whereas when they are both negative, then y^2 would be positive while z^3 would be negative. Hence this is insufficient.

2) IMO A.

Out of 89, 45th box is going to be the median. The heaviest box of shelf J is the 45th term.

1) IMO B.
Given statement B, 3^x5^y = 1125. Factorizing 1125 gives 9 and 125 that are 3^2 and 5^3 resp.
So x=2.