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(u/v)/w and (x/y)/z

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(u/v)/w and (x/y)/z

by sanju09 » Wed Feb 25, 2009 5:14 am
What is the probability that (u/v)/w and (x/y)/z are recirocal fractions?

(1) v, w, y, and z are each randomly chosen from the first 100 positive integers.

(2) The product (u) (x) is the median of 100 consecutive integers.
The mind is everything. What you think you become. -Lord Buddha



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Source: — Data Sufficiency |
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Re: (u/v)/w and (x/y)/z

by Stuart@KaplanGMAT » Wed Feb 25, 2009 1:27 pm
sanju09 wrote:What is the probability that (u/v)/w and (x/y)/z are recirocal fractions?

(1) v, w, y, and z are each randomly chosen from the first 100 positive integers.

(2) The product (u) (x) is the median of 100 consecutive integers.
Tough question!

Let's start by rewriting it:

(u/v)/w = u/vw
(x/y)/z = x/yz

Question: what's the probability that u/vw = yz/x

or:

what's the probability that ux = vwyz?

(1) nothing about u or x.. insufficient.
(2) nothing about v, w, y or z.. insufficient.

Together:

From (1), we know that v, w, y and z are all integers. Therefore, vwyz is an integer.

From (2), we know that ux is the median of 100 consecutive integers, therefore ux is NOT an integer (the median of an even number of terms is the average of the two middle terms; the average of two consecutive integers is going to end in .5).

Since vwyz IS an integer and ux is NOT an integer, the probability that ux=vwzy is 0... sufficient, choose (C).
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Re: (u/v)/w and (x/y)/z

by sanju09 » Thu Feb 26, 2009 1:26 am
Stuart Kovinsky wrote:
sanju09 wrote:What is the probability that (u/v)/w and (x/y)/z are recirocal fractions?

(1) v, w, y, and z are each randomly chosen from the first 100 positive integers.

(2) The product (u) (x) is the median of 100 consecutive integers.
Tough question!

Let's start by rewriting it:

(u/v)/w = u/vw
(x/y)/z = x/yz

Question: what's the probability that u/vw = yz/x

or:

what's the probability that ux = vwyz?

(1) nothing about u or x.. insufficient.
(2) nothing about v, w, y or z.. insufficient.

Together:

From (1), we know that v, w, y and z are all integers. Therefore, vwyz is an integer.

From (2), we know that ux is the median of 100 consecutive integers, therefore ux is NOT an integer (the median of an even number of terms is the average of the two middle terms; the average of two consecutive integers is going to end in .5).

Since vwyz IS an integer and ux is NOT an integer, the probability that ux=vwzy is 0... sufficient, choose (C).
:) HATS OFF! This is my explanation, word by word; mind-boggling Stuart Kovinsky!
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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