Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56300

What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
30 Oct 2014, 09:46
Question Stats:
37% (01:45) correct 63% (01:50) wrong based on 281 sessions
HideShow timer Statistics
Tough and Tricky questions: Probability. What is the probability that u/v/w and x/y/z are reciprocal 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Manager
Joined: 02 Jul 2012
Posts: 186
Location: India
GPA: 2.6
WE: Information Technology (Consulting)

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
30 Oct 2014, 18:20
Bunuel wrote: Tough and Tricky questions: Probability. What is the probability that u/v/w and x/y/z are reciprocal 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. I got this one wrong. Please correct my reasoning: 1  we only know about v,w,y,z. No information about u and x  Insufficient 2. u*x is the median of 100 consecutive integers. Firstly, the only information provided is about u and x and secondly, the the consecutive integers can start from anywhere. They may be negative positive, some negative some positive, so Insufficient Combining the two, We have some information about v,w,y,z but info about u and x is unclear So E
_________________
Give KUDOS if the post helps you...



Intern
Joined: 23 Sep 2012
Posts: 20
Concentration: Technology, Operations
GPA: 4
WE: Information Technology (Computer Software)

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
31 Oct 2014, 12:00
@ThoughttoSphere: the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ? Statement 1: really doesn't tells us anything except that v,w , y and z are integers chosen from the first 100 positive integers. Clearly not sufficient. Statement 2: tells us that product of x and u is the median of 100 consecutive integers. Now whatever be the set of 100 consecutive integers, the median is always going to be the mean of the 50th and 51st term and since the terms are consecutive, the median will be x.5 which is not an integer. Clearly not sufficient by itself. Combining 1 and 2  we can surely tell that since v,w,y and z are integers their product can never be equal to a fraction. Hence C.



Manager
Joined: 21 Sep 2012
Posts: 212
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
01 Nov 2014, 06:10
romitsn wrote: @ThoughttoSphere: the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ? Statement 1: really doesn't tells us anything except that v,w , y and z are integers chosen from the first 100 positive integers. Clearly not sufficient. Statement 2: tells us that product of x and u is the median of 100 consecutive integers. Now whatever be the set of 100 consecutive integers, the median is always going to be the mean of the 50th and 51st term and since the terms are consecutive, the median will be x.5 which is not an integer. Clearly not sufficient by itself. Combining 1 and 2  we can surely tell that since v,w,y and z are integers their product can never be equal to a fraction. Hence C. Please explain how did you get the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ?



Intern
Joined: 04 Jul 2014
Posts: 46

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
05 Nov 2014, 13:46
u/v/w & x/y/z are the two fractions given.
I can interpret it in two ways
1. [u/v] / w & [x/y] / z ....in this case ...we get the equation ux = yzvw ...and hence ans C.
2. But if i interpret the fraction as
U / [v/w] & X / [y/z] ....in this case....we get the equation uwxz = vy ...i.e. ux = vy / wz i.e. 50.5 = vy / wz [50.5 if we consider 1 to 100]
So this equation can give somenumber point five and no way we can determine if reciprocals or not . Hence here ans E.
@@bunuel ....should the question have more clarity . Please suggest.



Manager
Joined: 02 Jul 2015
Posts: 102

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
12 Sep 2015, 05:34
Not sure about the answer as the expression in the question is not clear. I could not get an expression which I could take as a premise. Different interpretations, different expressions. Bunuel.. a little help please!



Manager
Joined: 14 Oct 2012
Posts: 160

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
19 Oct 2016, 18:10
Question: (u*w)/v = y/(x*z) ? [1] v, w, y, z E {1,2, ... 99, 100} Not sufficient... [2] ux = (50+51)/2 = 101/2 = 50.5 Not sufficient... [1]+[2]: ux = 50.5 = (v/w)*(y/z) Now no pair of random values from 1 to 100 can form (101/2) as 101 is a prime number. So ux = (v/w)*(y/z) => (u*w)/v = y/(x*z) Sufficient. Answer C



Manager
Joined: 10 Apr 2015
Posts: 178
GPA: 3.31

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
26 Nov 2016, 04:24
This can be simplified as follows: What is the probability that ux/vy/wz = 1 ? What is the probability that ux/vy = wz ? Finally: What is the probability that ux = vywz ? Statement (1) tells us that vywz is an integer, since it is the product of integers. However, this gives no information about u and x and is herefore not sufficient to answer the question. Statement (2) tells us that ux is NOT an integer. This is because the median of an even number of consecutive integers is NOT an integer. (For example, the median of 4 consecut ive integers  9, 10, 11, 12  equals 10.5.) However, this gives us no information about vywz and is therefore not sufficient to answer the question. Taking both statements together, we know that vywz IS an integer and that ux is NOT an integer. Therefore vywz CANNOT be equal to ux. The probability that the fractions are reciprocals is zero. The correct answer is C: Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient
_________________
In case you find my posts helpful, give me Kudos. Thank you.



Manager
Joined: 09 Nov 2013
Posts: 63

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
28 Nov 2016, 11:03
if we know we can calculate the probability that all x,z,u,w,v, and y are integers, we get the answer.



Director
Joined: 26 Oct 2016
Posts: 626
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
06 Jan 2017, 07:06
Please find the solution of this problem attached.
Attachments
tough&tricky.PNG [ 27.35 KiB  Viewed 1863 times ]
_________________
Thanks & Regards, Anaira Mitch



Senior Manager
Joined: 02 Apr 2014
Posts: 473
Location: India
GPA: 3.5

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
22 Nov 2017, 12:43
anairamitch1804 wrote: Please find the solution of this problem attached. Hi anairamitch1804 Bunuel chetan2u can we call 1/2.5 as a fraction or fraction needs to be strictly reduced form of integers? Thanks



NonHuman User
Joined: 09 Sep 2013
Posts: 11709

Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
Show Tags
26 Nov 2018, 02:40
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
[#permalink]
26 Nov 2018, 02:40






