Last visit was: 27 Apr 2024, 13:10 It is currently 27 Apr 2024, 13:10

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10616 [3]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
avatar
Intern
Intern
Joined: 18 Nov 2017
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10616 [0]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10616 [0]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Expert Reply
Explanation

In Quantity A, if y = 4, then the numbers (arranged in increasing order) become \(x, x_2, x_3, x_4, x_4, x_6\); the mode is \(x_4\).


In Quantity B, if y = 5, then the numbers become \(x, x_2, x_3, x_4, x_4, x_6\). Usually, you’d need to take the average of the middle two numbers to find the median because there is an even number of values, but in this case they’re both \(x_4\).

The median, then, is \(x_4\), so the quantities are equal. Because x > 1, you don’t have to worry about special cases such as 0, 1, negatives, or fractions, and the correct answer is choice C.
avatar
Intern
Intern
Joined: 26 Aug 2017
Posts: 30
Own Kudos [?]: 47 [2]
Given Kudos: 0
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
2
How do you know that x > 1. And what does the subscript mean?
Verbal Expert
Joined: 18 Apr 2015
Posts: 28660
Own Kudos [?]: 33141 [0]
Given Kudos: 25178
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Expert Reply
Bump the discussion
avatar
Intern
Intern
Joined: 19 May 2021
Posts: 3
Own Kudos [?]: 2 [0]
Given Kudos: 2
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Can someone explain this question again?
Verbal Expert
Joined: 18 Apr 2015
Posts: 28660
Own Kudos [?]: 33141 [0]
Given Kudos: 25178
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Expert Reply
OE

In Quantity A, if y = 4, then the numbers (arranged in increasing order) become \(x,x^2,x^3,x^4,x^4,x^6\), and the mode is \(x^4\)

In Quantity B, if y = 5, then the numbers become \(x,x^2,x^3,x^4,x^4,x^5,x^6\)

Usually, you’d need to take the average of the middle two numbers to find the median because there is an even number of values, but in this case they’re both \(x^4\). The median, then, is \(x^4\), so the quantities are equal. Because x > 1, you don’t have to worry about special cases such as 0, 1, negatives, or fractions, and the correct answer is choice (C).
Manager
Manager
Joined: 22 Sep 2020
Posts: 74
Own Kudos [?]: 64 [0]
Given Kudos: 97
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
The explanation is not still clear to me, I mean after plugging the value of y = 4 and 5 there is a list given, but I don't understand how this list is inferred. Can anyone share a more detailed answer.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11691 [4]
Given Kudos: 136
Send PM
x,x2,xy,xy-1,x4,x6 [#permalink]
4
sandy wrote:
\(x>1\)

\(x,x^2,xy,xy^{-1},x^4,x^6\)

Quantity A
Quantity B
The mode of the numbers above when y = 4
The median of the numbers above when y =5


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

Drill 2
Question: 1
Page: 524

I don't like this question...

Given set: \({x,x^2,xy,xy^{-1},x^4,x^6}\)

If \(y = 4\), then the set becomes: \({x,x^2,4x,\frac{1}{4}x,x^4,x^6}\)
So, if x = 2, then the set becomes \({2,4,8,\frac{1}{2},16,64}\), in which case the mode is \(2,4,8,\frac{1}{2},16,64\)

Similarly, if \(y = 5\), then the set becomes: \(x,x^2,5x,\frac{1}{5}x,x^4,x^6\)
So, if x = 2, then the set becomes \(2,4,20,\frac{2}{5},16,64\), in which case the median is \(10\)

So we have:
QUANTITY A: \(2,4,8,\frac{1}{2},16,64\)
QUANTITY B: \(10\)

There's no way to compare these two quantities.

Answer: Bad question
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4432
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: x,x2,xy,xy-1,x4,x6 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne