Last visit was: 22 Apr 2026, 15:13 It is currently 22 Apr 2026, 15:13
Close
GMAT 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 GMAT 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
bilv23
Joined: 26 Nov 2023
Last visit: 18 Jul 2024
Posts: 1
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,754
Own Kudos:
Given Kudos: 105,823
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,754
Kudos: 810,663
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Sababa1561
Joined: 03 Dec 2017
Last visit: 04 Jan 2025
Posts: 4
Own Kudos:
Given Kudos: 125
Posts: 4
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
saransh2797
Joined: 23 May 2021
Last visit: 28 Jan 2025
Posts: 11
Given Kudos: 10
Posts: 11
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,
what confused me in the solution is that if -x^2 and -x^3 are -1/4 and -1/8, then how are we again treating them as x individually. The way question is written it seems that the set has 3 numbers, x, -x^2, and -x^3

Please help me out with this, thanks :)
Bunuel
5. For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)


A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:


\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:


\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:


\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,754
Own Kudos:
810,663
 [1]
Given Kudos: 105,823
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,754
Kudos: 810,663
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
saransh2797
Hi Bunuel,
what confused me in the solution is that if -x^2 and -x^3 are -1/4 and -1/8, then how are we again treating them as x individually. The way question is written it seems that the set has 3 numbers, x, -x^2, and -x^3

Please help me out with this, thanks :)
Bunuel
5. For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)


A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:


\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:


\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:


\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D.

The set's rule means each number generates new elements based on the pattern. Here, -x^2 and -x^3 are treated as elements in the set, not individual variables. They follow the same rule to generate further numbers.
User avatar
notsobigshaq
Joined: 18 Nov 2024
Last visit: 16 Mar 2026
Posts: 1
Given Kudos: 6
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
11. Certain bowl contains 5 red marbles and 3 blue marbles only. One by one, every marble is drawn at random and without replacement. What is the probability that the seventh marble drawn is NOT blue?

A. 7/8
B. 3/4
C. 2/3
D. 5/8
E. 3/8

Total arrangements possible: 8!/(5!*3!) = 56

Arrangements with Blue at 7th: 7!(5!*2!) = 21 (Position for 1 Blue Marble is fixed, arrange the remaining marbles)
Arrangements with Blue NOT at 7th: 56-21 = 35

Probability: 35/56 = 5/8

Answer: D

Is this a good approach? Or was it a fluke that I got it right with this method??
   1   2   3   4 
Moderators:
Math Expert
109754 posts
Tuck School Moderator
853 posts