Last visit was: 19 Nov 2025, 21:08 It is currently 19 Nov 2025, 21:08
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
aeros232
Joined: 04 Sep 2009
Last visit: 22 Jul 2013
Posts: 31
Own Kudos:
327
 [28]
Given Kudos: 9
Posts: 31
Kudos: 327
 [28]
2
Kudos
Add Kudos
26
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,390
 [8]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,390
 [8]
4
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
General Discussion
User avatar
aeros232
Joined: 04 Sep 2009
Last visit: 22 Jul 2013
Posts: 31
Own Kudos:
Given Kudos: 9
Posts: 31
Kudos: 327
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mydreammba
Joined: 28 Jul 2011
Last visit: 06 Dec 2013
Posts: 224
Own Kudos:
Given Kudos: 16
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE:Accounting (Commercial Banking)
Posts: 224
Kudos: 1,625
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
aeros232
A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper?

(1 yard = 3 Ft)

A. 8
B. 9
C. 18
D. 27
E. 36


Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).

The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet.

Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)).

So, it must be true that \(6a^2\leq{63}\).

If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube.
If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet.

Answer: D.

If we won't limit the value of a side to integers only, then: \(6a^2={63}\) --> \(a=\sqrt{10.5}\) --> \(volume=a^3=(\sqrt{10.5})^3\).

Hope it's clear.

Bunnel

Can you please let me know Why you didn't consider a=2? then the volume would be 8
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,390
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mydreammba
Bunuel
aeros232
A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper?

(1 yard = 3 Ft)

A. 8
B. 9
C. 18
D. 27
E. 36


Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).

The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet.

Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)).

So, it must be true that \(6a^2\leq{63}\).

If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube.
If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet.

Answer: D.

If we won't limit the value of a side to integers only, then: \(6a^2={63}\) --> \(a=\sqrt{10.5}\) --> \(volume=a^3=(\sqrt{10.5})^3\).

Hope it's clear.

Bunnel

Can you please let me know Why you didn't consider a=2? then the volume would be 8

We need the largest cube that could be packed using this paper. Now, the larger the side of a cube larger the volume and since a can be 3, no need to consider smaller lengths.

Hope it's clear.
User avatar
mydreammba
Joined: 28 Jul 2011
Last visit: 06 Dec 2013
Posts: 224
Own Kudos:
Given Kudos: 16
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE:Accounting (Commercial Banking)
Posts: 224
Kudos: 1,625
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Bunnel i din't read the Q properly and can you please respond the private message on statistics which i have sent you yesterday?

and for the post the link is here

clarification-on-statistics-sd-140907.html
User avatar
mydreammba
Joined: 28 Jul 2011
Last visit: 06 Dec 2013
Posts: 224
Own Kudos:
Given Kudos: 16
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE:Accounting (Commercial Banking)
Posts: 224
Kudos: 1,625
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mydreammba
Bunuel
aeros232
A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper?

(1 yard = 3 Ft)

A. 8
B. 9
C. 18
D. 27
E. 36


Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).

The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet.

Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)).

So, it must be true that \(6a^2\leq{63}\).

If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube.
If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet.

Answer: D.

If we won't limit the value of a side to integers only, then: \(6a^2={63}\) --> \(a=\sqrt{10.5}\) --> \(volume=a^3=(\sqrt{10.5})^3\).

Hope it's clear.



Thanks Bunnel i din't read the Q properly and can you please respond the private message on statistics which i have sent you yesterday?

and for the post the link is here

clarification-on-statistics-sd-140907.html
avatar
Franzhel
Joined: 19 Jul 2018
Last visit: 26 Oct 2024
Posts: 32
Own Kudos:
65
 [1]
Given Kudos: 18
Location: France
Concentration: Economics, Nonprofit
Schools: ESSEC '22
GMAT 1: 680 Q47 V35
GMAT 2: 710 Q49 V38
WE:Analyst (Manufacturing)
Schools: ESSEC '22
GMAT 2: 710 Q49 V38
Posts: 32
Kudos: 65
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In my opinion the question is wrongly formulated. It asks "What is the volume in Ft3 of the largest cube that could be packed using this paper?". As said in by Bunuel, the largest one is sqrt(10.5)^3. Option D is not the largest cube that could be packed.

Either it should change the information given (say that the sides are integers) or change the question.
avatar
M1ke
Joined: 03 Dec 2019
Last visit: 17 May 2020
Posts: 18
Own Kudos:
13
 [1]
Given Kudos: 13
Posts: 18
Kudos: 13
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Easier approach to this question:

Wrapping paper yard to foot conversion: 1*3=3 and 7*3=21 so the total space on the wrapping paper is 3 * 21 = 63 foot.
Factorize 63 => 3 * 3 * 7.

\(x^3\) is what we need for a cubed object, the maximum size therefore is the smallest factor of 63 which is 3. This is the biggest number which we can cube within our wrapping paper.

So the answer is \(3^3=27\)
avatar
ankur2510
Joined: 27 Apr 2018
Last visit: 31 May 2021
Posts: 15
Own Kudos:
Given Kudos: 188
Posts: 15
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Experts,

Can anyone tell me if my logic is correct here?

Since we are asked the volume of the largest cube out of the given option, and we know that the volume of the cube is a^3, that means in the answer we need to check which one is the perfect cube and the side could be between 3 and 7, we need to see the the options which is the perfect cube. Out of the options it is 27.

Please help me clear my understanding on this.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,589
Own Kudos:
Posts: 38,589
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Math Expert
105390 posts
Tuck School Moderator
805 posts