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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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: \(6a^2={63}\) --> \(a=\sqrt{10.5}\) --> \(volume=a^3=(\sqrt{10.5})^3\).

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.

Great, thanks Bunuel
_________________

Aeros "Why are you trying so hard to fit in when you were born to stand out?" "Do or do not. There is no 'try'..."

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
_________________

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.

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?

Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]

Show Tags

20 Aug 2014, 22:41

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: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]

Show Tags

29 Apr 2016, 00:05

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: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]

Show Tags

12 Sep 2017, 06:53

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.
_________________