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A wrapping paper has dimensions 1 yard by 7 yards. What is the volume

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

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New post 17 Oct 2012, 19:21
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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
[Reveal] Spoiler: OA

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"Do or do not. There is no 'try'..."

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

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New post 17 Oct 2012, 20:59
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aeros232 wrote:
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

[Reveal] Spoiler:
D


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

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New post 17 Oct 2012, 21:11
Bunuel wrote:
aeros232 wrote:
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

[Reveal] Spoiler:
D


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

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New post 19 Oct 2012, 09:42
Bunuel wrote:
aeros232 wrote:
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

[Reveal] Spoiler:
D


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

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New post 19 Oct 2012, 09:45
mydreammba wrote:
Bunuel wrote:
aeros232 wrote:
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

[Reveal] Spoiler:
D


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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 19 Oct 2012, 09:53
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
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]

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New post 19 Oct 2012, 09:54
mydreammba wrote:
Bunuel wrote:
aeros232 wrote:
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

[Reveal] Spoiler:
D


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
_________________

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

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

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New post 29 Apr 2016, 01:05
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]

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New post 12 Sep 2017, 07:53
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).

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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume   [#permalink] 12 Sep 2017, 07:53
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