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 350,000 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:

The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 13:48

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

73% (02:27) correct
27% (01:37) wrong based on 73 sessions

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the rectangular solid, what the ratio of the volume of the cube to the volume within the rectangular solid that is not occupied by the cube? (A) 2:3 (B) 2:5 (C) 5:16 (D) 25:7 (E) 32:25

The question is today's question by Jeff Sackmann . In his answer he mentioned:

"Since the cube shares one of the dimensions of the rectangular solid, it must have a side of 4, 5, or 8. However, if its side is 5 or 8, it won't fit entirely within the solid."

Why ? could any body please clarify this statement further ? Thanks in advance

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 14:58

1

This post received KUDOS

We have a rectangular solid and a cube, the first has three different values for each side, the second has the same value of each side. rectangular: 4,5,8 ( as in our case) cube: x,x,x Imagine now this rectangular solid, how would you "cut" it in order to obtain a cube? Or even better: what is the biggest cube you can obtain from it? From the rectangular (4,5,8) you could extract a cube 1x1x1, but this isn't clearly the biggest possible, try now with a 2x2x2 and so on... you'll reach 4x4x4 and this is our biggest value. Why? Because if you try to "cut" a 4,1x4,1x4,1 cube from a rectangular 4x5x8, how can you obtain a 4,1 side (for the cube) from a side of 4? it's impossible! From 8 you can cut a side of 4,1; from 5 you can cut a side of 4,1; but you cannot do it from a side of 4! That's why in this problem the cube must have a side of 4.

So if you want to obtain a cube from a rectangular solid, the max value the side of the cube can have is the min value of the side of the rectangular. _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 15:19

1

This post received KUDOS

TheNona wrote:

hitman5532 wrote:

(it must be 4, becuase if it is 5 or 8 the cube will be longer than the shortest side of the rectangular solid (\(4\)) and therefore not fit in the box. Remember, a cube is the same dimension on each of the three sides)

This is exactly the part I do not understand ... why it cannot be longer than the shorter side of the solid ? still cannot imagine

If i told you to put a 12 inch tall product in a shipping box which is only 4" x 4" x 4", would it fit? No.

Why?

Because 12" is greater than 4" and therefore you could not close the box to ship the item, right?

This follows the same idea. If the smallest side of an item is larger than the smallest side of a box within which it is to fit, you could not close to box. With a cube, all three sides are the same size.

If you are still struggling with this, I suggest drawing or constructing two box of the given measurements in the problem and to a look at what is occuring

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 14:28

TheNona wrote:

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the rectangular solid, what the ratio of the volume of the cube to the volume within the rectangular solid that is not occupied by the cube? (A) 2:3 (B) 2:5 (C) 5:16 (D) 25:7 (E) 32:25

The question is today's question by Jeff Sackmann . In his answer he mentioned:

"Since the cube shares one of the dimensions of the rectangular solid, it must have a side of 4, 5, or 8. However, if its side is 5 or 8, it won't fit entirely within the solid."

Why ? could any body please clarify this statement further ? Thanks in advance

Volume of rectangular solid =\(4*5*8 = 160\)

Volume of cube = \(4*4*4 = 64\) (it must be 4, becuase if it is 5 or 8 the cube will be longer than the shortest side of the rectangular solid (\(4\)) and therefore not fit in the box. Remember, a cube is the same dimension on each of the three sides)

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 14:36

TheNona wrote:

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the rectangular solid, what the ratio of the volume of the cube to the volume within the rectangular solid that is not occupied by the cube? (A) 2:3 (B) 2:5 (C) 5:16 (D) 25:7 (E) 32:25

The question is today's question by Jeff Sackmann . In his answer he mentioned:

"Since the cube shares one of the dimensions of the rectangular solid, it must have a side of 4, 5, or 8. However, if its side is 5 or 8, it won't fit entirely within the solid."

Why ? could any body please clarify this statement further ? Thanks in advance

If the cube has a length of 4, it will fit in a 4x5x8 cube; because the length of the cube is equal to the shortest side of the solid. Even if our cube has a length of 4,1 it will not fit in such rectangular solid: in the faces 5x8 or 8x5 fits, but in a face 4x8 or 4x5 does not ( the cube `s side is longer than 4, so 0,1 will be outside of the face, and so outside of the solid)

Hope it is clear _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 14:37

hitman5532 wrote:

(it must be 4, becuase if it is 5 or 8 the cube will be longer than the shortest side of the rectangular solid (\(4\)) and therefore not fit in the box. Remember, a cube is the same dimension on each of the three sides)

This is exactly the part I do not understand ... why it cannot be longer than the shorter side of the solid ? still cannot imagine _________________

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
29 Mar 2013, 14:41

Zarrolou wrote:

TheNona wrote:

The dimensions of a rectangular solid are 4 inches, 5 inches, and 8 inches. If a cube, a side of which is equal to one of the dimensions of the rectangular solid, is placed entirely within the rectangular solid, what the ratio of the volume of the cube to the volume within the rectangular solid that is not occupied by the cube? (A) 2:3 (B) 2:5 (C) 5:16 (D) 25:7 (E) 32:25

The question is today's question by Jeff Sackmann . In his answer he mentioned:

"Since the cube shares one of the dimensions of the rectangular solid, it must have a side of 4, 5, or 8. However, if its side is 5 or 8, it won't fit entirely within the solid."

Why ? could any body please clarify this statement further ? Thanks in advance

If the cube has a length of 4, it will fit in a 4x5x8 cube; because the length of the cube is equal to the shortest side of the solid. Even if our cube has a length of 4,1 it will not fit in such rectangular solid: in the faces 5x8 or 8x5 fits, but in a face 4x8 or 4x5 does not ( the cube `s side is longer than 4, so 0,1 will be outside of the face, and so outside of the solid)

Hope it is clear

thanks for the kind care but I would appreciate if you clarify more , please _________________

Re: The dimensions of a rectangular solid are 4 inches, 5 inches [#permalink]
24 Oct 2013, 08:33

Here's the full Official Explanation is case anyone was interested.

Answer: A Since the cube shares one of the dimensions of the rectangular solid, it must have a side of 4, 5, or 8. However, if its side is 5 or 8, it won't fit entirely within the solid. Since one of the lengths of the solid is 4, all of the lengths of the cube must be 4 or shorter. Thus the side of the cube is 4, and the volume of the cube is 4^3 = 64.

The volume of the solid is the product of the dimensions: (4)(5)(8) = 160. We're looking for the ratio of the volume of the cube (64) to the volume of the solid that is not occupied by the cube-- that is, 160 - 64 = 96. The ratio of 64 to 96 can be simplied by dividing both terms by 32. The result is 2 to 3, choice (A).

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...