GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 23 Oct 2019, 05:58

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

Corners are sliced off a unit cube so that the six faces each become

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Corners are sliced off a unit cube so that the six faces each become  [#permalink]

Show Tags

New post 14 Mar 2019, 04:31
10
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

20% (01:55) correct 80% (02:35) wrong based on 20 sessions

HideShow timer Statistics

Director
Director
avatar
D
Joined: 19 Oct 2018
Posts: 997
Location: India
Premium Member CAT Tests
Corners are sliced off a unit cube so that the six faces each become  [#permalink]

Show Tags

New post 17 May 2019, 04:04
total number of tetrahedron=8 from each corner
let s be the side of the regular octagon, that left
edge of cube =\(s/\sqrt{(2)} + s + s/\sqrt{(2)} = 1\)
[\({\sqrt{(2)} +1}\)]s=1
\(s=1/{\sqrt{(2)} +1}=\sqrt{(2)} -1\)

Area of base which is equilateral triangle= \((\sqrt{3}/4)*s^2\)

height of tetrahedron= \(\sqrt{[(slant edge of tetrahedron)^2 - (circumcenter of base triangle)^2]}\)
height of tetrahedron= \(\sqrt{[(s^2/2)^-(s^2/3)]}\)= \(s/\sqrt{(6)}\)

Volume of tetrahedron= 1/3 * base area * height

Volume of tetrahedron =\(1/3*(\sqrt{3}/4)*s^2*s/\sqrt{(6)}= s^3/12*\sqrt{(2)}\)

\(s^3= {\sqrt{(2)} -1}^3\)= \({5*\sqrt{2} -7}\)

Volume of tetrahedron= \({5*\sqrt{2} -7}/12*\sqrt{(2)}\)
Total cutoff area= \(8*{5*\sqrt{2} -7}/12*\sqrt{(2)}\)
= \((10-7*\sqrt{2})/3\)

Enough Mathematics for today..... Took me more than 30 minutes to solve it
Bunuel wrote:
________________________
BUMPING FOR DISCUSSION.

Attachments

im.png
im.png [ 18.51 KiB | Viewed 233 times ]

GMAT Club Bot
Corners are sliced off a unit cube so that the six faces each become   [#permalink] 17 May 2019, 04:04
Display posts from previous: Sort by

Corners are sliced off a unit cube so that the six faces each become

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne