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

 It is currently 19 Jan 2019, 02:24

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

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

Events & Promotions

Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• Free GMAT Strategy Webinar

January 19, 2019

January 19, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
• FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

M25-08

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

16 Sep 2014, 00:23
31
00:00

Difficulty:

65% (hard)

Question Stats:

55% (01:10) correct 45% (01:17) wrong based on 457 sessions

HideShow timer Statistics

A sphere is inscribed in a cube with an edge of $$x$$ centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. $$x(\sqrt{3}-1)$$
B. $$\frac{x}{2}$$
C. $$x(\sqrt{2}-1)$$
D. $$\frac{x}{2}(\sqrt{3}-1)$$
E. $$\frac{x}{2}(\sqrt{2}-1)$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

16 Sep 2014, 00:23
7
13
Official Solution:

A sphere is inscribed in a cube with an edge of $$x$$ centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. $$x(\sqrt{3}-1)$$
B. $$\frac{x}{2}$$
C. $$x(\sqrt{2}-1)$$
D. $$\frac{x}{2}(\sqrt{3}-1)$$
E. $$\frac{x}{2}(\sqrt{2}-1)$$

Say $$x=10$$ centimeters.

Then, since a sphere is inscribed in cube then the edge of the cube equals to the diameter of a sphere, thus $$Diameter=10$$.

Next, diagonal of a cube equals to $$Diagonal=\sqrt{10^2+10^2+10^2}=10\sqrt{3}$$.

Now half of Diagonal minus Diameter is gap between the vertex of a cube and the surface of the sphere, thus $$gap=\frac{Diagonal-Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)$$.

Since $$x=10$$ then $$5(\sqrt{3}-1)=\frac{x}{2}(\sqrt{3} - 1)$$.

_________________
Intern
Joined: 07 May 2014
Posts: 27
Concentration: Strategy, Technology
Schools: Anderson '17
GMAT 1: 730 Q50 V40
GPA: 3.1
WE: Analyst (Consulting)

Show Tags

08 Oct 2014, 04:50
Cube length = x cm

therefore, diagonal of the cube - a = x*sqrt(3)
diameter of the sphere inscribed - b = x * sqrt(2)

distance left is a - b. this distance is the distance between both parts which are left, either side of the diameter of the sphere. so the distance between the point is half the distance. hence, 1/2 * (a-b)
Intern
Joined: 08 Oct 2014
Posts: 4

Show Tags

08 Oct 2014, 13:39
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)-x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)-1).
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

08 Oct 2014, 13:56
temoalta wrote:
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)-x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)-1).

We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face.
_________________
Intern
Joined: 08 Oct 2014
Posts: 4

Show Tags

08 Oct 2014, 14:16
Bunuel wrote:
temoalta wrote:
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)-x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)-1).

We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face.

I do understand we have a right triangle. I guess I don't see what is wrong with my logic.
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

08 Oct 2014, 14:25
temoalta wrote:
Bunuel wrote:
temoalta wrote:
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)-x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)-1).

We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face.

I do understand we have a right triangle. I guess I don't see what is wrong with my logic.

You are considering wrong right triangle.
_________________
Intern
Joined: 08 Oct 2014
Posts: 4

Show Tags

09 Oct 2014, 07:03
Isn't that formula used to calculate the greatest distance? The question is asking for the shortest possible distance. What is the correct triangle we should we use here?
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

09 Oct 2014, 07:16
Intern
Joined: 08 Oct 2014
Posts: 4

Show Tags

09 Oct 2014, 07:27
I understand now. I thought that vertices were the edges of the cube. Thanks.
Intern
Joined: 19 Sep 2014
Posts: 22
GPA: 3.96

Show Tags

04 Mar 2015, 00:32
1
I think this question is good and helpful.
Could anyone explain the above with the help of a diagram for better understanding?
Math Expert
Joined: 02 Sep 2009
Posts: 52294

Show Tags

04 Mar 2015, 02:52
schak2rhyme wrote:
I think this question is good and helpful.
Could anyone explain the above with the help of a diagram for better understanding?

Check the links in my post above.
_________________
Intern
Joined: 20 Dec 2015
Posts: 2

Show Tags

30 Dec 2015, 18:59
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 24 May 2013
Posts: 79

Show Tags

25 Mar 2016, 20:35
2
The sphere and cube arrangement will be as shown in figure.

1/2(Diagonal-Diameter) = 1/2 (a sqrt3 -a)= a/2 (sqrt3-1)
Attachments

Sphere.png [ 9.32 KiB | Viewed 6915 times ]

Intern
Joined: 12 May 2016
Posts: 7
Location: India
GPA: 3.93

Show Tags

18 Jan 2017, 08:43
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 16 Oct 2016
Posts: 192
Location: India
Concentration: General Management, Healthcare
GMAT 1: 640 Q40 V38
GMAT 2: 680 Q48 V35
GPA: 3.05
WE: Pharmaceuticals (Health Care)

Show Tags

22 May 2018, 07:56
"half of Diagonal minus Diameter is gap between the vertex of a cube and the surface of the sphere"

Shouldn't it be 0.5 (Diagonal) - Radius of the sphere

As the diagonal will have two segments which are not part of the diameter?
_________________

_____________________
Chasing the dragon

Intern
Joined: 12 Jan 2017
Posts: 35

Show Tags

09 Jun 2018, 08:58
temoalta wrote:
Bunuel wrote:
temoalta wrote:
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)-x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)-1).

We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face.

I do understand we have a right triangle. I guess I don't see what is wrong with my logic.

Why it is a 30:60:90 right triangle and not a 45:45:90 right triangle is because of the 3-D plane. A normal square would have a diagonal of $$x\sqrt{2}$$, but since this is a cube the diagonal is $$x\sqrt{3}$$.
Re: M25-08 &nbs [#permalink] 09 Jun 2018, 08:58
Display posts from previous: Sort by

M25-08

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.