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

 It is currently 08 Dec 2019, 16:19

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

The greatest possible (straight line) distance, in inches, between any

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59590
The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

07 Oct 2016, 02:33
1
8
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:26) correct 38% (02:31) wrong based on 113 sessions

HideShow timer Statistics

The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

_________________
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3300
Location: India
GPA: 3.12
The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

07 Oct 2016, 05:35
1
1
The greatest distance(straight line) between two points in a cube is its diagonal.
Here the length of the digaonal is 10.
Therefore, Side*$$\sqrt{3} = 10$$
Side = $$\frac{10}{\sqrt{3}}$$

Since the length and breadth doubles, and height remains the same.
The new box is a cuboid with length = breadth = $$\frac{20}{\sqrt{3}}$$ and height is $$\frac{10}{\sqrt{3}}$$

The straight line greatest distance is the diagonal.
Diagonal = $$\sqrt{length^2 + breadth^2 + height^2} = \sqrt{300} = 10*\sqrt{3}$$(Option B)
_________________
You've got what it takes, but it will take everything you've got
Manager
Joined: 06 Oct 2017
Posts: 50
Schools: Schulich "22 (A)
GPA: 3.6
Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

19 Feb 2018, 17:55
pushpitkc wrote:
The greatest distance(straight line) between two points in a cube is its diagonal.
Here the length of the digaonal is 10.
Therefore, Side*$$\sqrt{3} = 10$$
Side = $$\frac{10}{\sqrt{3}}$$

Since the length and breadth doubles, and height remains the same.
The new box is a cuboid with length = breadth = $$\frac{20}{\sqrt{3}}$$ and height is $$\frac{10}{\sqrt{3}}$$

The straight line greatest distance is the diagonal.
Diagonal = $$\sqrt{length^2 + breadth^2 + height^2} = \sqrt{300} = 10*\sqrt{3}$$(Option B)

Can you please explain how you got the square root of 300? I am so lost at that point?
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3300
Location: India
GPA: 3.12
Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

19 Feb 2018, 19:32
1
300 is 10*10*3 or $$3 * 10^2$$

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!
_________________
You've got what it takes, but it will take everything you've got
Manager
Joined: 06 Oct 2017
Posts: 50
Schools: Schulich "22 (A)
GPA: 3.6
The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

19 Feb 2018, 19:35
pushpitkc wrote:
300 is 10*10*3 or $$3 * 10^2$$

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!

But if the new box is a cuboid with length = breadth = 20/√3 and height is 10/√3 then where are the 10*10*3 coming from? Shountld you be multiplying the two 20/√3 with the 10/√3 ?

Sorry but for some reason this is literally going over my head....
Math Expert
Joined: 02 Aug 2009
Posts: 8290
Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

19 Feb 2018, 20:15
1
aanjumz92 wrote:
pushpitkc wrote:
300 is 10*10*3 or $$3 * 10^2$$

The square root of 300 is 10*sqft(3) as 10 comes outside the square

Hope that helps!

But if the new box is a cuboid with length = breadth = 20/√3 and height is 10/√3 then where are the 10*10*3 coming from? Shountld you be multiplying the two 20/√3 with the 10/√3 ?

Sorry but for some reason this is literally going over my head....

Diagonal = $$\sqrt{length^2 + breadth^2 + height^2} = \sqrt{(\frac{20}{sqroot3})^2 + (\frac{20}{sqroot3})^2 + (\frac{10}{sqroot3})^2}= \sqrt{(\frac{400}{3}) + (\frac{400}{3}) + (\frac{100}{3})}= \sqrt{\frac{900}{3} =} \sqrt{300} = 10*\sqrt{3}$$(Option B)
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9855
Location: Pune, India
The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

19 Feb 2018, 22:54
1
Bunuel wrote:
The greatest possible (straight line) distance, in inches, between any two points on a certain cube is 10. If the cube is modified so that its length is doubled and its width is doubled while its height remains unchanged, then what is the greatest possible (straight line) distance, in inches, between any two points on the modified box?

A. 10√2
B. 10√3
C. 20
D. 30
E. 30√3

Attachment:

Unknown.png [ 1.69 KiB | Viewed 852 times ]

The greatest possible distance between any two points on a cube is the diagonal (D) which is also the hypotenuse of the triangle made by the the shorter diagonal (d) and a. The shorter diagonal (d) is the hypotenuse of sides a and a.
So $$d^2 = a^2 + a^2$$
and $$D^2 = d^2 + a^2 = 3a^2$$
$$D = \sqrt{3}a = 10$$

If length and width are doubled, New $$d^2 = (2a)^2 + (2a)^2 = 8a^2$$
New $$D^2 = 8a^2 + a^2 = 9a^2$$
Then New $$D = 3a = 10\sqrt{3}$$

_________________
Karishma
Veritas Prep GMAT Instructor

Non-Human User
Joined: 09 Sep 2013
Posts: 13727
Re: The greatest possible (straight line) distance, in inches, between any  [#permalink]

Show Tags

18 Aug 2019, 10:04
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: The greatest possible (straight line) distance, in inches, between any   [#permalink] 18 Aug 2019, 10:04
Display posts from previous: Sort by