Last visit was: 24 Apr 2026, 01:21 It is currently 24 Apr 2026, 01:21
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
User avatar
wbricker3
Joined: 27 Jun 2015
Last visit: 19 Apr 2018
Posts: 1
Own Kudos:
9
 [9]
Given Kudos: 26
Location: United States
GMAT 1: 730 Q48 V42
GPA: 3.69
GMAT 1: 730 Q48 V42
Posts: 1
Kudos: 9
 [9]
1
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
User avatar
Skywalker18
User avatar
Retired Moderator
Joined: 08 Dec 2013
Last visit: 15 Nov 2023
Posts: 1,973
Own Kudos:
Given Kudos: 171
Status:Greatness begins beyond your comfort zone
Location: India
Concentration: General Management, Strategy
GPA: 3.2
WE:Information Technology (Consulting)
Products:
Posts: 1,973
Kudos: 10,167
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
generis
User avatar
Senior SC Moderator
Joined: 22 May 2016
Last visit: 18 Jun 2022
Posts: 5,258
Own Kudos:
Given Kudos: 9,464
Expert
Expert reply
Posts: 5,258
Kudos: 37,728
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,800
Own Kudos:
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,800
Kudos: 6,235
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Formula needed:
Area of equilateral triangle = \((\frac{√3}{4})* x^2\) where x is the area of an equilateral triangle.

Total surface area of a cube = \(6*z^2\) where z is the side of a cube.

We are given that the areas of equilateral triangle = total surface area of cube.

\((\frac{√3}{4})* x^2 = 6*z^2\)
\((\frac{√3}{4})* x^2 = 6*z^2\)
\(x^2 = \frac{6*4*√3*z^2}{√3*√3}\) (Multiplying and dividing the fraction by √3)
\(x^2 = \frac{6*4*√3*z^2}{3} = 8*√3*z^2\)

The answer has to be Option B.
User avatar
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
Posts: 608
Own Kudos:
Given Kudos: 276
Location: India
Concentration: Technology, Strategy
WE:Information Technology (Consulting)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
wbricker3
The area of an equilateral triangle with side length x is the same as the surface area of a cube with edge length z. If both x and z are integers, then what is x^2 in terms of z^2 ?

A. Z^2 (6√3)
B. Z^2 (8√3)
C. 72 Z^2
D. Z^2 (72√3)
E. 192 Z^2


Can anyone help explain?
Formula for Area of equilateral triangle with side \(x = \frac{x^2 \sqrt{3}}{4}\)

Formula for Surface area of cube with edge length \(z = 6z^2\)

Given Area of equilateral triangle and Surface area of cube are equal. Therefore,

\(\frac{x^2 \sqrt{3}}{4} = 6z^2\)

\(x^2 = \frac{(z^2)(6 * 4) }{\sqrt{3}}\)

\(x^2 = \frac{(z^2)(6 * 4)(\sqrt{3})}{\sqrt{3}*\sqrt{3}}\) --------- (Multiplying numerator and denominator by \(\sqrt{3}\))

\(x^2 = \frac{(z^2)(6 * 4)(\sqrt{3})}{3}\)

\(x^2 = (z^2)(2 * 4) (\sqrt{3})\)

\(x^2 = (z^2)(8 \sqrt{3})\)

Answer (B)...

I am also getting answer as B...

_________________
Please Press "+1 Kudos" to appreciate. :)
User avatar
shashankism
Joined: 13 Mar 2017
Last visit: 19 Feb 2026
Posts: 608
Own Kudos:
Given Kudos: 88
Affiliations: IIT Dhanbad
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE:Engineering (Energy)
Posts: 608
Kudos: 712
Kudos
Add Kudos
Bookmarks
Bookmark this Post
wbricker3
The area of an equilateral triangle with side length x is the same as the surface area of a cube with edge length z. If both x and z are integers, then what is x^2 in terms of z^2 ?

A. Z^2 (6√3)
B. Z^2 (8√3)
C. 72 Z^2
D. Z^2 (72√3)
E. 192 Z^2


Can anyone help explain?

\(\sqrt{3}\)/4 * x^2 = 6 * z^2
-> x^2 = (24/\(\sqrt{3}\))* z^2
-> x^2 =(8*\(\sqrt{3}\))*z^2

So both x and z can be integers only if x = z = 0 .

So, I think there is no such option and questions needs a reframing.. Bunuel for your reference..
User avatar
Basshead
Joined: 09 Jan 2020
Last visit: 07 Feb 2024
Posts: 906
Own Kudos:
Given Kudos: 431
Location: United States
Posts: 906
Kudos: 323
Kudos
Add Kudos
Bookmarks
Bookmark this Post
wbricker3
The area of an equilateral triangle with side length x is the same as the surface area of a cube with edge length z. what is x^2 in terms of z^2 ?

A. Z^2 (6√3)
B. Z^2 (8√3)
C. 72 Z^2
D. Z^2 (72√3)
E. 192 Z^2


Can anyone help explain?

\(\frac{\sqrt{3}}{4}x^2 = 6z^2\)

\(x^2 = \frac{24z^2}{\sqrt{3}}\)

\(x^2 = \frac{24z^2}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}}\)

\(x^2 = 8\sqrt{3}z^2\)

Answer is B.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,965
Own Kudos:
Posts: 38,965
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109802 posts
Tuck School Moderator
853 posts