Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Geometry question

Expert replies
by prernamalhotra » Fri May 30, 2014 6:35 am
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

1)10(√3 – 1)

2) 5

3) 10(√2 – 1)

4) 5(√3 – 1)

5) 5(√2 – 1)

Thank you,
Prerna
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Fri May 30, 2014 7:09 am
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A.10(√3 – 1)
B.5
C.10(√2 – 1)
D.5(√3 – 1)
E.5(√2 – 1)
Image

The formula for the diagonal of a cube = √(3e²).

In the figure above, x = the distance between the cube and the surface of the sphere.
The diagonal of the cube = 2x + the diameter of the sphere.
Thus, x = (diagonal of the cube - diameter of the sphere)/2.

The diagonal of the cube = √(3e²) = √(3*10²) = 10√3.
The diameter of the sphere = the edge of the cube = 10.
Thus, x = (10√3 - 10)/2 = 5√3 - 5 = 5(√3 - 1).

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by [email protected] » Fri May 30, 2014 10:48 am
Hi Prerna,

Mitch's solution correctly explains the math behind this question, so I won't rehash it here. I will add a few details to what Mitch did with his approach though.

First, when dealing with a "rectangular solid" (a box or cube), the longest distance between any two points on the surface of the rectangular solid is actually found by going THROUGH the solid.

eg. Upper-Right-Back corner to Lower-Left-Front corner

You can calculate that distance with this formula:

Sq. Root(Length^2 + Width^2 + Height^2)

-------------
In this question, all 3 dimensions of the cube are 10, so the diagonal through the cube = Sq.Root(10^2 + 10^2 + 10^2) = Sq.Root(300) = 10√3.

Since the sphere was inscribed in the cube, the middle of the sphere is also the middle of the cube. Also, the diameter of the sphere = side length of the cube = 10. Thus, the sphere's radius = 5.

So we can subtract the radius of the sphere (5) from from half the diagonal line (5√3) to find the length from the corner to the sphere. The answer choices expect us to factor out a 5.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Join the discussion