BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
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
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Cube Problem

Expert replies
by paritosh_b » Sat Dec 04, 2010 7:08 am
Not able to visualise it.Pls help me out

A large cube 10cm on a side is composed of smaller cubes 1 cm on a side. The large cube is printed red on its outside surfaces and then all the cubes are mixed in a drum. If one cube is selected randomly, which of the following is the approximate probability that the cube will have at least one red face?
OA:
[spoiler]1:2[/spoiler]
Join the discussion
Source: — Data Sufficiency |

by Bharat » Sat Dec 04, 2010 11:04 am
Hi Paritosh,

The original cube comprises 1000 unique 1-unit-cubes of which 488 unique 1-unit-cubes are on its outer surface.
-> The drum has 1000 cubes, of which 488 are painted red on one more surfaces.
-> Probability of desired event = 488/1000 = 1/2 (approximately)

Explanation: If you peel the surface only cubes from a cube with an edge length of n units, you will be left with a cube of edge length of n-2 units

Count of surface only cubes = difference in volumes of the two cubes
= 10^3 - 8^3
= 488 unit-cubes

PS: Note that the 8 unit-cubes on vertices are painted on 3 surfaces, the remaining 104 unit cubes on the edge are painted on 2 surfaces

Let me know if you have further questions.

Regards,
Bharat.
Join the discussion

by paritosh_b » Sat Dec 04, 2010 8:13 pm
Got it :)
Thanks.
Join the discussion

by goyalsau » Sat Dec 04, 2010 8:41 pm
Bharat wrote:
Explanation: If you peel the surface only cubes from a cube with an edge length of n units, you will be left with a cube of edge length of n-2 units AWESOME

I Never knew that we have formulas for these kind of questions as Well. :(

According to me,

64 *6 = 384 Cubes with one color FACE { Square on the face of a Cube with length 8 * 8 }

8 * 12 = 96 Cubes with two color on Its FACE { ON every Edge of Cube 8 Cube Excluding Corners of Cube }

8 Cubes with three color FACE { On every Cube 8 Corners are there }

384 + 96 + 8 = 488 Cubes with At least one FACE Colored.

Bharat wrote: PS: Note that the 8 unit-cubes on vertices are painted on 3 surfaces, the remaining 104 unit cubes on the edge are painted on 2 surfaces
I think you miscounted,
Saurabh Goyal
[email protected]
-------------------------


EveryBody Wants to Win But Nobody wants to prepare for Win.
Join the discussion

by Bharat » Sat Dec 04, 2010 9:28 pm
Hi goyalsau, thanks for the response & noticing the error.

You are correct, my count has a typographical error. Below is correction with detail.
Total surface area of cube = 600 (=6*10*10) sq. units
8 vertex cubes are painted on 3 surfaces, hence each of these is counted thrice (total 3*8) in the surface area, so need to discount 2*8 for removing duplicates.
96 cubes are colored on two surfaces, hence each of these is counted twice (total 2*96) in the surface area, so need to discount 1*96 for these:

so the final answer is: 600 - 16 - 96 = 488.

Regards,
Bharat.
Join the discussion

by junegmat221 » Sat Dec 04, 2010 10:29 pm
8 * 12 = 96 Cubes with two color on Its FACE { ON every Edge of Cube 8 Cube Excluding Corners of Cube }
@ Goyalsu,
I can only visually imagine this.
But it becomes harder when the cube is not 1000 in volume or if the smaller cubes are not 1cm in each of its side.
It is a little bit hard in exams when the units are not easily representable.
If anyone could visually represent the same data by means of formula with a diagram offcourse ,it should be really helpful.
Join the discussion