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
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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.

Number of actual directors "lost" Sets problem

Expert replies
Source: — Problem Solving |

by raijonney » Mon Sep 29, 2008 10:42 pm
I think the answer is 3
Attachments
beatgmat_directorQ.JPG
Join the discussion

this is not the answer

by ashish1354 » Tue Sep 30, 2008 10:40 am
nope! this is not the answer!
Join the discussion

Oops..I think the answer should be 2.
Join the discussion

Oops..I think the answer should be 2.
Join the discussion

by raijonney » Wed Oct 01, 2008 9:56 am
figure for answer
Attachments
beatgmat_directorQ.JPG
Join the discussion

voila! here's the answer!

by ashish1354 » Wed Oct 01, 2008 10:32 pm
answer is 13
Join the discussion

voila! here's the answer!

by raijonney » Thu Oct 02, 2008 7:26 am
Okay .. did i read it wrong? .. does the question mean that how many are distinct in total? then it is definitely 13.
Join the discussion

could you explain how?

by ashish1354 » Thu Oct 02, 2008 7:29 am
i could'nt understand how are 13 people distinct. Please explain if you got it.
Join the discussion

voila! here's the answer!

by raijonney » Thu Oct 02, 2008 8:09 am
if you add all the numbers from the latest figure, 2+2+2+1+1+1+4 =13

i.e. there are 13 distint persons satisfying all the conditions and serving on the 3 boards:
condition 1. 4 serve on 3 boards each (lets say, board a, b and c)
condition 2. each pair i.e. ab, bc, ca has 5 in common, and we have 4 from condition 1. so we need 1 person common between each pair to make it total 5.
condition 3. since each board has 8 persons in total, hence there must be 2 persons on each board who work only on that board.

hence 13 different persons in total are serving on 3 boards under above conditions.
Join the discussion

Re: voila! here's the answer!

by dally_gmat » Fri Oct 03, 2008 10:42 am
If we need to apply following equation here...how do we apply here??

For 3 sets A, B, and C: P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

I am not able to understand how we can make use of this equation?

Thanks in advance for your reply and time..

raijonney wrote:if you add all the numbers from the latest figure, 2+2+2+1+1+1+4 =13
conditions.
Join the discussion

voila! here's the answer!

by raijonney » Fri Oct 03, 2008 12:10 pm
In terms of :
P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

P(A), P(B),P(C) is the number of people in each board i.e. 8
P(AnB), P(AnC), P(BnC) is intersection i.e. common person in 2 boards i.e. 5
P(AnBnC) is intersection of all boards, i.e. common person in all 3 boards i.e. 4

hence the equation will be => 8 + 8 + 8 - 5 - 5 - 5 + 4 = 13, hence the answer.
Join the discussion