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.

Set Theory challenge

Expert replies

by Rabeea » Tue Mar 01, 2011 11:02 am
why we have taken cheese the largest possible value for the range?
Join the discussion

by fskilnik@GMATH » Tue Mar 01, 2011 11:30 am
Rabeea wrote:why we have taken cheese the largest possible value for the range?
Because we were looking for Xmax!

What I mean? If you have 20 people that like magazine A and 30 people that like magazine B, what is the maximum number of people that may like both magazines? 20. Why? Because 21 or more people could not be (why?) and 19 or less people is not the maximum possible, because ALL people that like mag A could also like mag B.

I hope you got it.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion

by Rabeea » Tue Mar 01, 2011 11:43 am
sir there is still confusion,
For example we have three sets,
say it set A , set B and Set C
20 people in A, 20 in B and 26 in C
How can we find the maximum number of people that is AuBuC
Rj
Join the discussion

by fskilnik@GMATH » Tue Mar 01, 2011 1:11 pm
Rabeea wrote:sir there is still confusion,
For example we have three sets,
say it set A , set B and Set C
20 people in A, 20 in B and 26 in C
How can we find the maximum number of people that is AuBuC
Hi there, Rabeea.

You must understand (and memorize) the following very-useful "formula":

AuBuC = A+B+C - (Sum of Exactly 2 of them) - 2* (Exactly 3 of them)

(The reason for why this is true is REALLY easy. Try to find by yourself "counting" the regions of the Venn diagram...)

Therefore AuBuC = 20+20+26 - (sum of exactly 2) - 2(exactly 3).

From the fact that (sum of exactly 2) and (exactly 3) are positive or zero (always), to maximize AuBuC the "ideal" would be to put zero in each one of them! The question is: is it possible to have A = 20, B = 20, C = 26 and all intersections mentioned equal to zero?

The answer is: it is! Please note that 20+20+26 is less or equal than 100 (percent), therefore we could have A = 20 = only A, B = 20 = only B and C = 26 = only C and the rest would go to neither A, nor B, nor C.

In this scenario AuBuC = 20+20+26 = 66 and this would be the maximum value for AuBuC (with the numbers you provided).

Is this what you were looking for (as an answer) ?

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion

by Rabeea » Tue Mar 01, 2011 1:46 pm
Actual question was
Each of the 49 students in the class understands equations,excersiceregularly or loves litreture.20 understand equations, 20 excercise regularly and 26 loves litreture.
36 understands equations or excersice regularly
38 understands equation or loves litreture
40 excersice regularly or loves litreture
Suppose you dont know the no.of students in the class,What is the largest possible no.of students in class?
Rj
Join the discussion

by fskilnik@GMATH » Tue Mar 01, 2011 2:06 pm
Rabeea wrote:Actual question was
Each of 49 students in the class understands equations,exercises regularly or loves literature.
20 understand equations, 20 excercise regularly and 26 loves litreture.
36 understands equations or excersice regularly
38 understands equation or loves litreture
40 excersice regularly or loves litreture
Suppose you dont know the no.of students in the class,What is the largest possible no.of students in class?
Total = (at least one thing) + (None), that is, Total = AuBuC + Remainder.

A = understand equations
B = exercise regularly
C = loves literature

From the question stem we know that:

AuBuC = 49
A = 20
B = 20
C = 26

AuB = 36
AuC = 38
BuC = 40

Imagine (Exactly 3) = x students

From what I mentioned we have:

49 = AuBuC = 20 + 20 + 26 - [(36-x) + (38-x) + (40-x)] - 2x therefore x = 97 (check that).

That means ABSURD. The data given in the question stem is impossible to be satisfied!

The red number means that I guess you typed it wrong.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Join the discussion

by Rabeea » Tue Mar 01, 2011 11:54 pm
this 49 was in my question may be absurd.
But we have to suppose that this 49 is not mentioned now we have to find the largest possible number of students.

from your concept if i get AuBuC 66
then by applying inclusion exclusion principle i can find the largest possible number of students.
that is to say
A inter B inter C
Rj
Join the discussion