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.

elements in common

Expert replies
by mberkowitz » Sat Sep 13, 2008 3:48 pm
Sets A, B, and C have some elements in common. If 16 elements are
in both A and B, 17 elements are in both A and C, and 18 elements are
in both B and C, how many elements do all three of the sets A, B, and
C have in common?
(1) Of the 16 elements that are in both A and B, 9 elements are also in
C.
(2) A has 25 elements, B has 30 elements, and C has 35 elements.

OA is A
Join the discussion
Source: — Data Sufficiency |

Re: elements in common

by parallel_chase » Sat Sep 13, 2008 11:46 pm
mberkowitz wrote:Sets A, B, and C have some elements in common. If 16 elements are
in both A and B, 17 elements are in both A and C, and 18 elements are
in both B and C, how many elements do all three of the sets A, B, and
C have in common?
(1) Of the 16 elements that are in both A and B, 9 elements are also in
C.
(2) A has 25 elements, B has 30 elements, and C has 35 elements.

OA is A
The answer is indeed A.

Use this formula

P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

We just have to find the value of P(AnBnC), Statement I gives us exactly that.
Join the discussion

by mberkowitz » Sun Sep 14, 2008 7:23 am
how do we know p(a) + p(b) + p(c) given sI?
Join the discussion

by tendays2go » Sun Sep 14, 2008 9:00 am
option(1) =>

n(A u B) = 16 out of which 9 are also there in C, thus these 9 elements are common to all A,B & C, hence the answer and therefore this is sufficient to answer.

option(2) =>

N(A) = 25
N(B) = 30
N(C) = 35

But, we don't know the other distribution...so we can't find the common elements to all A, B and C

Hence solution is (A)
Join the discussion

by Thouraya » Sun May 29, 2011 6:45 am
Didnt get why B is not sufficient. Thanks!
Join the discussion