Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 1810:00 AM PDT
-11:00 AM PDT
Join us in a live GMAT practice session and solve 25 challenging GMAT questions with other test takers in timed conditions, covering GMAT Quant, Data Sufficiency, Data Insights, Reading Comprehension, and Critical Reasoning questions.
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products. Ends 6/1
May 1501:00 PM IST
-11:00 AM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.- May 19
12:00 PM PDT
-01:00 PM PDT
Scoring 329 on the GRE is not always about using more books, more courses, or a longer study plan. In this episode of GRE Success Talks, Ashutosh shares his GRE preparation strategy, study plan, and test-day experience, explaining how he kept his prep.... - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
21 Apr 2026, 02:40
Kudos
Bookmarks
Video solution to this question:
09 May 2026, 13:15
Kudos
Bookmarks
The simplest way to solve it is algebraically:
We know that:
75+80+55−a−2b=100(1)
where a represents the number of people with exactly 2 devices and b the number of people with exactly 3 devices.
From (1), we have:
a+2b=110(2)
To maximize b, we set a=0.
Thus, b max =55
We also know that:
c+a+b=100(3)
where c represents the number of people with exactly 1 device.
From (2), we know that in order to minimize b, we want to maximize a.
To maximize a, we set c=0 in (3).
Therefore,
a+2b=110
and
a+b=100
Solving the linear system, we obtain:
b=10
Therefore, b min =10
Answer:
55−10=45.
We know that:
75+80+55−a−2b=100(1)
where a represents the number of people with exactly 2 devices and b the number of people with exactly 3 devices.
From (1), we have:
a+2b=110(2)
To maximize b, we set a=0.
Thus, b max =55
We also know that:
c+a+b=100(3)
where c represents the number of people with exactly 1 device.
From (2), we know that in order to minimize b, we want to maximize a.
To maximize a, we set c=0 in (3).
Therefore,
a+2b=110
and
a+b=100
Solving the linear system, we obtain:
b=10
Therefore, b min =10
Answer:
55−10=45.
09 May 2026, 13:27
Kudos
Bookmarks
Also notice that the maximum triple intersection is limited by the smallest set, which in this case is 55.
juliansorre
