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.

Weird?!?

Expert replies
by Deepthi Subbu » Wed Aug 03, 2011 4:16 am
Is this question right ?

Of the 250 students in a certain class, each student who majors
in mathematics also majors in computer science, and 90 of the
students major only in biology. If no student majors in all
three subjects, how many of the students major in two of the
three subjects?
(1) 50 of the students major only in computer science.
(2) 40 of the students do not major in any of the three
subjects.


In the question stem it says - no student majors in all
three subjects . But in statement B it says 40 of the students do not major in any of the three
subjects . Arent they contrary to each other ?

Also nothing is mentioned about students taking only math . Does it mean we dont have the necessary info or it means none has taken only math ?

To my surprise the answer is C .
Join the discussion
Source: — Data Sufficiency |

by Frankenstein » Wed Aug 03, 2011 5:58 am
Hi,
In the question stem it says - no student majors in all
three subjects . But in statement B it says 40 of the students do not major in any of the three
subjects . Arent they contrary to each other ?
No, they are not contrary. The question is just fine.
If we represent majors in Maths,Computer Science and Biology as M,C,B then
no student majors in all three subjects -> n(M n C n B) = 0
40 of the students do not major in any of the three subjects -> n( M U C U B) = 250 - 40.
Coming to the problem,
Each student who majors in mathematics also majors in computer science -> M is a subset of C i.e. set M lies entirely in C, when we draw a Venn diagram.
So, n(M n C) = n(M) and n(M U C) = n(C)
So, n(M U C U B) = n(C U B)
As no student majors in all three subjects, there should be any elements common to M and B.
So, n(M n B) = 0
Let n(M n B) = x.
Number of students student who major in 2 of the 3 subjects is given by
n(M n B) + n(M n C) + n(B n C) = 0 + n(M) + n(B n C)
So, we need to find n(M) + n(B n C)
Given that 90 of the students major only in biology -> n(only B) = 90
From(1):
n(Only C) = 50
This means n(C) - n(M) - n(C n B) = 50.
So, n(M) + n(B n C) = n(C) - 50.

From(2):
n(C U B U M) = 210.
So, n(C) + n(only B) = 210.
So, n (C) = 210 - 90 = 120.

From(1) and (2):
n(M) + n(B n C) = n(C) - 50 = 120-50 = 70

Hence, C

Hey I couldn't draw this as a Venn diagram and upload this. Please draw a Venn diagram, then it will be easy to understand.
Draw it in such a way, M lies entirely in C and there will be no common part for M and B although C and B can overlap.
Cheers!

Things are not what they appear to be... nor are they otherwise
Join the discussion

by gmatboost » Wed Aug 03, 2011 8:57 am
A good starting diagram of the situation could look like this

Image

We must find x + y.

You must remember that "None" is a possibility.
The 0 in the center and the 90 come from the question prompt directly, and the other two 0's come from the fact that every Math major MUST also be a CS major. So, no one can be Math only, and no one can be just Math and Bio.

To answer the original question: The prompt says that no one is in the very center, but Statement 2 is addressing the "None" region.

The answer is C because we need both the CS-only number and the "None" number in order to get everything but x+y. Once we have that total of everything else (90 + 50 + 40 = 180), we can get x + y = 250 - 180 = 70.

Hope this helps.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.


Also, check out the most useful GMAT Math blog on the internet here.
Join the discussion