Last visit was: 02 May 2026, 00:19

It is currently 02 May 2026, 00:19

Toolkit

Practice Tests

Data Insights Questions

Data Sufficiency (DS)

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

12 Days of Christmas GMAT Competition 2025 - Day 2: A survey of 90

1 2 3 4 5

prepapr

Joined: 06 Jan 2025

Last visit: 29 Apr 2026

Posts: 92

Own Kudos:

[1]

Given Kudos: 6

GMAT Focus 1: 615 Q85 V80 DI77 Verified score

Products:

GMAT Focus 1: 615 Q85 V80 DI77 Verified score

Posts: 92

Kudos: 82

[1]

16 Dec 2025, 10:14

Kudos

Bookmarks

Total number of students is 90. Every student belong to Debate or Math club.
Let number of students in debate club and math club be D and M respectively. Let the number of students in both clubs be B.
We have to find if D>M

Statement A:
3/5 participants in only Math club.
This shows greater than 50% students belong to only math Club. [(3/5)*90 = 54]
Hence M-B = 54.
Whatever the number of D, D will have to be less than M as
M-B is over 50% of total. So D-B<M-B. Or D<M
This is sufficient to get an answer as No

Statement B:
B = (2/9)*90 = 20
This does not give values for D or M.
Insufficent to compare D and M

hence answer is A

Bunuel

A survey of 90 students found that each student participates in at least one of two activities: Debate Club or Math Club. Is the number of students who participate in Debate Club greater than the number who participate in Math Club?

(1) 3/5 of the students participate in only Math Club.
(2) 2/9 of the students participate in both Debate Club and Math Club.

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

chasing725

Joined: 22 Jun 2025

Last visit: 13 Jan 2026

Posts: 176

Own Kudos:

173

[1]

Given Kudos: 5

Location: United States (OR)

Schools: Stanford

Posts: 176

Kudos: 173

[1]

16 Dec 2025, 10:14

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

We can plot in a 2 * 2 matrix

None = 0

Statement 1

Only math = 3/5 * 90 = 54

As 54 is more than half of the strength, we can conclude that number of students in the debate club cannot exceed the number of students in the math club.

This is just not possible.

S1 is sufficient.

Statement 2

2/9 = 20

Assume only math = only debate = 25

In that case the strength in both the clubs are equal. Also, we can adjust the numbers so that to arrive at different results. One in which the number of students in the math club club can be greater than the number of students in the debate club and vice versa.

Hence, this statement alone is not sufficient.

Option A

JiriNovacek

Joined: 14 Dec 2025

Last visit: 07 Feb 2026

Posts: 17

Own Kudos:

Posts: 17

Kudos: 9

16 Dec 2025, 10:18

Kudos

Bookmarks

Both statements are needed and sufficient
90 students, only Math 54 students, Both 20students => 16 students only Debating club, no, in Math club there are more students

Swanoff

Joined: 04 Aug 2021

Last visit: 07 Apr 2026

Posts: 27

Own Kudos:

Given Kudos: 92

Location: Uzbekistan

Concentration: Strategy, General Management

Schools: Olin

WE:Education (Non-Profit)

Products:

Schools: Olin

Posts: 27

Kudos: 34

16 Dec 2025, 10:26

Kudos

Bookmarks

Okay important thing is here we have 0 in neither category

Neither=0

Question: D>M ?

Statement 1: Math only = 3/5*90= 54
90= 54+D+Neither-Both
90=54+D+0-Both two unknowns we can't solve not sufficient

Statement 2: Both = 2/9*90=20

90= M+D-20+0
Again one equation with two unknowns not sufficient
(1) and (2) combo works because we needed Both in the first statement. Answer is C

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

gchandana

Joined: 16 May 2024

Last visit: 01 May 2026

Posts: 194

Own Kudos:

142

[1]

Given Kudos: 170

Location: India

Products:

Posts: 194

Kudos: 142

[1]

16 Dec 2025, 10:31

Kudos

Bookmarks

This is an overlapping sets question.

Total is 90, None = 0 (as everyone participates in at least one of the clubs).
We want to know if Debate club members > Math club members.

(1) Gives 54 students in the Math club, the maximum debate club can take is 34 (considering no overlap at all). So this alone is sufficient as Math club members> Debate Club members.

(2) Gives 20 students in both clubs. Now remaining 70, we don't know how they are distributed. It could be 35-35, making them equal, or 20-50 or 40-30. Not sufficient.

So, option A.

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

harishg

Joined: 18 Dec 2018

Last visit: 09 Apr 2026

Posts: 176

Own Kudos:

174

[1]

Given Kudos: 31

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Products:

GMAT Focus 1: 695 Q88 V84 DI81 Verified score

Posts: 176

Kudos: 174

[1]

16 Dec 2025, 10:37

Kudos

Bookmarks

Statement 1 says that 54 students participate only in math club. It means that 36 students are distributed between both math and debate club and only debate club. No matter what the distribution is, the number of debate club members can never be more than math club members. Therefore, statement 1 is sufficient.

Statement 2 says 20 students participate in both the clubs. The distribution of 70 among only math and only debate clubs will determine the answer. Therefore, this statement is not sufficient as we do not know this distribution.

Therefore, Option A

Rahilgaur

Joined: 24 Jun 2024

Last visit: 26 Jan 2026

Posts: 162

Own Kudos:

125

Given Kudos: 47

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Products:

GMAT Focus 1: 575 Q81 V82 DI72 Verified score

Posts: 162

Kudos: 125

16 Dec 2025, 10:38

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Lets Number of students who are participating in Debate Club = x, Math Club = y, Both = Z.

x+y-z= 90 ---

is x>y?

AD ---> y-z=3*90/5=270/5=54, Y-Z=54 ---X=36 ----> Insufficient

B ---> z=2*90/9 = 20 ----> x+y=110 -----> Insufficient

C ----> x=36, z=20, y= 74 ---x>y No Answer C

Abhishake04

Joined: 08 Jan 2025

Last visit: 01 May 2026

Posts: 12

Own Kudos:

Products:

Posts: 12

Kudos: 9

16 Dec 2025, 10:38

Kudos

Bookmarks

For statement 1 it says, 3/5 of 90 i.e. 54 are only in math club so student in debate Club [i.e. only in debate club + both club] is 36. Hence, Debate Club < Math Club. Statement 1 is sufficient.

For Statement 2 it says 2/9 of 90 i.e. 20 student belong to both club so remaining 70 can equally be divided or can be divided in any way so no unique result.

ikan444

Joined: 31 May 2025

Last visit: 28 Apr 2026

Posts: 18

Own Kudos:

[1]

Given Kudos: 23

Posts: 18

Kudos: 12

[1]

16 Dec 2025, 10:45

Kudos

Bookmarks

S1:

3/5 of the students participate in only maths clubs.
Only maths = (3/5)*90 = 54
This means 90-54 = 36, is the number of students who did not participate in only maths.
They either belong to only debate, both debate and maths, or are part of both and only.

Whatever the case may be, the number of students participating in debate can't be more than the number participating in maths (as maths only is already 54). We get a definite answer. Sufficient.

S2:

Students participating in both debate and math club = (2/9)*90 = 20
We have no idea about the number of students who participated in maths or the number who participated in debate. It can be any two numbers whose sum gives us 90-20 = 70. That means we can get yes as well as a know for the main question. Insufficient.

A (S1 alone is sufficient).

MANASH94

Joined: 25 Jun 2025

Last visit: 01 May 2026

Posts: 89

Own Kudos:

[1]

Given Kudos: 16

Location: India

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

GPA: 2.9

Products:

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

Posts: 89

Kudos: 63

[1]

16 Dec 2025, 10:49

Kudos

Bookmarks

It should be A because if On;y Math Category has majority which is 3/5 of the students, that means even with both and only debate will not have greater number of students.
Statement 1 is sufficent.

Now looking at statement 2: It only gives the information of the overlapping part but not the individual instances of participation of each club.
So Statement 2 is not sufficient hence:
A is the answer.

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

gemministorm

Joined: 26 May 2025

Last visit: 29 Apr 2026

Posts: 143

Own Kudos:

112

[1]

Given Kudos: 57

GMAT Focus 1: 565 Q82 V79 DI73 Verified score

GMAT Focus 2: 605 Q84 V83 DI73 Verified score

Products:

GMAT Focus 2: 605 Q84 V83 DI73 Verified score

Posts: 143

Kudos: 112

[1]

16 Dec 2025, 10:50

Kudos

Bookmarks

Venn diagram question: weather DC member > MC member
Since both is common to DC and MC, we only need to find weather only DC > only MC -> Yes/No

Statement 1 alone: MC only = 3/5*90 = 54
remaining 90 - 54 = 36 => of which some will be for both and some for DC.
we can conclude Yes MC > DC

Statement 2 alone: 2/9*90 = 20 both -- can't say anything about MC or DC.

hence, A.

aka007

Joined: 15 Nov 2019

Last visit: 19 Mar 2026

Posts: 42

Own Kudos:

[1]

Given Kudos: 84

Posts: 42

Kudos: 37

[1]

16 Dec 2025, 10:52

Kudos

Bookmarks

D= only Debate
M= only Maths
B= Both

D+B+M=90 students

question is D>M???

Statement 1. Only math = 3/5 x 90=54 54 students. 54 is greater than half of 90. Hence definitely be greater than D + B combined. Hence M>D sufficient.

Statement 2. Both is 2/9 x 90=20 20 students. however D+M=70. we don't know how much is D or M individually. hence not sufficient.

ANSWER is A

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Xdsa

Joined: 07 Jul 2025

Last visit: 01 May 2026

Posts: 57

Own Kudos:

[1]

Given Kudos: 1

Products:

Posts: 57

Kudos: 45

[1]

16 Dec 2025, 10:54

Kudos

Bookmarks

option a : 3/5 in math club only. so math club only will have 54 members and the rest can be in either debate club or both. which will be definitely less than math club. hence statement 1 is sufficient

option b: 2/9 are in both club.
so the rest 7/9=70 can be in either debate only or math only. it is not possible to check who is higher in number. statement 2 is not sufficient
hence statement 1 alone is sufficient

linnet

Joined: 11 Dec 2025

Last visit: 22 Jan 2026

Posts: 81

Own Kudos:

Given Kudos: 1

Posts: 81

Kudos: 42

16 Dec 2025, 10:55

Kudos

Bookmarks

Statement 1 alone is sufficient, Statement 2 alone is not sufficient

vikramadityaa

Joined: 28 Jul 2025

Last visit: 23 Dec 2025

Posts: 55

Own Kudos:

Given Kudos: 1

Posts: 55

Kudos: 41

16 Dec 2025, 10:56

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Total Student = 90
We're asked: D>M? (Yes/No question)

Statement 1: (3/5)*90 = 54
Remaining = 36 (these are either debate-only or both)
Too many possibilities, can't decide. Hence, Option A & Option D eliminated.

Statement 2: (2/9)*90 = 20 (Both)
Remaining: 70 (Split between math and debate, but we don't know how both are split.)
Hence, Option B is eliminated.

Combining both:
Only Math = 54; Both = 20; Math Total = 74
Debate = 90 - 54 = 36
Debate < Math.

Hence, OPTION C.

Sumimasen

Joined: 21 Jan 2024

Last visit: 02 May 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 11

Products:

Posts: 36

Kudos: 33

[1]

16 Dec 2025, 10:56

Kudos

Bookmarks

Total participants = 90

From statement I
Only Math club participants = 3/5*90 = 54
Total participants of Math club will be >= 54
Total participants of Debate club will be <= 36 which is (90-54).
So, participants of debate club will be lesser than math club students.
Sufficient to Answer.

From statement II
We only know the number of common participants in both debate & math club which is 2/9*90 = 20
We have not distribution of remaining 70 students.
Not sufficient to answer.

Correct answer is A

poojaarora1818

Joined: 30 Jul 2019

Last visit: 01 May 2026

Posts: 1,634

Own Kudos:

786

Given Kudos: 3,837

Location: India

Concentration: General Management, Economics

Schools: Duke '27 Haas '27 Kelley '27 Kellogg '27 McDonough '27

GPA: 3

WE:Human Resources (Real Estate)

Products:

Schools: Duke '27 Haas '27 Kelley '27 Kellogg '27 McDonough '27

Posts: 1,634

Kudos: 786

16 Dec 2025, 10:58

Kudos

Bookmarks

Solution:

Statement 1 Insufficient

	maths Club	No maths club	Total
Debate Club
No Debate Club	54
			90

Statement 2 Insufficient

	Maths Club	No Maths Club	Total
Debate Club	20
No Debate Club
			90

Even if we combine both statements, we won't get a conclusive answer

	Maths Club	No Maths Club	Total
Debate Club	20
No Debate Club	54
	74	16	90

Hence, option E

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

LunaticMaestro

Joined: 02 Jul 2023

Last visit: 01 May 2026

Posts: 44

Own Kudos:

[1]

Given Kudos: 106

Posts: 44

Kudos: 26

[1]

16 Dec 2025, 11:15

Kudos

Bookmarks

Clue1: tells (M - D) = 60% and remaining is 40%; now consider the boundaries that 1% in only D club then D < M; consider 40% in only D club then D < M;
Sufficient to make conclusion

Clue2: tells MnD ~ 22% remaining of the students can fall on either side so can't make a certain conclusion.

Answer: A; Only Clue 1

flippedeclipse

Joined: 26 Apr 2025

Last visit: 01 May 2026

Posts: 105

Own Kudos:

[1]

Given Kudos: 37

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Products:

GMAT Focus 1: 655 Q80 V87 DI80 Verified score

Posts: 105

Kudos: 73

[1]

16 Dec 2025, 11:18

Kudos

Bookmarks

Bunuel

Gift

12 Days of Christmas Competition

This question is part of our holiday event

Win $40,000 in prizes: courses, tests, and more

Stem is a Yes or No question; we need to be able to answer definitively for each statement.

Statement 1:

If 3/5 of the students only participate in Math Club, its a straightforward calculation.

$\frac{3}{5}*90 = 54$

This is a majority over any participants in debate club, so we can say definitively that yes, there are less students in Debate Club. Sufficient.

Statement 2:

$90 = Only Debate + Only Math - Both$

$Both = 90*\frac{2}{9}=20$

$90-20 = 70 = Only Debate + Only Math$

We can't discern the values of either Debate or Math at this point. Not sufficient.

Therefore A is the answer.

vasu1104

Joined: 10 Feb 2023

Last visit: 02 May 2026

Posts: 415

Own Kudos:

246

[1]

Given Kudos: 667

Location: Canada

Products:

Posts: 415

Kudos: 246

[1]

16 Dec 2025, 11:19

Kudos

Bookmarks

Say total in both - x
People in debate- x
People in only debate- a-x
People in maths - b
People in only maths- b-x

1) b-x = 54
a-x+x+b-x= 90
A - 36
Sufficient

2) x -20
a+ b - 110

Not sufficient