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

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Bunuel

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16 Dec 2025, 10:00

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GMAT Club Official Explanation:

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.

3/5 of 90 participate in only Math Club, so only Math = 54. Since every student is in at least one club, the remaining 36 students must be those in Debate Club only plus those in both clubs. Therefore, the total number in Debate Club is 36. The total number in Math Club is only Math + both = 54 + both, which is at least 54. So Math Club (at least 54) has more participants than Debate Club (36) for sure. Sufficient.

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

2/9 of 90 participate in both, so both = 20. The remaining 70 are split between only Math and only Debate. We cannot tell which of those two groups is larger. Not sufficient.

Answer: A.

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paragw

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16 Dec 2025, 09:26

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Let:
D = number in Debate
M = Number in Math

Is D>M

Statement I:
3/5 of the students participate in only Math.

3/5 of 90 = 54
So far: Only Math = 54
Remaining students = 90-54 = 36

These 36 students must be either
Only Debate or
Both Debate and Math

Now count totals:
Debate participants = (only Debate) + (both)
Math participants = 54 + (both)

No matter how we split 36

Math total is at least 54

M >> D
So definite NO
Sufficient

Statement II:
2/9 of the students participate in both clubs
2/9 of 90 = 20 students in both

Remining students = 90-20 =>70

These 70 are split between only debate and only math
Possible cases:
a. If all 70 in Debate -> D = 90, M= 20 => Yes D>M
b. If all 70 are in Math -> D = 20, M=90 => No m>D
We get different answers.
Statement II is not sufficient

Answer: A

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16 Dec 2025, 09:28

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Using table analysis, is x>y
D ND
M y
NM 0
x 90 (total)
St 1: Gives x, no information on y. NS
St 2: Given information on D&M (cell 1), nothing on X and y
two together: (I did not solve completely) but gives you both x and y values. Hence sufficient.
C

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16 Dec 2025, 09:29

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Bunuel

Deconstructing the Question
Total students = 90.
Constraint: Each student participates in at least one activity (No student does neither).
Let $D$ = Total Debate Club members.
Let $M$ = Total Math Club members.
We can break the groups down into three mutually exclusive sets:
1. Only Debate ($O_D$)
2. Only Math ($O_M$)
3. Both ($B$) $\\
\\
Total = O_D + O_M + B = 90$

Question: Is $D > M$?
Since $D = O_D + B$ and $M = O_M + B$, the question simplifies to:
Is $O_D + B > O_M + B$?
Is $O_D > O_M$?

Analyze Statement (1)
"3/5 of the students participate in only Math Club."
$O_M = \frac{3}{5} \times 90 = 3 \times 18 = 54$.

We know that $O_D + B + O_M = 90$.
Substituting $O_M$:

$O_D + B + 54 = 90$

$O_D + B = 36$

Notice that $O_D + B$ is exactly the definition of the Total Debate Club ($D$).
So, $D = 36$.

Now look at Math Club ($M$):
$M = O_M + B = 54 + B$.

Compare $D$ and $M$:
Is $36 > 54 + B$?
Since $B \ge 0$, $54 + B$ is at least 54.
36 is definitely NOT greater than a number that is at least 54.

The answer is a definitive NO. SUFFICIENT

Analyze Statement (2)
"2/9 of the students participate in both Debate Club and Math Club."

$B = \frac{2}{9} \times 90 = 20$.

Using the total equation:

$O_D + O_M + 20 = 90$

$O_D + O_M = 70$

We need to know if $O_D > O_M$.

We have infinite integer pairs that sum to 70.

Case 1: $O_D = 60, O_M = 10 \implies D > M$ (YES).
Case 2: $O_D = 10, O_M = 60 \implies D < M$ (NO). INSUFFICIENT

Answer: A

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16 Dec 2025, 09:34

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Let number of students be M and D in Maths and Debate clubs
Is D > M?

S1
3/5 = 60% are in Math only
That means D total can be a max of 40%
So M > D
Answer is No
Sufficient

S2
2/9 is the overlap but we need D and M values to know the relation b/w D and M
Insufficient

Answer A

Bunuel

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Number of students	Debate Club	No Debate Club	Total
Math Club			y
No Math Club		0
Total	x		90

Is x > y ?

(1) 3/5 of the students participate in only Math Club

Number of students	Debate Club	No Debate Club	Total
Math Club		3/5*90 = 54	y>54
No Math Club		0
Total	x=90-54=36	54	90

x = 36
y > 54
x < y
x is NOT > y
Sufficient

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

Number of students	Debate Club	No Debate Club	Total
Math Club	2/9*90=20	y-20	y
No Math Club	90-y	0	90-y
Total	x = 20+90-y = 110-y	y-20	90

x = 110 - y

20 < y < 90

Case 1: y = 30; x = 80; x > y
Case 2: y = 80; x = 30; x < y

Not sufficient

IMO A

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16 Dec 2025, 09:38

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to simplify the calculations assume there are 45 students (lcm of 5,9)
M = Math club
D = Debate club
(1) alone
we know 3/5 * 45 = 27 students are in Only M
the remaining 18 students could be in Only D or Both.
Whatever the distribution of those 18 students between only D or Both, M is always greater than D.
Let's consider 2 extreme cases (all other possibilities will fall in between these 2 scenarios)
Case 1: Only M=27 Only D=0 Both=18
M = 27 + 18
D = 18
M > D
Case 2: Only M=27 Only D=18 Both=0
M = 27
D = 18
M > D

(2) alone
Both = 2/9 * 45 = 10
If we consider this information alone we can have both situations where M > D and M < D.
Case 1: Both = 10 Only M = 35 Only D = 0
M = 45 > D = 10
Case 2: Both = 10 Only M = 0 Only D = 35
M = 10 < D = 45

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16 Dec 2025, 09:40

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s1 : clearly qualifes that 60% are only math club
so math club people > debate club

s2 : 20 student participate in both clubs
lets take 2 hypothetical situattion

case 1 -
only m = 10
then only d = 60
so debate club = 80 (60 + 20) , math club (20 + 10)= 30

now reverse this
only m = 60
only d = 10
so debate club = 30, math club = 80

we cant conclusively deicde based on s2.

hence ans is A

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16 Dec 2025, 09:43

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Let, number of students in Debate Club = D and in Math Club= M
Total students = 90
Everyone is in atleast one club.

number of students in D> number of students in M ?

(1) only M= (3/5)*90=54
No info about Debate club students
Insufficient

(2) In both clubs = (2/9)*90=20
No info about students in only maths or debate club
Insufficient

(1)&(2)
Math only = 54 and Both= 20
Total students in Maths Club= 54+20=74
Remaining students are only in Debate club= 90-74=16
Total students in debate club = 16+20=36
D=36 and M=74
Sufficient

C

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16 Dec 2025, 09:45

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M = n(Math)
D = n(Debate)

(1) 3/5 of 90(total students) = 54 is M_only.
Hence D < = 90-54 = 36 < 54
Using this we can say that M which is > = M_only is greater than D

(2) Math and Debate = 2/9of 90 = 20. This doesnt tell us anything about Math or debate or their comparison with each other.

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16 Dec 2025, 09:45

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St1 3/5×90=54- only Math
Not sufficient
St2 2/9×90=20 both
Not sufficient
St1+st2
90-74=16
Both sufficient

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16 Dec 2025, 09:46

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the question is asking for is number of students in debate club grater than number of student in math club.
Given: Total is 90
Only math =M
Only Debate =D
Debate & Math=x
is M+x>d+x
M>D yes no answer is good
1)3/5*90=54 for only math that is more than d (sufficient)

2)x=2/9*90=20
this does not say anything about M and D
M and D can take any number from 20-90 there not sufficient
There A is the answer

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16 Dec 2025, 09:46

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	D	not D	Total
M
not M		0
Total			90

Using statement 1:

	D	not D	Total
M	x	54	54+
not M		0
Total	36	54	90

We can see that even if x=0, Number of students in Debate club will be lesser than students in Math club
Statement 1 is sufficient alone.

Statement 2 is clearly not sufficient alone.

Ans: A

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16 Dec 2025, 09:49

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3/5 out of 90 participate only in math = 54 out of 90 sufficient
2/9 out of 90 participate in both = 20 out of 90 Not sufficient
1 alone is sufficient as we can get to know that more than 50% are participating only in math
2 alone is not sufficient as we only know that 20 participate in both, which is not enough evidence as the other 70 could be divided in any manner and this 20 doesn't help as 20 are present for both which evens out.

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16 Dec 2025, 09:54

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The answer should be C.

Statement 1 tells that only Math= 3/5*90= 54
Statement 2 tells that both Math and Debate is= 2/9*90= 20

Combining both, we can find only Debate= 90-54-20 = 16
which is sufficient to answer whether there were more students in math or debate club.

Bunuel

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16 Dec 2025, 09:57

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Let D & M be the number of students in respective clubs.
Total = 90
Is D > M ?

Statement (1):
M(only) = 3/5 * 90 = 54

Remaining = 90 - 54 = 36

If 54 students participate only in math club, then the remaining 36 are participating in D(only) OR D/M.
If all the remaining 36 participate in D(only), D will still not be great than M.

Hence, Statement (1) is Sufficient

Statement (2):
D/M = 2/9 * 90 = 20
Remaining = 90 - 20 = 70

No further information is given to find the exact number in each of the clubs.

Hence Statement (2) is not sufficient

Option A

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16 Dec 2025, 10:04

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90 students can be divided into:-
1. Only Debate Club
2. Only Math Club
3. Math Club and Debate Club.

(1) 3/5th students participate only in Math Club.
Only Math Club= 90*3/5= 54
To maximise only Debate club, we can assume both Debate Club and Math Club as 0.
Only Debate Club= 90-54= 36

Even then,
Number of students in Debate Club< No of students in Math Club.

So, answer is no.

So we can eliminate options B,C and E.
It has to be either A or D.

(2) 2/9th students participate in both Debate Club and Math Club.
Both Math and Debate Club= 90*2/9= 20 Students.
Remaining= 90-20= 70 are divided among Only debate club and only math club.

As we do not have any more information, we can assume any division. So, we cannot get a definite yes or no.

So, D can be eliminated.

So, answer is Option A.

Bunuel

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16 Dec 2025, 10:11

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IMO A

St1 : 3/5 of 90 = 54 only maths which leaves D or D+M = 36, so definitely we can answer if D>M -----> sufficient
St2: 2/9 of 90 = 20 both D & M ------------> insufficient

Hence A

Bunuel