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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 02:50
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The Answer is E, Neither is sufficient
A- Tells us the value of Both, From this we can deduce the value of Both to be 4. Hence in math club the total number of people is 4 (Both) + Only Math and for total people in Science= 4(Both) + Only science.
This alone is insufficient to answer the number of teams that can be formed in which at least one student belongs to both clubs
B -Tells us the value of neither =9
Total number of people= Group Science + Group Math -Both +Neither (9). This is also insufficient
Taking both statements together we have
Total number of people= 4+only science+ 4+only Math -4 +9
Total =12+only science+only math
We still cannot find the number of teams that can be formed in which at least one student belongs to both clubs because we will need the value of only science and only math.
The answer should include- only math +Both + only science+both + Both science and Math
Which we cannot find with the available information
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 03:41
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At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
We know that there are 4 groups of students: Both M&S, Only M, Only S, and Not M&S.
Both M&S + Only M >5
Both M&S + Only S >5


(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(Only M) x (only S) = 9. The possible combinations can be (9,1), (1,9), (3,3). We can't answer the question

(2) The number of teams that can be formed in which both students belong to both clubs is 6.
Since 4C2 = 6, we can infer that Both M&S = 4. We can't answer the question

Both (1) & (2): From the above info, Only M and Only S must be 3 since the group M or group S must be larger than 5. The answer is: 4C1 x 6C1 = 24.

Final answer: C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 03:44
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Given,
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club.
  • 1 from the math club
· 1 from the science club
M: Number of students in math club
S: Number of students in science club
B: Number of students in both clubs
Then:
  • Students only in math club = M - B
  • Students only in science club = S – B
Total number of possible teams = M* S

To find
If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
Statement 1:
The number of teams that can be formed in which neither student belongs to both clubs is 9.
(M- B)*(S- B) = 9
3 unknowns. Not enough to find a single answer.

Not sufficient

Statement 2:
The number of teams that can be formed in which both students belong to both clubs is 6.
B*(B – 1) = 6
B = 3
M & S unknowns , not possible to solve the question.

Not sufficient

Using both statements

  • From 2: B = 3
  • From 1: ( M – 3 )* ( S – 3) = 9
Using cases, we get
M = 6, S = 6
M*S = 36

Now we will verify how many teams have at least one student in both:
Total – neither in both = 36 – 9 = 27
Sufficient together
Ans: C


Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 04:21
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Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
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Let number of students from math club only = x
number of students from science club only = y
number of students from both club = z

No. of teams that can be formed in which at least one student belongs to both clubs = zC1(xC1 +yC1) + zC2 =z(x+y) + z(z-1)/2= ???

1. The number of teams that can be formed in which neither student belongs to both clubs = 9 = xC1*yC1 + xC2 +yC2 = x*y + x(x-1)/2 + y(y-1)/2 ---Insufficient.

2. The number of teams that can be formed in which both students belong to both clubs = 6 = zC2 = z(z-1) = 3*2 ---> z=3 however, x & y not available -- Insufficient

Combining both -- z=3,
x*y + x(x-1)/2 + y(y-1)/2=9 --> two variables one equation --Insufficient E.
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:11
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Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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Let M= number of students in math club
S = number of students in science club
B = number of students who belong to both clubs

We are looking for the Number of teams where at least one student is in both clubs = Total teams−Teams where neither student belongs to both clubs
=> M.S - (M-B)(S-B)

Stmt (1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(M-B)(S-B) = 9, We can have a combination 1,9 or 3,3 or 9,1

We cannot have a single value of M, N here.

Hence, stmt 1 is insufficient.

Stmt (2) The number of teams that can be formed in which both students belong to both clubs is 6.

B(B-1) = 6, B can be 3. But we don't have any info about M,N.

Hence, stmt 2 is insufficient.

Let's combine both stmt 1 & stmt 2:

So we have B=3 and (M-B)(S-B) = 9

So (M-3)(S-3) = 9

M cannot be 4, As per given condition. So only possible value of M,S are 6,6.

M.S - (M-B)(S-B) = 36 - 9 = 27.

Hence IMO C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:24
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Let’s say:

Total Math Club students = M
Total Science Club students = S
Number of students who belong to both clubs = B

So:
Only Math students = M – B
Only Science students = S – B

Total possible teams = M × S
(1 from Math club, 1 from Science club)
Teams with neither from both clubs
Means: pick from only Math and only Science= (M – B) × (S – B)
Teams with at least one from both =
Total teams – Teams with neither from both= M × S – (M – B)(S – B)

Statement(1):
Teams where neither student belongs to both = 9
==>(M – B)(S – B) = 9
Possible values:
(M – B, S – B) could be: (1,9), (3,3), (9,1)
So many values possible,so M and S not fixed
Statement(1) not sufficient

Statement(2)
Teams where both students are from both clubs = 6
==>B × B = 6, So B = \(\sqrt{6}\), not an integer,Not valid
Statement(2) alone not sufficient

Both combined:
Since B=\(\sqrt{6}\) invalid so(M-B) not defined properly
Still not sufficient

So Option E is correct
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:34
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S1 Insufficient this only tells us that each group has 3 students in each group who do not belong to the other
S2 Insufficient based on xC2=6 meaning X=4 . 4 students belong to both groups but no info about students that belong to only one group
S1+S2 Both are sufficient because we now have all the numbers that we need to compute for teams in which at least one student belongs to both clubs
ANS C
Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:38
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Option E is the correct answer.

Lets understand the information mentioned in the question before trying to solve it.

So the question starts by telling us that "At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students". And then asks us how many such teams can be formed in which at least one student belongs to both clubs.

Now lets take a look at the statements and see whether we can answer the question with the help of them or not.

Statement 1: "The number of teams that can be formed in which neither student belongs to both clubs is 9". Now this statement tells us that the number of pairs which can be formed by the students who attend only only of the clubs is 9. So from here we can say the ratio of Math:Science could be 1:9, 3:3 or 9:1 and the question tells us that their are more than 5 students in each club so from these information we can not answer the question. Not Sufficient


Statement 2: "The number of teams that can be formed in which both students belong to both clubs is 6". Now this statement tells us the number of pairs formed by the students who attend both the clubs is 6. Again here also the ration of Math:Science could be 2:3, 3:2, 1:6 or 6:1 and the question tells us that their are more than 5 students in each club so from these information we can not answer the question. Not Sufficient


So as check above both the statements alone can not answer the question so lets try by Combining both of them.

The first statement tells us that the ratio of student who attend only one club i.e. ratio of Math:Science could be could be 1:9, 3:3 or 9:1 and from the second statement we get that ratio of students attending both clubs i.e. Math:Science could be could be 2:3, 3:2, 1:6 or 6:1. After combing these two ration and considering the condition that "each club has more than 5 students" we can get the ratio of total number of students attending Match:Science club to be 6:14, 9:9, 8:17, 11:11. So this also gives us multiple values. So from here with even solving any further we can tell that even Combination is not enough to answer the question.


So based on the above solution and calculations we can say that as neither statement alone or combined is unable to answer the question that's why Option E is our answer.

Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:42
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At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?
Math club M has more than 5 students
Science club S has more than 5 students
One student belongs to both club
M>5, S>5 Let B be the students of both club
Teams that can be formed from atleast one student from each club=MxS
Teams that can be formed from students neither from both clubs= (M-B)(S-B)
Teams that can be formed from students from both clubs = Bx(B-1) = B2-B
Teams from students where at least one student belong to each group = Total teams- teams of students of no group
=MS-MS+MB+BS-B2= MB+BS-B2

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
MS-BS-MB+B2=9
MB+BS-B2=?
M,S and B are not known
So statement is insufficient

(2) The number of teams that can be formed in which both students belong to both clubs is 6.
B2-B=6 so B=3
3M+3S-9 =?
Still M & S are not known

Combine both statements
MS-3S-3M=0
MS= 3(S+M)
M=6
S=6
So solution=18+18-9=27
So combine statements are sufficient.
Answer is C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:44
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Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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Option C is my answer


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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:45
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We have to find how many such teams can be formed in which at least one student belongs to both clubs

Total Math club = M (>5)
Total Science club = S (>5)
Math & Science overlap = X.

No. of teams that can be formed = M * S.
No. of teams that can be formed where neither is in both clubs = (M–X)*(S–X)
No. of teams that can be formed where both are in both clubs = X * (X-1)

(1) (M–X) * (S–X) = 9.
We still can't find total no. of teams from this. Insufficient

(2) X*(X–1)=6 -> X=3.
But we still can't find total no. of teams from this. Insufficient

Both 1 & 2 :
-> X=3
-> (M–3) * (S–3)=9.
The only way with M>5 and S >5 is M – 3 = 3, S – 3 = 3
-> M=S=6.
Since we now know both M, S & X , we can find the answer

Answer: C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:54
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Given,

Two student committee: 1 from MC and 1 from SC

MC/SC> 5

In a Venn-diagram,
let a = no. of students in only MC
b = no. of students in only SC
c = no. of students in both a&b

Q: no. of committee in which at least 1 student is from c.?

Three scenarios are possible:
(1) 01 student from a and 01 from c
(2) 01 student from b and 01 from c
(3) both students from C

Total possible teams = MC(a+c)*SC(b+c)


So, no. of committee in which at least 1 student is from c = Total - Neither of students belong to both(c)

To find this, we need the values of a,b individually or (a+b) and c

St1: We have the info on Neither student belonging to both club(c) = 9.
since both a,b>5 one possibility is a=3 b= 3
however, we still don't know about the total no. of committee. Insufficient

St2: Gives the value of scenario (3).
If a=4 and c=3, Total teams that can be formed = 12
If a=3 and c=4, Total teams that can be formed = 12

Still we don't know the exact values or total of a,b and c. Insufficient

Combining St1 and St2:

From st1: a=3 b= 3
From St2: c=4 as a/b*c = 12

So, we have the values of all the three a, b and c. Sufficient.

IMO Option C


Bunuel
At a school, a two-student project team is to be formed by selecting one student from the math club and one from the science club. If each club has more than 5 students, how many such teams can be formed in which at least one student belongs to both clubs?

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.


 


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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 05:56
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onlyM + MandS > 5
onlyS + MandS > 5

Calculate the number of teams required.
both students from MandS: (MandS)C2 = MandS*(MandS-1)/2
one student from onlyM and the other from MandS: onlyM * MandS
one student from onlyS and the other from MandS: onlyS * MandS

Teams = MandS*(MandS-1)/2 + MandS*(onlyM + onlyS)

(1)
onlyM*onlyS=9

Multiple solutions for Teams

Statement (1) alone is insufficient.

(2)
MandS*(MandS-1)/2=6

Multiple solutions for Teams

Statement (2) alone is insufficient.

(1)+(2)
onlyM*onlyS=9 -> 1*9, 9*1 or 3*3, but only 3*3 is valid because in the other two onlyM + MandS=5 or onlyS + MandS=5
MandS*(MandS-1)/2=6 -> MandS*(MandS-1)=12 -> MandS=4

onlyM=3
onlyS=3
MandS=4

Teams=6+4*6=30

Statement (1) and (2) together are sufficient

Answer C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 06:31
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Two-student project team is to be formed by selecting one student from the math club and one from the science club.
Each club has more than 5 students


Let total students in Math club = M
Let total students in Science club = S
Let total students in both clubs = B

=> Total number of teams = M*S
=> Total number of teams where neither student belongs to both clubs = ( M - B ) * ( S - B)
=> Total number of teams with at least one student in both clubs = M * S - ( M - B ) * ( S - B)

Given statements,

(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.

=> ( M - B ) * ( S - B) = 9

Here, multiple values of M,S and B satisfy the equation

=> Statement (1) is not sufficient

(2) The number of teams that can be formed in which both students belong to both clubs is 6.

=> Total number of teams where both students belong to both clubs = B * B = B^2 = 6
=> B = 6^0.5 which cant be possible since not an integer

=> Statement (2) is not sufficient

Combining statements (1) and (2),
( M - B ) * ( S - B) = 9
B = 6^0.5

We wont get a valid integer solution for B.

=> E. Statements (1) and (2) together are not sufficient
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 07:11
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maths = (only maths) + both
science = (only science) + both

maths > 5
science > 5

teams at least one student in both = (only maths)*both + (only science)*both + both(both-1)/2

(1)
(only maths)*(only science)=9

Insufficient because both is unknown

Insufficient

(2)
both(both-1)/2=6

Insufficient because (only maths) and (only science) are unknown

Insufficient

(1) and (2)
both(both-1)/2=6
both(both-1)=12
both=4

(only maths)*(only science)=9

(only maths)=1 and (only science)=9, no because maths = (only maths) + both = 5 is not >5
(only maths)=9 and (only science)=1, no because science = (only science) + both = 5 is not >5

(only maths)=3
(only science)=3
both=4

teams at least one student in both=30

Sufficient

Correct answer is C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 07:25
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Math=M, Science=S, both=B , only math=M-B, Only science=S-B
total team=MS
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9. -Insufficient , we do not know the total number of team as we do not know M and S
(2) The number of teams that can be formed in which both students belong to both clubs is 6. - Insufficient , we do not know only M OR S
Combining both:
Still insufficient as root6 will not give us any integer
IMO:E
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LastHero
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 07:26
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m=only maths
s=only scince
b=both

teams = b*m + b*s + bC2

It depends on knowing m, s, and b

(1)
m*s=9
m and s can be:
- 1 and 9
- 9 and 1
- 3 and 3

Need to know b too

Statement is insufficient

(2)
bC2=6
b(b-1)/2=6
b(b-1)=12
b=4

Need to know m and s too

Statement is insufficient

(1)+(2)
b=4
m and s can be:
- 1 and 9
- 9 and 1
- 3 and 3

But m+b>5 only if m>1
But m+s>5 only if s>1

The only possibility is b=4, m=3, s=3

teams=12+12+6=30

Both statements are sufficient

The right answer is C
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GMAT Club Olympics 2025 (Day 10): At a school, a two-student project 12 Jul 2025, 07:32
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oM = only maths
oS = only science
MS = maths and science

M = oM+MS > 5
S = oS+MS > 5

number of teams with at least one student in MS = MS*(oM + oS) + MS(MS-1)/2

To calculate this number it's necessary to know oM, oS and MS

(1)
oM*oS=9
(oM,oS) can be (1,9), (3,3) or (9,1)
No information about MS.

INSUFFICIENT

(2)
MS(MS-1)/2 = 6
MS(MS-1) = 12
MS = 4
No information about oM and oS.

INSUFFICIENT

(1)+(2)
(oM,oS) can be (1,9), (3,3) or (9,1)
MS = 4

As oM+MS=oM+4 must be greater than 5, option (oM,oS) = (1,9) is discarded
As oS+MS=oS+4 must be greater than 5, option (oM,oS) = (9,1) is discarded

oM=3, oS=3, MS=4
number of teams = 4(3+3) + 4*3/2 = 24 + 6 = 30

SUFFICIENT

IMO C
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