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11 Jul 2025, 08:00
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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?
(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.
(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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11 Jul 2025, 08:30
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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?
Let:
Total number of 2-person project teams where one student is from the math club and the other from the science club is xy + z(x + y) + zC2. Where:
We are asked: how many such teams have at least one student who is in both clubs?
So we want: z(x + y) + zC2
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
This implies xy = 9, which allows multiple (x, y) pairs such as (1, 9), (3, 3), or (9, 1). In addition, we have no information about z, so this is not sufficient.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
This implies zC2 = 6, which gives z!/(2!(z - 2)!) = 6, or z(z - 1)/2 = 6. Solving gives z = 4. But we don’t know x or y, so not sufficient.
(1)+(2) From (1), we found that (x, y) can be (1, 9), (3, 3), or (9, 1), and from (2), we know that z = 4. However, if (x, y) is (1, 9) or (9, 1), then the number of members in one of the clubs, math or science, becomes 1 + z = 5, which violates the condition that each club has more than 5 students. So (x, y) must be (3, 3). We know the values of each unknown, and thus we can answer the question. Sufficient.
Answer: C.
GMAT Club Official Explanation:
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?
Let:
- x = number of students in only the math club
- y = number of students in only the science club
- z = number of students in both clubs
Total number of 2-person project teams where one student is from the math club and the other from the science club is xy + z(x + y) + zC2. Where:
- xy = both students are in only one club each (math-only * science-only)
- z(x + y) = one student from both clubs and the other from only one club
- zC2 = both students are from both clubs
We are asked: how many such teams have at least one student who is in both clubs?
So we want: z(x + y) + zC2
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
This implies xy = 9, which allows multiple (x, y) pairs such as (1, 9), (3, 3), or (9, 1). In addition, we have no information about z, so this is not sufficient.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
This implies zC2 = 6, which gives z!/(2!(z - 2)!) = 6, or z(z - 1)/2 = 6. Solving gives z = 4. But we don’t know x or y, so not sufficient.
(1)+(2) From (1), we found that (x, y) can be (1, 9), (3, 3), or (9, 1), and from (2), we know that z = 4. However, if (x, y) is (1, 9) or (9, 1), then the number of members in one of the clubs, math or science, becomes 1 + z = 5, which violates the condition that each club has more than 5 students. So (x, y) must be (3, 3). We know the values of each unknown, and thus we can answer the question. Sufficient.
Answer: C.
General Discussion
Heix
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11 Jul 2025, 08:34
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Let's redefine the problem:
* M = number of students in math club only
* S= number of students in science club only
* B = number of students in both clubs
The total number of teams with at least one student belonging to both clubs includes:
* One student from math-only and one from both clubs: M × B teams
* One student from science-only and one from both clubs: S × B teams
* Both students from the group that belongs to both clubs: B(B-1)/2 teams (combination of 2 students from B)
So the answer we seek is: M×B + S×B + B(B-1y/2
Statement (i): The number of teams with neither student in both clubs is 9.
This means MxS = 9. Insufficient alone.
Statement (2): The number of teams with both students in both clubs is 6.
This means B(B-1)/2 = 6, so B = 4. Insufficient alone.
Combining (1) and (2):
We know B= 4 and MxS = 9
The number of teams with at least one student in both clubs is:
MxB + SxB + B(B-1)/2 = 4M+ 45 + 6
We can simplify: 4M + 4S = 4(M + S)
Now, we need to find M+S. From the total number of students in each club:
* Math club has M+ B students = M+4
* Science club has S + B students = S + 4
Since we know MxS = 9, and we're told each club has more than 5 students, we have:
* M+ 4>5, 5 so M>1
* S+4>5, 5 so >1
With MxS = 9 and both M and S being positive integers greater than 1, the only possibility is M = 3 and S = 3.
Therefore M+S = 6, and the number of teams with at least one student in both clubs is:
4(M + S) + 6= 4(6) + 6=24 + 6=30
With both statements together, we can determine the answer. The correct answer is C) Both statements together are sufficient, but neither alone is sufficient.
* M = number of students in math club only
* S= number of students in science club only
* B = number of students in both clubs
The total number of teams with at least one student belonging to both clubs includes:
* One student from math-only and one from both clubs: M × B teams
* One student from science-only and one from both clubs: S × B teams
* Both students from the group that belongs to both clubs: B(B-1)/2 teams (combination of 2 students from B)
So the answer we seek is: M×B + S×B + B(B-1y/2
Statement (i): The number of teams with neither student in both clubs is 9.
This means MxS = 9. Insufficient alone.
Statement (2): The number of teams with both students in both clubs is 6.
This means B(B-1)/2 = 6, so B = 4. Insufficient alone.
Combining (1) and (2):
We know B= 4 and MxS = 9
The number of teams with at least one student in both clubs is:
MxB + SxB + B(B-1)/2 = 4M+ 45 + 6
We can simplify: 4M + 4S = 4(M + S)
Now, we need to find M+S. From the total number of students in each club:
* Math club has M+ B students = M+4
* Science club has S + B students = S + 4
Since we know MxS = 9, and we're told each club has more than 5 students, we have:
* M+ 4>5, 5 so M>1
* S+4>5, 5 so >1
With MxS = 9 and both M and S being positive integers greater than 1, the only possibility is M = 3 and S = 3.
Therefore M+S = 6, and the number of teams with at least one student in both clubs is:
4(M + S) + 6= 4(6) + 6=24 + 6=30
With both statements together, we can determine the answer. The correct answer is C) Both statements together are sufficient, but neither alone is sufficient.
