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11 Jul 2025, 12:15
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Bunuel
Let M denote the number of students in Math club.
Let S denote the number of students in science club.
Let x denote the number of students who belong to both clubs.
Given that each club has more than 5 students. Thus, M ,S > 5.
How many teams can be formed with at least 1 student from each club. ?
Statement 1:
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
(M - x) ( S- x) = 9
since, we don’t know the values of M, S and x. This is Insufficient
Statement 2:
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
x .x = 6 . Hence, x ^2 = 6 .
Hence, Insufficient
Without knowing the value of x, even combing statements 1 and 2, won’t give us a clear answer.
Hence, Insufficient
Option E
11 Jul 2025, 12:36
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We have to make a 2 member team.
Math club has >5 students.
Science club has >5 students
How many such teams can be formed in which at least one student belongs to both clubs ?
Let number of students in Math club be x, and number of students in Science club be y and number of students in both be c.
We need to find: c(c-1) + c(y-c-1) + c(x-c-1).
Statement 1:
Number of teams that can be formed in which neither student belongs to both clubs is 9.
This means (x-c)(y-c) = 9. This is not sufficient to get the value we need.
Statement 2:
The number of teams that can be formed in which both students belong to both clubs is 6.
This means c(c-1) = 6
This is not sufficient.
Combining the statements:
c(c-1) = 6
=> c = 3
(x-c)(y-c) = 9
The only way this is possible is if x and y are both 6. We have all the values now.
Answer is C.
Math club has >5 students.
Science club has >5 students
How many such teams can be formed in which at least one student belongs to both clubs ?
Let number of students in Math club be x, and number of students in Science club be y and number of students in both be c.
We need to find: c(c-1) + c(y-c-1) + c(x-c-1).
Statement 1:
Number of teams that can be formed in which neither student belongs to both clubs is 9.
This means (x-c)(y-c) = 9. This is not sufficient to get the value we need.
Statement 2:
The number of teams that can be formed in which both students belong to both clubs is 6.
This means c(c-1) = 6
This is not sufficient.
Combining the statements:
c(c-1) = 6
=> c = 3
(x-c)(y-c) = 9
The only way this is possible is if x and y are both 6. We have all the values now.
Answer is C.
Bunuel
11 Jul 2025, 12:50
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Given data,
We're forming a team with two students: one from the Math Club and one from the Science Club.
lets understand the condition here, "two-student project team" usually means the two students selected must be different people. If you pick John from the Math Club and then pick John again from the Science Club (because he's in both), that's just one person, not a "two-student" team. So, the two selected students must be distinct.
We want to find teams where at least one student belongs to both clubs = ?
Let b be the number of students in the Both group (Both Clubs - Math Club and the Science Club)
Let m be the number of students in the M_only group (Math Club only)
Let s be the number of students in the S_only group(Science Club only)
The total number of students in the Math Club = m + b.
The total number of students in the Science Club = s + b.
each club has more than 5 students = m + b > 5 and s + b > 5.
The teams looking for (at least one student from the Both group) can be=
student from M_only and student from Both => Number of teams = m * b
student from Both and student from S_only =>Number of teams: b * s
Two students, both from the Both group, but they must be different people =>Number of teams = b * (b-1))
So, the total number of teams with at least one student from Both = (m * b) + (b * s) + (b * (b-1)). This is what we need to find.
Now lets check the statements:
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
"Neither student belongs to both clubs" means one student is from M_only AND the other student is from S_only. (These students will always be different because the "only" groups don't overlap).
So, m * s = 9.
Possible pairs for (m, s) that multiply to 9 = (1, 9), (3, 3), (9, 1).
However, we don't know the value of b from this statement. The number of teams we want to find depends on b, m, and s. Since b can be different, we could get different answers. Since we get different answers, Statement (1) alone is NOT enough.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
"Both students belong to both clubs" means you pick one student from the Both group for the Math spot, AND you pick a different student from the Both group for the Science spot.
If there are b students in the Both group, the number of ways to pick two different students from this group is b * (b-1).
So, b * (b-1) = 6.
=> b^2 - b - 6 = 0. This factors to (b - 3)(b + 2) = 0.
Since b must be a number of students (cannot be negative), b = 3.
So, this statement tells us there are 3 students who belong to both clubs.
However, this statement doesn't tell us m (Math only) or s (Science only). So we can't find the total number of teams we want.Therefore, Statement (2) alone is NOT enough.
Combining Both Statements (1) and (2):
From Statement (2), we know b = 3.
From Statement (1), we know m * s = 9.
Now, let's use the rule that each club has more than 5 students:
Math Club students = m + b > 5 -> m + 3 > 5 -> m > 2.
Science Club students = s + b > 5 -> s + 3 > 5 -> s > 2.
So we need m * s = 9, where m is greater than 2 and s is greater than 2.
Let's look at the pairs for m * s = 9 again:
(1, 9): m=1 is NOT greater than 2. (No)
(3, 3): m=3 is greater than 2, and s=3 is greater than 2. (Yes!)
(9, 1): s=1 is NOT greater than 2. (No)
This means the only possible numbers are m = 3, s = 3, and b = 3.
Now that we know the exact numbers for m, s, and b, we can calculate the number of teams where at least one student belongs to both clubs:
Number of teams = (m * b) + (b * s) + (b * (b-1))
= (3 * 3) + (3 * 3) + (3 * (3-1))
= 9 + 9 + (3 * 2)
= 9 + 9 + 6
= 24
Since combining both statements gives us a single, unique answer, both statements together are sufficient.
We're forming a team with two students: one from the Math Club and one from the Science Club.
lets understand the condition here, "two-student project team" usually means the two students selected must be different people. If you pick John from the Math Club and then pick John again from the Science Club (because he's in both), that's just one person, not a "two-student" team. So, the two selected students must be distinct.
We want to find teams where at least one student belongs to both clubs = ?
Let b be the number of students in the Both group (Both Clubs - Math Club and the Science Club)
Let m be the number of students in the M_only group (Math Club only)
Let s be the number of students in the S_only group(Science Club only)
The total number of students in the Math Club = m + b.
The total number of students in the Science Club = s + b.
each club has more than 5 students = m + b > 5 and s + b > 5.
The teams looking for (at least one student from the Both group) can be=
student from M_only and student from Both => Number of teams = m * b
student from Both and student from S_only =>Number of teams: b * s
Two students, both from the Both group, but they must be different people =>Number of teams = b * (b-1))
So, the total number of teams with at least one student from Both = (m * b) + (b * s) + (b * (b-1)). This is what we need to find.
Now lets check the statements:
(1) The number of teams that can be formed in which neither student belongs to both clubs is 9.
"Neither student belongs to both clubs" means one student is from M_only AND the other student is from S_only. (These students will always be different because the "only" groups don't overlap).
So, m * s = 9.
Possible pairs for (m, s) that multiply to 9 = (1, 9), (3, 3), (9, 1).
However, we don't know the value of b from this statement. The number of teams we want to find depends on b, m, and s. Since b can be different, we could get different answers. Since we get different answers, Statement (1) alone is NOT enough.
(2) The number of teams that can be formed in which both students belong to both clubs is 6.
"Both students belong to both clubs" means you pick one student from the Both group for the Math spot, AND you pick a different student from the Both group for the Science spot.
If there are b students in the Both group, the number of ways to pick two different students from this group is b * (b-1).
So, b * (b-1) = 6.
=> b^2 - b - 6 = 0. This factors to (b - 3)(b + 2) = 0.
Since b must be a number of students (cannot be negative), b = 3.
So, this statement tells us there are 3 students who belong to both clubs.
However, this statement doesn't tell us m (Math only) or s (Science only). So we can't find the total number of teams we want.Therefore, Statement (2) alone is NOT enough.
Combining Both Statements (1) and (2):
From Statement (2), we know b = 3.
From Statement (1), we know m * s = 9.
Now, let's use the rule that each club has more than 5 students:
Math Club students = m + b > 5 -> m + 3 > 5 -> m > 2.
Science Club students = s + b > 5 -> s + 3 > 5 -> s > 2.
So we need m * s = 9, where m is greater than 2 and s is greater than 2.
Let's look at the pairs for m * s = 9 again:
(1, 9): m=1 is NOT greater than 2. (No)
(3, 3): m=3 is greater than 2, and s=3 is greater than 2. (Yes!)
(9, 1): s=1 is NOT greater than 2. (No)
This means the only possible numbers are m = 3, s = 3, and b = 3.
Now that we know the exact numbers for m, s, and b, we can calculate the number of teams where at least one student belongs to both clubs:
Number of teams = (m * b) + (b * s) + (b * (b-1))
= (3 * 3) + (3 * 3) + (3 * (3-1))
= 9 + 9 + (3 * 2)
= 9 + 9 + 6
= 24
Since combining both statements gives us a single, unique answer, both statements together are sufficient.
