A High School has 400 students 1/2 attend the arithmetic club, 5/8 attend the Biology club, 3/4 attend the Chemistry club, and 3/8 attend all 3 clubs. If every student attends at least one club, how many students attend exactly 2 clubs?
A. 50
B. 75
C. 150
D. 200
E. 300
The OA is A.
I solved this PS question as follows,
Total=A+B+C-(Exactly 2 Groups) - 2*All three
Exactly 2 Groups = X
400=200+250+300-X-2*150
X=50
Has anyone another approach to solve this PS question? Regards!
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
A high School has 400 students 1/2 attend the arithmetic
This topic has expert replies
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi All,
We're told that a High School has 400 students: 1/2 attend the arithmetic club, 5/8 attend the Biology club, 3/4 attend the Chemistry club, and 3/8 attend all 3 clubs. If every student attends at least one club, we're asked for the number of students who attend EXACTLY 2 clubs. 3-Group Overlapping Sets questions are relatively rare on the Official GMAT (you likely will NOT see this version of Overlapping Sets on Test Day). However, there is a formula that you can use to solve it.
Total = (1st group) + (2nd group) + (3rd group) - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd) - 2(all 3 groups).
In overlapping sets questions, any person who appears in more than one group has been 'counted' more than once. When dealing with groups of people, you're not supposed to count any individual more than once, so the formula 'subtracts' all of the extra times that a person is counted. For example, someone who is in BOTH the 1st group and the 2nd group will be counted twice....that's why we SUBTRACT that person later on [in the (1st and 2nd) group].
In this prompt, we're given the Total, a fraction (which we can convert to a number) for each of the 3 individual groups and the number of people who appear in all 3 groups. The equation would look like this...
400 = 200 + 250 + 300 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)- 2(150)
400 = 450 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)
0 = 50 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)
(1st and 2nd) + (1st and 3rd) + (2nd and 3rd) = 50
Since the prompt asks for the total number of students that are in exactly 2 classes, we have our answer.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
We're told that a High School has 400 students: 1/2 attend the arithmetic club, 5/8 attend the Biology club, 3/4 attend the Chemistry club, and 3/8 attend all 3 clubs. If every student attends at least one club, we're asked for the number of students who attend EXACTLY 2 clubs. 3-Group Overlapping Sets questions are relatively rare on the Official GMAT (you likely will NOT see this version of Overlapping Sets on Test Day). However, there is a formula that you can use to solve it.
Total = (1st group) + (2nd group) + (3rd group) - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd) - 2(all 3 groups).
In overlapping sets questions, any person who appears in more than one group has been 'counted' more than once. When dealing with groups of people, you're not supposed to count any individual more than once, so the formula 'subtracts' all of the extra times that a person is counted. For example, someone who is in BOTH the 1st group and the 2nd group will be counted twice....that's why we SUBTRACT that person later on [in the (1st and 2nd) group].
In this prompt, we're given the Total, a fraction (which we can convert to a number) for each of the 3 individual groups and the number of people who appear in all 3 groups. The equation would look like this...
400 = 200 + 250 + 300 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)- 2(150)
400 = 450 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)
0 = 50 - (1st and 2nd) - (1st and 3rd) - (2nd and 3rd)
(1st and 2nd) + (1st and 3rd) + (2nd and 3rd) = 50
Since the prompt asks for the total number of students that are in exactly 2 classes, we have our answer.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7305
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We can create the equation:AAPL wrote:A High School has 400 students 1/2 attend the arithmetic club, 5/8 attend the Biology club, 3/4 attend the Chemistry club, and 3/8 attend all 3 clubs. If every student attends at least one club, how many students attend exactly 2 clubs?
A. 50
B. 75
C. 150
D. 200
E. 300
Total = A + B + C - (Exactly 2 clubs) - 2 x (All 3 clubs) + (None of the 3 clubs)
400 = 1/2(400) + 5/8(400) + 3/4(400) - d - 2 x 3/8(400) + 0
400 = 200 + 250 + 300 - d - 2(150)
400 = 750 - d - 300
d = 50
Answer: A
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
-
deloitte247
- Legendary Member
- Posts: 2214
- Joined: Fri Mar 02, 2018 2:22 pm
- Followed by:5 members
How many students attend exactly 2 clubs?
Arithmetic club = 1/2 of total students = $$\frac{1}{2}\cdot400=200$$
Biology club = 5/8 of total student= $$\frac{5}{8}\cdot400=250$$
Chemistry club = 3/4 of total student = $$\frac{3}{4}\cdot400=300$$
Total = 750 students
Using equation for over lapping set
A+B+C = n+x+2y
where A = Arithmetic Club
B = Biology Cub
C = Chemistry club
n = number of students
x = number of student in two clubs
y = number of student in three clubs
y = $$\frac{3}{8}\cdot400=150$$
A+B+C = n+x+2y
200+250+300 = 400+x+2(150)
750 = 400+x+300
750 = 700+x
750-700 = x
50 = x
Option A is the answer
Arithmetic club = 1/2 of total students = $$\frac{1}{2}\cdot400=200$$
Biology club = 5/8 of total student= $$\frac{5}{8}\cdot400=250$$
Chemistry club = 3/4 of total student = $$\frac{3}{4}\cdot400=300$$
Total = 750 students
Using equation for over lapping set
A+B+C = n+x+2y
where A = Arithmetic Club
B = Biology Cub
C = Chemistry club
n = number of students
x = number of student in two clubs
y = number of student in three clubs
y = $$\frac{3}{8}\cdot400=150$$
A+B+C = n+x+2y
200+250+300 = 400+x+2(150)
750 = 400+x+300
750 = 700+x
750-700 = x
50 = x
Option A is the answer
