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A high School has 400 students 1/2 attend the arithmetic

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by AAPL » Thu Apr 26, 2018 5:05 am

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Answers

A

B

C

D

E

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Difficulty—

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!
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by [email protected] » Thu Apr 26, 2018 11:50 am
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

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by Scott@TargetTestPrep » Fri Apr 27, 2018 9:29 am
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
We can create the equation:

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

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by deloitte247 » Sun Apr 29, 2018 11:35 am
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
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