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Students in the tenth grade at a certain high school must take at least one science course: chemistry, physics, or biology. Each of these classes contains 20 students, and any two classes
have the same number of students in common. Five students are taking all three classes simultaneously.

Using the information given, identify a possible number of students in the tenth grade taking only one science class and the corresponding number of students common to any pair of classes.
The numbers must be consistent with each other. Make only two selections, one in each column.

Only One Science___Common to A Pair of Classes_______Number
_____________ ____________________________________2
_____________ ____________________________________7
_____________ ____________________________________9
_____________ ____________________________________24
_____________ ____________________________________33
_____________ ____________________________________39


Lets consider : Number of student studying only one science = 33.

Each class will have 11 students studying only 1 subject. 5 students study all subject. so total students studying only one and three subjects in
each class = 16.

Now we know that each class has 20 students . 20-16 = 4 will be the students in each class that should study, if selected option is correct, two
subjects. Each class has 4 students that should study two subjects and as it is evenly distributed, out of these 4 , 2 students should study one of the three subjects...

5 students will are common for all class will also be common for any two class. Therefore, 2+5 = 7.

33 , 7 is the correct answer. Check will other option it want satisfy the condition
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Hi melaos.,

Check out the below link for clarification

https://gmatclub.com/forum/advanced-over ... 44260.html

melaos
MacFauz
Only in one group = x
Students common = y

x = 20 + 20 + 20 - 2(y) + 3(5)
x = 60 + 15 -2(y) = 75 - 2y

So values should satisfy the equation

x = 75 - 2y

don't quite understand how you get the numbers.

can pls elaborate?

Had a chance to look at the link -- -which formula are you using ?

Total=A+B+C−(sum of 2−group overlaps)+(all three)+NeitherTotal=A+B+C−(sum of 2−group overlaps)+(all three)+Neither.

OR

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+NeitherTotal=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither.
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Hi,

I don't get the same answer.

I get :
- 7 students chose one discipline only in each class
- 9 students for each pair of classes (with 4 students with 2 disciplines and 5 students with 3 disciplines)
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sir correct answer must be exactly1=33, exactly2=6, which is not provided in the choices
since number of students in Chem+Phy+Bio=Exactly1+2*Exactly2+3* Exactly3
60=e1+2*e2+3*5
e1+2*e2=45
e1=33 & e2=7 satisfies the condition.
Please guide me I m incorrect
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pappal
sir correct answer must be exactly1=33, exactly2=6, which is not provided in the choices
since number of students in Chem+Phy+Bio=Exactly1+2*Exactly2+3* Exactly3
60=e1+2*e2+3*5
e1+2*e2=45
e1=33 & e2=7 satisfies the condition.
Please guide me I m incorrect
­You are looking at Exactly one and common to any pair, and not exactly one and exactly 2.

 
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Can someone please clearly explain this question using venn diagram?

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Appreciable guidance

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The question says each class has 20 students, and among these classes there are students in common, thus the total students are less than 60.

with previously given answers, if total students with only 1 class is 33 and students with only 2 classes is 7 then for any class, total students will be (33/3)+7+7+5 = 30, which does not satisfy the 20 student per class criteria.

The answer is, 2 students common between only 2 class, and total 33 students with only one class: here total students in a class is (33/3)+2+2+5=20.
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