Students in the tenth grade at a certain high school must take at

Question

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.

chetan2u · Accepted Answer

See if this helps.

MacFauz · Answer

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

dhirajdas53 · Answer

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

jabhatta2 · Answer

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.

agueye · Answer

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)

pappal · Answer

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

chetan2u · Answer

You are looking at  Exactly one  and  common to any pair,  and not exactly one and exactly 2.

Nidhibatra · Answer

Can someone please clearly explain this question using venn diagram?

Posted from my mobile device

pappal · Answer

Appreciable guidance

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Arzeyone · Answer

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.

cumquead · Answer

I feel like the venn diagram made a lot of assumptions.. why is x equal in each common area? Why is y equal? There's not reason they should be. Is this definitely the answer?

chetan2u · Answer

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.

Go through the statements above and you will get your answer.

Adit_ · Answer

Question sounds confusing but the point is usual:
Draw the venn diagram.

You will quickly realise that the number of students common to any 2 classes is 5+x for ALL 3 classes. Only then will it be equal.
Now, we need to find the value for x and y  that satisfy the rules.

Common to two classes cant be 2 as we have 5+x.
Start with 7 then, meaning x=2. 
From venn diagram one circle comprising of 20 students has 
Single+5+2x = Single+5+2*2 = Single+9=20
Thus Single = 11.
Now Only single classes again is uniform throughout the venn diagram as number of students is 20 in each.
Thus 11 in each meaning 11*3=33 students.
We do have 33 in the list.
Thus 33,7 is our answer.

Answer:  33,7

Adit_ · Answer

No assumptions involved.
They said common between any 2 classes is same.
We have all 3 classes to be 5, thus in the venn diagram 5+x is what is common to any 2 classes!
If you have 5+y for one and 5+x it doesnt matter both must be equal so 5+x=5+y or x=y anyway.
This way we conclude that Only 2 items at a time is x throughout.

Like somebody already pointed out best way to solve is using the equation method.

Consider x = only 2 items, y = only 1 item
60-2(5)-3x = Total
y+5+3x = Total
Equating the 2:

60-10-3x=y+5+3x
50-3x=y+5+3x
45=6x+y
To find common for 2 classes we need to add x+5. So substract 5 from the second column and fit it into equation and u will get the answer.

Students in the tenth grade at a certain high school must take at

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Prep Toolkit

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