help me to solve one prob I saw in a test

[Andijansky]

heres the details
40 students in a class, 37 of them know English, 31 French, 28 Spanish, 27 German, so how many of them know all of the languages?
a)1
b)2
c)3
d)4
e)5
if Im not mistaken, answers were like that, help me on this type, pls

[goyalsau]

This is the first time I have seen a problem With Four Overlapping SETS..............

What is the source of the problem.............

[rishab1988]

Dude this problem is crazy.Drawing Venn diagram for this is like showing 4 dimensions on a page.

haha..

Anyway,I doubt you'll ever see such question ever show on GMAT.Maybe when GMAT is scaled upto 60.

I have browsed almost entire internet and I couldn't fnd a solution to that.

[oddball]

Do you remember the question properly?
I think some data are missing. Only with the informatin you provided it is not possible to solve the problem.

[rishab1988]

SOLVED THE QUESTION.

Okay this problem can be solved.I crunched my brain hard and devised a way.This can be done using combinatorics and sets.

Yep you are right combinatorics in sets!

Now ,when there are 2 sets (say A and B).There is just one intersection.You know how to calculate this?

I'll show.

In intersection of elements are two places.

No of ways for choosing first element : 2 [you could either choose A or B]
Now of ways for choosing second element: 1 [after you have chosen ,say A,you are left with just B]

So no of ways = 2*1=2.But wait A intersection B is same as B intersection A,so divide by 2.
No of intersections =1

Similarly in a 3 element set Venn diagram.

No of intersections of 2 elements : (3*2)/2! =3
No of intersections of 3 elements : (3*2*1)3!=1

Now for a 4 element set:

No of intersections of 2 elements : (4*3)/2!=6
No of intersections of 3 elements : (4*3*2)/3!=4
No of intersections of 4 elements : (4*3*2*1)/4!=1

Now,

In a 2 element set

Let A denote only A
B denote only B
and a denote intersection of A and B.
Then total no elements = A+B+a

Similarly in a 3 element set,

Total no of elements : A+B+C+a+b+c+d [ where a,b,and c denote intersection of 2 elements and d denotes intersection of all 3 elements]

Now IN A 4 element set

Total = A+B+C+D+ [a+b+c+d+e+f]+[g+h+i+j]+[k] [ arranged order of intersection]=40

a= A intersection B
b= A and C
c= A and D
d= B and C
e= B and D
f = C and D

g= A and B and C
h= A and B and D
i = A and C and D
j = B and C and D

k= A and B and C and D

Question is what is k?

Now stay here only if you want your brains to be FRIED!

k = 40 - [a+b+c+d+e+f] - [g+h+i+j] - [A+B+C+D]

A+a+b+c+g+h+i+k =37 [English] [Include all those that have A]

B+a+d+e+g+h+j+k = 31 [Spanish]

C+b+d+f+g+i+j+k = 28 [Spanish]

D+b+e+f+h+i+j+k =27 [German]


Adding all of them!!!!!!!

A+B+C+D+2[a+b+c+d+e+f]+3[g+h+i+j]+4k = 123

Substituting:

A+B+C+D+[a+b+c+d+e+f]+[g+h+i+j]+k =40 in the above equation:

[a+b+c+d+e+f]+2[g+h+i+j+k]+3k = 83

WHAT THE HECK!

YOU DIDN'T PROVIDE COMPLETE INFO.

I NEED no students in 2 classes and 3 classes!

I assume no of 2 students = 34
No of 3 students =17

[I WON'T LET MY EFFORT GO WASTE]

34+34+3k=83
3k=15
k=5

or no of students who study all classes = 5

THIS IS NOT A GMAT QUESTION.

THIS IS ATLEAST A 5-10 MINUTE QUESTION

[Andijansky]

Do you remember the question properly?
I think some data are missing. Only with the informatin you provided it is not possible to solve the problem.

the thing is I saw it in a test for job application and thought I have to be able to solve being a student who's going to take GMAT,anyway, thank you all!:)

[goyalsau]

Dude this problem is crazy.Drawing Venn diagram for this is like showing 4 dimensions on a page.

haha..

Great Work Rishab, I doubt will i able to do one like this by myself.,

but i must say Your 4 Dimension Thought Could may Give the Great Einstein some sleeping Nightmares,

What to say buddy , I just can't stop Laughing.....................

[goyalsau]

the thing is I saw it in a test for job application and thought I have to be able to solve being a student who's going to take GMAT,anyway, thank you all!:)

Buddy, I have a free suggestion to make, Don't Join that Company , If they ask such question on the Test , Just Wonder what they will do to you when you will join them,,,,,,,,,

[Andijansky]

thnx!:) I'll keep that in mind, and if I get accepted, will convey the message sure:)