SOLUTION: In a class of 36 students 25 study chemistry and 22 study mathematics and 25 study physics ,17 study physics and mathematics,18 study physics and chemistry,15 study one of the th

CMP
CM
CP
MP
C
M
P

CMP     x
CM      y
CP      18-x
MP      17-x
C       7-y
M       5-y
P       x-10
-------------
total   37-y

CMP     15
CM       1
CP       3
MP       2
C        6
M        4
P        5
-------------
total   36

I originally posted this solution with an error.  I reposted it under my
other pseudonym "AnlytcPhil".

Edwin

Using the same interpretation as tutor @greenestamps, and borrowing the nice Venn diagram code from tutor @Edwin_M, and adding my own lettering:









a = Chem only

b = Chem & Math only

c = Math only

d = all 3 subjects

e = Chem & Phys only

f = Math & Phys only

g = Phys only





There are 7 equations in 7 unknowns:



1. a+c+g = 15

2. e+d = 18

3. f+d = 17

4. a+b+d+e = 25

5. c+b+d+f = 22

6. d+e+f+g = 25

7. a+b+c+d+e+f+g = 36







Which has the solution (courtesy of http://www.bluebit.gr/matrix-calculator/solve.aspx):



a = 6

b = 1     <<< Taking Math & Chem only

c = 4

d = 15    <<< Taking all 3 subjects

e = 3

f = 2

g = 5  





This agrees with tutor @greenestamps answer.  Tutor @Edwin_McCrary has a at least one setup problem that I noticed: (referring to his diagrams)  e + g + i = 15 should be d + f + j = 15.



Best wishes.

 

 

We can solve this, but it contains a very badly-worded ambiguous statement.

The word "one" could either mean "EXACTLY one" or "ONE or more".

could give one answer and

could give another.  However, luckily we can tell which it is in this problem.
We can tell that it means "EXACTLY one" because 15 happens to be too small for
"ONE or more" to make sense, because, for instance, there are more than 15 in,
say, the 25 that study chemistry.  So luckily we can solve the problem, but you
should point out to your teacher not to just write "one" ever, but always
"EXACTLY one" or "ONE or more".  For if you have $5, you have $1, right?

Anyway:

 
d+e+f+g+h+i+j+k = 36

  
d+e+g+h = 25

  
e+f+h+i = 22

  
g+h+i+j = 25

 
h+i = 17

 
g+h = 18

  
d+f+j = 15

So we line up the equations:

1   d+e+f+g+h+i+j+k = 36
2   d+e  +g+h       = 25
3     e+f  +h+i     = 22
4         g+h+i+j   = 25
5           h+i     = 17
6         g+h       = 18
7   d  +f      +j   = 15

This is an "under-determined system" for it has 8 unknowns
but only 7 equations.  The determining thing will be the fact
that the solutions must all be non-negative integers.

We use matrix feature of TI-83 or 84.  And we get the
answers in terms of k, the number studying none of the three 
subjects.

d = 6-k
e = 1+k
f = 4-k
g = 3-2k
h = 15+2k
i = 2-2k
j = 5+2k

Since i = 2-2k ≥ 0
           -2k ≥ -2  
             k ≤ 1

So there are two solutions, k=0 and k=1

Substituting k = 1 in the equations above gives



Substituting k = 0 in the equations above gives



Find a)Those who study all three subject 

That's h, which is 15 for k=0, or 17 for k=1

b) only mathematics and chemistry 

That's g, which is 3 for k=0, or 1 for k=1. 

The second solution above with k=0 agrees with the other tutors.  But I do
consider the first case k=1 to be a legitimate solution as well.

Edwin

We are given a universal set of 36 students (the entire class).


We are also given info about three its subsets : 

    C (students enrolled to Chemistry)  of n(C) = 25 elements;  
    M (Mathematics)                     of n(M) = 22 elements, and 
    P (Physics)                         of n(P) = 25 elements.


We are also given the following info about their in-pair intersections:

    PM (studying Physics and Math)      of n(PM) = 17 elements;
    PC (studying Physics and Chemistry) of n(PC) = 18 elements;
    MC (studying Math    and Chemistry) of n(MC) = unknown number of elements.

 
We also should find the number of elements in the the triple intersection

    PMC (studying Physics, Math and Chem) of n(PMC) (unknown number of elements).


We assume (from the context) that the union of subsets P, M and C  is the entire set of the 36 students in the class.


In this situation the following formula works


    n(P U M U C) = n(P) + n(M) + n(C) - n(P n M) - n(P n C) - n(M n C) + n(P n M n C).     (1)


Substitute the given data. You will get


    36 = 25 + 22 + 25 - 17 - 18 - n(MC) + n(PMC),   or

    n(MC) - n(PMC) = 25 + 22 + 25 - 17 - 18 - 36 = 1.


This difference, n(MC) - n(PMC), is exactly the number of those who study only Math and Chemistry,

so it is just answer to question b) of the problem.


Now we will work to answer question a) and to find the unknown n(PMC).

For it, I should create the other equation, based on the second given condition.


The number of those who study only Math      is  22-n(PM)-n(MC)+n(PMC) = 22-n(PM)-(n(MC)-n(PMC)) =
                        replace here n(PM) by 17 and replace (n(MC)-n(PMC)) by 1, as we just found, to continue the line
                                                                                                 = 22-17-1 = 4.

The number of those who study only Physics   is  25-n(PM)-n(PC)+n(PMC) = 25-17-18+n(PMC) = -10+n(PMC).

The number of those who study only Chemistry is  25-n(PC)-n(MC)+n(PMC) = 25-n(PC)-(nMC)-n(PMC) = 
                        replace here n(PC) by 18 and replace (n(MC)-n(PMC)) by 1, as we just found, to continue the line                   
                                                                                               = 25-18-1 = 6.

Now the equation "the sum of those who study only one of the three subjects" is

    4 - 10+n(PMC) + 6 = 15,  which implies  n(PMC) = 15 - 4 + 10 - 6 = 15.


ANSWER to (a) : 15.    ANSWER to (b) : 1.

    It is totally clear to you why I add the first three addends in the formula (1).

    But when I add them, I count twice every term in each in-pair intersection.

    Therefore, I subtract the numbers of terms in each in-pair intersection.

    Next, when I add three first addends, I count thrice each term in the triple intersection;

    and when I subtract in-pair intersections, I cancel these terms thrice.

    Therefore, I must add the number of terms in the triple intersection one more time to restore the balance.

    Probably, the formulation is not perfect; probably, it could be better.

    But as formulated, it does not leave any doubt/doubts

        - that the condition covers and involves all 36 students of the class, and
        - what does it mean "15 study one of the three subject".


     It is a STANDARD WAY to formulate such circumstances in this sort of problems,
     and I saw tens of them in this forum, formulated in this way.



     Regarding matrix solutions by Edwin and by @Math_helper, I do not think that it is the expected way for the school students
     to get the solution.


     It is why I produced my own solution and placed it here.