Question 1149313
<pre>We can solve this, but it contains a very badly-worded ambiguous statement.</pre>>>15 study one of the three subjects<<<pre>The word "one" could either mean "EXACTLY one" or "ONE or more".</pre>>>15 study EXACTLY one of the three subjects<<<pre>could give one answer and</pre>>>15 study one OR MORE of the three subjects<<<pre>
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:
</pre>
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 three subjects. Find a)Those who study all three
subjects b) only mathematics and chemistry 

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,P),
locate(-.45,-1,"j=?"),
locate(.6,.4,"i=?"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,C),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,M),
locate(-1.5,.5,"g=?"),
locate(-.4,2.3,"e=?"),
locate(1.8,2,"f=?"),
locate(-.4,1.1,"h=?") )}}}

</pre>In a class of 36 students<pre> 
d+e+f+g+h+i+j+k = 36
</pre>25 study chemistry<pre>  
d+e+g+h = 25
</pre>and 22 study mathematics<pre>  
e+f+h+i = 22
</pre>and 25 study physics<pre>  
g+h+i+j = 25
</pre>17 study physics and mathematics<pre> 
h+i = 17
</pre>18 study physics and chemistry<pre> 
g+h = 18
</pre>15 study EXACTLY one of the three subjects<pre>  
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

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=5"),
locate(-3.5,-2,"k=1"),
locate(0,-2.7,P),
locate(-.45,-1,"j=7"),
locate(.6,.4,"i=0"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,C),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,M),
locate(-1.5,.5,"g=1"),
locate(-.4,2.3,"e=2"),
locate(1.8,2,"f=3"),
locate(-.4,1.1,"h=17") )}}}

Substituting k = 0 in the equations above gives

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=6"),
locate(-3.5,-2,"k=0"),
locate(0,-2.7,P),
locate(-.45,-1,"j=5"),
locate(.6,.4,"i=2"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,C),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,M),
locate(-1.5,.5,"g=3"),
locate(-.4,2.3,"e=1"),
locate(1.8,2,"f=4"),
locate(-.4,1.1,"h=15") )}}}

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</pre>