&lt;p&gt;in a secondary school of 60 teachers, 30 teach maths, 27 teach physics, 21 teach chemistry, 12 teach maths andphysics, but none teaches both maths and chemistry&lt;/p&gt; &lt;p&gt;1) how many teach physics and chemistry?&lt;/p&gt; &lt;p&gt;​2) how many teach only physics&lt;/p&gt; &lt;p&gt;​&lt;/p&gt; &lt;p&gt;​&lt;/p&gt; &lt;p&gt;​&lt;/p&gt; &lt;p&gt;​&lt;/p&gt; &lt;p&gt;​&lt;/p&gt; &lt;p&gt;​&lt;/p&gt;

If you want good help, go to http://brainly.com.

Help is always 100% free there.

answered Dec 21, 2016 by Level 10 User (55,380 points)

The problem can be solved using the above diagram. The three interlocking circles represent the subject teachers (maths, physics, chemistry). The letters are regions created by the circles. The solution consists of relating the lettered regions to facts given in the problem.

The large circle is all teachers. The regions are as follows (some regions may be zero):

A Teachers of other subjects

B Only maths

C Only physics

D Only chemistry

E Maths & chemistry but not physics

F Physics & chemistry but not maths

G Maths & physics but not chemistry

H All 3 subjects

Now we use the numbers in the problem:

A+B+C+D+E+F+G+H=60 (all teachers)

C+F+G+H=27 (physics)

B+E+G+H=30 (maths)

D+E+F+H=21 (chemistry)

G+H=12 (maths & physics & maybe chemistry)

E=0 (no-one teaches maths and chemistry)

H=0 because E=0, so no-one can teach all 3 subjects

So G=12, C+F=15, D+F=21, B=18

A+18+15+D+12=60, A+D=15. Assume A=0 (no teachers of other subjects) then D=15 and F=6 (physics and chemistry) and C=9 (only physics).

answered Dec 21, 2016 by Top Rated User (425,260 points)