In a class of 40 students, 16 take maths, 21 take chem., 18 take phy., 9 take math & chem., 5 take math & phy., # of students who take...maths only & (chem. + phy.) only are equal &...

In a class of 40 students, 16 take maths, 21 take chem., 18 take phy., 9 take math & chem., 5 take math & phy., # of students who take...

maths only & (chem. + phy.) only are equal & 3 students take all three subjects. Find the number of students who take maths only.

Expert Answers
crmhaske eNotes educator| Certified Educator

Information provided:

16/40 students take math
21/40 students take chemistry
18/40 students take physics

9/40 take math and chemistry
5/40 take math and physics
3/40 take all three subjects

This doesn't add up because this would imply that 9+5+3 students take math, but not only math; however, that equals 17 students?

Edit: because I assumed that the three groups were mutually exclusive.  ie. the 3 taking all three subjects were not accounted for in the math/chemisty or math/physics group.

If they are not mutually exclusive then it would be be 16 - (9+5-3) = 5 students taking math only.

To solve for the number of students taking physics and chemisty sum the number of students taking physics and the num of students taking chemistry first: 21+18 = 39

Subtract from it all other possible combinations: 39 - 5 (math only) - 9 (math and chemistry) - 5 (math and physics) - 7 (physics only) - 8 (chemistry only) + 3 (all three) = 8

neela | Student

The entire set of students = 40

n(M) =  16 , number of students taking maths,

n(MC) = 9 number of students who take maths and chemistry

n(MP) = 5 number of students who take maths and physics.

n(MPC) = 3

Therefore  the number of students who take maths only are: n(MP'C') = n(M) - {n(MP)+n(MC)-n(MPC)}, where  '  indicate not taking.

= 16- {5+9-3}

=16-14+3

n(MP'C') = 5, is the number who take only maths.

 

n(M) =16 , n(P) = 18,  n(C) = 21

n(MP) = 5, n(MC) = 9, n(PC) = 8 not given but calculated.

n(MPC) = 3

 

Tally :  n(U) = n(M)+n(P)+n(C) - {n(MP)+n(MC)+n(PC)}+2 n(MPC) + n(MPC) + n(M'P'C')

n(U)=40 =  16+18+21 -(5+9+8)+2(3)+ n(M'P'C')

40 = 55-22+6 +n(M'P'C')

40 = 39 +nM'P'C").

Therefore  n(M'P'C') = 40 -39 =1

n(M'P'C') = 1. So there is one cadidate who do not take any of the 3 given subjects and takes the subject other than M or P or C.