In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics,

In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

Answer

M : mathematics

P  : physics

C :  chemistry 

We have the formula for arithmetic series

an = a+(n-1)d here

an = last term

 a= first term

 n = number of term and

d = difference

subtract a both side and divide by (n-1)

we get                 

number of term  n= (an- a)/d  + 1

use this formula to get the number of student

Student numbered for physics                  ={2,4,6,8,……….120}

Number of student                         n(P)       =(120- 20)/6 +1 = 60

 Student numbered for chemistry            ={5,10,15,20………..120}

Number of student                         n(C)       = (120 – 5)/5 +1=24

Student number for mathematics            = {7,14,21,28………119}

Number of student                         n(M)      =(119-7)/7 +1 =17

Student of Physics and chemistry will have number which will be divisible by  2*5 = 10

Hence student for the physics and chemistry will have the number = {10,20,30,40…….120 }

Number of student                         n(P∩C)                 = (120-10)/10 +1 = 12

Student of Physics and mathematics will have number which will be divisible by  of 2*7 = 14

Hence student for the physics and mathematics will have the number = {14,28,42…….112 }

Number of student                         n(P∩C)                 = (112-14)/10 +1 = 8

Student of mathematics and chemistry will have number which will be divisible by  7*5 = 35

Hence student for the mathematics and chemistry will have the number = {35,70,105 }

Number of student                         n(M∩C)               = 3

Student of mathematics, physics and chemistry will have the number which will be divisible by  2*5*7 = 70

Hence Student of mathematics, physics and chemistry will have the number ={70}

There are only 1 such student hence

n(M∩P∩C) = 1

We have the formula.

n (AUBUC) = n(A) + n(B) +n(c) – n(A∩B) – n(B∩C) –n(C∩A)  + n(A∩B∩C) 

replace A,B and C by C,P and M respectively and plug the value we get

n(PUCUM)          = 60 + 24+17 -12-3-8 + 1 

                                                                                                =79

Number of student which non of above subject   = total -   n(CUPUM)

                                                                                                = 120-79

                                                                                                =41 students