Wednesday, April 9, 2014

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(PC)                 = (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(PC)                 = (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(MC)               = 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(MPC) = 1
We have the formula.
n (AUBUC) = n(A) + n(B) +n(c) – n(AB) – n(BC) –n(CA)  + n(ABC) 
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


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