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
