Question

In a class of 140 students numbered from 1 to 140, all even number students opted mathematics course, those whose number is divisible by 3 opted physics course and those whose number is divisible by 5 opted chemistry course. Then the number of students who did not opt for any of the courses is:

A. 102

B. 42

C. 1

D. 38

Answer 1

Hint: Solve the question with the help of set theory. First find the number of students who have opted any of the 3 courses, i.e. find the value of \[n\left( A\cup B\cup C \right).\] Thus find the number of students who did not opt any subjects from the total of 140 students.


Complete step-by-step answer:

Given that the total number of students in a class is 140.

It is said that out of 140, even numbered students opted for the mathematics course.
\[\therefore \dfrac{140}{2}=70\] students.

Let n(A) represent the students who took mathematics courses.
\[\therefore n(A)=70.\]

Out of 140 students, whose number is divisible by 3, opted Physics course.

Let us consider n(B) as the students who opted for the Physics course, i.e. out of 140, 138 is the highest number divisible by 3.
\[\therefore n(B)=\dfrac{138}{3}=46\] students.

Now the students whose number is divisible by 5 opted chemistry course.

Let n(C) be the number of students who opted for the chemistry course.
\[\therefore n(C)=\dfrac{140}{5}=28\] students.

Thus we found n(A) = 70 students, n(B) = 46 students, n(C) = 28 students.

Now we need to find \[n\left( A\cap B \right)\], which means that the term should be divisible by both 2 and 3.

Let us take LCM of 2 and 3 \[=2\times 3=6\], i.e. the number should be divisible by 6.
So out of 140, the number divisible by 6 would be,
6, 12, 18,….,36 = 23 terms
\[\therefore n(A\cap B)=\dfrac{136}{6}=23\].

Similarly \[n(B\cap C)\] should be divisible by both 3 and 5.
LCM = 15

The numbers are 15, 30, 45…., 135 = 9 terms.
\[\therefore n(B\cap C)=9\].

Similarly we need to find \[n(A\cap C)\]where both should be divisible by 2 and 5.
LCM = 10

Thus the number will be 10, 20, 30…..140 = 14 terms.
So, \[n(A\cap C)=14\].

Now let us find \[n(A\cap B\cap C)\] where the number should be divisible by 2, 3 and 5.
LCM \[=2\times 3\times 5=30\]

The numbers are 30, 60, 90…..,120, out of 140 terms = 4 terms.
\[\therefore n(A\cap B\cap C)=4\].

Thus we have n(A) = 70, n(B) = 9, n(C) = 14, \[n(A\cap B\cap C)\]= 4.

We need to find the number of students who did not opt for anything of the three courses.

We know \[n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cup C)+n(A\cap B\cap C)\]
\[\therefore n(A\cup B\cup C)=70+46+28-23-9-14+4=102.\]

Thus the total number of students who opted for the courses are \[n(A\cup B\cup C)\] is equal to 102 students.
So, the number of students who did not opt for any course:
= Total number of students – number of students who opted for the courses
= Total students - \[n(A\cup B\cup C)\]
\[=140-102=38\]

Thus, 38 students didn’t opt for any courses.

Option D is the correct answer.

Note:
Remember as these are 3 values, n(A), n(B) and n(C), to get all the opted students, you need to find the union of these sets, i.e. \[n(A\cup B\cup C)\]. If you get the number of students who have opted for the courses, then it’s easier to find the number of students who didn’t opt for any course from the total number of students.