In a class of 120 students, numbered 1 to 120, all even numbered students opt for physics, those whose numbers are divisible by 5 opt for chemistry and those whose numbers are divisible by 7 opt for maths. how many opt for none of the three subjects?
Answers
There are 12 divisible numbers by 5
There are 8 divisible no. by 7
So,60+12+8=80
Answer:
Approach: Let us find the number of students who took at least one of the three subjects and subtract the result from the overall 120 to get the number of students who did not opt for any of the three subjects.
Number of students who took at least one of the three subjects can be found by finding out n(A U B U C), where A is the set of students who took Physics, B is the set of students who took Chemistry and C is the set of students who opted for Math.
Now, n(A∪B∪C) = n(A) + n(B) + n(C) - {n(A ∩ B) + n(B ∩ C) + n(C ∩ A)} + n(A ∩ B ∩ C)
A is the set of those who opted for Physics = 1202 = 60 students
B is the set of those who opted for Chemistry = 1205 = 24
C is the set of those who opted for Math = 1207 = 17.
Number of students who opted for Physics and Chemistry
Students whose numbers are multiples of 2 and 5 i.e., common multiples of 2 and 5 would have opted for both Physics and Chemistry.
The LCM of 2 and 5 will be the first number that is a multiple of 2 and 5. i.e., 10 is the first number that will be a part of both the series.
The 10th, 20th, 30th..... numbered students or every 10th student starting from student number 10 would have opted for both Physics and Chemistry.
Therefore, n(A ∩ B) = 12010 = 12
Number of students who opted for Physics and Math
Students whose numbers are multiples of 2 and 7 i.e., common multiples of 2 and 7 would have opted for both Physics and Math.
The LCM of 2 and 7 will be the first number that is a multiple of 2 and 7. i.e., 14 is the first number that will be a part of both the series.
The 14th, 28th, 42nd..... numbered students or every 14th student starting from student number 14 would have opted for Physics and Math.
Therefore, n(C ∩ A) = 12014 = 8
Number of students who opted for Chemistry and Math
Students whose numbers are multiples of 5 and 7 i.e., common multiples of 5 and 7 would have opted for both Chemistry and Math.
The LCM of 5 and 7 will be the first number that is a multiple of 5 and 7. i.e., 35 is the first number that will be a part of both the series.
The 35th, 70th.... numbered students or every 35th student starting with student number 35 would have opted for Chemistry and Math.
Therefore, n(B ∩ C) = 12035 = 3
Number of students who opted for all 3 subjects
Students whose numbers are multiples of 2, 5, and 7 i.e., common multiples of 2, 5, and 7 would have opted for all 3 subjects.
The LCM of 2, 5, and 7 will be the first number that is a multiple of 2, 5, and 7. i.e., 70 is the first number that will be a part of all 3 series.
70 is the only multiple of 70 in the first 120 natural numbers. So, the 70th numbered student is the only one who would have opted for all three subjects.
Therefore, n(A∪B∪C) = 60 + 24 + 17 - (12 + 8 + 3) + 1 = 79.
Number of students who opted for none of the three subjects = 120 - 79 = 41.