In a class of 140 students numbered 1 to 140, all even numbered 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 three courses is

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In a class of 140 students numbered 1 to 140, all even numbered 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 three courses is

102

$n (p) = [\frac{140}{5}] = 46$

$n (C) = [\frac{140}{5}] = 28$

$n (M) = [\frac{140}{2}] = 70$

$N (P \cup C \cup M) = n (P) + n (C) + n (M) - n (P \cap C) - n (C \cap M) - n (M \cap P) + n (P \cap M \cap C)$

$= 46 + 28 + 70 - [\frac{140}{15}] - [\frac{140}{10}] - [\frac{140}{6}] + [\frac{140}{30}]$

= 144 – 9 – 14 – 23 + 4 = 102

So required number of student = 140 – 102 = 38

This problem involves finding the number of students who did not opt for any of three courses (Mathematics, Physics, Chemistry) in a class of 140 students, based on their roll numbers. We'll use the principle of inclusion-exclusion to solve it.

Step 1: Define the Sets

Let:

M be the set of students who opted Mathematics (even numbers).
P be the set of students who opted Physics (divisible by 3).
C be the set of students who opted Chemistry (divisible by 5).

Step 2: Find the Number of Students in Each Set

Number of students in M: Even numbers from 1 to 140. There are 140/2 = 70 students.
$n (M) = 70$

Number of students in P: Numbers divisible by 3. The largest multiple ≤140 is 138 = 3×46, so there are 46 students.
$n (P) = ⌊\frac{140}{3}⌋ = 46$

Number of students in C: Numbers divisible by 5. The largest multiple ≤140 is 140 = 5×28, so there are 28 students.
$n (C) = ⌊\frac{140}{5}⌋ = 28$

Step 3: Find Pairwise Intersections

Students in M and P: Numbers divisible by LCM(2,3)=6. Largest multiple ≤140 is 138=6×23, so 23 students.
$n (M \cap P) = ⌊\frac{140}{6}⌋ = 23$

Students in M and C: Numbers divisible by LCM(2,5)=10. Largest multiple ≤140 is 140=10×14, so 14 students.
$n (M \cap C) = ⌊\frac{140}{10}⌋ = 14$

Students in P and C: Numbers divisible by LCM(3,5)=15. Largest multiple ≤140 is 135=15×9, so 9 students.
$n (P \cap C) = ⌊\frac{140}{15}⌋ = 9$

Step 4: Find the Triple Intersection

Students in all three sets: Numbers divisible by LCM(2,3,5)=30. Largest multiple ≤140 is 120=30×4, so 4 students.
$n (M \cap P \cap C) = ⌊\frac{140}{30}⌋ = 4$

Step 5: Apply Inclusion-Exclusion Principle

Number of students who opted for at least one course:
$n (M \cup P \cup C) = n (M) + n (P) + n (C) - n (M \cap P) - n (M \cap C) - n (P \cap C) + n (M \cap P \cap C)$
$= 70 + 46 + 28 - 23 - 14 - 9 + 4 = 102$

Step 6: Find Students Who Opted for None

Total students = 140
Students who opted for at least one course = 102
Students who did not opt for any course = Total - (At least one) = 140 - 102 = 38

Final Answer: 38

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