Question

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 :

• A. 102
• B. 42
• C. 1
• D. 38

Answer 1

The Correct Option is D

To solve this problem, we need to determine the number of students who did not opt for any of the three courses: Mathematics, Physics, and Chemistry. Let's analyze step-by-step:

1. Students opting for Mathematics are those with even numbers. The total number of even numbers from 1 to 140 can be calculated by:
   • The sequence of even numbers is 2, 4, 6, ..., 140. This is an arithmetic sequence where the first term \(a = 2\) and the last term \(l = 140\), with a common difference \(d = 2\).
   • The number of terms \(n\) in this sequence is given by the formula: n = \frac{l - a}{d} + 1 = \frac{140 - 2}{2} + 1 = 70
2. Students opting for Physics are those whose numbers are divisible by 3.
   • The sequence of these numbers is 3, 6, 9, ..., 138.
   • Using the arithmetic sequence formula: n = \frac{l - a}{d} + 1 = \frac{138 - 3}{3} + 1 = 46
3. Students opting for Chemistry are those whose numbers are divisible by 5.
   • The sequence of these numbers is 5, 10, 15, ..., 140.
   • Using the arithmetic sequence formula: n = \frac{l - a}{d} + 1 = \frac{140 - 5}{5} + 1 = 28
4. Now, we apply the principle of inclusion-exclusion to find the number of students who opted for at least one course:
   • \( |A| = 70 \) where \( A \) is students who opted for Mathematics.
   • \( |B| = 46 \) where \( B \) is students who opted for Physics.
   • \( |C| = 28 \) where \( C \) is students who opted for Chemistry.
   • \( |A \cap B| \) is the number of students who opted for both Mathematics and Physics, i.e., divisible by 6: = \frac{138 - 6}{6} + 1 = 23
   • \( |A \cap C| \) is the number of students who opted for both Mathematics and Chemistry, i.e., divisible by 10: = \frac{140 - 10}{10} + 1 = 14
   • \( |B \cap C| \) is the number of students who opted for both Physics and Chemistry, i.e., divisible by 15: = \frac{135 - 15}{15} + 1 = 9
   • \( |A \cap B \cap C| \) is the number of students who opted for all three courses, i.e., divisible by 30: = \frac{120 - 30}{30} + 1 = 4
   • Total number of students opting for at least one course is given by: = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| = 70 + 46 + 28 - 23 - 14 - 9 + 4 = 102
5. So, the number of students who did not opt for any of the three courses is: = 140 - 102 = 38

Therefore, the correct answer is 38.