Let \( A = \{ n \in [100, 700] \cap \mathbb{N} : n \text{ is neither a multiple of 3 nor a multiple of 4} \} \).
Then the number of elements in \( A \) is:
Then the number of elements in \( A \) is:
- 300
- 280
- 310
- 290
The Correct Option is A
Solution and Explanation
Step 1: Find the total number of elements in [100, 700]:
\[ \text{Total} = 700 - 100 + 1 = 601. \]
Step 2: Find the number of multiples of 3 in [100, 700]:
Multiples of 3: \( 102, 105, 108, \ldots, 699 \).
This is an arithmetic progression (AP) with: \[ a = 102, \, d = 3, \, \text{and } l = 699. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 699 = 102 + (n - 1)3. \] Simplify: \[ 597 = 3(n - 1) \implies n = 200. \] Thus, \( n(3) = 200 \).
Step 3: Find the number of multiples of 4 in [100, 700]:
Multiples of 4: \( 100, 104, 108, \ldots, 700 \).
This is an AP with: \[ a = 100, \, d = 4, \, \text{and } l = 700. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 700 = 100 + (n - 1)4. \] Simplify: \[ 600 = 4(n - 1) \implies n = 151. \] Thus, \( n(4) = 151 \).
Step 4: Find the number of multiples of both 3 and 4 (i.e., multiples of 12):
Multiples of 12: \( 108, 120, 132, \ldots, 696 \).
This is an AP with: \[ a = 108, \, d = 12, \, \text{and } l = 696. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 696 = 108 + (n - 1)12. \] Simplify: \[ 588 = 12(n - 1) \implies n = 50. \] Thus, \( n(3 \cap 4) = 50 \).
Step 5: Use the inclusion-exclusion principle to find \( n(3 \cup 4) \):
\[ n(3 \cup 4) = n(3) + n(4) - n(3 \cap 4). \] Substitute values: \[ n(3 \cup 4) = 200 + 151 - 50 = 301. \]
Step 6: Find the number of elements in \(A\) (neither multiples of 3 nor 4):
\[ n(A) = \text{Total} - n(3 \cup 4). \] Substitute values: \[ n(A) = 601 - 301 = 300. \]
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