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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Sequences and Series

Let A=\ n ∈[100,700] ℕ: n is neither a multiple of 3 nor a multiple...

Let A={n∈[100,700]∩N:nA=\{n \in[100,700] \cap \mathbb{N}: n is neither a multiple of 3 nor a multiple of 4.Thenthenumberofelementsin. Then the number of elements inA$ is

Options

  1. A290
  2. B280
  3. C300
  4. D310

Correct answer

C. 300

Step-by-step solution

n(3)⇒ multiple of 3102,105,108,…..,699Tn=699=102+(n−1)(3)n=200n(3)=200∵n(4)⇒ multiple of 4\begin{aligned} & \mathrm{n}(3) \Rightarrow \text{ multiple of } 3 \\ & 102,105,108, \ldots . ., 699 \\ & \mathrm{ T}_{\mathrm{n}}=699=102+(\mathrm{n}-1)(3) \\ & \mathrm{n}=200 \\ & \mathrm{n}(3)=200 \\ & \because \mathrm{n}(4) \Rightarrow \text{ multiple of } 4\end{aligned}100,104,108,….,700Tn=700=100+(n−1)(4)n=151n(4)=151n(3∩4)⇒ multiple of 34 both 108,120,132,….,696Tn=696=108+(n−1)(12)n=50n(3∩4)=50n(3∪4)=n(3)+n(4)−n(3∩4)=200+151−50=301n(3∪4‾)= Total −n(3∪4)= neither a multiple  of 3 nor a multiple of 4=601−301=300\begin{aligned} & 100,104,108, \ldots ., 700 \\ & T_n=700=100+(n-1)(4) \\ & n=151 \\ & n(4)=151 \\ & n(3 \cap 4) \Rightarrow \text{ multiple of } 3 & 4 \text{ both } \\ & 108,120,132, \ldots ., 696 \\ & T_n=696=108+(n-1)(12) \\ & n=50 \\ & n(3 \cap 4)=50 \\ & n(3 \cup 4)=n(3)+n(4)-n(3 \cap 4) \\ & \quad=200+151-50 \\ & \quad=301 \\ & n(\overline{3 \cup 4})=\text{ Total }-n(3 \cup 4)=\text{ neither a multiple } \\ & \text{ of } 3 \text{ nor a multiple of } 4 \\ & =601-301=300\end{aligned}$
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