Question

If a set $A=\{3,7,11,...,407\}$ and a set $B=\{2,9,16,...,709\}$ then $n(A\cap B)=$

• A. $13$
• B. $14$
• C. $15$
• D. $16$

Answer 1

The correct option is B

$14$

Step1. The given sets :

Set $A=\{3,7,11,...,407\}$ and the set $B=\{2,9,16,...,709\}$

For$A$, common difference, $d=4$

For$B$, common difference, $d=7$

First-term of common series $=23$

Common difference $=4\times 7=28$

Common series is given by $=23,71,99,127\dots$

Step2. Find the value of $\mathit{n}\mathbf{(}\mathit{A}\mathbf{}\mathbf{\cap }\mathbf{}\mathit{B}\mathbf{)}$:

$23+(n-1)28\le 407$ (the last term of the common series will be less than or equal to 407)

$$\begin{gathered} \Rightarrow 28n-28\le 384 \\ \Rightarrow 28n\le 412 \\ \Rightarrow n\le 412/28 \\ \Rightarrow n\le 14.71 \end{gathered}$$

So $n=14$

Hence the correct option is B.