• # question_answer Match the following. Column - I Column - II (P) If $213x27$ is divisible by 9, then $x=$ (i) 2 (Q) If $2415x$is divisible by 6, then $x=$ (ii) 8 (R) If $22135x$ is divisible by 4 and 3, then $x=$ (iii) 3 (S) If $7251x93$ is divisible by 11, then $x=$ (iv) 6 A)   (P)$\to$(iii); (Q)$\to$(ii); (R)$\to$(iv); (S)$\to$(i)B)  (P)$\to$(ii); (Q)$\to$(iv); (R)$\to$(i); (S)$\to$(iii)C)  (P)$\to$(iii); (Q)$\to$(iv): (R)$\to$(i); (S)$\to$(ii)D)  (P)$\to$(ii); (Q)$\to$(iii); (R)$\to$(i); (S)$\to$(iv)

(P) Since, $213\times 27$ is divisible by 9. So, 2+1+3+x+2+7=15+xis divisible by 9. $\therefore$ x=3 (Q) $2415x$ is divisible by 6, if it is divisible by both 2 and 3. $\therefore x=6$ (R) $23245x$ is divisible by 4 and 3 $\Rightarrow 5x$ is divisible by 4 $\therefore$ Possible values of x are 2, 6             Also, $2+3+2+4+5+x=16+x$ is divisible by 3. $\therefore$ Possible values of $x$ is 2. (S) We have, $7251x\text{ }93$ is divisible by 11 $\therefore [(7\text{ }+\text{ }5\text{ }+\text{ }x\text{ }+\text{ }3)\text{ }-\text{ }(2\text{ }+\text{ }1\text{ }+\text{ }9)]$ is divisible by 11 $\Rightarrow 15+x-12=3+x$ is divisible by 11 $\therefore x$ can be equal to 8.