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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sets
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If $ A=\{x\;/\;x\in\:N, \: $ and $\:x$ is a multiple of $3$ $\leq100\}$ and $B=\{x/x\in\:N\:$ and $\:x$ is multiple of $5$ $\leq\:100\}$, then no. of elements in $(A\times\:B)\cap\:(B\times\:A)\:is$

$\begin{array}{1 1} 36 \\ 33 \\ 20 \\ 6\end{array}$

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  • If $n(A)=n,\:then\:n(A\times A)=n^2$
Ans: (A) 36
$A=\{3,6,9,.....................99\}$ n(A)=33
$B=\{5,10,15,..................100\}$ n(B)=20
$A\times B=\{(3,5),(3,10),...............\}$
$B\times A=\{(5,3),(5,6),(5,9)...............\}$
$\Rightarrow\:(A\times B)\cap (B\times A)=\{(x,y)\:\}$ where $x\:and\:y$ both should be multiple of 3 and 5.
i.e., multiple of $15\leq 100$
$\Rightarrow\:x,y\in\:\:C=\{15,30,45,60,75,90\}$
$n(C)=6$
$\Rightarrow\:(A\times B)\cap (B\times A)=C\times C$
$\Rightarrow\:$ no. of elements in $ (A\times B)\cap (B\times A)=6\times6=36$
answered Apr 29, 2013 by rvidyagovindarajan_1
 
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