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1.Set Theory
hard
Let $A =\{1,2,3, \ldots \ldots, 10\}$ and $B=\left\{\frac{m}{n}: m, n \in A, m < n \text { and } \operatorname{gcd}(m, n)=1\right\} . $Then $n(B)$ is equal to :
A$31$
B$36$
C$37$
D$29$
(JEE MAIN-2025)
Solution
$A=\{1,2, \ldots .10\}$
$\begin{array}{l} B \left\{\frac{ m }{ n }= m , n \in A , m < n , \operatorname{gcd}( m , n )=1\right\} \\ n ( B )\end{array}$
$n=2 \quad\left\{\frac{1}{2}\right\}$
$n=3 \quad\left\{\frac{1}{3}, \frac{2}{3}\right\}$
$n=4 \quad\left\{\frac{1}{4}, \frac{3}{4}\right\}$
$n=5 \quad\left\{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right\}$
$n=6 \quad\left\{\frac{1}{6}, \frac{5}{6}\right\}$
$n=7 \quad\left\{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right\}$
$n=8 \quad\left\{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right\}$
$n=9 \quad\left\{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right\}$
$n=10 \quad\left\{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right\}$
$n(B)=31$
$\begin{array}{l} B \left\{\frac{ m }{ n }= m , n \in A , m < n , \operatorname{gcd}( m , n )=1\right\} \\ n ( B )\end{array}$
$n=2 \quad\left\{\frac{1}{2}\right\}$
$n=3 \quad\left\{\frac{1}{3}, \frac{2}{3}\right\}$
$n=4 \quad\left\{\frac{1}{4}, \frac{3}{4}\right\}$
$n=5 \quad\left\{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right\}$
$n=6 \quad\left\{\frac{1}{6}, \frac{5}{6}\right\}$
$n=7 \quad\left\{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right\}$
$n=8 \quad\left\{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right\}$
$n=9 \quad\left\{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right\}$
$n=10 \quad\left\{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right\}$
$n(B)=31$
Standard 11
Mathematics