JEE Mathematics 2025 Question with Solution

Given: $A = \{1, 2, 3, \dots, 10\}$ and $B = \left\{ \frac{m}{n} : m, n \in A, m < n \text{ and } \gcd(m, n) = 1 \right\}$.

Find: $n(B)$.

For each fixed denominator $n$ with $2 \le n \le 10$, we count the admissible numerators $m$ such that $1 \le m < n$ and $\gcd(m,n)=1$. This count is $\varphi(n)$, where $\varphi$ is Euler's totient function.

Therefore,

Now compute:

Adding these values,

Therefore, $n(B)=31$ and the correct option is A.

Given: $A = \{1, 2, 3, \dots, 10\}$ and $B = \left\{ \frac{m}{n} : m, n \in A, m < n \text{ and } \gcd(m, n) = 1 \right\}$.

Find: $n(B)$.

List the valid fractions for each denominator.

For $n=2$: $\left(\frac{1}{2}\right)$

For $n=3$: $\left(\frac{1}{3}, \frac{2}{3}\right)$

For $n=4$: $\left(\frac{1}{4}, \frac{3}{4}\right)$

For $n=5$: $\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right)$

For $n=6$: $\left(\frac{1}{6}, \frac{5}{6}\right)$

For $n=7$: $\left(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right)$

For $n=8$: $\left(\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right)$

For $n=9$: $\left(\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right)$

For $n=10$: $\left(\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right)$

Counting all these valid fractions gives

Therefore, $n(B)=31$, so the correct option is A.