1. We need to count the number of distinct elements in $B=\left\{\frac{m}{n}: m,n\in A=\{1,2,\dots,10\},\ m<n,\ \gcd(m,n)=1\right\}.$

Because $\gcd(m,n)=1$, each fraction $\frac{m}{n}$ is already in lowest terms. Also, two different coprime pairs cannot give the same reduced fraction. So $n(B)$ is exactly the number of pairs $(m,n)$ such that

• $1\le m<n\le 10$,
• $\gcd(m,n)=1$.

So we count, for each denominator $n$, how many numerators $m<n$ are coprime to $n$. This is precisely Euler's totient function $\varphi(n)$.

Thus, $n(B)=\sum_{n=2}^{10}\varphi(n).$

2. Now compute $\varphi(n)$ for $n=2$ to $10$:

• $\varphi(2)=1$
• $\varphi(3)=2$
• $\varphi(4)=2$
• $\varphi(5)=4$
• $\varphi(6)=2$
• $\varphi(7)=6$
• $\varphi(8)=4$
• $\varphi(9)=6$
• $\varphi(10)=4$

3. Add them: $1+2+2+4+2+6+4+6+4=31.$

Therefore, $n(B)=31.$

4. Option check:

• A: $29$ ❌
• B: $31$ ✅
• C: $37$ ❌
• D: $36$ ❌

Hence the correct option is B.