We are given \( A = \{1, 2, 3, \dots, 10\} \) and the set \( B = \left\{ \frac{m}{n} : m, n \in A, m <n \text{ and } \gcd(m, n) = 1 \right\} \).
To find \( n(B) \), we list the pairs \( (m, n) \) where \( m <n \) and \( \gcd(m, n) = 1 \).
We compute this for each \( n \in A \):
For \( n = 2 \), the valid pairs are \( \left( \frac{1}{2} \right) \).
For \( n = 3 \), the valid pairs are \( \left( \frac{1}{3}, \frac{2}{3} \right) \).
For \( n = 4 \), the valid pairs are \( \left( \frac{1}{4}, \frac{3}{4} \right) \).
For \( n = 5 \), the valid pairs are \( \left( \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \right) \).
For \( n = 6 \), the valid pairs are \( \left( \frac{1}{6}, \frac{5}{6} \right) \).
For \( n = 7 \), the valid pairs are \( \left( \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \right) \).
For \( n = 8 \), the valid pairs are \( \left( \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8} \right) \).
For \( n = 9 \), the valid pairs are \( \left( \frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9} \right) \).
For \( n = 10 \), the valid pairs are \( \left( \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10} \right) \).
Counting all the valid pairs, we get \( n(B) = 31 \). Thus, the answer is \( \boxed{31} \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).