Let A=\ 1,2,3, , 10\ and B=\ (m)/(n): m, n ∈ A, m n. and .gcd(m, n)=1\ . Then n(B) is equal to :

Correct answer: B. 31 aligned & A=\ 1,2, , 10\ & B\ m n =m, n ∈ A, m < n, gcd(m, n)=1\ & n(B) aligned n=2 \ (1)/(2)\ n=3 \ (1)/(3), (2)/(3)\ array ll n=4 & \ (1)/(4), (3)/(4)\ n=5 & \ (1)/(5), (2)/(5), (3)/(5), (4)/(5)\ array array ll n=6 & \ (1)/(6), (5)/(6)\ n=7 & \ (1)/(7), (2)/(7), (3)/(7), (4)/(7), (5)/(7), (6)/(7)\ n=8 & \ (1)/(8), (3)/(8), (5)/(8), (7)/(8)\ array array ll n=9 & \ (1)/(9), (2)/(9), (4)/(9), (5)/(9), (7)/(9), (8)/(9)\ n=10 & \ (1)/(10), (3)/(10), (7)/(10), (9)/(10)\ n(B)=31 array

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Sets and Relations

Let A=\ 1,2,3, , 10\ and B=\ (m)/(n): m, n ∈ A, m n. and .gcd(m,...

Let

A=\{1,2,3, \ldots, 10\} and

B=\left\{\frac{m}{n}: m, n \in A, m \lt n\right. and \left.\operatorname{gcd}(m, n)=1\right\}

. Then

n(B) is equal to :

Options

A $36$
B $31$
C $37$
D $29$

Correct answer

B. $31$

Step-by-step solution

\begin{aligned} & \mathrm{A}=\{1,2, \ldots, 10\} \\ & \mathrm{B}\left\{\frac{\mathrm{m}}{\mathrm{n}}=\mathrm{m}, \mathrm{n} \in \mathrm{A}, \mathrm{m} < \mathrm{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\right\} \\ & \mathrm{n}(\mathrm{B}) \end{aligned}

\mathrm{n}=2 \quad\left\{\frac{1}{2}\right\}

\mathrm{n}=3 \quad\left\{\frac{1}{3}, \frac{2}{3}\right\}

\begin{array}{ll}\mathrm{n}=4 & \left\{\frac{1}{4}, \frac{3}{4}\right\} \\ \mathrm{n}=5 & \left\{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right\}\end{array}

\begin{array}{ll}\mathrm{n}=6 & \left\{\frac{1}{6}, \frac{5}{6}\right\} \\ \mathrm{n}=7 & \left\{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right\} \\ \mathrm{n}=8 & \left\{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right\}\end{array}

\begin{array}{ll}\mathrm{n}=9 & \left\{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right\} \\ \mathrm{n}=10 & \left\{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right\} \\ \mathrm{n}(\mathrm{B})=31\end{array}

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