Question

Let $A=(x|x$ is a prime number and $x<300$ > the number of different rational numbers, whose numerator and denominator belong to $A$ is:

• A. 91
• B. 84
• C. 106
• D. None of these

Answer 1

The correct option is A 91

Set A contains 10 elements. 2 different numbers for numerator and denominator from these can be obtained in $10\times 9=90$ ways and each permutation will form a unique rational number different from one. In addition, one will be formed when numerator and denominator are same. So required number of numbers $90+1=91$