If first $100$ natural numbers are divided in groups of ten each, then the minimum number of primes are between:

$1 t o 10$

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$51 t o 60$

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$81 t o 90$

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$91 t o 100$

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Solution

The correct option is D
$91 t o 100$

Explanation for the correct option:
Option (D) $91 t o 100$
Prime number between $91 t o 100$ is $97$ . There is only one prime number. and thus it is the minimum number of prime number among the given options.
Explanation for the incorrect options:
Option (A) $1 t o 10$
Prime numbers between $1 t o 10$ are $2$ , $3$ , and $7$ . There are only three prime numbers between $1 t o 10$ .
Option (B) $51 t o 60$
Prime numbers between $51 t o 60$ are $53$ , and $59$ . There are two prime numbers between $51 t o 60$ .
Option (C) $81 t o 90$
Prime numbers between $81 t o 90$ are $83$ , and $89$ . There are two three prime numbers between $81 t o 90$
Hence, the correct answer is option (D) $91 t o 100$

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