Question

Find the value of $99\times 101$ using standard identity.

• A. $9999$
• B. $9989$
• C. $9979$
• D. $1009$

Answer 1

The correct option is A $9999$

Given, $99\times 101=(100-1)(100+1)$.

We know, $(a+b)(a-b)=a^2-b^2$.

Then,

$99\times 101=(100-1)(100+1)$
$=(100{)}^2-1^2$

$=10000-1$

$=9999$.

Therefore, option $A$ is correct.