Question

Find the H.C.F of three monomials $56m^{2}n,12m^{2}n,10m^3{n}^2$.

• A. $2m^{2}n$
• B. $2mn$
• C. $2$
• D. $2m^3{n}^2$

Answer 1

The correct option is A $2m^{2}n$

$56mn,12m^{2}n,10m^3{n}^2$
$56mn=2\times 4\times 7\times m^{2}n$
$12m^{2}n=2\times 2\times 3\times m^{2}n$
$10m^3{n}^2=2\times 5\times m^3{n}^2$
So, the common factor is $2m^{2}n$.
Therefore, H.C.F is $2m^{2}n$.