Question

Solve L.C.M. for the monomial $7x{y}^2,35z{y}^3,40x^{2}z$

• A. $180x^2{y}^{3}z$
• B. $280x^2{y}^{3}z$
• C. $280x^2{y}^3{z}^2$
• D. $280x^2{y}^{2}z$

Answer 1

The correct option is B $280x^2{y}^{3}z$

To find the L.C.M. of two or more monomials is the product of the L.C.M. of their numerical coefficients and the L.C.M. of their literal coefficients.
The L.C.M. of numerical coefficients $=$ The L.C.M. of $7,35$ and $40$.
$7=1\times 7$
$35=5\times 7$
$40=2x4\times 5$
Therefore, L.C.M. of $7,35$ and $40=7\times 5\times 2\times 4=280$
The L.C.M. of literal coefficients $=$ The L.C.M. of $x{y}^2,z{y}^3,x^{2}z=x^2{y}^{3}z$
So, the L.C.M. of $7x{y}^2,35z{y}^3,40x^{2}z=280x^2{y}^{3}z$