Question

The H.C.F. of $16a^2{b}^3{x}^4{y}^5$, $40a^3{b}^2{x}^3{y}^4$ and $24a^5{b}^5{x}^6{y}^4$ is .

• A. $16a^2{b}^2{x}^4{y}^4$
• B. $8a^2{b}^2{x}^3{y}^4$
• C. $24a^5{b}^5{x}^6{y}^4$
• D. $40a^3{b}^2{x}^3{y}^4$

Answer 1

The correct option is B $8a^2{b}^2{x}^3{y}^4$

The H.C.F. of 16, 40 and 24 is 8.
a, b, x and y are the common elementary factors and $a^2,b^2,x^3,y^4$ are their highest powers that exactly divide the given quantities