Question

If H.C.F of $x^3{y}^4{z}^2$, $x^6{y}^8{z}^4$ is 'a' and H.C.F of $x^5{y}^6{z}^3$, $x^6{y}^7{z}^4$ is 'b' then $a/b=$ ___

• A. $x^2{y}^4{z}^4$
• B. $x^{-2}{y}^{-2}{z}^{-1}$
• C. $x^{-3}{y}^{-4}{z}^2$
• D. $x^3{y}^4{z}^2$

Answer 1

The correct option is B

$x^{-2}{y}^{-2}{z}^{-1}$

HCF of $x^3{y}^4{z}^2$ and $x^6{y}^8{z}^4$ = $x^3{y}^4{z}^2=a$

HCF of $x^5{y}^6{z}^3$ and $x^6{y}^7{z}^4$ = $x^5{y}^6{z}^3=b$

$⟹$$\frac{a}{b}=x^{-2}{y}^{-2}{z}^{-1}$