H.C.F of $x^{3} y^{4} z^{2}$ and $x^{6} y^{8} z^{4}$ is 'a' and H.C.F of $x^{5} y^{6} z^{3}$ and $x^{6} y^{7} z^{4}$ is 'b'. What is $\frac{a}{b}$

$x^{2} y^{4} z^{4}$

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

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$x^{- 3} y^{- 4} z^{2}$

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$x^{3} y^{4} z^{2}$

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Solution

The correct option is B
$x^{- 2} y^{- 2} z^{- 1}$

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

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

Therefore,

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

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Similar questions

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 $a / b =$ ___

x^{3} y^{4} z^{2}

x^{6} y^{8} z^{4}

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H.C.F of $a^{2} b^{3}$ and $a^{3} b^{2} d^{7}$ is $a^{2} b^{2}$ . What is their H.C.F

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