Question

the hcf and lcm of 2 numbers are 6a and 360a^2 b^3 respectively. if one of the number is 120b^2, find the other

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given:

• HCF = 6a
• LCM = 360a²b³
• One Number = 120b²

To find:

• The other number = x

Solution:

We know that,

$\boxed{\tt{\red{HCF\times LCM=Product\:of\:both\:numbers}}}$

So,

$\leadsto\tt{6a\times 360{a}^{2}{b}^{3}=120{b}^{2}\times x}$

$\leadsto\tt{2160 {a}^{3}{b}^{3}=120{b}^{2}\times x}$

$\leadsto\tt{x=\dfrac{2160{a}^{3}{b}^{3}}{120{b}^{2}}}$

$\huge\leadsto{\tt{\green{x=18 {a}^{3}b}}}$

So, the other number is = 18a³b.

$\rule{200}2$

Answer 2

$\large{\underline{\bold{\red{\rm{Answer-}}}}}$

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The other number is 18a³b.

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$\large{\underline{\bold{\red{\rm{Explanation-}}}}}$

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$\begin{lgathered}\bold{\pink{Given}} \begin{cases}\sf{\blue{LCM\:of\:two\:numbers\:=\:360a^2\:b^2}} \\ \sf{\blue{HCF\:of\:two\:numbers\:=\:6a}}\\ \sf{\blue{One\:number\:=120b^2}}\end{cases}\end{lgathered}$

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We have to find the other number.

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Let the other number be k.

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We know that,

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$\large{\boxed{\rm{HCF\:\times\:LCM\:=\:Product\:of\:two\:numbers}}}$

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Putting the values,

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$\implies$ 6a × 360a²b² = 120b² × k

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$\implies$ 2160a³b² = 120b² × k

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$\implies$ k = $\sf{\cancel{\dfrac{2160a^3\:b^2}{120b^2}}}$

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$\implies$ k = 18a³b

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Hence, the other number is 18a³b.