Question

hcf and lcm of two number are 48 and 384 respectively.if one of the number is 96 . find the other number.​

Answer 1

$\bf{\underline{\underline{Given}}}$

$\mathsf{\star\: \: HCF\: of\:two\:no's = 48}$

$\mathsf{\star\: \: LCM\: of\:two\:no's = 384}$

$\mathsf{\star\: \: One\:number = 96}$

$\bf{\underline{\underline{To\:find}}}$

$\mathsf{\star\:\: The \:other\: number}$

$\bf{\underline{\underline{Solution}}}$

Let the two numbers be a and b respectively

where a = 96 (Given)

We know that

$\mathtt{HCF × LCM = Product\: of \: a \: and \: b}$

$\mathtt{\implies\: 48 × 384 = a × b}$

$\mathtt{\implies\: 48 × 384 = 96 × b}$

$\mathtt{\implies\: \frac{48 × 384}{96} = b}$

$\mathtt{\implies\: \green{b = 192}}$

∴ The other number (b) = 192

Answer 2

$\huge{\tt{\underline{\pink{\: Given:-}}}}$

HCF of the two number= 48

LCM of the two number = 384

The one number = 96

${\tt{\red{\: To \: Find:-}}}$

The Other number =?

$\huge{\tt{\pink{\underline{\: solution:-}}}}$

We know the formula to solve,

${\tt{ HCF × LCM = product \: of \: a \: and \: b}}$

${\implies{\tt{ 48 × 384= a×b}}}$

${\implies{\tt{ 48 × 384= 96×b}}}$

${\implies{\tt{\: or, \: b = \cancel\frac{48× 384} {96}}}}$

${\implies{\tt{ \: b = 192}}}$

$\implies\boxed {\green{ \: 192}}$