Question

d
9. The HCF of two numbers is 23 and their
LCM is 1449. If one of the numbers is 161
what is the other?
(a

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The H.C.F of two numbers is 23 and their L.C.M is 1449.If one of the numbers is 161.

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

The other number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the other number be r

We know that formula as;

$\boxed{\bf{Product\:of\:two\:numbers=H.C.F.\times L.C.M.}}}}$

A/q

$\longrightarrow\sf{161\times r=23\times 1449}\\\\\\\longrightarrow\sf{161r=33327}\\\\\\\longrightarrow\sf{r=\cancel{\dfrac{33327}{161} }}\\\\\\\longrightarrow\sf{\pink{r=207}}$

Thus;

The other number will be r = 207 .

Answer 2

$\black{\huge{\mathtt{\underline{\underline{ Given : }}}}}$

$\red{\tt{\:\:}}$

• $\red{\tt{\: HCF\:of\:two\: number = 23}}$

• $\red{\tt{\: LCM\:of\:two\: number = 1449}}$

• $\red{\tt{\: One\: number = 161}}$

$\red{\tt{\:\:}}$

$\black{\huge{\mathtt{\underline{\underline{ To\: Find :}}}}}$

$\red{\tt{\:\:}}$

• $\red{\tt{\: Other\: number }}$

$\red{\tt{\:\:}}$

$\black{\huge{\mathtt{\underline{\underline{ Solution : }}}}}$

$\red{\tt{\:\:}}$

$\green{\tt{Let,\: the\: other\:no.\:be\:x}}$

$\red{\tt{\:\:}}$

$\green{\tt{Product\: of\:two\: no. = LCM \times HCF}}$

$\green{\tt{\implies 161 \times x = 1449 \times 23}}$

$\green{\tt{\implies x = \dfrac{1449 \times 23}{161} }}$

$\green{\tt{\implies x = \dfrac{1449 \times \cancel{23}}{\cancel{161}} }}$

$\green{\tt{\implies x = \dfrac{1449 \times 1}{7} }}$

$\green{\tt{\implies x = \dfrac{1449}{7} }}$

$\green{\tt{\implies x = \dfrac{\cancel{1449}}{\cancel{7}} }}$

$\green{\tt{\implies x = 207 }}$

$\red{\tt{\:\:}}$

$\blue{\bf{ Other\: number= 207 }}$