Question

9 The HCF of two numbers is 58 and their LCM is 54810. If one of the numbers
1566. find the other number.​

Answer 1

Answer:

HCF × LCM =one number × other number ( be x)

58 × 54810 = 1566× x

(58×54810)/1566= x

2030 = x

hence, other number = x = 2030

Answer 2

Given:

• HCF of two numbers is 58
• LCM of those numbers is 54810
• One number is 1566

To find :

• Other number?

Solution:

Here, we have given that, there are two numbers (Let suppose one is $\small{\tt{X}}$ and another is $\small{\tt{Y}}$) whose LCM is 54810 and HCF is 58. And,One number is (Let suppose it is $\small{\tt{X}}$) 1566.

We know that :

• $\large{\boxed{\sf{LCM×HCF\:=\:{\tt{X × Y}}}}}$

Where,

• LCM = 54810
• HCF = 58
• $\small{\tt{X}}$ = 1566
• $\small{\tt{Y}}$= ?

Let put known value :-

LCM × HCF = $\small{\tt{X×Y}}$

→ 54810 × 58 = 1566 × $\small{\tt{Y}}$

→ 3178980 = 1566 × $\small{\tt{Y}}$

→ $\small{\tt{Y}}$ = 3178980/1566

→ $\small{\tt{Y}}$ = 2030

Therefore,

• $\large{\tt{Another\: number\: is\:2030}}$