Question

please help me today is my exam​

Answer 1

Answer:

Option(3)

Step-by-step explanation:

Let the LCM be x and HCF be y.

(i)

Sum of Squares of LCM and HCF is 3609.

x² + y² = 3609

(ii)

LCM is 57 times more than their HCF.

x = y + 57

Substitute (ii) in (i), we get

⇒ (y + 57)² + y² = 3609

⇒ y² + 3249 + 114y + y² = 3609

⇒ 2y² + 114y - 360 = 0

⇒ y² + 57y - 180 = 0

⇒ y² + 60y - 3y - 180 = 0

⇒ y(y + 60) - 3(y + 60) = 0

⇒ (y - 3)(y + 60) = 0

⇒ y = 3, -60{It cannot be negative}

⇒ y = 3

Substitute y = 3 in (ii), we get

⇒ x = y + 57

⇒ x = 60

Now,

Product of two numbers = x * y

                                         = 60 * 3

                                   = 180.

Hope it helps!

Answer 2

given that the LCM of two numbers is 57 more than their HCF.

Let their LCM be x

then their HCF would be x-57

 according to the qstn,

x² +(x-57)² = 3609

x²+x²+3249-114x = 3609

2x²-114x -360 = 0

⇒ x²-57x -180 = 0

on solving we get

x= 60 or 3

3 is not possible as the LCM of any two numbers ia always greater than their HCF.

∴LCM = 60 and HCF = 60-57 = 3

LCM*HCF= product of two numbers

60*3 = 180