The least common multiple of two numbers is 168 and highest
common factor of them is 12. If the difference between the numbers
is 60, what is the sum of the numbers?
no spammming
give explanation
Answers
Answer:
108
Step-by-step explanation:
HCF = 168
LCM = 12
Difference = 60
Let the numbers be 'X' and 'Y'
So, X - Y = 60
⇒ X = 60 + Y
Wkt, HCF x LCM = X x Y
⇒ 168 x 12 = ( 60 + Y )( Y ) [Plugging in all the values]
⇒ 2016 = 60Y + Y²
⇒ Y² + 60Y - 2016 = 0
Now, by using quadratic formula
⇒ Y = [-b ± √(b² - 4ac)] / 2a
⇒ Y = [ -60 ± √( 3600 + 8064) ] / 2
⇒ Y = [ -60 ± √( 11664) ] / 2
⇒ Y = [ -60 ± 108 ] / 2
By taking 'Y = [ -60 + 108 ] / 2', we have
⇒ Y = [ -60 + 108 ] / 2
⇒ Y = [ 48 ] / 2
⇒ Y = 24
By taking 'Y = [ -60 - 108 ] / 2', we have
⇒ Y = [ -60 - 108 ] / 2
⇒ Y = [ -168 ] / 2
⇒ Y = -84
Now we have to find the value of X
X = 60 + Y
⇒ X = 60 + 24 or X = 60 - 84
⇒ X = 84 or -24
So the values of X and Y could be 84 & 24 or -24 & -84
Since we usually don't choose negative values, we should take the first pair, i.e, X = 84 and Y = 24
So the sum will be 84 + 24 = 108
Hope this helped! Sorry if it is complicated
Answer:
108
Step-by-step explanation:
HCF = 168
LCM = 12
Difference = 60
Let the numbers be 'X' and 'Y'
So, X - Y = 60
⇒ X = 60 + Y
Wkt, HCF x LCM = X x Y
⇒ 168 x 12 = ( 60 + Y )( Y ) [Plugging in all the values]
⇒ 2016 = 60Y + Y²
⇒ Y² + 60Y - 2016 = 0
Now, by using quadratic formula
⇒ Y = [-b ± √(b² - 4ac)] / 2a
⇒ Y = [ -60 ± √( 3600 + 8064) ] / 2
⇒ Y = [ -60 ± √( 11664) ] / 2
⇒ Y = [ -60 ± 108 ] / 2
By taking 'Y = [ -60 + 108 ] / 2', we have
⇒ Y = [ -60 + 108 ] / 2
⇒ Y = [ 48 ] / 2
⇒ Y = 24
By taking 'Y = [ -60 - 108 ] / 2', we have
⇒ Y = [ -60 - 108 ] / 2
⇒ Y = [ -168 ] / 2
⇒ Y = -84
Now we have to find the value of X
X = 60 + Y
⇒ X = 60 + 24 or X = 60 - 84
⇒ X = 84 or -24
So the values of X and Y could be 84 & 24 or -24 & -84
Since we usually don't choose negative values, we should take the first pair, i.e, X = 84 and Y = 24
So the sum will be 84 + 24 = 108
Hope this helped! Sorry if it is complicated