Question

The LCM of two numbers is 168 and hcf of them is 12. if the difference between the no's is 60.What is the sum of the numbers.

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Answer 1

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LCM of a, b = 168

HCF of a, b = 12

a - b = 60

a + b =??

= (a+b) ² = (k) ² = a² + b² +2ab = k²

= (a + b) ² = (a + b) =a² +ab + ba + b² = a²+2ab +b²

= (a- b)² = (60)²

= a² + b² - 2ab = 3600

= a²+b² = 3600 + 2ab [(a+b) ² = (k) ²]

=k² = 3600+2ab+2

=k² = 3600+4ab

=k²= 3600 + 4×168×12[LCM of a, b = 168, HCF of a,

b=12]

=k² = 11664

$= k = \sqrt{11664 } \\ = k = 108$

Answer 2

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The LCM of two numbers is 168 and hcf of them is 12. if the difference between the no's is 60.What is the sum of the numbers.

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Lcm ×hcf = product of two numbers

Hence 168×12=x×y

2016=xy----(1)

Given  x-y=60----(2)

Hence x×(x-60)=2016

So x^2-60x=2016

Hence x^2-60x-2016=0

So x= 60+/-(3600+8064)^0.5/2

So x=(60+/-108)/2

Hence x=84 or -24 so y=24 or y=-84

Hence the sum of numbers =108

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