The sum of the lcm and hcf of two numbers is 1884 and their difference is 1860. If the difference between the numbers is 12, then determine the sum of the numbers.
Answers
Answered by
1
*Plz try to solve by this method...:*
Let LCM(a,b)=x and HCF(a,b)=y
First number=a
Second number=b
x+y=2196
x-y=2172
Adding the above two equations,
2x=4368
x=2184
So,
y=12
LCM(a,b)=2184,
HCF(a,b)=12
a-b=12
a+b=?
We know that LCM(a,b)*HCF(a,b)=Product of a & b
2184*12=ab
(a-b)^2 + 4ab= (a+b)^2
(12)^2 + 4*2184*12 =(a+b)^2
(a+b)^2=104976
(a+b)=324
So,the sum of numbers will be 324.
Let LCM(a,b)=x and HCF(a,b)=y
First number=a
Second number=b
x+y=2196
x-y=2172
Adding the above two equations,
2x=4368
x=2184
So,
y=12
LCM(a,b)=2184,
HCF(a,b)=12
a-b=12
a+b=?
We know that LCM(a,b)*HCF(a,b)=Product of a & b
2184*12=ab
(a-b)^2 + 4ab= (a+b)^2
(12)^2 + 4*2184*12 =(a+b)^2
(a+b)^2=104976
(a+b)=324
So,the sum of numbers will be 324.
Similar questions