​The sum and difference of the LCM and HCF of two numbers are 682 and 638 respectively. If the sum of the two numbers is 286, find the numbers. ​

Correct option is B

Given:

The sum and difference of the LCM and HCF of two numbers are 682 and 638 respectively.

If the sum of the two numbers is 286.

Formula used:

Product of Numbers = LCM × HCF

Solution:

LCM + HCF = 682

LCM - HCF = 638

Now add,

2LCM = 1320

LCM = 660

660 + HCF = 682

HCF = 682 - 660

HCF = 22

Let 1^st number = X

2^nd number = Y

then,

Product of Numbers = LCM × HCF

X × Y = 660 × 22

X × Y = 14,520

We have, 

X + Y = 286

Now quadratic equation becomes:

​$${t^2} - (\text{X+Y})t +(\text{X×Y})$$   .......(1)

Put the values in equation (1)

​$${t^2} - (\text{286})t +14520$$ = 0

​​$${t^2} - (\text{220+66})t +14520$$ = 0 

​$${t^2} - 220t-66t +14520$$ = 0 

​$$t(t-220) -66(t-220)$$ = 0

​​$$(t-220) (t-66)$$ = 0

​$$t = 220, t = 66$$ 

Thus, the numbers are 220 and 66.

Thus, option (b) is correct.