The LCM of two numbers is 126 times their HCF. If their HCF is 7 and the difference between the two numbers is five times the HCF, then the sum of the two numbers is:

Correct option is C

Given:

The HCF of the two numbers is 7.

The LCM of the two numbers is 126 times their HCF.

The difference between the two numbers is 5 times the HCF.

We are required to find the sum of the two numbers.

Concept Used:

The relationship between the HCF, LCM, and the two numbers N1 and N2 is as follows:

LCM × HCF = N1 × N2

Solution: 

Let the two numbers be N1 = 7x  and N2 = 7y, where x and y are co-prime.

Now, 

LCM = 126 × HCF = 126 × 7 = 882

we know that

N1 − N2 = 5 × HCF = 5 × 7 = 35

Then:

LCM = 7xy and N1 − N2 = 7(x − y)

Given:

882 = 7xy

xy = $$\frac{882}{7}$$​ = 126

and:

35 = 7(x - y)

x - y = $$\frac{35}{7}$$​ = 5

We need to find two co-prime numbers xx and yy such that:

xy = 126 and x − y = 5

The factor pairs of 126 are:

(1, 126) → Difference = 125

(2, 63) → Difference = 61

(3, 42) → Difference = 39

(6, 21) → Difference = 15

(9, 14) → Difference = 5 (This pair satisfies the condition).

Thus, x = 14 and y = 9

the two numbers:

N1 = 7x =7 × 14 = 98

N2 = 7y = 7 × 9 = 63

N1 + N2 = 98 + 63 = 161

Thus, the sum of the two numbers is 161.