The sum and difference of the L.C.M. and H.C.F. of two numbers are 592 and 518 respectively. If the sum of the numbers is 296 , find the numbers.

Correct option is B

Solution:

Given:
1. $$\text{L.C.M.} + \text{H.C.F.} = 592$$​
2. $$\text{L.C.M.} - \text{H.C.F.} = 518$$​
3. a + b = 296 , where a and b are the two numbers.

Step-by-Step Solution:

Step 1: Solve for L.C.M. and H.C.F.

$$\text{L.C.M.} = \frac{(\text{Sum} + \text{Difference})}{2} = \frac{592 + 518}{2} = 555$$​


$$\text{H.C.F.} = \frac{(\text{Sum} - \text{Difference})}{2} = \frac{592 - 518}{2} = 37$$​

Step 2: Use the product relationship of L.C.M. and H.C.F.

$$\text{L.C.M.} \times \text{H.C.F.} = a \times b$$​

Substitute L.C.M. = 555 and H.C.F. = 37 :

$$a \times b = 555 \times 37 = 20535$$​

Step 3: Solve for the numbers a and b :
Let a and b be the roots of the quadratic equation:

$$x^2 - (\text{Sum of numbers})x + (\text{Product of numbers}) = 0$$​

Substitute a + b = 296 and $$a \times b$$​ = 20535 :

$$x^2 - 296x + 20535 = 0$$​

Step 4: Solve the quadratic equation:
Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$​

Here, a = 1 , b = -296 , and c = 20535 :

$$x = \frac{-(-296) \pm \sqrt{(-296)^2 - 4(1)(20535)}}{2(1)}$$​


$$x = \frac{296 \pm \sqrt{87616 - 82140}}{2}$$​


$$x = \frac{296 \pm \sqrt{5476}}{2}$$​


$$x = \frac{296 \pm 74}{2}$$​
Solve for both roots:

x = $$\frac{296 + 74}{2} = 185, \quad x = \frac{296 - 74}{2} = 111$$​

Final Answer:

The two numbers are 185 and 111 .