The HCF of two numbers is 18 and their product is 5832. If the numbers lie between 30 and 200, then the sum of their reciprocals is: 

Given:

HCF of two numbers = 18

Product of two numbers = 5832

The numbers lie between 30 and 200.

Formula Used:

For two numbers 'a' and 'b', Product (a × b) = HCF(a, b) × LCM(a, b)

If HCF(a, b) = H, then a = Hx and b = Hy, where x and y are coprime integers.

Sum of reciprocals = \(\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}\)

Calculation:

Let the two numbers be 'a' and 'b'.

We are given HCF(a, b) = 18 and a × b = 5832.

a × b = HCF(a, b) × LCM(a, b)

5832 = 18 × LCM(a, b)

⇒ LCM(a, b) = \(\frac{5832}{18}\)

⇒ LCM(a, b) = 324

Since HCF(a, b) = 18, we can write the numbers as:

a = 18x

b = 18y

where x and y are coprime integers (HCF(x, y) = 1).

Substitute these into the product equation:

(18x) × (18y) = 5832

324xy = 5832

⇒ xy = \(\frac{5832}{324}\)

⇒ xy = 18

Now, find pairs of coprime integers (x, y) whose product is 18:

If x = 1, y = 18. HCF(1, 18) = 1. (Valid pair)

If x = 2, y = 9. HCF(2, 9) = 1. (Valid pair)

If x = 3, y = 6. HCF(3, 6) = 3. (Not coprime, so not valid)

Let's check the numbers 'a' and 'b' for each valid (x, y) pair against the condition that they lie between 30 and 200.

Case 1: (x, y) = (1, 18)

a = 18 × 1 = 18

b = 18 × 18 = 324

Here, a = 18 is not > 30, and b = 324 is not < 200. So, this pair is not suitable.

Case 2: (x, y) = (2, 9)

a = 18 × 2 = 36

b = 18 × 9 = 162

Here, a = 36 (30 < 36 < 200) and b = 162 (30 < 162 < 200). This pair satisfies all conditions.

So, the two numbers are 36 and 162.

Sum of reciprocals = \(\frac{1}{a} + \frac{1}{b}\)

⇒ Sum of reciprocals = \(\frac{1}{36} + \frac{1}{162}\)

To add these fractions, find the LCM of 36 and 162.

36 = 2² × 3² = 4 × 9

162 = 2 × 3⁴ = 2 × 81

LCM(36, 162) = 2² × 3⁴ = 4 × 81 = 324

⇒ Sum of reciprocals = \(\frac{9}{324} + \frac{2}{324}\)

⇒ Sum of reciprocals = \(\frac{9 + 2}{324}\)

⇒ Sum of reciprocals = \(\frac{11}{324}\)

∴ The sum of their reciprocals is \(\frac{11}{324}\).