The sum of two numbers is 36 and the HCF and LCM of these numbers are 3 and 105 respectively.  Find the sum of their reciprocals.

\(\frac{4}{35}\)

Product of other factors = \(\frac{LCM}{HCF}\) = \(\frac{105}{3}\) = 35

Sum of other factors = \(\frac{Sum\;of\;numbers}{HCF}\) = \(\frac{36}{3}\) = 12

Then, only 7, 5 satisfy the condition.

Numbers are: -

7 × 3 = 21

5 × 3 = 15

Reciprocal of numbers = \(\frac{1}{21}\) + \(\frac{1}{15}\) = \(\frac{5 + 7}{105}\) = \(\frac{12}{105}\) = \(\frac{4}{35}\)