The sum of two numbers is 36 and their H.C.F . and L.C.M. are 3 and 105 respectively . The sum of the peciprocals of two numbers is

Options

A. `(2)/(35)`
B. `(3)/(25)`
C. `(4)/(35)`
D. `(2)/(25)`

Answer

To solve the problem step by step, we need to find the two numbers whose sum is 36, and their H.C.F. (Highest Common Factor) is 3 and L.C.M. (Lowest Common Multiple) is 105. We will also find the sum of their reciprocals.

### Step 1: Set Up the Equations
Let the two numbers be \( x \) and \( y \). According to the problem:
1. \( x + y = 36 \)
2. \( \text{H.C.F.}(x, y) = 3 \)
3. \( \text{L.C.M.}(x, y) = 105 \)

### Step 2: Use the Relationship Between H.C.F. and L.C.M.
We know that:
\[ \text{H.C.F.} \times \text{L.C.M.} = x \times y \]
Substituting the values we have:
\[ 3 \times 105 = x \times y \]
\[ 315 = x \times y \]

### Step 3: Express \( y \) in Terms of \( x \)
From the first equation, we can express \( y \) as:
\[ y = 36 - x \]

### Step 4: Substitute \( y \) in the Product Equation
Now substitute \( y \) in the equation \( x \times y = 315 \):
\[ x(36 - x) = 315 \]
Expanding this gives:
\[ 36x - x^2 = 315 \]
Rearranging it leads to:
\[ x^2 - 36x + 315 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we will solve the quadratic equation \( x^2 - 36x + 315 = 0 \) using the factorization method. We need two numbers that multiply to \( 315 \) and add to \( 36 \).

After testing factors of \( 315 \), we find:
- \( 15 \) and \( 21 \) work because \( 15 + 21 = 36 \) and \( 15 \times 21 = 315 \).

Thus, we can factor the equation as:
\[ (x - 15)(x - 21) = 0 \]

### Step 6: Find the Values of \( x \) and \( y \)
Setting each factor to zero gives us:
1. \( x - 15 = 0 \) → \( x = 15 \)
2. \( x - 21 = 0 \) → \( x = 21 \)

Thus, the two numbers are \( 15 \) and \( 21 \).

### Step 7: Calculate the Sum of the Reciprocals
Now, we need to find the sum of the reciprocals of \( x \) and \( y \):
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{15} + \frac{1}{21} \]

### Step 8: Find a Common Denominator
The least common multiple of \( 15 \) and \( 21 \) is \( 105 \). Therefore, we can rewrite the fractions:
\[ \frac{1}{15} = \frac{7}{105}, \quad \frac{1}{21} = \frac{5}{105} \]
Now, adding these gives:
\[ \frac{7}{105} + \frac{5}{105} = \frac{12}{105} \]

### Step 9: Simplify the Result
Now we simplify \( \frac{12}{105} \):
\[ \frac{12}{105} = \frac{4}{35} \]

### Final Answer
The sum of the reciprocals of the two numbers is:
\[ \frac{4}{35} \]

---