The harmonic mean of two quantities a and b is defined as `(2ab)/(a+b)`. If the harmonic mean of the quantities x and 12 is one less than the average of the same two quantities, which of the following can be the value of the sum of digits in x?

Options

A. 1
B. 2
C. 4
D. 5

Answer

To solve the problem, we need to find the value of \( x \) such that the harmonic mean of \( x \) and \( 12 \) is one less than the average of \( x \) and \( 12 \).

### Step 1: Write the formulas for harmonic mean and average
The harmonic mean (HM) of two quantities \( a \) and \( b \) is given by:
\[ HM = \frac{2ab}{a+b} \]
The average (A) of two quantities \( a \) and \( b \) is given by:
\[ A = \frac{a+b}{2} \]

### Step 2: Substitute the values into the formulas
For our specific case, we have \( a = x \) and \( b = 12 \). Thus, the harmonic mean becomes:
\[ HM = \frac{2 \cdot x \cdot 12}{x + 12} = \frac{24x}{x + 12} \]
And the average becomes:
\[ A = \frac{x + 12}{2} \]

### Step 3: Set up the equation based on the problem statement
According to the problem, the harmonic mean is one less than the average:
\[ HM = A - 1 \]
Substituting the expressions for HM and A, we get:
\[ \frac{24x}{x + 12} = \frac{x + 12}{2} - 1 \]

### Step 4: Simplify the right-hand side
First, simplify the right-hand side:
\[ \frac{x + 12}{2} - 1 = \frac{x + 12 - 2}{2} = \frac{x + 10}{2} \]
Now we have the equation:
\[ \frac{24x}{x + 12} = \frac{x + 10}{2} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 24x \cdot 2 = (x + 10)(x + 12) \]
This simplifies to:
\[ 48x = x^2 + 12x + 10x + 120 \]
\[ 48x = x^2 + 22x + 120 \]

### Step 6: Rearrange the equation
Rearranging the equation gives:
\[ 0 = x^2 + 22x + 120 - 48x \]
\[ 0 = x^2 - 26x + 120 \]

### Step 7: Factor the quadratic equation
We need to factor \( x^2 - 26x + 120 \). We look for two numbers that multiply to \( 120 \) and add to \( -26 \):
\[ (x - 20)(x - 6) = 0 \]

### Step 8: Solve for \( x \)
Setting each factor to zero gives:
\[ x - 20 = 0 \quad \Rightarrow \quad x = 20 \]
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]

### Step 9: Find the sum of the digits in \( x \)
Now we need to find the sum of the digits for both possible values of \( x \):
- For \( x = 20 \): The sum of digits is \( 2 + 0 = 2 \).
- For \( x = 6 \): The sum of digits is \( 6 \).

### Conclusion
The possible values for the sum of digits in \( x \) are \( 2 \) and \( 6 \). Since the problem asks for which of the following can be the value of the sum of digits in \( x \), the answer is:
\[ \text{Sum of digits can be } 2 \text{ or } 6. \]