The frequency of `B` is `3%` greater than that of `A` . The frequency of `C is 2%` less than that of `A` . If `B and C` produce `8 beats//s`, then the frequency of `A` is

Options

A. `136 Hz`
B. `168 Hz`
C. `164 Hz`
D. `160 Hz`

Explanation

To solve the problem, we need to find the frequency of A given the relationships between the frequencies of A, B, and C, and the number of beats produced by B and C.

### Step-by-Step Solution:

1. **Define the Frequencies**:
Let the frequency of A be \( f_A \).
- The frequency of B is 3% greater than A:
\[ f_B = f_A + 0.03 f_A = 1.03 f_A \]
- The frequency of C is 2% less than A:
\[ f_C = f_A - 0.02 f_A = 0.98 f_A \]

2. **Understand the Beat Frequency**:
The beat frequency produced by two waves is given by the absolute difference of their frequencies:
\[ \text{Beat frequency} = |f_B - f_C| \]
According to the problem, B and C produce 8 beats per second:
\[ |f_B - f_C| = 8 \]

3. **Substituting the Frequencies**:
Substitute the expressions for \( f_B \) and \( f_C \) into the beat frequency equation:
\[ |1.03 f_A - 0.98 f_A| = 8 \]

4. **Simplifying the Equation**:
Simplify the left side:
\[ |(1.03 - 0.98) f_A| = 8 \]
This simplifies to:
\[ |0.05 f_A| = 8 \]

5. **Solving for \( f_A \)**:
Since frequency cannot be negative, we can drop the absolute value:
\[ 0.05 f_A = 8 \]
Now, solve for \( f_A \):
\[ f_A = \frac{8}{0.05} = 160 \text{ Hz} \]

### Final Answer:
The frequency of A is \( 160 \text{ Hz} \).