Three quantities A, B, C are such that AB = KC, where K is a constant. When A is kept constant, B varies directly as C, When B is kept constant, A varies directly as C and when C is kept constant, A varies inversely as B.
Initially, A was at 5 and A : B : C was 1 : 3 : 5. Find the value of A when B equals 9 at constant C.

Options

A. 8
B. `8.33`
C. 9
D. `9.5`

Answer

To solve the problem step by step, we will follow the relationships given in the question and use the initial values provided.

### Step 1: Understand the Ratios
We are given that the initial ratios of A, B, and C are 1:3:5. Since A is 5, we can find the values of B and C.

- Let A = 1x, B = 3x, and C = 5x.
- If A = 5, then:
- 1x = 5 ⇒ x = 5
- B = 3x = 3 * 5 = 15
- C = 5x = 5 * 5 = 25

So, we have:
- A = 5
- B = 15
- C = 25

### Step 2: Apply the Relationships
We know that:
- When C is constant, A varies inversely as B. This means:
\[ A \cdot B = k \quad \text{(where k is a constant)} \]

Using the initial values:
\[ A_1 \cdot B_1 = 5 \cdot 15 = 75 \]

### Step 3: Find A when B = 9
Now, we need to find the value of A when B = 9 and C is constant. We can use the relationship derived above:
\[ A_2 \cdot B_2 = A_1 \cdot B_1 \]
Substituting the known values:
\[ A_2 \cdot 9 = 75 \]

### Step 4: Solve for A
Now, we can solve for A:
\[ A_2 = \frac{75}{9} = \frac{25}{3} \approx 8.33 \]

### Final Answer
Thus, the value of A when B equals 9 at constant C is:
\[ A \approx 8.33 \]