The rate of flow of water (in litre per min) of three pipes are 2, N and 3, where `2 lt N lt 3`. The lowest and the highest flow rates are both decreased by a certain quantity `x`, while the intermediate rate is left unchanged. If the reciprocals of the three flow rates, in the order given above, are in arithmetic progression both before and after the change, then what is the quantity x(in litre per min)? (Negative flow rates indicate that the pipes act as emptying pipes instead of filling pipes.

Options

A. A) 2.6
B. B) `-3.2`
C. C) 3.8
D. D) `-2.6`

Answer

To solve the problem, we need to determine the quantity \( x \) under the given conditions. Let's break it down step by step.

### Step 1: Define the flow rates
We have three flow rates:
- Pipe 1: \( 2 \) liters/min
- Pipe 2: \( n \) liters/min (where \( 2 < n < 3 \))
- Pipe 3: \( 3 \) liters/min

### Step 2: Write the reciprocals of the flow rates
Before any changes, the reciprocals of the flow rates are:
- \( \frac{1}{2} \)
- \( \frac{1}{n} \)
- \( \frac{1}{3} \)

### Step 3: Establish the condition for arithmetic progression (AP)
For three numbers \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]
Applying this to our reciprocals:
\[ 2 \cdot \frac{1}{n} = \frac{1}{2} + \frac{1}{3} \]

### Step 4: Simplify the equation
Calculating the right side:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]
So we have:
\[ \frac{2}{n} = \frac{5}{6} \]

### Step 5: Solve for \( n \)
Cross-multiplying gives:
\[ 2 \cdot 6 = 5n \]
\[ 12 = 5n \]
\[ n = \frac{12}{5} = 2.4 \]

### Step 6: Write the new flow rates after decreasing by \( x \)
After decreasing the lowest and highest flow rates by \( x \):
- New Pipe 1: \( 2 - x \)
- Pipe 2: \( n = 2.4 \) (unchanged)
- New Pipe 3: \( 3 - x \)

### Step 7: Write the new reciprocals
The new reciprocals are:
- \( \frac{1}{2 - x} \)
- \( \frac{1}{2.4} \)
- \( \frac{1}{3 - x} \)

### Step 8: Establish the condition for the new reciprocals to be in AP
Using the same condition for AP:
\[ 2 \cdot \frac{1}{2.4} = \frac{1}{2 - x} + \frac{1}{3 - x} \]

### Step 9: Simplify the new equation
Calculating the left side:
\[ 2 \cdot \frac{1}{2.4} = \frac{2}{2.4} = \frac{5}{6} \]

Now, we need to simplify the right side:
\[ \frac{1}{2 - x} + \frac{1}{3 - x} = \frac{(3 - x) + (2 - x)}{(2 - x)(3 - x)} = \frac{5 - 2x}{(2 - x)(3 - x)} \]

### Step 10: Set the two sides equal
Now we have:
\[ \frac{5}{6} = \frac{5 - 2x}{(2 - x)(3 - x)} \]

### Step 11: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 5(2 - x)(3 - x) = 30 \]

### Step 12: Expand and simplify
Expanding the left side:
\[ 5(6 - 5x + x^2) = 30 \]
\[ 30 - 25x + 5x^2 = 30 \]
Subtracting 30 from both sides:
\[ 5x^2 - 25x = 0 \]

### Step 13: Factor the equation
Factoring out \( 5x \):
\[ 5x(x - 5) = 0 \]

### Step 14: Solve for \( x \)
Setting each factor to zero gives:
1. \( 5x = 0 \) → \( x = 0 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

Since \( x \) must be less than both flow rates (2 and 3), we discard \( x = 5 \).

### Conclusion
Thus, the only valid solution is:
\[ x = 0 \]