Find the appropriate relation for quantity 1 and quantity 2 in the following question :
An artificial kund is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the kund at the same time during which the kund is filled by the third pipe alone. The second pipe fills the kund 5 hours faster than the first pipe and 4 hours slower than the third pipe.
Quantity 1 : The time required by the first pipe ?
Quantity 2 : Time taken by all three pipes to fill the kund simultaneously

Options

A. A)Quantity 1 `gt` Quantity 2
B. B)Quantity 1 `ge` Quantity 2
C. C)Quantity 1 `lt` Quantity 2
D. D)Quantity 1 `le` Quantity 2

Answer

To solve the problem step by step, we will denote the time taken by each pipe to fill the kund as follows:

1. Let the time taken by the third pipe (Pipe C) be \( T \) hours.
2. The second pipe (Pipe B) fills the kund 4 hours slower than Pipe C, so the time taken by Pipe B is \( T + 4 \) hours.
3. The first pipe (Pipe A) fills the kund 5 hours slower than Pipe B, so the time taken by Pipe A is \( T + 4 + 5 = T + 9 \) hours.

Now, according to the problem, the time taken by the first two pipes (A and B) working together is equal to the time taken by the third pipe (C) alone.

### Step 1: Determine the rates of filling for each pipe
The rate of filling for each pipe can be expressed as:
- Rate of Pipe A = \( \frac{1}{T + 9} \)
- Rate of Pipe B = \( \frac{1}{T + 4} \)
- Rate of Pipe C = \( \frac{1}{T} \)

### Step 2: Set up the equation based on the given condition
Since Pipes A and B together fill the kund in the same time that Pipe C fills it alone, we can write the equation:
\[ \frac{1}{T + 9} + \frac{1}{T + 4} = \frac{1}{T} \]

### Step 3: Solve the equation
To solve this equation, we can find a common denominator and simplify:
\[ \frac{(T + 4) + (T + 9)}{(T + 9)(T + 4)} = \frac{1}{T} \]
This simplifies to:
\[ \frac{2T + 13}{(T + 9)(T + 4)} = \frac{1}{T} \]

Cross-multiplying gives:
\[ T(2T + 13) = (T + 9)(T + 4) \]
Expanding both sides:
\[ 2T^2 + 13T = T^2 + 13T + 36 \]
Simplifying:
\[ 2T^2 + 13T - T^2 - 13T - 36 = 0 \]
This simplifies to:
\[ T^2 - 36 = 0 \]
Factoring gives:
\[ (T - 6)(T + 6) = 0 \]
Thus, \( T = 6 \) (we discard \( T = -6 \) since time cannot be negative).

### Step 4: Calculate the time for each pipe
Now we can find the time taken by each pipe:
- Time taken by Pipe C (T) = 6 hours
- Time taken by Pipe B = \( T + 4 = 10 \) hours
- Time taken by Pipe A = \( T + 9 = 15 \) hours

### Step 5: Calculate Quantity 1 and Quantity 2
- **Quantity 1**: Time required by the first pipe (Pipe A) = 15 hours.
- **Quantity 2**: Time taken by all three pipes to fill the kund simultaneously.

To find the time taken by all three pipes together, we calculate their combined rate:
\[ \text{Combined Rate} = \frac{1}{15} + \frac{1}{10} + \frac{1}{6} \]
Finding a common denominator (30):
\[ \text{Combined Rate} = \frac{2}{30} + \frac{3}{30} + \frac{5}{30} = \frac{10}{30} = \frac{1}{3} \]
Thus, the time taken by all three pipes together is:
\[ \text{Time} = \frac{1}{\frac{1}{3}} = 3 \text{ hours} \]

### Conclusion
Now we compare Quantity 1 and Quantity 2:
- Quantity 1 = 15 hours
- Quantity 2 = 3 hours

Thus, we conclude that:
**Quantity 1 is greater than Quantity 2.**