Question

The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to

• A. \(76\)
• B. \(78\)
• C. \(73\)
• D. \(71\)

Hint:

In pipes and cisterns problems:
Convert ratios into actual rates using any one known filling time.
Add rates (not times) when pipes work together.
Do all calculations in hours first, and convert to minutes only at the end to avoid mistakes.

Answer 1

The Correct Option is C

To solve this problem, we first need to determine the individual water flow rate of each pipe and then calculate the combined flow rate to find out how long it will take to fill the tank when all three pipes are used simultaneously.

1. Given that the rates of water flow through pipes A, B, and C are in the ratio 4:9:36. 
2. Let the water flow rates of A, B, and C be \(4x\), \(9x\), and \(36x\) respectively.
3. Pipe A can fill the tank in 15 hours, which implies:

Pipe A's rate

=

\(\frac{1}{15}\) of the tank per hour

1. Since the flow rate of A is \(4x\), equate the amount of water filled in one hour by A to its rate: \(4x = \frac{1}{15}\).
2. Solving for \(x\), we get: \(x = \frac{1}{60}\)
3. Now, calculate the flow rates for pipes B and C:

Pipe B's rate	=	\(9x = 9 \times \frac{1}{60} = \frac{9}{60} = \frac{3}{20}\) of the tank per hour
Pipe C's rate	=	\(36x = 36 \times \frac{1}{60} = \frac{36}{60} = \frac{3}{5}\) of the tank per hour

1. Calculate the combined rate of all three pipes:
   • Combined rate = \(4x + 9x + 36x = 49x\)
   • Substitute \(x = \frac{1}{60}\):

Combined rate

=

\(49 \times \frac{1}{60} = \frac{49}{60}\)of the tank per hour

1. Time required to fill the tank when using all three pipes: \(\frac{1}{\frac{49}{60}} = \frac{60}{49}\)hours.
2. Convert hours into minutes:
   • \(\frac{60}{49} \times 60 = \frac{3600}{49} \approx 73\) minutes.

Hence, the nearest time required to fill up the entire tank using all three pipes is 73 minutes.

Answer 2

Step 1: Convert the rate ratio into actual rates. Let the rates of flow (in tank/hour) of pipes A, B and C be: \[ \text{A : B : C} = 4 : 9 : 36. \] Suppose the common factor is \(k\). Then: \[ \text{Rate of A} = 4k. \] Given that pipe A alone can fill the empty tank in 15 hours, its rate is: \[ \text{Rate of A} = \frac{1}{15} \text{ tank/hour}. \] So: \[ 4k = \frac{1}{15} \quad \Rightarrow \quad k = \frac{1}{60}. \] Therefore: \[ \text{Rate of A} = 4k = \frac{4}{60} = \frac{1}{15}, \] \[ \text{Rate of B} = 9k = \frac{9}{60} = \frac{3}{20}, \] \[ \text{Rate of C} = 36k = \frac{36}{60} = \frac{3}{5}. \]
Step 2: Find the combined rate of all three pipes. \[ \text{Combined rate} = \frac{1}{15} + \frac{3}{20} + \frac{3}{5}. \] Take LCM of 15, 20 and 5, which is 60: \[ \frac{1}{15} = \frac{4}{60}, \quad \frac{3}{20} = \frac{9}{60}, \quad \frac{3}{5} = \frac{36}{60}. \] So: \[ \text{Combined rate} = \frac{4}{60} + \frac{9}{60} + \frac{36}{60} = \frac{49}{60} \text{ tank/hour}. \]
Step 3: Calculate the total time to fill the tank. \[ \text{Time taken} = \frac{1}{\text{Combined rate}} = \frac{1}{\frac{49}{60}} = \frac{60}{49} \text{ hours}. \] Convert this into minutes: \[ \frac{60}{49} \times 60 = \frac{3600}{49} \text{ minutes}. \] Approximate: \[ 49 \times 73 = 3577,\quad 49 \times 74 = 3626. \] So: \[ \frac{3600}{49} \approx 73.47 \text{ minutes}, \] which is closest to \(73\) minutes. \[ \boxed{73} \]