Question

A water tanker can be filled by 2 pipes A & B separately in 16 min & 32 min respectively. The outlet of the tanker is partially open and it can empty the full tanker completely in 1 hour 4 min. Pipes A & B were opened simultaneously for 9 min to fill the tanker but the partially open outlet was not closed. After 9 min the pipes A & B were closed and the tanker then went to Mohan's house, 6 km away to deliver water. If the tanker moved at a constant speed of 36 km/hr, approximately what percentage of the tanker was full, when it reached Mohan's house? 
1. 55%
2. 63%
3. 58%
4. None of the options

Answer 1

Correct Answer - Option 1 : 55%

Calculation:

Consider the outlet of the tanker as pipe C that works negatively and empties the tank in 1 hour 4 min or 64 minutes. Now, take the LCM of time taken by pipes A, B, and C to complete the work.

LCM(16, 32, 64) = 64

Let the total capacity of the tank be 64 litres. Then, the efficiency of tank A will be 64/16 = 4 ltr/min. Similarly, the efficiencies of tanks B and C will be 2 ltr/min and -1 ltr/min respectively. 

Now, when all three pipes are open the tank will be filled up to 5 litres in a minute. It is given that the pipe A, B, and C were opened for 9 minutes. So, the tank was filled up to 45 litres in 9 minutes. Calculate the time taken by the truck to cover 6 km.

⇒ T = distance/speed

⇒ T = 6/36 hours

⇒ T = 1/6 hours

⇒ T = 10 minutes

So, for the 10 minutes of the journey, only pipe C is opened which will drain out 10 litres of water in that period of time. So, when the tanker will reach Mohan's house it will contain 45 - 10 = 35 litres of water.

Calculate the percentage of tanker that was full.

⇒ Percentage of tanker that was full = (35/64) × 100

⇒ Percentage of tanker that was full = 54.64

⇒ Percentage of tanker that was full ≈ 55%

The total work done by a person can be obtained by multiplying the period of time for which it works with the efficiency with which he works for that period of time.