Hose A runs at a constant rate and can fill a 11,000 gallon

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Hose A runs at a constant rate and can fill an 11,000-gallon pool in 44 hours. How much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates?

1) Both Hose A and Hose B can fill the same fraction of the pool in one hour.
2) It takes Hose B twice as long to fill the pool as it takes Hose A and Hose B running simultaneously to fill the pool.

The OA is D
Source: — Data Sufficiency |

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by deloitte247 » Sun Nov 17, 2019 11:51 am
Volume of pool = 11,000 gallon
Rate of Hose A = 11,000 gallon per 44 hours
in 1 hour, it will be = 11,000 / 44 = 250 gallons per hour.
Now, how much less time would it take to fill the pool if Hose A and Hose B ran simultaneously at their respective constant rates.
Statement 1: Both Hose A and Hose B can fill the same fraction of the pool in 1 hour.
If hose A can fill 250 gallons in 1 hour, hose B will also fill 250 gallons in 1 hour.
Both hoses working simultaneously will fill 250+250 = 500 gallons in 1 hour.
If both hoses fill 500 gallons in 1 hour, the time taken to fill 11,000 gallons = 11,000 / 500 = 22 hours
Hose A working alone = 44 hours
Both hoses = 22 hours
Time difference = 44 - 22 = 22 hours.
Therefore, it takes 22 hours less for both hoses to fill the pool. Thus, statement 1 is SUFFICIENT.

Statement 2: It takes hose B twice as long to fill the pool as it takes hose A and hose B running simultaneously.
The time taken by hose B = 2 * time taken by hose A and B.
If hose A and B filled up the pool in 'x' hours, hose B alone will fill it up in 2x hours.
This means that the workdone by hose B when both hoses are working simultaneously = 1/2 0f total workdone.
Hence, hose A filled half of the gallon and hose B filled the remaining half.
If hose A fills 250 gallons per hour, then hose B also fills 250 gallons per hour.
This leads back to statement 1 which tells us that it takes 22 hours less for both hoses to fill the pool. Hence, statement 2 is also SUFFICIENT.

Answer = option D