shreerajp99 wrote:

Hose A runs at a constant rate and can fill a 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 concepts being tested here are:

1. Rates are additive. Rate of work of A and B working together is the sum of rates of A and B. i.e.

Rate of work of A = R_A

Rate of work of B = R_B

Rate of work of both A and B working together = R_A + R_B

2. Work done = Rate * Time taken

So if A and B do the same work, R_A*T_A = R_B*T_B

R_A/R_B = T_B/T_A

Ratio of their rates will be inverse of ratio of their time taken.

We are given the work done and time taken for hose A so we can find the rate of work for Hose A. (which is 11000/44 = 250 gallons/hr just for clarity)

To find the time taken when both hoses work together, we need to find their combined rate of work. Hence we need to know the rate of work of Hose B too.

1.Both Hose A and Hose B can fill the same fraction of the pool in one hour.

This tells us that their rate of work is the same. Rate of work of hose B = 250 gallons/hr too. When they work together, they will take half the usual time so they will take 22 hrs i.e. 22 hrs less. Sufficient.

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.

Time taken by B = 2*Time taken together

Rate of B = 1/2 * Rate of working together (since ratio of rates is inverse)

So Rate of B = Rate of AThis boils down to statement 1 and hence is sufficient too.

Answer (D)

I have 1 doubt how Rb = 1/2 (Ra + Rb) converts to Rb = Ra in next step. (I dont think any where Rb+Ra= 2 Ra is given anywhere in prompt).