Pump A can empty a pool in A minutes, and pump B can empty

Question

Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins. Beginning from the time pump A starts, how many minutes will it take to empty the pool?

A. \(\frac{A+B-1}{2}\)

B. \(\frac{A(B+1)}{A+B}\)

C. \(\frac{AB}{(A+B)}\)

D. \(\frac{AB}{(A+B)} -1\)

E. \(\frac{A(B-1)}{(A+B)}\)

A.  \frac{A+B-1}{2}

B.  \frac{A(B+1)}{A+B}

C.  \frac{AB}{(A+B)}

D.  \frac{AB}{(A+B)} -1

E.  \frac{A(B-1)}{(A+B)}

Bunuel · Accepted Answer

The rate of pump A is  \frac{1}{A}  pool/minute;
The rate of pump B is  \frac{1}{B}  pool/minute.

Their combined rate is  \frac{1}{A} + \frac{1}{B} = \frac{A+B}{AB}  pool/minute.

In one minute that A works alone it does (rate)*(time) =  \frac{1}{A} * 1= \frac{1}{A}  part of the job, thus  1-\frac{1}{A}=\frac{A-1}{A}  part of the job is remaining to be done by A and B together.

To do  \frac{A-1}{A}  of the job both pumps working together will need (time) = (job)/(combined rate) =  \frac{A-1}{A}*\frac{AB}{A+B}=\frac{(A-1)B}{A+B} .

So, the total time is  1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B} .

Answer: B.

KarishmaB · Answer

Taking numbers is a great strategy but try to take easier numbers. Say, 1, 0 etc as long as you meet all conditions mentioned. 
I would take A = B = 1 (since there are no symmetric options in A and B such as A/(A+B) and B(A + B), I can take both A and B equal)
Now I need to do no calculations. A starts and works for a minute. In that time, the pool is empty. B joins but has nothing to do. Total time taken to empty the pool is 1 min. 
Only option (B) gives 1.

Bunuel · Answer

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jlgdr · Answer

I did this question using smart numbers

A = 6
B = 3

Therefore in 1 minute A did 1/6 of the job. 
5/6 of the job remains

Together A and B work at a rate of 1/2.

Thus 1/2 (x) = 5/6

x = 5/3 minutes

Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes

Replacing we can see that in (B)

(6)(4) / 9 = 8/3

Therefore it is the correct answer choice

Gimme some freaking Kudos!
Cheers
J

GyanOne · Answer

Method 1: 
The time taken by both pumps working together to do the work = AB/(A+B) minutes
Now, since Pump A has already worked for 1 minute, 1*(1/A) of the work has already been done. 
Time taken to do the remaining work =  *(1-1/A) = B(A-1)/(A+B) minutes
Total time taken from the start =   + 1 = A*(B+1)/(A+B)
Option (B).

Method 2: 
Let t be the time taken from the start. Then pump A works for t minutes to finish the work and pump B for (t-1) minutes
=> t(1/A) + (t-1)(1/B) = 1 
=> t = A*(B+1)/(A+B)
Option (B).

Method 3: 
We know that the time taken to do the work by both pumps working together right from the beginning is AB/(A+B) minutes.
Lets flip the question around to say that the two pumps worked together to do the work, except pump B was off in the first minute. So we must add the time it would take the two pumps to do the work that pump B would have done in this one minute to the total time. In this one minute, pump B would have done 1*(1/B) of the work = 1/B of the work. 
Total time taken is therefore = AB/(A+B) + AB*(1/B)/(A+B) 
= A*(B+1)/(A+B) 
Option (B).

shashankism · Answer

Since A empties a pool in A minutes. In 1 minute A empties 1/A fraction of the pool.
B empties a pool in B minutes. In 1 minute B empties 1/B fraction of the pool.

So, 1/A + (1/A + 1/B)n = 1 where is n is the time in minutes for which A and B worked together.
(A+b)/AB *n = 1- 1/A
n = AB/(A+B) - B /(A+B)= (AB- B)/(A+B)

SO, total time taken to empty a pool = 1+ (AB- B)/(A+B) = (A+B+ AB-B)/(A+B) = A(B+1)/(A+B)

Answer B

kannu44 · Answer

The fastest time in which tank can be emptied is by working together which is AB /(A+B)

But in this case it would be greater than AB/(A+B) since they have not worked together all time.

out of option B , C , D & E only option "B" is greater than AB /(A+B)

Amlu95 · Answer

Pump A rate = 1/A
Pumb B rate = 1/B

rate*time = work

The question says A begun 1 minute prior to ( A+B) working together.

So, 1/A + (A+B/AB) ( t-1) = 1

Which gives us , t = AB+A/A+B

mm229966 · Answer

If we mark time used as T.
Then A works for T minutes and B for (T-1) minutes => T*(1/A) + (T-1)*(1/B) = 1 => BT + A(T-1) = AB; =>
BT + AT - A = AB => T(B+A) - A = AB; => T = A*(B+1)/(A+B)

jkkids12 · Answer

I did this question slightly different, so let me know if it makes sense.

The rate for pump A is  \frac{(1\ pool)}{(A\ minutes)}

The rate for pump B is  \frac{(1\ pool)}{(B\ minutes)}

we can say  1\ pool   =  \frac{(1\ pool)}{(A\ minutes)}(t \ minutes) +  \frac{(1\ pool)}{(B\ minutes)} (T - 1\ minutes)

We can then simplify this as 1 =  \frac{(T)}{(A)}  +  \frac{(T - 1)}{(B)}

1  =  \frac{(BT)}{(AB)}  +  \frac{(AT - A)}{(AB)}

1  =  \frac{(BT + AT - A)}{(AB)}

AB  =  BT + AT - A

AB + A  =  BT + AT

AB + A  =  (B + A)T

\frac{(AB + A)}{(A + B)}  = T

\frac{A(B + 1)}{(A + B)}  = T

Kinshook · Answer

Pump A can empty a pool in A minutes, and pump B can empty the same pool in B minutes. Pump A begins emptying the pool for 1 minute before pump B joins.

Beginning from the time pump A starts, how many minutes will it take to empty the pool?

Part of pool emptied by Pump A in 1 minute = 1/A

Remaining fraction of pool to be emptied by both pumps = 1 - 1/A = (A-1)/A

Work done by both pumps in a minutes = 1/A + 1/B = (A+B)/AB

Time taken by both pumps after pump B starts =  {(\frac{A-1)}{A}}/{(\frac{A+B)}{AB}} = \frac{B(A-1)}{(A+B)}

Time taken to empty the pool after pump A starts =  1 + \frac{B(A-1)}{(A+B)} = \frac{(A + B + AB - B)}{(A+B)} = \frac{A(1+B)}{(A+B)}

IMO B

Pump A can empty a pool in A minutes, and pump B can empty

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Pump A can empty a pool in A minutes, and pump B can empty

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