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# 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  [#permalink]

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Updated on: 24 Jan 2014, 06:19
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Difficulty:

75% (hard)

Question Stats:

62% (02:31) correct 38% (02:31) wrong based on 380 sessions

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

Originally posted by gmatgambler on 24 Jan 2014, 06:13.
Last edited by Bunuel on 24 Jan 2014, 06:19, edited 2 times in total.
Edited the question.
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Posts: 50543
Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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24 Jan 2014, 06:29
6
6
gmatgambler wrote:
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. $$\frac{A(B-1)}{(A+B)}$$

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

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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24 Jan 2014, 06:35
1
1
Bunuel wrote:
gmatgambler wrote:
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. $$\frac{A(B-1)}{(A+B)}$$

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

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Hope this helps.
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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30 Mar 2014, 14:32
2
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
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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30 Mar 2014, 19:24
jlgdr wrote:
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

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.
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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01 Apr 2014, 09:32
1
2
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 = [AB/(A+B)]*(1-1/A) = B(A-1)/(A+B) minutes
Total time taken from the start = [B(A-1)/(A+B)] + 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).
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Solve Work Problem using Numerical approach  [#permalink]

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Updated on: 18 Apr 2014, 00:25
Q. 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) A+B-1/2

(b) A(B+1)/A+B

(c) AB/A+B

(d) AB/A+B -1/1

(e) A(B-1)/A+B

-----------------------------------------------

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

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Originally posted by royQV on 17 Apr 2014, 09:57.
Last edited by royQV on 18 Apr 2014, 00:25, edited 1 time in total.
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Re: Solve Work Problem using Numerical approach  [#permalink]

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17 Apr 2014, 18:05
1
royQV wrote:
Q. 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?

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

Can you please post the OA?

Rate of Pump A $$= \frac{1}{a}$$

Rate of Pump B $$= \frac{1}{b}$$

Work done by Pump A in 1 min $$= \frac{1}{a} * 1 = \frac{1}{a}$$

Work remaining (To be done by both pumps) $$= 1 - \frac{1}{a} = \frac{a-1}{a}$$

Combined rate of Pump A & B$$= \frac{1}{a} + \frac{1}{b}$$

Time required to complete the remaining work = t

$$\frac{a+b}{ab} * t = \frac{a-1}{a}$$

$$t = \frac{b(a-1)}{a+b}$$

Total time required

= $$\frac{b(a-1)}{a+b} + 1$$

$$= \frac{a(b-1)}{a+b}$$
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Re: Solve Work Problem using Numerical approach  [#permalink]

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18 Apr 2014, 01:30
royQV wrote:
Q. 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) A+B-1/2

(b) A(B+1)/A+B

(c) AB/A+B

(d) AB/A+B -1/1

(e) A(B-1)/A+B

-----------------------------------------------

METHOD ONE: Algebraic approach

This approach to this question involves some tricky algebra.

Pump A works at a rate of 1/A and pump B, at a rate of 1/B (these rates are given in units of "pools/minute"). For the time they are working together, we add rates. That's a HUGE idea in work problems - when two machines or people work together, we add the rates.

In the first minute, pump A works alone and drains an amount of 1/A (that is, one "A-th" of a pool). This leaves an amount of

The time it will take the two pumps, working at the combined rate, to drain this, is:

That's the time from when the two pumps start working together, which is 1 minute after pump A starts. To get the total time, we need to add 1 to this (this is the trickiest algebra in the whole problem!)

METHOD TWO: Numerical approach

Let's say that Pump A can drain a pool in A = 6 minutes, and pump B can drain a pool in B = 3.

Pump A works for a minute, draining 1/6 of the pool, and leaving 5/6 of the pool left.

Then pump B kicks in --- A & B work at the combined rate of 1/6 + 1/3 = 1/2. How long does it take the two pumps, working at a rate of 1/2, to drain 5/6 of a pool?

Add the first minute for total time.

If we plug in the starter values A = 6 and B = 3, how many of the answers will yield this answer of 8/3 as the total time?

Merging similar topics. Please refer to the discussion above.

Hope it helps.

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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07 Aug 2017, 22:24
gmatgambler wrote:
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. $$\frac{A(B-1)}{(A+B)}$$

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)

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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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17 Aug 2017, 06:49
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)
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Re: Pump A can empty a pool in A minutes, and pump B can empty  [#permalink]

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08 Sep 2018, 04:46
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Re: Pump A can empty a pool in A minutes, and pump B can empty &nbs [#permalink] 08 Sep 2018, 04:46
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