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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] New post 24 Jan 2014, 06:13
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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. \(\frac{A(B-1)}{(A+B)}\)
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Jan 2014, 06:19, edited 2 times in total.
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Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink] New post 24 Jan 2014, 06:29
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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 \(1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}\).

Answer: B.
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Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink] New post 24 Jan 2014, 06:35
Expert's post
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 \(1+\frac{(A-1)B}{A+B}=\frac{A(B+1)}{A+B}\).

Answer: B.


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

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Re: Pump A can empty a pool in A minutes, and pump B can empty [#permalink] New post 30 Mar 2014, 19:24
Expert's post
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!
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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] New post 01 Apr 2014, 09:32
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] New post 17 Apr 2014, 09:57
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?

Answer Choices:

(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!)

Answer = B

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?

Only answer (B) works, so that the correct answer.
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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] New post 17 Apr 2014, 18:05
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!)

Answer = B

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?

Only answer (B) works, so that the correct answer.



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] New post 18 Apr 2014, 01:30
Expert's post
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?

Answer Choices:

(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!)

Answer = B

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?

Only answer (B) works, so that the correct answer.


Merging similar topics. Please refer to the discussion above.

Hope it helps.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1, 3, and 8. Thank you.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Solve Work Problem using Numerical approach   [#permalink] 18 Apr 2014, 01:30
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