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Machines A, B, and C can either load nails into a bin or

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Machines A, B, and C can either load nails into a bin or [#permalink] New post 29 Oct 2009, 18:01
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A
B
C
D
E

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  25% (medium)

Question Stats:

83% (02:42) correct 17% (02:57) wrong based on 198 sessions
Machines A, B, and C can either load nails into a bin or unload nails from that bin. Each machine works at a constant rate that is the same for loading and for unloading, although the individual machines may have different rates. Working together to load at their respective constant rates, machines A and B can load the bin in 6 minutes. Likewise, working together to load at their respective constant rates, machines B and C can load the bin in 9 minutes. How long will it take machine A to load the bin if machine C is simultaneously unloading the bin?

(A) 12 minutes
(B) 15 minutes
(C) 18 minutes
(D) 36 minutes
(E) 54 minutes
[Reveal] Spoiler: OA

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Re: Unusual work problem. [#permalink] New post 29 Oct 2009, 18:20
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It doesn't look very difficult, but I need to confirm my approach by somebody else.

So, compile standard working problem equations

1. \frac{1}{A}+\frac{1}{B}=\frac{1}{6}

2. \frac{1}{B}+\frac{1}{C}=\frac{1}{9}

and now we see that all we have to do is just substract the second equation from the first one:

\frac{1}{A}+\frac{1}{B}-\frac{1}{B}-\frac{1}{C}=\frac{1}{6}-\frac{1}{9}

\frac{1}{A}-\frac{1}{C}=\frac{3}{18}-\frac{2}{18}=1/18

therefore, answer should be C.

KUDOS if you find it useful;)
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Re: Unusual work problem. [#permalink] New post 29 Oct 2009, 18:31
(\frac{1}{A}+\frac{1}{B})-(\frac{1}{B}+\frac{1}{C})=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}
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Re: Unusual work problem. [#permalink] New post 30 Oct 2009, 05:55
yes, 18 is what I got tooo, same method as above two...
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Re: Unusual work problem. [#permalink] New post 30 Oct 2009, 08:30
18 for me too.
I think this is the correct answer as I've seen this Q in a recent mock test (forgot which). Solution mentioned above nails it right...
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Re: Unusual work problem. [#permalink] New post 30 Oct 2009, 12:48
gmattokyo wrote:
18 for me too.
I think this is the correct answer as I've seen this Q in a recent mock test (forgot which). Solution mentioned above nails it right...


Yes..its 18 min. This question is in Kaplan -Advanced book.
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Re: Unusual work problem. [#permalink] New post 02 Sep 2010, 19:04
yes it is 18 ....
P S guys I am shaping up .
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Re: Unusual work problem. [#permalink] New post 02 Sep 2010, 21:19
I too got 18 minutes, but did a lot of unnecessary calculations.... need to keep the eye on the target..... was looking all over the question.

Vyacheslav wrote:
It doesn't look very difficult, but I need to confirm my approach by somebody else.

So, compile standard working problem equations

1. \frac{1}{A}+\frac{1}{B}=\frac{1}{6}

2. \frac{1}{B}+\frac{1}{C}=\frac{1}{9}

and now we see that all we have to do is just substract the second equation from the first one:

\frac{1}{A}+\frac{1}{B}-\frac{1}{B}-\frac{1}{C}=\frac{1}{6}-\frac{1}{9}

\frac{1}{A}-\frac{1}{C}=\frac{3}{18}-\frac{2}{18}=1/18

therefore, answer should be C.

KUDOS if you find it useful;)


This is was the quickest approach..... +1 from me.
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Re: Unusual work problem. [#permalink] New post 02 Sep 2010, 23:08
yup!! the above method helps..
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Re: Unusual work problem. [#permalink] New post 03 Sep 2010, 03:17
Answer is 18 minutes. There can be no other answer. Here is the reason.

Suppose there are 54 units of work.

then rate of A and B together is 9 units/min and rate of B and C working together is 6 units/min.

If you write down all possible combinations of rates for A, B and C, then you'll see the following relationship:

Rate of : A B C
Case 1: 8 1 5
Case 2: 7 2 4
Case 3: 6 3 3
Case 4: 5 4 2
Case 5: 4 5 1

If you now notice, you'll realise that Rate of A - Rate of C is always equal to 3. Hence, the time taken will always be 54/3 = 18 minutes.

Hope that helps.
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Re: Unusual work problem. [#permalink] New post 03 Sep 2010, 03:46
I am getting 18 mins
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Re: Unusual work problem. [#permalink] New post 11 Sep 2010, 11:41
I.
( A + B ) * 6 -> 100%
Ib.
( A + B ) * 9 -> 150%
II.
( B + C ) * 9 -> 100%
?
( A - C ) * x -> 100%

From Ib-II:

( A - C ) * 9 -> 50%

( A - C ) * 18 -> 100%
x = 18
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Re: Unusual work problem. [#permalink] New post 15 Jan 2011, 12:32
i'll use a similar approach to Vyacheslav's approach

\frac{1}{A} + \frac{1}{B} = \frac{1}{6}

\frac{1}{B} + \frac{1}{C} = \frac{1}{9}

if A is loading and C is unloading then we need to get rid of B from the equations.

\frac{1}{A} + (\frac{1}{9} - \frac{1}{C}) = \frac{1}{6}

\frac{1}{A} - \frac{1}{C} = \frac{1}{18}

Answer: C

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Re: Unusual work problem. [#permalink] New post 10 Apr 2012, 00:41
I feel percentages approach in solving Work Rate problem is better than other approaches.

Let's have a look at the solution for this question.

A&B take 6 mins to load the bin and B&C take 9 mins for the same work.

In one minute, A+B complete 16.66% (1/6) of work and B+C complete 11.11% (1/9) of work.
A+B=16.66
B+C=11.11

Subtracting the 2 equations above:
(A+B) - (B+C) = 16.66 - 11.11 = 5.55 %

A - C= 5.55%
That means A's loading and C's unloading together complete 5.55% of work in a minute.

100% of work will take 18 mins (100/5.55).


PS - I have been solving all Work Rate problems in this forum since morning with percentage approach and I found each one of them quite simple to solve using percentages.
I don't have time else I would have submitted solutions for each problem.

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Re: Unusual work problem. [#permalink] New post 10 Apr 2012, 02:26
palsays wrote:
I feel percentages approach in solving Work Rate problem is better than other approaches.

Let's have a look at the solution for this question.

A&B take 6 mins to load the bin and B&C take 9 mins for the same work.

In one minute, A+B complete 16.66% (1/6) of work and B+C complete 11.11% (1/9) of work.
A+B=16.66
B+C=11.11

Subtracting the 2 equations above:
(A+B) - (B+C) = 16.66 - 11.11 = 5.55 %

A - C= 5.55%
That means A's loading and C's unloading together complete 5.55% of work in a minute.

100% of work will take 18 mins (100/5.55).


PS - I have been solving all Work Rate problems in this forum since morning with percentage approach and I found each one of them quite simple to solve using percentages.
I don't have time else I would have submitted solutions for each problem.

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You dont really have to convert fractions into percentages as it can be extremely time consuming.
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Re: Unusual work problem. [#permalink] New post 10 Apr 2012, 04:19
stevie1111 wrote:
palsays wrote:
I feel percentages approach in solving Work Rate problem is better than other approaches.

Let's have a look at the solution for this question.

A&B take 6 mins to load the bin and B&C take 9 mins for the same work.

In one minute, A+B complete 16.66% (1/6) of work and B+C complete 11.11% (1/9) of work.
A+B=16.66
B+C=11.11

Subtracting the 2 equations above:
(A+B) - (B+C) = 16.66 - 11.11 = 5.55 %

A - C= 5.55%
That means A's loading and C's unloading together complete 5.55% of work in a minute.

100% of work will take 18 mins (100/5.55).


PS - I have been solving all Work Rate problems in this forum since morning with percentage approach and I found each one of them quite simple to solve using percentages.
I don't have time else I would have submitted solutions for each problem.

_________



Ravender Pal Singh


You dont really have to convert fractions into percentages as it can be extremely time consuming.


I have just given detailed explanation for your understanding. There is no sense in calculating 100/5.55 as it is pretty obvious that it would be slightly less than 20 (range in between 16.66 - 20).
The best thing about GMAT is that one need not do the complete calculation to solve the question. Once you get the equation, one can easily guess the answer as GMAT has answer options in a good range.
Instead of having equations in inverted ratios, if we can have linear equations at one go, solution becomes very simple to correctly guess.
I have solved so many questions till now and none of them took more than 1-2 mins.

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Re: Machines A, B, and C can either load nails into a bin or [#permalink] New post 15 Nov 2012, 22:52
\frac{1}{A}+\frac{1}{B}=\frac{1}{T}

It takes A units of time for A to do it alone.
It takes B units of time for B to do it alonge.
BUT this means it takes T units of time for A and B to accomplish the work where A>T and B>T.

My Solution:
\frac{1}{Amin}+\frac{1}{Bmin}=\frac{1}{6min}
\frac{1}{Bmin}+\frac{1}{Cmin}=\frac{1}{9min}

Combine the two equations:
\frac{1}{Amin}+\frac{1}{Bmin}-\frac{1}{Bmin}-\frac{1}{Cmin}=\frac{1}{6}-\frac{1}{9}

\frac{1}{Amin}-\frac{1}{Cmin}=1/18min

This means A would do 18 minutes of loading while C would do 18 minutes of unloading to complete the task.

Answer: 18
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Re: Machines A, B, and C can either load nails into a bin or [#permalink] New post 30 Aug 2013, 05:58
Another method ---- Rate(AB) = 1/6 and Rate(BC) = 1/9

Then Rate(AB) - Rate(BC) = Rate A - Rate C = (1/6) - (1/9) = 1/18

Hence Rate of A - Rate of C = 18 mins ----- Time to fill the bin
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Re: Machines A, B, and C can either load nails into a bin or [#permalink] New post 25 Sep 2013, 20:25
This took a lot of time.

We can use RTW chart, rate * time = work

Combined rate of A/B => 1/A + 1/B = (A+B)/AB

Rate * time = Work

(A+B)/AB * tab = 1 => tab = AB/(A+B) = 6

Getting A in terms of B => A = 6B/(B-6) ..... (1)

Similarly BC/(B+C) = 9

Getting C in terms of B

C = 9B/(B-9)..... (2)

A is loading and C is unloading hence

1/A - 1/C => tac = AC/(C-A)

Substituting values from (1) and (2)
we get 18
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Re: Machines A, B, and C can either load nails into a bin or [#permalink] New post 04 Dec 2013, 06:08
1 Job = (A+B) *6 => A+B = 1/6
1 Job = (B+C) *9 => B+C = 1/9

Now, if A+B would load and B+C would unload => (A+B) - (B+C) = A-C = 1/6 - 1/9 = 3/18 - 2/18 = 1/18

so rate for A is 1/18 => 1Job = 1/18 * t => 1Job / 1/18 = 18

Answer C.
Re: Machines A, B, and C can either load nails into a bin or   [#permalink] 04 Dec 2013, 06:08
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