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Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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29 Oct 2009, 18:01
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84% (01:44) correct 16% (02:37) wrong based on 410 sessions
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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
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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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02 Sep 2010, 21:19
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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03 Sep 2010, 03:17
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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 IbII:
( A  C ) * 9 > 50%
( A  C ) * 18 > 100% x = 18



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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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10 Apr 2012, 00:41
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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: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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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Ravender Pal Singh You dont really have to convert fractions into percentages as it can be extremely time consuming.



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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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10 Apr 2012, 04:19
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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 12 mins.  Ravender Pal Singh



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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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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 unload nails [#permalink]
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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 unload nails [#permalink]
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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/(B6) ..... (1) Similarly BC/(B+C) = 9 Getting C in terms of B C = 9B/(B9)..... (2) A is loading and C is unloading hence 1/A  1/C => tac = AC/(CA) 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 unload nails [#permalink]
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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) = AC = 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.



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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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16 Dec 2014, 21:39
Bunuel wrote: 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 1/A + 1/ B = 1/6 1/B + 1/C = 1/9 1/A+1/B  1/B1/C = 1/61/9 1/A1/C = 1/18 = 18 minutes = Ans C
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Re: Machines A, B, and C can either load nails into a bin or unload nails [#permalink]
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29 Jan 2018, 14:50
Hi All, This question can be solved in a couple of different ways (and they all require a certain amount of 'math work', so this question will likely take you at least 23 minutes to solve it regardless of how you approach it). Here's a way to approach it that involves rates and TESTing VALUES. We're told that it takes Machines A and B, working together, to fill the bin in 6 minutes. Conceptually, it's easiest if those 2 Machines have the same rate, so let's TEST: Machine A = 12 minutes to fill the bin alone Machine B = 12 minutes to fill the bin alone Thus, in 6 minutes, each of them will fill half the bin. Next, we're told that it takes Machines B and C, working together, to fill the bin in 9 minutes. Since we've set Machine B's rate, we have to mathematically determine Machine C's rate. In 9 minutes, Machine B will fill 3/4 of the bin. Thus, in those 9 minutes, Machine C has to fill the other 1/4 of the bin. 9 minutes = (1/4)(Full) 36 minutes = Full Machine C = 36 minutes to fill the bin alone Now that we've established the rates for Machines A and C, we can calculate how long it takes to fill the bin when Machine A is FILLING the bin and Machine C is EMPTYING the bin. In 1 minute, Machine A 'fills' 1/12 of the bin. In that same minute, Machine C 'empties' 1/36 of the bin... 1/12  1/36 = 3/36  1/36 = 2/36 1/18 Thus, every minute, 1/18 of the bin is filled. Knowing that, it takes 18 minutes to fill the bin under these conditions. Final Answer: GMAT assassins aren't born, they're made, Rich
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