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If Machine B and Machine C work together at their constant individual
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27 Jan 2015, 07:41
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64% (02:01) correct 36% (01:56) wrong based on 493 sessions
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If Machine B and Machine C work together at their constant individual rates to produce x widgets, what percent of the widgets will be produced by Machine B? (1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours. (2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours. Kudos for a correct solution.
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Re: If Machine B and Machine C work together at their constant individual
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27 Jan 2015, 09:18
ans E... i think i should be correct although not 100% sure.. both the statements are insufficient as they deal with only one machine B or C.. combined it just tells us that B is faster than C... but till we know either of three machines speed or no of widgets produced ans cannot be given
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Re: If Machine B and Machine C work together at their constant individual
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27 Jan 2015, 11:26
Quote: (1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
we search for: hours/rate(b) > percentage of rate b and hours/rate(c) > percentage rate c (1) > gives us nothing but 1/A+1/B=1/9 for x=100 (2) > gives us nothing but 1/A+1/C=1/12 for x=100 1/91/B=1/121/C 4+36/C=3+36/B 1+36/C=36/B 36/B36/C=1 B=C/2 B does 2/3 of the work > C



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Re: If Machine B and Machine C work together at their constant individual
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27 Jan 2015, 12:29
Hi Eddyki, I am having a slight trouble understanding your procedure. Please could you explain 1/91/B=1/121/C To 4+36/C=3+36/B How did you approach the above step. eddyki wrote: Quote: (1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
we search for: hours/rate(b) > percentage of rate b and hours/rate(c) > percentage rate c (1) > gives us nothing but 1/A+1/B=1/9 for x=100 (2) > gives us nothing but 1/A+1/C=1/12 for x=100 1/91/B=1/121/C 4+36/C=3+36/B 1+36/C=36/B 36/B36/C=1 B=C/2 B does 2/3 of the work > C
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Kudos to you, for helping me with some KUDOS.



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Re: If Machine B and Machine C work together at their constant individual
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27 Jan 2015, 13:26
Bunuel wrote: If Machine B and Machine C work together at their constant individual rates to produce x widgets, what percent of the widgets will be produced by Machine B?
(1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
Kudos for a correct solution. I think the answer is E: let A, B and C be the respective times for machine A, B and C from 1: AB/(A+B) = 9 or 1/A+1/B = 1/9> no info on C NSF from 2: AC/(A+C) = 12 or 1/A+ 1/C = 1/12> no info on B NSF 1+ 2 1/A= 1/9 1/B from 1 1/A = 1/121/C from 2 equating these expressions 1/91/B = 1/121/C => 1/B1/C = 1/36 we can just tell that B is faster than C and nothing more from this expression. so NSF E



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Re: If Machine B and Machine C work together at their constant individual
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28 Jan 2015, 09:10
hi shriramvelamuri1/91/B=1/121/C ................ *36 36/936/B=36/1236/C ................ +36/C +36/B shriramvelamuri wrote: Hi Eddyki, I am having a slight trouble understanding your procedure. Please could you explain 1/91/B=1/121/C To 4+36/C=3+36/B How did you approach the above step. eddyki wrote: Quote: (1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
we search for: hours/rate(b) > percentage of rate b and hours/rate(c) > percentage rate c (1) > gives us nothing but 1/A+1/B=1/9 for x=100 (2) > gives us nothing but 1/A+1/C=1/12 for x=100 1/91/B=1/121/C 4+36/C=3+36/B 1+36/C=36/B 36/B36/C=1 B=C/2 B does 2/3 of the work > C



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Re: If Machine B and Machine C work together at their constant individual
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31 Jan 2015, 05:54
eddykihow did u reach from 36/b  36/c =1 to b=c/2 ?



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Re: If Machine B and Machine C work together at their constant individual
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02 Feb 2015, 02:47
Bunuel wrote: If Machine B and Machine C work together at their constant individual rates to produce x widgets, what percent of the widgets will be produced by Machine B?
(1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:E. While the setup of this problem  asking for a percentage and not an actual number  may make it seem as though both statements together can be sufficient, the algebra demonstrates that you cannot get a ratio of B to C given the information provided. One way to definitively prove the information to be insufficient is to plug in a couple numbers to get different results: Say that the rate of A were 1job/18hours, where you can call x widgets equal to just "one job". Based on the fact that rates are additive, that would get you to: Solve for B: A+B=1/9hours 1/18 + B = 1/9 B = 1/9 − 1/18 B = 1/18 Solve for C: A + C = 1/12hours 1/18 + C = 1/12 C = 1/12 − 1/18 C=1/36 So B works twice as fast as C, meaning that B would do 2/3 of the work. You can try it again with a different number to see if you get a different relationship, and you do if you try something like A = 1/36 (taking care to use numbers divisible by 12 and 9, the numbers in the problem). Here you can solve that the same way as the above to find that B=1/12 and C=1/18, a case in which B does not work twice as fast and therefore does not do the same proportion of work. Because even with both pieces of information together you cannot solve for a proportion, the correct answer must be E.
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If Machine B and Machine C work together at their constant individual
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29 Jul 2016, 11:29
Hello Bunuel,
I used the below method for solving the equation:
If we take x =36 then we get for statements 1 and 2 we get two equations:
1) A+B=4 widgets produced per hour 2) A+C=3widgets produced per hour
Combining these two equations we get multiple values of B and C.
For example for A=1, B= 3 and C=2thus ratio would become 3/560% but for A=2,B=2 and C=1thus ratio would become 2/366%
Can you please confirm whether taking an arbitrary value is the best way to go about such questions?
Thanks Akash



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Re: If Machine B and Machine C work together at their constant individual
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30 Jul 2016, 02:15
AkashKashyap wrote: Hello Bunuel,
I used the below method for solving the equation:
If we take x =36 then we get for statements 1 and 2 we get two equations:
1) A+B=4 widgets produced per hour 2) A+C=3widgets produced per hour
Combining these two equations we get multiple values of B and C.
For example for A=1, B= 3 and C=2thus ratio would become 3/560% but for A=2,B=2 and C=1thus ratio would become 2/366%
Can you please confirm whether taking an arbitrary value is the best way to go about such questions?
Thanks Akash Yes, this could be another way to solve this question.
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Re: If Machine B and Machine C work together at their constant individual
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10 Dec 2016, 08:12
I feel the answer hint is in question stem itself. It gives you how much work is done by B & C and not the time taken by both.
So this means X = (BC/(B+C))* t > Saytime taken is t
from 1. X = (AB/(A+B))*9 from 2. X = (AC/(A+C))*12
Unless we know t we can not find work done by B and its percentage.
Experts please comment if this approach is correct....



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If Machine B and Machine C work together at their constant individual
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23 Oct 2018, 13:08
Bunuel wrote: If Machine B and Machine C work together at their constant individual rates to produce x widgets, what percent of the widgets will be produced by Machine B?
(1) Machine A and Machine B, working together at their constant individual rates, can produce x widgets in 9 hours.
(2) Machine A and Machine C, working together at their constant individual rates, can produce x widgets in 12 hours.
\(?\,\,\,:\,\,\,{\text{in}}\,\,B \cup C\,\,\left( {{\text{any}}} \right)\,\,{\text{widget}}\,\,{\text{production}}\,,\,\,{\text{% }}\,\,{\text{done}}\,\,{\text{by}}\,\,{\text{B}}\) Let´s present a BIFURCATION for (1+2), that is, two EXPLICIT VIABLE scenarios, each one giving a different answer to our FOCUS! One possible scenario is the following: A does x/2 widgets in 9 hours (hence x/6 in 3 hours and 2x/3 in 12 hours) , B does x/2 widgets in 9 hours and C does x/3 widgets in 12 hours. Conclusion: in 12h, B does 2x/3 widgets and C does x/3 widgets , hence our FOCUS is 2x/3 divided by x (=2x/3+x/3), that is , 2/3. Another possible scenario is the following: A does x/4 widgets in 9 hours (hence x/12 in 3 hours and x/3 in 12 hours) , B does 3x/4 widgets in 9 hours (hence x widgets in 12h) and C does 2x/3 widgets in 12 hours. Conclusion: in 12h, B does x widgets and C does 2x/3 widgets , hence our FOCUS is x divided by 5x/3 (=x+2x/3), that is 3/5 (NOT 2/3). The correct answer is therefore (E). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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If Machine B and Machine C work together at their constant individual
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