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# If Machine B and Machine C work together at their constant individual

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Math Expert
Joined: 02 Sep 2009
Posts: 55228
If Machine B and Machine C work together at their constant individual  [#permalink]

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27 Jan 2015, 07:41
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Difficulty:

55% (hard)

Question Stats:

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  [#permalink]

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27 Jan 2015, 09:18
3
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  [#permalink]

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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

1/9-1/B=1/12-1/C
4+36/C=3+36/B
1+36/C=36/B
36/B-36/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  [#permalink]

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27 Jan 2015, 12:29
Hi Eddyki,

I am having a slight trouble understanding your procedure. Please could you explain

1/9-1/B=1/12-1/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

1/9-1/B=1/12-1/C
4+36/C=3+36/B
1+36/C=36/B
36/B-36/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  [#permalink]

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27 Jan 2015, 13:26
2
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 = 1/12-1/C from 2
equating these expressions
1/9-1/B = 1/12-1/C => 1/B-1/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  [#permalink]

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28 Jan 2015, 09:10
1
hi shriramvelamuri
1/9-1/B=1/12-1/C .......-......... *36
36/9-36/B=36/12-36/C .......-......... +36/C +36/B

shriramvelamuri wrote:
Hi Eddyki,

I am having a slight trouble understanding your procedure. Please could you explain

1/9-1/B=1/12-1/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

1/9-1/B=1/12-1/C
4+36/C=3+36/B
1+36/C=36/B
36/B-36/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  [#permalink]

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31 Jan 2015, 05:54
eddyki

how did u reach from
36/b - 36/c =1

to

b=c/2 ?
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Joined: 02 Sep 2009
Posts: 55228
Re: If Machine B and Machine C work together at their constant individual  [#permalink]

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02 Feb 2015, 02:47
1
1
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  [#permalink]

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29 Jul 2016, 11:29
1
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=3----widgets produced per hour

Combining these two equations we get multiple values of B and C.

For example for A=1, B= 3 and C=2----thus ratio would become 3/5---60%
but for A=2,B=2 and C=1-----thus ratio would become 2/3---66%

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  [#permalink]

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30 Jul 2016, 02:15
1
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=3----widgets produced per hour

Combining these two equations we get multiple values of B and C.

For example for A=1, B= 3 and C=2----thus ratio would become 3/5---60%
but for A=2,B=2 and C=1-----thus ratio would become 2/3---66%

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  [#permalink]

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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  [#permalink]

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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   [#permalink] 23 Oct 2018, 13:08
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