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

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.

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

hi shriramvelamuri
1/9-1/B=1/12-1/C .......-......... *36
36/9-36/B=36/12-36/C .......-......... +36/C +36/B

eddyki

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

to

b=c/2 ?

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.

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.

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....

\(?\,\,\,:\,\,\,{\text{in}}\,\,B \cup C\,\,\left( {{\text{any}}} \right)\,\,{\text{widget}}\,\,{\text{production}}\,,\,\,{\text{% }}\,\,{\text{done}}\,\,{\text{by}}\,\,{\text{B}}\)

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.

The answer is C.

Machine A means nothing when it comes to answering the question. They're asking for B and C working together.

For all we care, A makes zero widgets. (it is a constant for both scenarios and thus can be cancelled) it literally has zero influence on the final answer.

I can say Machine B makes x in 9 hours and C makes x in 12 hours. And there you have a rate for both and answer the question.

What we need:
x/(widgets produced by b)
= x/(b*x/(b+c)) {Assuming rate of B = b; rate of C = c; rate of A = a}
= (b+c)/b = 1 + c/b

(1) x/(a+b) = 9
a = (x/9) - b
(1) is insufficient

(2) x/(a+c) = 12
a = (x/12) - c
(2) is insufficient

Combining (1) and (2)
a= (x/12) - c = (x/9) - b
=> b-c = 1/36

Clearly, relation between b & c depends on x

But, what we need (highlighted in Blue) is independent of x
We cannot find the values of 3 variables with 1 equation. Thus, (E)

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