M28-41

Official Solution:


Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

Given that \(\frac{1}{A}+\frac{1}{B}=\frac{1}{6}\), where A is the time needed for machine A to complete the task working alone and B is the time needed for machine B to complete the task working alone.

(1) The average time A and B can complete the task working alone is 12.5 days. This statement implies that \(A+B=2*12.5=25\). Now, since we don't know which machine works faster then even if we substitute B with \(25-A\) (\(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\)) we must get two different answers for A and B: \(A \lt B\) and \(A \gt B\). Not sufficient.

(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task. \(A=B+5\), so we have that \(\frac{1}{A}+\frac{1}{A-5}=\frac{1}{6}\). From this we can find that \(A=2\) (not a valid solution since in this case B will be negative) or \(A=15\). Sufficient.


Answer: B
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Hi Bunuel,

Would you mind please explaining your thought process on how the equation of \(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\) yields two solutions for A and B while the equation \(\frac{1}{A}+\frac{1}{A-5}=\frac{1}{6}\) yields only one?

It seems that you outright solved the second (and eliminated the illogical solution). However for the first one, were you able to see that it had two solutions by inspection or did you actually solve the quadratic? I'm more curious because it seems to me that solving both quadratics would be too time consuming for ~2 mins

Thank you!

For the DS questions, we are only concerned with the sufficiency of the statement. In other words, we only need to figure out whether the statement provided to answer the question is sufficient or not. If you approach the DS Q's this way, you will save lots of time on the real exam.

Coming to your question,
Think of it this way:

We are provided with the average time, without any correlation between the two machines. For example, we do not know whether A works faster than B or B works faster than A (We don't know if Rate of Machine A > Rate of Machine B or not).

In the case that the average is provided, we have no way to determine whether one machine is faster than the other, or the same or vice versa. Hence, Statement A is insufficient to answer the question.

I hope this solves your doubts.

Please Bunuel,
may you expertise help here a little bit
most people need to solve the quadratic actually, does only comparison of the rates allow for an answer? Without solving ? How come ?
Thanks

The solution talks about actually solving the quadratics, which will be A^2 -17A + 30 = 0.
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Thanks Bunuel
You are the fastest answering tutor. Always to the rescue. I love Gmatclub.

A shortcut to eliminate S1 so you don't have to solve a quadratic is as follow:

(1/A + 1/B) * 6 = 1
Recognize that [(A+B)/AB]*6 = 1
S1 says that A + B = 25, plug that back in

(25*6)/AB = 1 or AB = 25*6; As one can see there are many solutions to what A and B can be so Insufficient.

Unfortunately for S2 I had to solve the quadratic, but recognizing what I did in S1 allowed me to gain the extra seconds I needed.

Hi Bunuel,
Can you spot my error

(A+B)/AB = 1/6.

Statement 1 : A + B = 25
AB = 150.

A-B = ((A+B)^2 - 4AB)^1/2
A-B = constant.
A+B = 25

Can't we get statement 1 enough.

???

From (1) we get two solutions: A = 10 and B = 15 OR A = 15 and B = 10.
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chetan2u Bunuel

Hi. Could someone please explain why \(\frac{1}{B}+\frac{1}{B+5}=\frac{1}{6}\) as a solution to st-2 is not a valid equation. Solving this yields b=3, but clearly \(\frac{1}{3}+\frac{1}{8}\) is not equal to \(\frac{1}{6}\), which led me to believe I had done sth else wrong or was applying the incorrect methodology altogether. Appreciate your help please.

Hi..
You are going wrong in finding solution..
The solutions will be 10 and -3..
(1/b)+(1/(b+5))=1/6..
6(b+5+b)=b^2+5b...12b+30=b^2+5b...
b^2-7b-30=0....(b-10)(b+3)=0..
So b=10 and -3

Hope it helps

Posted from my mobile device
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How do I move fwd with statement 2 in this?

Rate of A = A units/day
Rate of B= B units / day

total work= 6 (A+B)

So using statement 2
6 (A+B) /A = 6(A+B)/B + 5


Can we use this method? if yes how to proceed I was unable to simplify this.

First define some variables
A - rate of A
B - rate of B
x - time that it takes machine A working alone
y - time that it takes machine B working alone

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

Given in statement 2 It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task.
This means \(x > y\)
\(x = y + 5\)

Equation 1:
\(A + B = \frac{1}{6}\)

\(Rate = \frac{Work}{Time}\)

\(A = \frac{1}{x}\)
\(x = y + 5\)

\(A = \frac{1}{y+5}\)

\(B = \frac{1}{y}\)

\(A + B = \frac{1}{6}\)
\(\frac{1}{y+5} + \frac{1}{y} = \frac{1}{6}\)

\(\frac{y + y + 5}{y*(y+5)} = \frac{1}{6}\)

\(\frac{2y + 5}{y^2+5y} = \frac{1}{6}\)

\(12y+30=y^2+5y\)

\(y^2-7y-30 = 0\)

\((y-10)(y+3)=0\)

\(y = 10\) or \(y = -3\) reject y = -3 because it is implicitly assumed that time cannot be negative

\(x = y + 5\)
\(x = 10+5\)
\(x= 15\)

Thus statement 2 is sufficient

chetan2u GMATinsight Bunuel
how are we getting this slution for st 1? We are getting A=10 and -5 . Why would we randomly consider other variations

For (1) we have 1/A + 1/B = 1/6 and A + B = 25. After substituting B = 25 - A into 1/A + 1/B = 1/6 we get 1/A + 1/(25 - A) = 1/6, which gives A^2 -25A + 150 = 0, which gives A = 10 or A = 15. But as explained in the solution, you don't really need to do all that if you recognize that for (1) we have no way of distinguishing machines A and B from one another and thus we have to get two solutions for A and B: \(A \lt B\) and \(A \gt B\).
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