Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

70% (01:27) correct 30% (01:22) wrong based on 64 sessions

HideShow timer Statistics

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?

(1) The average time A and B can complete the task working alone is 12.5 days.

(2) It would take machine A 5 more days to complete the task alone than it would take machine B to 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.

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

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

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

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.

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

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.

I like your reasoning for statement 1, however is there a similar reasoning to apply to statement 2? I'm hoping there is a shortcut, but as I understood it I had to actually solve the quadratic for statement 2 to determine that A was 2 or 15, and then realize that the first solution gave a negative time for B and thus wasn't an actual solution.

A shortcut to solving \(\frac{1}{X}+\frac{1}{X-N}=\frac{1}{M}\) for x when N and M are provided would significantly trim problems like these down quite a bit.

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

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.
_________________