May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55271

Question Stats:
72% (01:51) correct 28% (01:48) wrong based on 100 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re M2841
[#permalink]
Show Tags
16 Sep 2014, 01:31
Official Solution: 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 \(25A\) (\(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\)) 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 \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\). 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
_________________



Intern
Joined: 12 Jul 2015
Posts: 5

Re: M2841
[#permalink]
Show Tags
05 Aug 2015, 04:10
Bunuel wrote: Official Solution:
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 \(25A\) (\(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\)) 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 \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\). 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 \(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\) yields two solutions for A and B while the equation \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\) 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!



Intern
Joined: 07 Dec 2012
Posts: 9

Re: M2841
[#permalink]
Show Tags
19 Apr 2016, 18:39
CountClaud wrote: Bunuel wrote: Official Solution:
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 \(25A\) (\(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\)) 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 \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\). 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 \(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\) yields two solutions for A and B while the equation \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\) 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.



Current Student
Joined: 23 Nov 2016
Posts: 72
Location: United States (MN)
GPA: 3.51

Re: M2841
[#permalink]
Show Tags
10 Feb 2017, 22:05
herein wrote: CountClaud wrote: Bunuel wrote: Official Solution:
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 \(25A\) (\(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\)) 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 \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\). 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 \(\frac{1}{A} + \frac{1}{25A} = \frac{1}{6}\) yields two solutions for A and B while the equation \(\frac{1}{A}+\frac{1}{A5}=\frac{1}{6}\) 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 \(\frac{1}{X}+\frac{1}{XN}=\frac{1}{M}\) for x when N and M are provided would significantly trim problems like these down quite a bit.



Intern
Joined: 04 Sep 2016
Posts: 19

Re: M2841
[#permalink]
Show Tags
20 Jun 2017, 00:16
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



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: M2841
[#permalink]
Show Tags
20 Jun 2017, 01:23
alexlovesgmat wrote: 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.
_________________



Intern
Joined: 04 Sep 2016
Posts: 19

Re: M2841
[#permalink]
Show Tags
20 Jun 2017, 04:36
Thanks Bunuel You are the fastest answering tutor. Always to the rescue. I love Gmatclub.



Intern
Joined: 18 Oct 2013
Posts: 5

Re: M2841
[#permalink]
Show Tags
12 Oct 2017, 07:14
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.



Director
Joined: 26 Aug 2016
Posts: 618
Location: India
Concentration: Operations, International Business
GMAT 1: 690 Q50 V33 GMAT 2: 700 Q50 V33 GMAT 3: 730 Q51 V38
GPA: 4
WE: Information Technology (Consulting)

Re: M2841
[#permalink]
Show Tags
22 May 2018, 22:16
Hi Bunuel, Can you spot my error (A+B)/AB = 1/6. Statement 1 : A + B = 25 AB = 150. AB = ((A+B)^2  4AB)^1/2 AB = constant. A+B = 25 Can't we get statement 1 enough. ???



Math Expert
Joined: 02 Sep 2009
Posts: 55271

Re: M2841
[#permalink]
Show Tags
23 May 2018, 06:02
Nightmare007 wrote: Hi Bunuel, Can you spot my error (A+B)/AB = 1/6. Statement 1 : A + B = 25 AB = 150. AB = ((A+B)^2  4AB)^1/2 AB = 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.
_________________



Manager
Joined: 06 Jul 2013
Posts: 61
Location: United States

Re: M2841
[#permalink]
Show Tags
02 Oct 2018, 09:11
Bunuel wrote: Nightmare007 wrote: Hi Bunuel, Can you spot my error (A+B)/AB = 1/6. Statement 1 : A + B = 25 AB = 150. AB = ((A+B)^2  4AB)^1/2 AB = 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. Hi Bunuel, How do you get to 10 and 15 with A^2 25A +150? I keep getting solutions 30 and 5.



Manager
Joined: 22 Jun 2017
Posts: 178
Location: Argentina

Re M2841
[#permalink]
Show Tags
18 Feb 2019, 11:47
I think this is a poorquality question and I agree with explanation. I didn't get correctly the first statement.. 1st means that A starts working, then stops and since then B starts working on the task?
_________________
The HARDER you work, the LUCKIER you get.










