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M28-41

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M28-41 [#permalink]

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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.

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Re M28-41 [#permalink]

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New post 16 Sep 2014, 01:31
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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 \(25-A\) (\(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\)) 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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Re: M28-41 [#permalink]

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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 \(25-A\) (\(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\)) 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!
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Re: M28-41 [#permalink]

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New post 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 \(25-A\) (\(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\)) 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.
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Re: M28-41 [#permalink]

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New post 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 \(25-A\) (\(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\)) 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.
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Re: M28-41 [#permalink]

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New post 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
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Re: M28-41 [#permalink]

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New post 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.
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Re: M28-41 [#permalink]

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New post 20 Jun 2017, 04:36
Thanks Bunuel
You are the fastest answering tutor. Always to the rescue. I love Gmatclub.
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Re: M28-41 [#permalink]

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New post 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.
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Re: M28-41 [#permalink]

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New post 22 May 2018, 22:16
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.

???
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Re: M28-41 [#permalink]

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New post 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.

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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M28-41   [#permalink] 23 May 2018, 06:02
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