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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
M28-41  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 72% (01:51) correct 28% (01:48) wrong based on 100 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re M28-41  [#permalink]

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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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Intern  Joined: 12 Jul 2015
Posts: 5
Re: M28-41  [#permalink]

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1
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!
Intern  Joined: 07 Dec 2012
Posts: 9
Re: M28-41  [#permalink]

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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.
Current Student B
Joined: 23 Nov 2016
Posts: 72
Location: United States (MN)
GMAT 1: 760 Q50 V42 GPA: 3.51
Re: M28-41  [#permalink]

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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.
Intern  B
Joined: 04 Sep 2016
Posts: 19
Re: M28-41  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 55271
Re: M28-41  [#permalink]

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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  B
Joined: 04 Sep 2016
Posts: 19
Re: M28-41  [#permalink]

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Thanks Bunuel
You are the fastest answering tutor. Always to the rescue. I love Gmatclub.
Intern  B
Joined: 18 Oct 2013
Posts: 5
Re: M28-41  [#permalink]

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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  P
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: M28-41  [#permalink]

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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.

???
Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: M28-41  [#permalink]

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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.
_________________
Manager  B
Joined: 06 Jul 2013
Posts: 61
Location: United States
GMAT 1: 720 Q49 V38 Re: M28-41  [#permalink]

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Bunuel wrote:
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.

Hi Bunuel,

How do you get to 10 and 15 with A^2 -25A +150? I keep getting solutions 30 and -5.
Manager  G
Joined: 22 Jun 2017
Posts: 178
Location: Argentina
Schools: HBS, Stanford, Wharton
GMAT 1: 630 Q43 V34 Re M28-41  [#permalink]

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I think this is a poor-quality 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. Re M28-41   [#permalink] 18 Feb 2019, 11:47
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