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A qualified worker digs a well in 5 hours. He invites 2 appr

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A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 04 Oct 2006, 07:14
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A
B
C
D
E

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Question Stats:

51% (02:57) correct 48% (02:07) wrong based on 128 sessions
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?

A. 1:24
B. 1:34
C. 1:44
D. 1:54
E. 2:14
[Reveal] Spoiler: OA
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 [#permalink] New post 04 Oct 2006, 09:24
Q rate/day = 1/5
A=(3/4)(1/5)=(3/20)(2)=3/10
T=(1/5)(1/5)=(1/25)(2)=1/50

t(1/5+3/10+1/50)=1 = 29/50 –-> t= 50/29
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 [#permalink] New post 04 Oct 2006, 13:34
X & Y wrote:
Q rate/day = 1/5
A=(3/4)(1/5)=(3/20)(2)=3/10
T=(1/5)(1/5)=(1/25)(2)=1/50

t(1/5+3/10+1/50)=1 = 29/50 –-> t= 50/29

I think it must be 2/25 instead of 1/50.
The explanation is valid though.
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 [#permalink] New post 04 Oct 2006, 13:44
anandsebastin wrote:
X & Y wrote:
Q rate/day = 1/5
A=(3/4)(1/5)=(3/20)(2)=3/10
T=(1/5)(1/5)=(1/25)(2)=1/50

t(1/5+3/10+1/50)=1 = 29/50 –-> t= 50/29

I think it must be 2/25 instead of 1/50.
The explanation is valid though.


You are right; since I did it on paper and then copied to the computer I missed that. Thanks..

It is 2/25
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 [#permalink] New post 05 Oct 2006, 00:33
I have another way.

A can do a work in 5 hours.

Now 2 persons whose efficiency is 3/4 of A joins him.
ie 1 person whose efficiency is 2x3/4 ie 3/2 times of A joins him.

Again 2 persons whose efficiency is 1/5 of A joins him.
ie 1 person whose efficiency is 2x1/5 ie 2/5 times of A joins him.

So finally the people involved in the work is
A+ 3/2(A) + 2/5(A)
ie 29/10 (A)

Since A alone takes 5 hours
29/10 (A) will take 10/29 x 5 i.e 50 /29 hours
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 [#permalink] New post 14 Jun 2007, 14:48
Q rate/day = 1/5
A=(3/4)(1/5)=(3/20)(2)=3/10
T=(1/5)(1/5)=(1/25)(2)=2/25

t(1/5+3/10+2/25)=1 = 29/50 --> t= 50/29 or 1:40
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 [#permalink] New post 01 Jul 2007, 10:57
I must be totally blind! I do not see how you can get 1.4 hours out of 50/29.

Here's how I see it: each hour 29/50 of the job is completed, therefore after the first hour, 29/50 is complete and 21/50 is still remaining to be completed. 21/50 is just slightly less than 29/50, which is inline with the 1.7 that Juaz got.

Can someone please explain what I am missing? :oops:

Thanks!
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 [#permalink] New post 01 Jul 2007, 11:11
Bluebird wrote:
I must be totally blind! I do not see how you can get 1.4 hours out of 50/29.

Here's how I see it: each hour 29/50 of the job is completed, therefore after the first hour, 29/50 is complete and 21/50 is still remaining to be completed. 21/50 is just slightly less than 29/50, which is inline with the 1.7 that Juaz got.

Can someone please explain what I am missing? :oops:

Thanks!


50/29 = 1.7 hrs

(.7)60 = approx 40 mins

so total time = 1 hour and 40 mins.

I think the answers are a bit confusing.
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 [#permalink] New post 01 Jul 2007, 11:19
Thank you salr15! I am quite embarrassed, but you're right, the answers are a bit confusing... 1.4 to me means 1.4 hours, not 1 hr and 40 minutes. Thank you for the clarification!
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 [#permalink] New post 16 Oct 2007, 01:13
Well, technically --- the question asks:
"how much time does the team need to finish the job?"

After the work calculation, we are left with 50/29 hours. This is equivalent to 1 hour and 21/29*60 minutes.

21/29 roughly yields "0.7241" which multiplied by 60 is roughly equivalent to 43.4 minutes.

Therefore, if we give the team 1:40 minutes, they will NOT be able to complete the job. --- Amongst the available choices, the team NEEDS 1:50 minutes to complete the job.

Is my logic wrong on this?
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A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 04 Oct 2013, 22:54
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?
A 1:24
B 1:34
C 1:44
D 1:54
E 2:14
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Re: A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 05 Oct 2013, 01:36
Stiv wrote:
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?
A 1:24
B 1:34
C 1:44
D 1:54
E 2:14


Your question sounds vague to me. The answer should be in terms of time. Anyway I just solved based on the data given.

Rate of the Work = 1 / 5

Rate of 2 apprentice = 2 * (1/5) * (3/4) = 6/20

Rate of 2 Trainees = 2 * (1/5) * (1/5) = 2/25

Total Rate of 5 people = (1/5) + (6/20) + (2/25) = 58/100

Time required to dig 1 Well = 1/ (58/100) = 100/58 hours
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Re: A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 05 Oct 2013, 01:57
Actually question is asking about the time they need. options are in hours:minutes.
Question is ok.

and answer is correct as you have calculated : 100/58 hrs that is 1.72 hrs - 1 hr and 44 minutes, so 1:44, hence, C.
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Re: A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 05 Oct 2013, 02:18
Chiranjeevee wrote:
Actually question is asking about the time they need. options are in hours:minutes.
Question is ok.

and answer is correct as you have calculated : 100/58 hrs that is 1.72 hrs - 1 hr and 44 minutes, so 1:44, hence, C.


Thanks for the info. I better get used to this type of time representation.
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Re: A qualified worker digs a well in 5 hours. He invites 2 appr [#permalink] New post 17 Oct 2013, 11:30
Q rate = 1/5
A rate = 1/5*3/4
T rate = 1/5*1/5

1/5+2*(1/5*3/4)+2*(1/5*1/5)=1/t

1/5+6/20+2/25=1/t

58/100=1/t

58t=100

t=100/58=1 21/29, which is slightly below 1 3/4, so it must be 1:44 for an answer
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Work/rate problem - was my approach incorrect? [#permalink] New post 02 Nov 2013, 20:57
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?

A) 1:24
B) 1:34
C) 1:44
D) 1:54
E) 2:14



I started with the original rate for the worker as 1/5.

Each apprentice would have a rate (1/5)*(3/4) = 3/20

Each trainee would have a rate (1/5) * (1/5) = 1/25

Combining all of it together 1/5 + 3/20 + 3/20 + 1/25 + 1/25 = 58/100 per hour

After one hour, the remaining work to be done is 42/100

Is it possible to use this approach to solve the problem?
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Re: Work/rate problem - was my approach incorrect? [#permalink] New post 03 Nov 2013, 03:30
undecidedonmba wrote:
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?

A) 1:24
B) 1:34
C) 1:44
D) 1:54
E) 2:14



I started with the original rate for the worker as 1/5.

Each apprentice would have a rate (1/5)*(3/4) = 3/20

Each trainee would have a rate (1/5) * (1/5) = 1/25

Combining all of it together 1/5 + 3/20 + 3/20 + 1/25 + 1/25 = 58/100 per hour

After one hour, the remaining work to be done is 42/100

Is it possible to use this approach to solve the problem?




Hi,

Your approach is perfectly correct but I do not see the answer options to be in sync with the question that's asked. The question is "hpw much time does the team need to finish the job" but answer options are in proportion/ratio format. Can you verify whether the question stem is correct?
According to me the answer should be 50/29 hours. I used the same approach as yours.
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Re: Work/rate problem - was my approach incorrect? [#permalink] New post 03 Nov 2013, 04:38
Expert's post
undecidedonmba wrote:
A qualified worker digs a well in 5 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?

A) 1:24
B) 1:34
C) 1:44
D) 1:54
E) 2:14



I started with the original rate for the worker as 1/5.

Each apprentice would have a rate (1/5)*(3/4) = 3/20

Each trainee would have a rate (1/5) * (1/5) = 1/25

Combining all of it together 1/5 + 3/20 + 3/20 + 1/25 + 1/25 = 58/100 per hour

After one hour, the remaining work to be done is 42/100

Is it possible to use this approach to solve the problem?


Merging similar topics.
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Re: Work/rate problem - was my approach incorrect?   [#permalink] 03 Nov 2013, 04:38
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