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# A carpenter worked alone for 1 day on a job that would take him 6 more

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Intern
Joined: 27 Jun 2011
Posts: 2
A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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Updated on: 02 Feb 2015, 05:44
2
10
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:10) correct 34% (02:32) wrong based on 375 sessions

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A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone?

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

Originally posted by kspice on 27 Jun 2011, 11:00.
Last edited by Bunuel on 02 Feb 2015, 05:44, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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31 Oct 2011, 02:42
3
3
kspice wrote:
Hello!

I just need help on setting up this work problem:

A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone.

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

Please include a detailed explanation as to how you arrived at the solution.

Thank you kindly!

~k

This is how I would deal with this problem:

Rate of work of first carpenter = 1/7 work/day
Both, working together, took 4 days to complete (6/7)th of the work.
So they will take 4*7/6 (= 14/3 days) to complete the full work.

Rate of work of both together = 3/14 work/day

Mind you, working alone, first carpenter does 1/7 = 2/14 work/day
Together, they complete 3/14 work/day. So the second carpenter must be doing 1/14 work/day.
So working alone, he will take 14 days.
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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27 Jun 2011, 20:10
3
2
it takes 7 days for carpenter 1 to complete the work.

=> rate of carpenter1 = 1/7

=> amount of work done by carpenter1 on first day = 1/7

rest of 6/7 work is done both by carpenter1 and carpenter2 in 4 days

=> 1/7+rate of carpenter2 = (6/7)/4
=> rate of carpenter2 = 1/14

=> carpenter2 alone can finish this work in 14 days.

##### General Discussion
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Posts: 1820
Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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27 Jun 2011, 11:11
4
kspice wrote:
Hello!

I just need help on setting up this work problem:

A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone.

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

Please include a detailed explanation as to how you arrived at the solution.

Thank you kindly!

~k

A carpenter worked only 1 day on something that takes him 6 MORE days.
Means;
Carpenter finishes his work in 7 days.
Let his buddy finish the same task in x days.

Respective rates per day:
1/7 AND 1/x

To complete 1 work:
First guy worked for 5 days @ rate=1/7 per day.
Second one worked for 4 days @ rate=1/x per day

Expression:
Days*Rate=Work
5*1/7+4*1/x=1
5x+28=7x
2x=28
x=14 days.

Ans: "D"
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Manager
Joined: 14 Mar 2011
Posts: 192
Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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27 Jun 2011, 13:39
2
First carpenter can finish the job in 7 days. After first day carpentar 2 joins and the work gets completed in 5 days.

Let the total work be 35 units. (LCM of 5 & 7) Rate of carpentar 1 = 35/7 = 5 units per day. He works for 5 days , so he does total of 25 units out of 35 units. That means carpentar 2 does 35-25 units = 10 units in 4 days. so rate of carpentar 2 - 10/4 = 2.5 units per day. So he will do complete work in 35/2.5 = 14 days.
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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27 Jun 2011, 16:37
fluke wrote:
kspice wrote:
Hello!
...

A carpenter worked only 1 day on something that takes him 6 MORE days.
Means;
Carpenter finishes his work in 7 days.
Let his buddy finish the same task in x days.

Respective rates per day:
1/7 AND 1/x

To complete 1 work:
First guy worked for 5 days @ rate=1/7 per day.
Second one worked for 4 days @ rate=1/x per day

Expression:
Days*Rate=Work
5*1/7+4*1/x=1
5x+28=7x
2x=28
x=14 days.

Ans: "D"

Nicely explained Fluke
Manager
Joined: 25 May 2011
Posts: 119
Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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30 Oct 2011, 11:28
2
1
the speed of first carpenter$$=\frac{1}{7}$$
the speed of second carpenter$$=\frac{1}{X}$$

$$1*\frac{1}{7}+4*(\frac{1}{7}+\frac{1}{X})=1$$

$$\frac{4}{7}+\frac{4}{X}=\frac{6}{7}$$

$$X=14$$
Intern
Joined: 30 Oct 2011
Posts: 2
Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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31 Oct 2011, 01:13
shahideh wrote:
the speed of first carpenter$$=\frac{1}{7}$$
the speed of second carpenter$$=\frac{1}{X}$$

$$1*\frac{1}{7}+4*(\frac{1}{7}+\frac{1}{X})=1$$

$$\frac{4}{7}+\frac{4}{X}=\frac{6}{7}$$

$$X=14$$

Fascinatingggggggggggg
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Joined: 07 Aug 2011
Posts: 538
GMAT 1: 630 Q49 V27
A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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27 Mar 2015, 23:27
2
kspice wrote:
A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone?

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

Carpenter takes 1+6 = 7days
When he was joined by his fellow carpenter i.e. after 1 day $$\frac{1 ^t^h}{7}$$ of the work was already done so remaining $$\frac{6^t^h}{7}$$ was done together in 4 days . setting this up on equation .

$$\frac{1}{7} + \frac{1}{X} = \frac{6}{7} * \frac{1}{4}$$
X=14
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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29 Jun 2016, 08:48
kspice wrote:
A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone?

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

On 1st day one carpenter did = 1/7th work (Total days required to finish work= 7 days

Remaining work = 6/7

In next 4 days 1st carpenter must have done 4/7 work, leaving 6/7-4/7= 2/7 for 2nd carpenter.

Now, if 2nd carpenter does 2/7 work in 4 days. he requires 4*7/2= 14 days to finish the whole task.

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A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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05 Jun 2017, 17:45
kspice wrote:
A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone?

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

1.(no.of days worked by the first person)/ time taken by him if working alone + (no.of days worked by the second person)/ time taken by him if working alone=1
2. (5/7)+ (4/x) = 1, So x=14
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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08 Jun 2017, 16:04
kspice wrote:
A carpenter worked alone for 1 day on a job that would take him 6 more days to finish. He and another carpenter completed the job in 4 more days. How many days would it have taken the second carpenter to do the complete job working alone?

A) 4 2/3
B) 7
C) 9
D) 14
E) 24

The rate of the first carpenter is 1/7.

Since the two carpenters can complete the remaining 6/7 of the job in 4 days, their combined rate is:

(6/7)/4 = 6/28 = 3/14

We can let x = the number of days it takes the second carpenter to complete the job alone; thus, his rate = 1/x. We can create the following equation:

1/7 + 1/x = 3/14

Multiplying the equation by 14x, we have:

2x + 14 = 3x

14 = x

The second carpenter could complete the job alone in 14 days.

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Re: A carpenter worked alone for 1 day on a job that would take him 6 more  [#permalink]

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31 Jul 2018, 15:45
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Re: A carpenter worked alone for 1 day on a job that would take him 6 more &nbs [#permalink] 31 Jul 2018, 15:45
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