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# A and B together can do a peace of work in 12 days,which B

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A and B together can do a peace of work in 12 days,which B [#permalink]

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05 Feb 2007, 11:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A and B together can do a peace of work in 12 days,which B and C together can do in 16 days.After A has been working at it for 5 Days and B for 7 Days,C finishes in 13 days.In how many days C alone will do the work?

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05 Feb 2007, 12:18
I Think this PS would be easily solved by changing the answers in the equations. Don't you have them?

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Re: A and B and C [#permalink]

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05 Feb 2007, 12:48
sravan_m444 wrote:
A and B together can do a peace of work in 12 days,which B and C together can do in 16 days.After A has been working at it for 5 Days and B for 7 Days,C finishes in 13 days.In how many days C alone will do the work?

Assuming everybody works at constant rate and independent from each other.

When A works for 5 days and B works for 7 days, we can conclude that A and B work together for 5 days and B works alone for 2 days. And

Work Rate(A+B) = 1/12 (work/day)

Thus, Work is done by A and B for 5 days together = 5 days x 1/12 (work/day) = 5/12 work

When B works for 7-5 = 2 days and C works for 13 day, we can conclude that B and C work together for 2 days and C work alone for 13 - 2 = 11 days.

Work Rate(B+C) = 1/16 (work/day)

Thus, Work is done by B and C for 2 days = 2 days x 1/16 (work/day) = 1/8 work

The remaining job = 1 - 5/12 - 1/8 = (24 - 10 - 3)/24 = 11/24 work

C can finish this part of work within 11 days. Therefore, the work rate for C = 11/24 (work) x 1/11(1/day) = 1/24 (work/day)

Or C can finish one work in = 24 days

Hope this is right.

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05 Feb 2007, 14:59
nice answer devilmirror, but i'm not sure if it accurately represents the stem:

"After A has been working at it for 5 Days and B for 7 Days", then C has 13 days left...

It seems to me that A,B and C work as in a sequence: first A, second B and then C.

ARe you sure that A and B can work together?

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05 Feb 2007, 15:16
I solved it like this

Let's say rate of A is R1 work/day,
rate of B is R2 work/day, and
rate of C is R3 work/day

Based on the information we have we know that

R1 + R2 = 1/12 ---------------------- 1
R2 + R3 = 1/16----------------------- 2
5R1 + 7R2 + 13R3 = 1 -------------- 3

Let's solve equation 3 for R2

5(1/12 - R2) + 7R2 + 13(1/16 - R2) = 1

5/12 - 5R2 + 7R2 + 13/16 - 13R2 = 1

5/12 + 13/16 -11R2 = 1

Therefore, 11R2 = 5/12 + 13/16
R2 = 1/11(5/12 + 13/16)
R2 = 0.11

We need to find R3 though, we know from 2 that
R3 = 1/16 - R2
R3 = 1/16 - 0.11
R3 = -0.0475

But rate cannot be negative hence
R3 = 0.0475 w/day
Hence in order to finish 1 work R3 will take 1/0.0475 = 21 days
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05 Feb 2007, 15:20
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05 Feb 2007, 15:30
This is kind of vague (at least for me). Are A,B and C working in succession or in unison?

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05 Feb 2007, 15:31
I solved it considering succession
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05 Feb 2007, 18:09
pau.sabria wrote:
nice answer devilmirror, but i'm not sure if it accurately represents the stem:

"After A has been working at it for 5 Days and B for 7 Days", then C has 13 days left...

It seems to me that A,B and C work as in a sequence: first A, second B and then C.

ARe you sure that A and B can work together?

Hi pau.sabria,

Altough, A, B, and C work as sequence but you can consider them working to gether on the same days and use the rate provide from the question. Just keep in mind that the time that they work will be 2 time faster.

Here the prove.

Assuming A's work rate = 1/a (work/day)
Assuming B's work rate = 1/b (work/day)

If A works for T day the work that A does = T/a work
If B also works for T day the work that B does = T/b work
Total work done = T/a + T/b

However, if A and B working together they will use T day to finish this job.
Work rate = (T/a + T/b)/T = (1/a + 1/b)
The equation above is the work rate of A + B combined!

Therefore, we can safely conclude that.

If A work on T day and B work on T day seperately, this will equal to A and B working together for T day.

Using the prove above to solve that question and the answer is 24.

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05 Feb 2007, 18:28
devilmirror, Can you look at my solution? I think the answer is 21
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05 Feb 2007, 18:33
amorpheus wrote:
I solved it like this

Let's say rate of A is R1 work/day,
rate of B is R2 work/day, and
rate of C is R3 work/day

Based on the information we have we know that

R1 + R2 = 1/12 ---------------------- 1
R2 + R3 = 1/16----------------------- 2
5R1 + 7R2 + 13R3 = 1 -------------- 3

Let's solve equation 3 for R2

5(1/12 - R2) + 7R2 + 13(1/16 - R2) = 1

5/12 - 5R2 + 7R2 + 13/16 - 13R2 = 1

5/12 + 13/16 -11R2 = 1

Therefore, 11R2 = 5/12 + 13/16
R2 = 1/11(5/12 + 13/16)
R2 = 0.11

We need to find R3 though, we know from 2 that
R3 = 1/16 - R2
R3 = 1/16 - 0.11
R3 = -0.0475

But rate cannot be negative hence
R3 = 0.0475 w/day
Hence in order to finish 1 work R3 will take 1/0.0475 = 21 days

Amorpheus, you set the equation correctly but you solved it wrong.

R1 + R2 = 1/12 ---------------------- 1
R2 + R3 = 1/16----------------------- 2
5R1 + 7R2 + 13R3 = 1 -------------- 3

5(1/12 - R2) + 7R2 + 13(1/16 - R2) = 1

5/12 - 5R2 + 7R2 + 13/16 - 13R2 = 1

5/12 + 13/16 -11R2 = 1

Therefore, 11R2 = 5/12 + 13/16 - 1 {you missed -1 this line}
11R2 = (20 + 39 - 48)/48 = 11/48
R2 = 1/48
Therefore R3 = C's work rate = 1/16 - 1/48 = (3-1)/48 = 2/48 = 1/24 (work/day)

C will use 24 days to finish this job.

I believe some people may like this method better than mine.

Last edited by devilmirror on 05 Feb 2007, 18:36, edited 1 time in total.

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05 Feb 2007, 18:35
oh bummer haste creates waste I always do that... I see it now... yeah I think my method is easy to understand and probably quicker also.

Thanks for pointing that out.
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05 Feb 2007, 18:40
amorpheus wrote:
oh bummer haste creates waste I always do that... I see it now... yeah I think my method is easy to understand and probably quicker also.

Thanks for pointing that out.

Nah-ah!

I think both methods can solve the question beautifully.

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05 Feb 2007, 19:16
Guys answer is 24.I liked both the ways

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05 Feb 2007, 19:16
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