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Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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03 Nov 2009, 12:21

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Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

B. 7

1/10 + 1/12 + 15 = 15/60 of the work done each day.

60 - 15/60-15/60 = 30/60 (abel then leaves)

Ben and carla working together finish 9/60 each day. Carla alone finished 4/60 each day so I figured the amount carla finishes alone will be a multiple of 4

30/60 - 9/60 = 21/60 (3 days total) 21/60-9/60 = 12/60 (4 days total; and coincidentally a multiple of 4) so assume carla works alone after this 12/60-4/60 = 8/60 (5 days) 8/60 - 4/60 = 4/60 (6 days) 4/60-4/60 = 0 (7 days)

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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19 Oct 2013, 09:35

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Expert's post

asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

Not to sound like a broken record from some of my earlier posts, but, worst case, you could always plug in answer choices for this problem.

Start with C and you get 2/10 of a job from Abel (which, notice, will always be the case), (8-3)/12 from Ben, and 8/15 from Carla.

LCM and add these up, and you get 12/60+25/60+32/60. Too much!

Do the same with B. Abel stays at 2/10, Ben is now 4/12, and Carla is 7/15. So, 12/60+20/60+28/60 = 60/60.

This approach could take longer in some circumstances, but it's always a default strategy where you have answer choices like these and no idea how to proceed. _________________

Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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03 Nov 2013, 12:37

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asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

Responding to a pm.

First 2 days all three of them worked together, thus they did 2*(1/10 + 1/12 + 1/15) = 1/2 of the work.

Last 3 days only Carla worked, thus she did 3/15 = 1/5 of the work.

1 - 1/2 - 1/5 = 3/10 of the work was done by Ben and Carla: (time)*(combined rate)=(job done) --> t*(1/12 + 1/15) = 3/10 --> t = 2 days. So, we have that Ben and Carla worked together for 2 days.

Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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18 Mar 2014, 07:30

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Let the work be completed in 't'.

Then again "Rate * Time = Work"

Rate(A)) = 1/10 Rate(B)=1/12 Rate(C)=1/15

Since A worked for 2 Days Work done by A= 2/10 Since B worked for 3 Days before work was completed work done B= (t-3)/12 Since C worked for full number of days = t/15

Adding them gives total work which is 1 unit.

2/10 + (t-3)/12 + t/15 = 1

Hence t=7 _________________

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Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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08 May 2014, 14:49

asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

Abel in the 2 days that he worked completed 1/5 of the job = 4/5 remains Then if Ben had to leave 3 days before the completion, this means that Carla had to work alone for these 3 days in which she completed 1/5 of the job.

Now together, Ben and Carla completed the job in (1/12 + 1/15)(t) = 3/5

3/20 (t) = 3/5 ---> t = 4

Therefore, these 4 days worked plus the 3 days that Carla had to work by herself add to 7 days

Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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06 Apr 2015, 07:53

Bunuel wrote:

asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

Responding to a pm.

First 2 days all three of them worked together, thus they did 2*(1/10 + 1/12 + 1/15) = 1/2 of the work.

Last 3 days only Carla worked, thus she did 3/15 = 1/5 of the work.

1 - 1/2 - 1/5 = 3/10 of the work was done by Ben and Carla: (time)*(combined rate)=(job done) --> t*(1/12 + 1/15) = 3/10 --> t = 2 days. So, we have that Ben and Carla worked together for 2 days.

Total days = 2 + 3 + 2 = 7.

Answer: B.

Hope it's clear.

Bunuel's answer is the quickest way to think about the structure of the solution. Just want to add a small tip: Because right off the bat, you notice you're adding 3 fractions. Instead of doing the usual cross multiply trick to add fractions, I would find the LCM immediately by sketching out a quick venn diagram:

10 = 2*5 12 = 2*2*3 15 = 3*5

2 is common, 3 is common, and 5 is common - leftover is just one 2. Therefore, the LCM is 2*3*5*2 = 60. Rewrite all of the fractions with a denominator of 60 and this problem can be solved in under 2 minutes.

All working together = 6/60 + 5/60 + 4/60 = 15/60 for 2 days = 30/60. B and C working together = 5/60 + 4/60 = 9/60 for (x) days. C working alone = 4/60 for 3 days = 12/60.

Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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06 Apr 2015, 21:07

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Expert's post

asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

Another option is to convert it to units of work if you don't want to work with fractions.

Say, the work involves 60 units (LCM of 10, 12 and 15). So Abel does 60/10 = 6 units a day, Ben does 60/12 = 5 units a day and Carla does 60/15 = 4 units a day. Together, they do 6+5+4 = 15 units a day.

In 2 days, they complete 15*2 = 30 units and are left with 30 units. Then only Ben and Carla are working and doing 5+4 = 9 units a day. The last 3 days Carla works alone and does 4*3 = 12 units of the 30 units so Ben and Carla together do 30 - 12 = 18 units. Hence Ben and Carla work together in the middle at the rate of 9 units per day for 18/9 = 2 days.

Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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06 Apr 2015, 21:43

asterixmatrix wrote:

Abel can complete a work in 10 days, Ben in 12 days and Carla in 15 days. All of them began the work together, but Abel had to leave after 2 days and Ben 3 days before the completion of the work. How long did the work last?

A. 6 B. 7 C. 8 D. 9 E. 10

speed of A=1.2B =1.5C .

combined speed of B and C = \(\frac{1}{B} +\frac{1}{C} = \frac{B+C}{BC} = \frac{3}{20}\) \(1-(\frac{1}{5} +\frac{1}{6}+ \frac{2}{15} + \frac{3}{15} )\) = work done together by B and C= \(\frac{9}{30}.\)

so time take by B and C together (in A's absence) = 2 days

answer 2+2+3=7 days _________________

Thanks, Lucky

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Re: Abel can complete a work in 10 days, Ben in 12 days and Carl [#permalink]

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19 Nov 2015, 07:17

Amount of work Carla completed + Amount of work Abel completed + Amount of work Ben completed = Total amount of work completed t/15 + 2/10 + (t-3)/12 = 1 t = 7

The hardest part of this problem (at least for me) was interpreting the amount of time Ben spent working.

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