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Difficult Math Problem #101 - Work

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Difficult Math Problem #101 - Work

by 800guy » Fri Feb 23, 2007 3:54 pm
John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

oa coming after some people respond. from the diff math doc.

have a good weekend!!
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Re: Difficult Math Problem #101 - Work

by ssekkeuny » Sat Feb 24, 2007 6:56 am
800guy wrote:John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

oa coming after some people respond. from the diff math doc.

have a good weekend!!
the answer is '3.5days', i think...

john's work rate = 1/20 jane's work rate = 1/12

together = 2/15

this means that it takes 7.5days to finish.

7.5 - 4 = 3.5 days.

am i right?
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by jayhawk2001 » Sat Feb 24, 2007 9:09 am
Lets consider x as the number of days John and Jane worked together.

John completes 1/20 of work in 1 day.
Jane completes 1/12 of work in 1 day.

(x+4)*1/20 + x*1/12 = 1

So, x = 6

So, total number of days in which the work was complete = 10 days

Cross-checking: 6 * 1/12 + 10 * 1/20 = 1
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by banona » Mon Feb 26, 2007 11:15 am
John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

John and Jane would complete the task in ( 20*12/ ( 12+20)) = 7.5 days
However, they only worked togeother for 3.5 days ( as Jane left 4 days before the task be comlpleted)
this means that, the fraction of the work done when Jane got indisposed is equal to : 3.5/7.5 = 7/15
CONSEQUENTLY, John would have to complete 8/15 of the work alone. As he can complete the task in 20 days, he would do 8/15 of the work in 32/3 days,
then, the task would be complete in 3.5 + 32/3 = 14 1/6 days
I appologize for my Frenchy-English.
I am working on it.
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oa

by 800guy » Mon Feb 26, 2007 12:41 pm
oa from the doc:

Together they do 1/20+1/12=4/30 of the task.

Jane and John started the work together, but only John finished the work because Jane gets sick. So let x be the number of days they worked together.

x*3/40+4*1/20=1

x4/30=4/5 and therefore x =6

So in total they worked 6 days on it together and John worked 4 days on it. So total days spent=10, but if the question is asking how many time did they spend working on the project together, then the answer is 6.
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by Tame the CAT » Sun Apr 08, 2007 6:18 pm
Rate John works = 1/20
Rate for Jane = 1/12

Rate Total = 2/15

Since John works solo for 4 days

RT=W
(1/20)*4 = 1/5

He completes 20 percent of the work alone and thus there is 80% or 4/5 left.

How much time (in days) does it take for BOTH to complete 80% of the work?

T=W*1/R
T=4/5*15/2 = 6

6+4 = 10 days
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by Cybermusings » Tue Apr 10, 2007 1:20 am
John - 20 days
Jane - 12 days
Together in 1 day - 1/20 + 1/12 = (3+5)/60 = 8/60th of the task
Thus task completed by John and Jane working together in 60/8 = 15/2 days
Jane leaves 4 days before the job is completed
Thus John works alone for these 4 days
In 4 days John completes ----- 4/20 ---- 1/5 of the job
Hence John + Jane had already completed 4/5 of the job working together
John + Jane complete 4/5 of the job in ------ 2/15 * 4/5 = 6 days
Hence they worked together for 6 days
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by mathrupradeep » Fri May 09, 2014 9:01 am
John - 20 days
Jane - 12 days

Jane quiets before 4 days of completion.

if it took x no of days to complete the project. Jane left when it was x-4 days

so

x-4/12 + x/20 = 1
by solving
x=10.

It took 10 days for the project to get completed.
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by GMATGuruNY » Fri May 09, 2014 10:17 am
800guy wrote:John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

oa coming after some people respond. from the diff math doc.

have a good weekend!!
Let the job = the LCM of 20 and 12 = 60 units.

Since John can complete the job in 20 days, John's rate = w/t - 60/20 = 3 units per day.
Since Jane can complete the job in 12 days, Jane's rate = w/t = 60/12 = 5 units per day.

After Jane leaves, John works on his own for 4 days to complete the job.
Work produced by John in the last 4 days = r*t = 3*4 = 12 units.

Of the 60 units that must be produced, John produces the last 12.
Thus, the first 48 units are produced by John and Jane together.
Combined rate for John and Jane = 3+5 = 8 units per day.
Time for John and Jane to produce 48 units = w/r = 48/8 = 6 days.

Total time for the job = (4 days for John alone) + (6 days for John and Jane together) = 10 days.
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