25 men reap a field in 20 days. when should 15 men leave the work if the whole field is to be reaped in 37-1/2 days after they leave the work?
I dont know where to start????
Thanks guys
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25 men reap a field in 20 days
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scoobydooby
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25 men take 20 days to complete a piece of work
=> 1 man takes 500 days to complete the piece of work.
=> 1 mans 1 day work=1/500 of the total work
=>10 mens 1 day work=1/500*10 of total work
=>10 mens 37.5 days work= 1/500*10*75/2=3/4 work
so 1/4 of the work must have been done by all 25 men before 15 of them left.
to do full work 25 men take 20 days
=>to do 1/4 work, 25 men will take 20/4=5 days.
so 15 men should leave after 5 days
(it looks clumsy, am sure theres a shorter/smarter solution somewhere)
=> 1 man takes 500 days to complete the piece of work.
=> 1 mans 1 day work=1/500 of the total work
=>10 mens 1 day work=1/500*10 of total work
=>10 mens 37.5 days work= 1/500*10*75/2=3/4 work
so 1/4 of the work must have been done by all 25 men before 15 of them left.
to do full work 25 men take 20 days
=>to do 1/4 work, 25 men will take 20/4=5 days.
so 15 men should leave after 5 days
(it looks clumsy, am sure theres a shorter/smarter solution somewhere)
I used this method to solve the problem.
but the result is different with the previous post could anyone tell me what is wrong?
let "t" be the time required by the first 25 men .
we know that 25 do the job in 20 days. their total production is 20*25
in the same way for the first part of job where 25 men were involved
25*t+(25-15)*((75/2)-t)=25*20
solving this equation yields
t=25/3.
but the result is different with the previous post could anyone tell me what is wrong?
let "t" be the time required by the first 25 men .
we know that 25 do the job in 20 days. their total production is 20*25
in the same way for the first part of job where 25 men were involved
25*t+(25-15)*((75/2)-t)=25*20
solving this equation yields
t=25/3.
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harshavardhanc
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it should be 5 days.clock60 wrote:agree with answer 5-days
let it be total job-500,and rate of given worker=1, x-number of days they worked all together
25*1*x+(25-15)*1*37 1/2=500
x=5
per person per day of work is 1/(25*20) = 1/500
see that the question says : 37.5 days after they leave the work. That means, for these 37.5 days, 10 workers are on the job who will do (37.5 *10)/500 = 3/4th of the total work. That means, 1/4th is already done by 25 people.
1/4 of the work will be done in 20/4 = 5 days.
HTH
in these
Regards,
Harsha
Harsha
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rbansal
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Hello,
When you got that 1 man takes 500 days to complete the task. I understand you just did 25 x 20 but can you please explain the rational behind it. I am getting very stuck on why 1 man would take 500 days to complete the task .
Thank you,
RB
When you got that 1 man takes 500 days to complete the task. I understand you just did 25 x 20 but can you please explain the rational behind it. I am getting very stuck on why 1 man would take 500 days to complete the task .
Thank you,
RB
scoobydooby wrote:25 men take 20 days to complete a piece of work
=> 1 man takes 500 days to complete the piece of work.
=> 1 mans 1 day work=1/500 of the total work
=>10 mens 1 day work=1/500*10 of total work
=>10 mens 37.5 days work= 1/500*10*75/2=3/4 work
so 1/4 of the work must have been done by all 25 men before 15 of them left.
to do full work 25 men take 20 days
=>to do 1/4 work, 25 men will take 20/4=5 days.
so 15 men should leave after 5 days
(it looks clumsy, am sure theres a shorter/smarter solution somewhere)
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GMATGuruNY
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Let rate per man = 1 unit per day.nhai2003 wrote:25 men reap a field in 20 days. when should 15 men leave the work if the whole field is to be reaped in 37-1/2 days after they leave the work?
I dont know where to start????
Thanks guys
Work completed by the 25 men each day = 25*1 = 25 units.
Over 20 days, total work = 20*25 = 500 units.
After the 15 men leave, rate for the 10 remaining men = 10 units per day.
Work completed by these 10 men over the last 37.5 days = 10 * 37.5 = 375 units.
Thus, work to be completed before the 15 men leave = 500-375 = 125 units.
Time for all 25 men to complete 125 units = w/r = 125/25 = 5 days.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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GmatKiss
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Thanks GMATGuru, u made the simplest amd the effective solution!!GMATGuruNY wrote:Let rate per man = 1 unit per day.nhai2003 wrote:25 men reap a field in 20 days. when should 15 men leave the work if the whole field is to be reaped in 37-1/2 days after they leave the work?
I dont know where to start????
Thanks guys
Work completed by the 25 men each day = 25*1 = 25 units.
Over 20 days, total work = 20*25 = 500 units.
After the 15 men leave, rate for the 10 remaining men = 10 units per day.
Work completed by these 10 men over the last 37.5 days = 10 * 37.5 = 375 units.
Thus, work to be completed before the 15 men leave = 500-375 = 125 units.
Time for all 25 men to complete 125 units = w/r = 125/25 = 5 days.
