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A contractor estimated that his 10-men crew (m07q06) [#permalink]
12 Feb 2009, 19:45

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A contractor estimated that his 10-men crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, after 5 days of rain, on the 61-st day, he hired 6 more people to finish the project early. If the job was done in 100 days, on how many days after day 60 was it raining?

Re: A contractor estimated that his 10-men crew [#permalink]
12 Feb 2009, 20:17

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xALIx wrote:

A contractor estimated that his 10-men crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, after 5 days of rain, on the 61-st day, he hired 6 more people to finish the project early. If the job was done in 100 days, on how many days after day 60 was it raining?

* 4 * 5 * 6 * 7 * 8

Total man-days needed = 10 X 110 = 1100

Man-days worked before 61st day =(60-5)X10 = 550

Man-days worked after 61st day = 16 X (40-x), where x is the number of days it rained after day 60.

1100 = 550 + 16(40-x)

16x = 90 ---> x = 5.6....

number of days it rained after day 60 = 5 ... option B.

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
18 Sep 2009, 05:05

well, I solved the question a little bit differently but I think it can help you about the explanation, it took 34,37 days for 16 workers to finish the left over job without any rainy days. the whole job was completed in 100 days, the first 10 worker group worked for 60 days (55+5rainy), so total time spent with 16 workers was 40 (100-60), so we are left with the days that people worked for 34,37 which we round up to 35 because even if they don't work the whole day they were able work in some part of that day so we understand it didn't rain that day, too. so the number of days it rained is 40-35=5

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
18 Sep 2009, 17:00

ya i was having trouble with the last part as i rounded it to 6..its pretty tricky..does it mean that if the 5.6 goes to 6 it means it rained for MORE days n' so the workers would take MORE than 100 days to complete?Thats the reason we are rounding down?
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Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
15 Jan 2010, 02:06

10 men can do a work in 110 days they worked for 55 days ( 60-5 days) so they worked for half of days required and hence completed half of the work. balance half was to be done by 16 men ( 10 + 6 new hires) 10 man can do in 55 days 16 man can do in X days.....(solving we get x = 34.37 ~ 35 days). 40 days - 35 days = 5 days

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
15 Jan 2010, 04:58

Hi,

If someone wants to avoid decimal/round-off approach then try this alternative

Total work= 1100 man-days Work done after 60 days (including 5 rainydays)= 550 man-days

After addition on 6 men into the crew total work per day = 16 man-days therefore, from day 61-100 total work (considering no rainy day in between)= 40*16=640 man-days. No. of man- hours lost due to rain between day 61-100= 640-550=90 man-days So, no. of rainy days between day 61-100=90/16= 5 days

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
19 Jan 2011, 05:56

B.

total work required for one person to complete the job= 110 * 10 = 1100 61st day .. 5 days of rain.. hence work done in number of days = 60-5 = 55 d half of the work remians... = 550 d of work for one person if 10 + 6 people are deployed to complete work then => 550/16 = 34.4 d we will consider this as 35 d as full day of work is counted even if you spent an hour working on the last day.

100 - ( 55 + 5d of rain + 35 d of work) = 100 - 95 = 5 d of rain

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
19 Jan 2011, 06:22

10 men completed work in 110 days. In 55 working days half of the work was done by 10 men. so, 1 man can do half of the work in 550 days. 16 men can do the same work in 550/16 days = 34.37 Total days to complete = 40 Rainy days = 40-34.37 = 5.63 ~ 5 days.

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
19 Jan 2011, 23:51

I explain: Call a man with a working day is Y Total working part to finish the construction : 10Y x 110 = 1100Y ( without rainny) According to the data above, It rains from 57th to 61st day-5days. So, the working is 56x 10Y=560 Y And, the working requirement is 1100Y So that, the rest working requirement is 1.100 Y- 560Y = 540 Y In the data, the construction finish with 100days. We have working day from 62 from 100days. That mean : 39 working day x 16Y=624Y The balance is 540 Y and 624Y : 84Y - a day with rain and no working with 16 man , 84Y : 16= 5.3 So, B is correct.

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
20 Jan 2011, 00:58

tejal777 wrote:

The OE states:

Quote:

x=5.625 Round the number of rainy days down to 5 as with 6 rainy days the crew would have worked less than 550 mh and not have completed the job.

Could somebody please explain this part in detail?I rounded it to 6 as this no. should have been in any other context..

i did the calculation and got the same result. it should rain for 5.625 day, which we can round off to 6(0.625 can be rounded to1). apart from this i didnt get the anyidea why OE gives such an explanation. if it rains for 6 day, then total men work will be = 16*34=544mendays work(which is less than required) but acc to problem it should be 550mendays work.

the answer should be 5.
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Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
23 Jan 2012, 10:14

powerka wrote:

Here's the answer using a RTW chart.

As the gmatclub explanation says,

Round the number of rainy days down to 5 as with 6 rainy days the 16 men crew would have completed less than 1/2 of the job.

Could you explain how you got 16/10 for the 2nd part of this table please. The chart does help out significantly in doing the problem I just don't understand that part nor the part where u got the equation below the chart. Also do you know where I can find a thread that discusses using a wrt table? In my Veritas Prep book they don't mention any tables for that equation just manipulating the equation to find the answers. Thanks!

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
23 Jan 2012, 18:43

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Expert's post

AzWildcat1 wrote:

powerka wrote:

Here's the answer using a RTW chart.

As the gmatclub explanation says,

Round the number of rainy days down to 5 as with 6 rainy days the 16 men crew would have completed less than 1/2 of the job.

Could you explain how you got 16/10 for the 2nd part of this table please. The chart does help out significantly in doing the problem I just don't understand that part nor the part where u got the equation below the chart. Also do you know where I can find a thread that discusses using a wrt table? In my Veritas Prep book they don't mention any tables for that equation just manipulating the equation to find the answers. Thanks!

Below approach might help to understand better the problem as well as the solution provided by powerka: A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? A. 4 B. 5 C. 6 D. 7 E. 8

Given: 10-man crew needs 110 days to complete the construction.

"On the 61-st day, after 5 days of rain ..." --> as it was raining for 5 days then they must have bee working for 55 days thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 men).

Then contractor "hired 6 more people" --> speed of construction increased 1.6 times, so the new 16-man crew needed 55/1.6=~34.4 days to complete the construction, but after they were hired job was done in 100-60=40 days --> so 5 days rained. (They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.)

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
23 Jan 2012, 22:46

Thanks Bunuel, that helped and made a lot more sense. I wish I could understand some of these harder problems more, it usually takes me a solid min or 90 seconds to sometimes comprehend and write out the equations or logic of the problems for these harder Q on the quant section

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
24 Jan 2012, 03:00

Hi My Solution is :

Total Work = 10 x 110 = 1100 men-days Work competed = (Days 10 men work for) = 10 x 55 = 550 (no. of days are 55 because 5 days before 61st day are 55 days) Work remainig = 1100 - 550 = 550. Now, if 16 men work to complete this remaining work they will take = 550/16 = 34. 3 = ~ 35 Days.

No of days actually taken = 100 - 61 + 1 (this is to calulate the numbers between 100 and 61) = 40 Total no days it rained = 40 - 35 = 5 [B] _________________

Thanks & Regards Yodee

‘A good plan violently executed now is better than a perfect plan executed next week.’ - General Georg S. Patton

Re: A contractor estimated that his 10-men crew (m07q06) [#permalink]
23 Jan 2013, 05:51

no. of men initially: 10 work to be done : 1 no. of days expected to finish using 10 men: 110 amount of work done in 1 day by 10 men :1/110----- (1)

hence it means 10 men complete 1 work in 110 days therefore 1 men will complete ds work in 1100 days------(2)

now 1100 days reqd to complete 1 work by 1 men therefore in 1 day amount of workdone by 1 man will be 1/1100

now till 60th day 5 days are lost so in 55 day work done by 10 men will be :55/110 using (1) and amount of work left :1-55/110=1/2

now we have 6 more men therefore in 1 day 16 men will complete 16/1100 work using (2) therefore no . of days more required to complete 1/2 work will be: (1/2)/(16/1100)= 35.625 days=35 days

therefore no. of days it rained again=(100-60)-35=5 days

no
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