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It takes 7 high school students, working at identical constant individ
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11 Nov 2015, 12:03
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It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM? A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM
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Re: It takes 7 high school students, working at identical constant individ
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11 Nov 2015, 21:06
shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM Hi, there can be various ways.. the simpler way would be to work on student hours.. 7 students can do a work in 10 hr, so we require 70 student hrs to finish the work..now till 1 pm student hrs(SH) spent=4*7=28.. remaining SH=7028=42.. each hr one student is added, so lets work out.. at 2pm SH left= 428.. at 3 pm SH left=349=25 at 4 pm SH left=2510=15 at 5 pm SH left=1511=4.. So here we see the SH from 5 to 6pm is 12 but we require only 4 ie 4/12 or they will take 1/3 of that hour=1/3*60=20 min.. ans 5:20 pm D
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It takes 7 high school students, working at identical constant individ
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Updated on: 11 Nov 2015, 23:09
It takes seven students working at x rate ten hours to paint 1 house.
\(7 * x * 10 = 1\) \(x = \frac{1}{70}\)
7 students paint from 9 AM to 1 PM (4 hours), resulting in \(\frac{4}{10}\) of the house being painted.
\(7 * \frac{1}{70} * 4 = \frac{4}{10}\)
This means there is \(\frac{6}{10}\) or \(\frac{42}{70}\) of the house left to paint.
At 2 PM, eight students paint: \(8 * \frac{1}{70} * 1 = \frac{7}{70}\)
At 3 PM, nine students paint: \(9 * \frac{1}{70} * 1 = \frac{8}{70}\)
You can see the pattern. By 5 PM, the total amount of the house painted is: \(\frac{8}{70} + \frac{9}{70} + \frac{10}{70} + \frac{11}{70} = \frac{38}{70}\)
By 6 PM, the amount of the house painted is \(\frac{50}{70}\)  this is more than what we needed painted. \(\frac{8}{70} + \frac{9}{70} + \frac{10}{70} + \frac{11}{70} + \frac{12}{70} = \frac{50}{70}\) So ultimately the house is finished between 5 and 6 PM, the only option of which is 5:20 PM.
Originally posted by chenm93 on 11 Nov 2015, 22:40.
Last edited by chenm93 on 11 Nov 2015, 23:09, edited 2 times in total.



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Re: It takes 7 high school students, working at identical constant individ
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11 Nov 2015, 22:47
chenm93 wrote: It takes seven students working at x rate ten hours to paint 1 house.
\(7 * x * 10 = 1\) \(x = \frac{1}{70}\)
7 students paint from 9 AM to 1 PM (4 hours), resulting in \(\frac{4}{10}\) of the house being painted.
\(7 * \frac{1}{70} * 4 = \frac{4}{10}\)
This means there is \(\frac{6}{10}\) or \(\frac{42}{70}\) of the house left to paint.
At 2 PM, seven students paint: \(7 * \frac{1}{70} * 1 = \frac{7}{70}\)
At 3 PM, eight students paint: \(8 * \frac{1}{70} * 1 = \frac{8}{70}\)
You can see the pattern. By 5 PM, the total amount of the house painted is: \(\frac{7}{70} + \frac{8}{70} + \frac{9}{70} + \frac{10}{70} = \frac{34}{70}\)
By 6 PM, the amount of the house painted is \(\frac{45}{60}\)  this is more than what we needed painted. \(\frac{7}{70} + \frac{8}{70} + \frac{9}{70} + \frac{10}{70} + \frac{11}{70} = \frac{45}{70}\) So ultimately the house is finished between 5 and 6 PM, the only option of which is 5:20 PM. Hi, had there been another value between 5 and 6 closer to 6, your answer would get you that answer... luckily there was no other choice between 5 and 6.. now the flaw in your approach.. the method is absolutely correct but you have taken the values wrong .. At 2pm there are 7+1=8 people working and not 7.. and so on for further hours...
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Re: It takes 7 high school students, working at identical constant individ
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11 Nov 2015, 23:06
Thanks! Good catch, let me fix that up.



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Re: It takes 7 high school students, working at identical constant individ
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26 Oct 2016, 07:25
shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM the easiest way I found is to suppose we have a job consisting 70 parts. in 1 hour, 7 students can do 7 parts. each additional student would do 1 part every hour. since 7 students start at 9 AM, they would do 7 parts till 10AM, 14 parts till 11AM, 21 parts till 12PM, and 28 parts till 1PM. starting 1PM, one more student is added. therefore, the total work done will increase with every hour by 1. 28+8 = 36  parts were done by 2PM 36+9 = 45  parts were done by 3PM 45+10 = 55  parts were done by 4PM 55+11 = 66  parts were done by 5 PM since we are left with 4 parts, we know for sure that it would be done in less than 1 hour. therefore, all but D are eliminated.
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Re: It takes 7 high school students, working at identical constant individ
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02 Nov 2016, 07:11
chenm93 wrote: It takes seven students working at x rate ten hours to paint 1 house.
\(7 * x * 10 = 1\) \(x = \frac{1}{70}\)
7 students paint from 9 AM to 1 PM (4 hours), resulting in \(\frac{4}{10}\) of the house being painted.
\(7 * \frac{1}{70} * 4 = \frac{4}{10}\)
This means there is \(\frac{6}{10}\) or \(\frac{42}{70}\) of the house left to paint.
At 2 PM, eight students paint: \(8 * \frac{1}{70} * 1 = \frac{7}{70}\)
At 3 PM, nine students paint: \(9 * \frac{1}{70} * 1 = \frac{8}{70}\)
You can see the pattern. By 5 PM, the total amount of the house painted is: \(\frac{8}{70} + \frac{9}{70} + \frac{10}{70} + \frac{11}{70} = \frac{38}{70}\)
By 6 PM, the amount of the house painted is \(\frac{50}{70}\)  this is more than what we needed painted. \(\frac{8}{70} + \frac{9}{70} + \frac{10}{70} + \frac{11}{70} + \frac{12}{70} = \frac{50}{70}\) So ultimately the house is finished between 5 and 6 PM, the only option of which is 5:20 PM. Your explanation is understandable as it follows all the concepts of Rate & work... Had there been another option between 56 PM what would be your answer??



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Re: It takes 7 high school students, working at identical constant individ
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02 Nov 2016, 07:58
shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM Let efficiency of each student be = 1 So, Total Work is 7*1*10 = 70 man day hours 9AM to 1 PM = 28 Man day hours 1PM to 2 PM = 8 Man day hours ( Total Work Completed is 36 Man day hours ) 2PM to 3 PM = 9 Man Day hours ( Total Work Completed is 45 Man day hours ) 3PM to 4 PM = 10 Man Day hours ( Total Work Completed is 55 Man day hours ) 4PM to 5 PM = 11 Man Day hours ( Total Work Completed is 66 Man day hours ) Efficiency is now 12 units/hour and only 4 units are left Time required is 4/12*60 Minutes = 20 minutes.. Hence total time required is 5PM + 20 minutes = 5:20 PM Thus answer will be (D) 5:20 PM
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Re: It takes 7 high school students, working at identical constant individ
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11 Nov 2016, 09:33
chetan2u wrote: shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM Hi, there can be various ways.. the simpler way would be to work on student hours.. 7 students can do a work in 10 hr, so we require 70 student hrs to finish the work..now till 1 pm student hrs(SH) spent=4*7=28.. remaining SH=7028=42.. each hr one student is added, so lets work out.. at 2pm SH left= 428.. at 3 pm SH left=349=25 at 4 pm SH left=2510=15 at 5 pm SH left=1511=4.. So here we see the SH from 5 to 6pm is 12 but we require only 4 ie 4/12 or they will take 1/3 of that hour=1/3*60=20 min.. ans 5:20 pm D I did this after I realized that the algebraic method was not gonna work.. until 1 PM 28 studenthours is done..so left > 70  28 = 42 Now an Arithmetic Progression would start from the next hour beginning with 8 studenthours, then 9 sh and so forth..got a quadratic to solve and was stuck! Please tell me how to decide when manual counting works better???!
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Re: It takes 7 high school students, working at identical constant individ
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15 Nov 2016, 18:30
Rate per worker = 1 house/70 hours @ 1pm: 5(7)(1/70)+(1/70) = 28/70 completed @ 2pm: (5)(7)(1/70)+(2/70)+(1/70) = 36/70 completed @ 3pm: (6)(7)(1/70)+(2/70)+(1/70)=45/70 completed @ 4pm: 7(7)(1/70)+(3/70)+(2/70)+(1/70)=55/70 completed 5pm: (8)(7)(1/70)+(4/70)+(3/70)+(2/70)+(1/70) = 66/70 completed We know from this point that we will need a few more minutes to reach 70/70  thus, D is the correct answer.



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Re: It takes 7 high school students, working at identical constant individ
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09 Jan 2017, 20:41
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It takes 7 high school students, working at identical constant individ
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16 Feb 2017, 23:22
shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM Official solution from Veritas Prep. If 7 students can complete a house in 10 hours, their combined hourly rate is \(\frac{1}{10}\). We can divide this by 7 to find the rate for one student: \(\frac{1}{70}\), 7 students start at 9:00AM and paint for 4 hours by themselves; at the end of this period (1:00PM), they have completed \(\frac{4}{70}\) (or \(\frac{28}{70}\)) of the house. During the next hour, 8 students complete \(\frac{8}{70}\) of the house, giving us a total of \(\frac{36}{70}\) completed at 2:00PM. During the next hour, 9 students paint \(\frac{9}{70}\) of the house, making the total \(\frac{45}{70}\) at 3:00PM. During the next hour, 10 students paint \(\frac{10}{70}\) of the house, bringing the total completed to \(\frac{55}{70}\) at 4:00PM. At this point, we can see that we will need more than 1 more hour but not 2 more hours; the finishing time must be between 5:00PM and 6:00PM. The correct answer is D.
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Re: It takes 7 high school students, working at identical constant individ
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04 May 2018, 02:51
OFFICIAL VERITAS SOLUTION If 7 students can complete a house in \(10\) hours, their combined hourly rate is \(\frac{1}{10}\). We can divide this by \(7\) to find the rate for one student: \(\frac{1}{70}\). \(7\) students start at 9:00AM and paint for \(4\) hours by themselves; at the end of this period (1:00PM), they have completed \(\frac{4}{10}\) or \(\frac{28}{70}\) of the house. During the next hour, \(8\) students complete \(\frac{8}{70}\) of the house, giving us a total of \(\frac{36}{70}\) completed at 2:00PM. During the next hour, \(9\) students paint \(\frac{9}{70}\) of the house, making the total \(\frac{45}{70}\) at 3:00PM. During the next hour, \(10\) students paint \(\frac{10}{70}\) of the house, bringing the total completed to \(\frac{55}{70}\) at 4:00PM. At this point, you can see that \(\frac{11}{70}\) of the house will be done in the next hour so \(\frac{66}{70}\) are done at 5pm. The finishing time must be between 5:00PM and 6:00PM as the next hour will take you beyond \(\frac{70}{70}\) and only one answer lies in that time frame. The correct answer is D.
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Re: It takes 7 high school students, working at identical constant individ
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07 May 2018, 10:24
shasadou wrote: It takes 7 high school students, working at identical constant individual rates, 10 hours to paint a certain house. At what time will the house be fully painted if 7 students start painting at 9 AM and one student, working at the same rate, is added per hour starting at 1 PM?
A. 3:00 PM B. 4:30 PM C. 5:00 PM D. 5:20 PM E. 6:20 PM Since the rate of 7 students is 1/10, the rate of one student is 1/70 and therefore the rate of any n students is n/70. By 1 PM ( 4 hours after 9 AM), the 7 students have completed 4 x 7/70 = 4 x 1/10 = 4/10 of the house. Since 1 more person is added at 1 PM, by 2 PM (1 hour after 1 PM), the 8 students have completed another 1 x 8/70 = 8/70 of the house, and thus 4/10 + 8/70 = 28/70 + 8/70 = 36/70 of the house has been painted.. Since 1 more person is added at 2 PM, by 3 PM (1 hour after 2 PM), the 9 students have completed another 1 x 9/70 = 9/70 of the house, and thus 36/70 + 9/70 = 45/70 of the house has been painted.. Since 1 more person is added at 3 PM, by 4 PM (1 hour after 3 PM), the 10 students have completed another 1 x 10/70 = 10/70 of the house, and thus 45/70 + 10/70 = 55/70 of the house has been painted. Since 1 more person is added at 4 PM, by 5 PM (1 hour after 4 PM), the 11 students have completed another 1 x 11/70 = 11/70 of the house, and thus 55/70 + 11/70 = 66/70 of the house has been painted. Since 1 more person is added at 5 PM, by 6 PM (1 hour after 4 PM), the 12 students have completed another 1 x 12/70 = 12/70 of the house and thus 66/70 + 12/70 = 78/70 of the house has been painted. However, this is more than one whole house. Thus, they must have finished the job some time before 6 PM (but after 5 PM). The only time that fits into this criterion is 5:20 PM. Thus, choice D is the answer. Answer: D
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