It takes 7 high school students, working at identical constant individ

Question

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

A. 3:00 PM
B. 4:30 PM
C. 5:00 PM
D. 5:20 PM
E. 6:20 PM

chetan2u · Accepted Answer

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=70-28=42..
each hr one student is added, so lets work out..
at 2pm SH left= 42-8..
at 3 pm SH left=34-9=25
at 4 pm SH left=25-10=15
at 5 pm SH left=15-11=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

chenm93 · Answer

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.

chetan2u · Answer

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...

chenm93 · Answer

Thanks! Good catch, let me fix that up.

mvictor · Answer

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.

suramya26 · Answer

Your explanation is understandable as it follows all the concepts of Rate & work...
Had there been another option between 5-6 PM what would be your answer??

Abhishek009 · Answer

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

ShashankDave · Answer

I did this after I realized that the algebraic method was not gonna work..
until 1 PM 28 student-hours is done..so left --> 70 - 28 = 42
Now an Arithmetic Progression would start from the next hour beginning with 8 student-hours, then 9 s-h and so forth..got a quadratic to solve and was stuck!
Please tell me how to decide when manual counting works better???!

law258 · Answer

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.

WilDThiNg · Answer

bb  - Points look quite cool on the user's DP! Unsure where else to post this comment, hence posting here

BillyZ · Answer

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.

AweG · Answer

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.

JeffTargetTestPrep · Answer

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

zebu · Answer

In this problem total working hour is 70. Let's assume that 140 is the number of work, then the rate of work is 2.
If we start the work at 9:00 am up to 1:00 pm, we will able to finish 7*2*4 =56 works. From 1:00 pm onwards the work completed will follows as (7+1)*2+(7+1+1)*2+ 9*2+10*2+11*2+ 12*2/3 = 86 works

So total hour is 8 hour and 20 minutes, The answer is 5:20 Pm

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