GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Aug 2019, 17:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)
It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

11 Nov 2015, 12:03
17
00:00

Difficulty:

95% (hard)

Question Stats:

53% (02:57) correct 47% (03:05) wrong based on 224 sessions

### HideShow timer Statistics

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

_________________
KUDO me plenty
Math Expert
Joined: 02 Aug 2009
Posts: 7755
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

11 Nov 2015, 21:06
7
4
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=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
_________________
##### General Discussion
Intern
Joined: 30 Oct 2015
Posts: 9
GPA: 3.89
WE: Analyst (Consulting)
It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

Updated on: 11 Nov 2015, 23:09
2
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.
Math Expert
Joined: 02 Aug 2009
Posts: 7755
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

11 Nov 2015, 22:47
2
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,
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...
_________________
Intern
Joined: 30 Oct 2015
Posts: 9
GPA: 3.89
WE: Analyst (Consulting)
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

11 Nov 2015, 23:06
Thanks! Good catch, let me fix that up.
Board of Directors
Joined: 17 Jul 2014
Posts: 2531
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

26 Oct 2016, 07:25
1
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.
_________________
Manager
Joined: 02 Jul 2016
Posts: 112
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

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...
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4573
Location: India
GPA: 3.5
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

02 Nov 2016, 07:58
1
1
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
_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Senior Manager
Joined: 03 Apr 2013
Posts: 268
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

11 Nov 2016, 09:33
chetan2u 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=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

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???!
_________________
Spread some love..Like = +1 Kudos
Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 368
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

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.
Manager
Joined: 29 Dec 2014
Posts: 66
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

09 Jan 2017, 20:41
bb - Points look quite cool on the user's DP! Unsure where else to post this comment, hence posting here
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1323
Location: Malaysia
It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

16 Feb 2017, 23:22
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.

_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Manager
Joined: 05 Dec 2016
Posts: 105
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

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.

_________________
lets all unite to reach our target together
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2820
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

07 May 2018, 10:24
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.

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Non-Human User
Joined: 09 Sep 2013
Posts: 12054
Re: It takes 7 high school students, working at identical constant individ  [#permalink]

### Show Tags

02 Aug 2019, 03:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: It takes 7 high school students, working at identical constant individ   [#permalink] 02 Aug 2019, 03:13
Display posts from previous: Sort by