Sep 27 08:00 PM PDT  09:00 PM PDT Working in collaboration with examPAL we will provide you with a unique online learning experience which will help you reach that higher score. Start your free 7 day trial today. Oct 02 08:00 PM PDT  09:00 PM PDT LEARN WITH AN EXPERT TEACHER—FOR FREE  Take a free practice test, learn content with one of our highestrated teachers, or challenge yourself with a GRE Workshop. Oct 07 07:30 AM PDT  08:30 AM PDT An overview of and discussion about the GRE argument essay. The first 20 minutes of this webinar will consist of a “presentation” on a specific topic, and the last 40 minutes consists of live Q&A covering presubmitted or live questions. Oct 07 09:00 PM PDT  11:00 PM PDT This admissions guide will help you plan your best route to a PhD by helping you choose the best programs your goals, secure strong letters of recommendation, strengthen your candidacy, and apply successfully.
Author 
Message 
TAGS:


Retired Moderator
Joined: 07 Jun 2014
Posts: 4803
WE: Business Development (Energy and Utilities)
Followers: 171
Kudos [?]:
2921
[0], given: 394

A cockroach population doubles every 3 days. In 30 days, by [#permalink]
02 Jun 2018, 03:39
Question Stats:
65% (01:13) correct
34% (00:59) wrong based on 44 sessions
A cockroach population doubles every 3 days. In 30 days, by what percent would a cockroach population increase? (A) 900% (B) 1,000% (C) 9,999% (D) 102,300% (E) 102,400%
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test




Active Member
Joined: 29 May 2018
Posts: 126
Followers: 0
Kudos [?]:
110
[2]
, given: 4

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
10 Jun 2018, 07:04
2
This post received KUDOS
sandy wrote: A cockroach population doubles every 3 days. In 30 days, by what percent would a cockroach population increase?
(A) 900% (B) 1,000% (C) 9,999% (D) 102,300% (E) 102,400% Given every 3 days population doubles and this continues for 30 days. This double increase overall happens for 10 times. (30/3) Now let's take the initial population is 10. Now for every 3 days population doubles and total 10 times. 10  20  40  80  160  320  640  1280  2560  5120  10240. Exactly after 10th time i.e. 30 days , population is 10240. Now take the percentage increase = 10240  10 / 10 = 10230 / 10 = 1023 = 102300%. Option D.



Retired Moderator
Joined: 07 Jun 2014
Posts: 4803
WE: Business Development (Energy and Utilities)
Followers: 171
Kudos [?]:
2921
[0], given: 394

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
11 Jul 2018, 13:36
ExplanationThe percent increase is the difference between the amounts divided by the original, converted to a percent. If the population doubles, mathematically the increase can be written as a power of 2. In the 30day interval, if the original population is 1, it will double to 2 after three days —so, 21 represents the population after the first increase, the second increase would then be 22 and so on. Since there are 10 increases, the final population would be 210 or 1,024. Therefore, the difference, 1,024 – 1, is 1,023. Use the percent change formula to calculate percent increase: Percent Change=\((\frac{Difference}{original} \times 100)\) Percent Change=\((\frac{1023}{1} \times 100)=102300 \%\) Note that the new number is 102,400% of the original, but that was not the question asked—the percent increase is 102,300%.
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Director
Joined: 22 Jun 2019
Posts: 517
Followers: 4
Kudos [?]:
104
[1]
, given: 161

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
17 Sep 2019, 04:55
1
This post received KUDOS
sandy wrote: Explanation
The percent increase is the difference between the amounts divided by the original, converted to a percent. If the population doubles, mathematically the increase can be written as a power of 2. In the 30day interval, if the original population is 1, it will double to 2 after three days —so, 21 represents the population after the first increase, the second increase would then be 22 and so on. Since there are 10 increases, the final population would be 210 or 1,024. Therefore, the difference, 1,024 – 1, is 1,023. Use the percent change formula to calculate percent increase:
Percent Change=\((\frac{Difference}{original} \times 100)\) Percent Change=\((\frac{1023}{1} \times 100)=102300 \%\)
Note that the new number is 102,400% of the original, but that was not the question asked—the percent increase is 102,300%. FROM 2ND LINE 21 represents the population after the first increase, the second increase would then be 22 This will be: \(2^1\) represents the population after the first increase, the second increase would then be \(2^2\)
_________________
New to the GRE, and GRE CLUB Forum? GRE: All About GRE  Search GRE Specific Questions  Download Vault Posting Rules: QUANTITATIVE  VERBAL
Questions' Banks and Collection: ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation.  ETS All Official Guides 3rd Party Resource's: All In One Resource's  All Quant Questions Collection  All Verbal Questions Collection  Manhattan 5lb All Questions Collection Books: All GRE Best Books Scores: Average GRE Score Required By Universities in the USA Tests: All Free & Paid Practice Tests  GRE Prep Club Tests Extra: Permutations, and Combination Vocab: GRE Vocabulary Facebook GRE Prep Group: Click here to join FB GRE Prep Group



Intern
Joined: 30 Aug 2020
Posts: 2
Followers: 0
Kudos [?]:
1
[0], given: 2

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
30 Aug 2020, 09:02
What I am confused by is the fact that if we use the equation of a geometric series, we get the value of the tenth term to be \(512*(\text{the original population})\). This is because we know that \(a_n=a_1*r^{n1}\). Using this formula, given that r=2, \(a_n=a_1*2^{101}=512*a_1\).
How is it that we cannot use the geometric series to find the 10th term in this setting?



GRE Instructor
Joined: 10 Apr 2015
Posts: 3835
Followers: 148
Kudos [?]:
4490
[2]
, given: 69

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
30 Aug 2020, 09:16
2
This post received KUDOS
dazedandconfused wrote: What I am confused by is the fact that if we use the equation of a geometric series, we get the value of the tenth term to be \(512*(\text{the original population})\). This is because we know that \(a_n=a_1*r^{n1}\). Using this formula, given that r=2, \(a_n=a_1*2^{101}=512*a_1\).
How is it that we cannot use the geometric series to find the 10th term in this setting? If \(a_1\) represents the INITIAL population, then \(a_2\) represents the population after 1 "doubling" So, \(a_2=a_1*2^{1}=a_1*(2)\) Likewise, \(a_3\) represents the population after 2 "doublings" So, \(a_3=a_1*2^{2}=a_1*(4)\) Likewise, \(a_4\) represents the population after 3 "doublings" So, \(a_4=a_1*2^{3}=a_1*(8)\) In general, \(a_n\) represents the population after n1 "doublings" So, \(a_n=a_1*r^{n1}\) 30/3 = 10 So, the population gets doubled 10 times. Since \(a_n\) represents the population after n1 "doublings," we need to find the value of \(a_{11}\) We get: \(a_{11}=a_1*2^{10}=a_1*1024\) Cheers, Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.com If you enjoy my solutions, you'll like my GRE prep course.
Sign up for GRE Question of the Day emails



Intern
Joined: 30 Aug 2020
Posts: 2
Followers: 0
Kudos [?]:
1
[1]
, given: 2

Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
30 Aug 2020, 09:27
1
This post received KUDOS
Hi Brent,
Just reread your response. This was very very helpful. Thanks so much!




Re: A cockroach population doubles every 3 days. In 30 days, by
[#permalink]
30 Aug 2020, 09:27





