NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 19:46 It is currently 25 Apr 2024, 19:46
Decision Tracker
My Rewards
New posts
New comers' posts

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If the population of a town is p in the beginning of any year then it

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619039 [10]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
If the population of a town is p in the beginning of any year then it [#permalink] New post   03 Apr 2020, 03:17
10
Bookmarks
Expert Reply
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

64% (02:45) correct 36% (02:35) wrong based on 88 sessions
Hide Show timer Statistics
If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be


A. \((997)2^{14} - 3\)

B. \((997)2^{14} + 3\)

C. \((1003)2^{15} − 3\)

D. \((997)^{15} − 3\)

E. \((1003)^{15} + 6\)


Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS


Are You Up For the Challenge: 700 Level Questions
ShowHide Answer
Official Answer
C
User avatar
L
CareerGeek
VP
VP
Joined: 20 Jul 2017
Posts: 1300
Own Kudos [?]: 3450 [3]
Given Kudos: 162
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 690 Q51 V30
WE:Education (Education)
Send PM
Re: If the population of a town is p in the beginning of any year then it [#permalink] New post   03 Apr 2020, 04:12
3
Kudos
Bunuel wrote:
If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

A. \((997)2^{14} - 3\)
B. \((997)2^{14} + 3\)
C. \((1003)2^{15} − 3\)
D. \((997)^{15} − 3\)
E. \((1003)^{15} + 6\)


2019 --> \(1000\)
After 1 year, Population = \(3 + 2*1000 = 3 + 2(1003 - 3) = 2(1003) - 3\)
After 2 years, Population = \(3 + 2[2(1003) - 3)] = 3 + 2^2(1003) - 6 = 2^2(1003) - 3\)
After 3 years, Population = \(3 + 2[2^2(1003) - 3)] = 3 + 2^3(1003) - 6 = 2^3(1003) - 3\)
.
.
.
After n years, Population = \(2^n(1003) - 3\)

In 2034, i.e, after 15 years, , Population = \(2^{15}(1003) - 3\)

Option C
avatar
B
prathyushaR
Intern
Intern
Joined: 14 Jan 2020
Posts: 26
Own Kudos [?]: 42 [3]
Given Kudos: 27
GMAT 1: 760 Q51 V41
Send PM
Re: If the population of a town is p in the beginning of any year then it [#permalink] New post   03 Apr 2020, 05:14
2
Kudos
1
Bookmarks
so the patter goes as:(let years from start is -n)

\(n=0: p\)

\(n=1:3+2p =3+2p\)

\(n=2:3+2*(3+2p)=3+3*2+2*2p = 3+3*2^1+2^1*2p=3(1+2^1)+2^2*p\)

\(n=3:3+2*(3+3*2+2*2p)=3+3*2+3*2*2+2*2*2p =3+3*2^1+3*2^2+2^2*p=3(1+2^1+2^2)+2^2*p\)

\(n=4:3+2*(3+3*2+3*2*2+2*2*2p)=3+3*2+3*2*2+3*2*2*2+2*2*2*2p=3+3*2^1+3*2^2+3*2^3+2^3*p=\)
\(= 3(1+2^1+2^2+2^3)+2^3*p\)


we can see it follows the pattern \(3(1+2^1+2^2+...2^{n-1} )+2^n*p\) which comes out to be\(3(2^n-1)+2^n p\) :GeometricProgression

at 2019 we have \(n=0 , p=1000\)

at 2034 we have\( n= 15 : 3*(2^{15}-1)+ 2^{15}*1000= 3*2^{15}-3+2^{15}*1000=2^{15}(1003)-3\)

ANSWER:C
User avatar
G
preetamsaha
Senior Manager
Senior Manager
Joined: 14 Oct 2019
Status:Today a Reader; Tomorrow a Leader.
Posts: 346
Own Kudos [?]: 344 [1]
Given Kudos: 127
Location: India
GPA: 4
WE:Engineering (Energy and Utilities)
Send PM
Re: If the population of a town is p in the beginning of any year then it [#permalink] New post   06 Apr 2020, 08:17
1
Kudos
Dillesh4096 3+2∗1000=3+2(1003−3)=2(1003)−3 how did u find this pattern? how will I able to make this pattern in the exam hall?
User avatar
S
gurmukh
Senior Manager
Senior Manager
Joined: 18 Dec 2017
Posts: 270
Own Kudos [?]: 203 [2]
Given Kudos: 20
Send PM
Re: If the population of a town is p in the beginning of any year then it [#permalink] New post   06 Apr 2020, 08:39
2
Kudos
This sum is not checking your maths.
Total there are 15 years
According to question Population after 1 year = 2003
According to option C
(1003)×2 -3 =2003
According to option D
997-3 = 994 ruled out
According to option A
(997)×2^15/2 - 3
997 -3
994 ruled out
Similarly other options are ruled out.
Answer is C, no conventional calculation required.

Posted from my mobile device
User avatar
G
reynaldreni
Manager
Manager
Joined: 07 May 2015
Posts: 80
Own Kudos [?]: 119 [2]
Given Kudos: 152
Location: India
Schools: Darden '21
GPA: 4
Send PM
If the population of a town is p in the beginning of any year then it [#permalink] New post   02 Sep 2020, 07:51
2
Kudos
Bunuel wrote:
If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be


A. \((997)2^{14} - 3\)

B. \((997)2^{14} + 3\)

C. \((1003)2^{15} − 3\)

D. \((997)^{15} − 3\)

E. \((1003)^{15} + 6\)


Project PS Butler


Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS


Are You Up For the Challenge: 700 Level Questions

This problem can be solved by analyzing patterns in the progressions. The Population increases as follows:
Year-0 (2019): P
Year-1 (2020): 2P+3
Year-2 (2021): 4P+9
Year-3 (2022): 8P+21
Year-4 (2023): 16P+45
Year-5 (2024): 32P+93
From the above the progression we see there are two parts (a) a Power of a 2 and (b) multiples of 3

Every subsequent year the (a) increases by 2^(n). That is in year 5 the part (a) is 2^5
So, for Year-15 (2034) the part (a) will 2^15P

We see similar patterns in the part (b),
Year 1: 3 = 3X1 = 3X(2^1 -1)
Year 2: 9 = 3X3 = 3X(2^2 -1)
Year 3: 21 = 3X7 = 3X(2^3 -1)
So, for Year 15 (2034): 3X(2^15-1)

combining (a)+(b), we get
= 2^15P + 3X(2^15 -1)
= 2^15P + 3*2^15 - 3
= 2^15(P+3) - 3
We know P=1000,
=2^15(1000+3) - 3
=2^15(1003) - 3

[Option C]
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32680
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: If the population of a town is p in the beginning of any year then it [#permalink] New post   07 Nov 2021, 23:00
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.
GMAT Club Bot
gmatclubot
Re: If the population of a town is p in the beginning of any year then it [#permalink]
07 Nov 2021, 23:00
Moderators:
Bunuel
Math Expert
92915 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions