Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2212:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
03 Apr 2020, 03:17
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (02:58) correct
35%
(02:29)
wrong
based on 158
sessions
History
Date
Time
Result
Not Attempted Yet
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. \((1003)^{15} + 6\)
Are You Up For the Challenge: 700 Level Questions
A. \((997)2^{14} - 3\)
B. \((997)2^{14} + 3\)
C. \((1003)2^{15} − 3\)
D. \((997)^{15} − 3\)
E. \((1003)^{15} + 6\)
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 04:12
Kudos
Bookmarks
Bunuel
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
03 Apr 2020, 05:14
Kudos
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
\(n=0: p\)
\(n=1:3+2p =3+2p\)
\(n=2:3+2*(3+2p)=3+3*2+2*2p = 3+3*2^1+2^1*2p=3(1+2^1)+2^2*p\)
\(n=3:3+2*(3+3*2+2*2p)=3+3*2+3*2*2+2*2*2p =3+3*2^1+3*2^2+2^2*p=3(1+2^1+2^2)+2^2*p\)
\(n=4:3+2*(3+3*2+3*2*2+2*2*2p)=3+3*2+3*2*2+3*2*2*2+2*2*2*2p=3+3*2^1+3*2^2+3*2^3+2^3*p=\)
\(= 3(1+2^1+2^2+2^3)+2^3*p\)
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
