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03 Apr 2020, 03:17
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C
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Difficulty:
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Question Stats:
66% (02:57) correct
34%
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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
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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
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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
