retagged by
902 views
2 votes
2 votes

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

  1. $(997)2^{14}+3$
  2. $(1003)^{15}+6$
  3. $(1003)2^{15}-3$
  4. $(997)^{15}-3$
retagged by

1 Answer

Best answer
3 votes
3 votes

Given that, the population of a town is $\text{‘P’}$ at the beginning of any year, then the population becomes $\text{‘3+2P’}$ at the beginning of the next year.  And the population at the beginning of $2019$ is $1000$.

Now, we can write each year's population in the following manner.

  • $\text{Year – 0:}$ Beginning of $ 2019  \longrightarrow \text{P} = 1000 = (1003)2^{0} – 3$
  • $\text{Year – 1:}$ Beginning of  $ 2020 \longrightarrow 3+2\text{P} = 3+2\times1000 = 2003 = (1003)2^{1} – 3$
  • $\text{Year – 2:}$ Beginning of  $ 2021 \longrightarrow 3+2\text{P} = 3+2\times2003 = 3 + 4006 = 4009 = (1003)2^{2} – 3$
  • $\text{Year – 3:}$ Beginning of  $ 2022 \longrightarrow 3+2\text{P} = 3+ 2 \times 4009 = 3+8018 = 8021 = (1003)2^{3}-3$
  • $\vdots \;\; \vdots \;\; \vdots$
  • $\text{Year – n: } \boxed {(1003)2^{n}-3}, \; n= \text{number of years}$

$\therefore\; \text{Year – 15:}$ Beginning of $ 2034 \longrightarrow (1003)2^{n} – 3 = (1003)2^{15} – 3$

Correct Answer $: \text{C}$

edited by
Answer:

Related questions

1 votes
1 votes
1 answer
4