Question:
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
- $(997) 2^{14}+3$
- $(1003)^{15}+6$
- $(1003) 2^{15}-3$
- $(997)^{15}-3$
Correct Answer: Option: 3
We can transform each of the options for ‘n’ years.
$(997){{2}^{14}}+3\equiv \left( p-3 \right){{2}^{n-1}}+3$
$(1003){{2}^{15}}+6\equiv \left( p+3 \right){{2}^{n}}+6$
$(1003){{2}^{15}}-3\equiv \left( p+3 \right){{2}^{n}}-3$
${{(997)}^{15}}-3\equiv {{\left( p-3 \right)}^{n}}-3$
As per the condition, in one year, the population ‘p’ becomes ‘3+2p’
Putting the value of n =1 in each option, and checking to get 3+2p, we have
$\left( p-3 \right){{2}^{n-1}}+3\equiv 3\ne 3+2p$
$\left( p+3 \right){{2}^{n}}+6\equiv \left( p+3 \right)2+6\ne 3+2p$
$(1003){{2}^{15}}-3\equiv \left( p+3 \right)2-3=3+2p$
${{\left( p-3 \right)}^{n}}-3\equiv \left( p-3 \right)-3\ne p-6$
Hence, the right answer is option 3.
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