CAT 2019 Quant Question with Solution 29

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. $(997)^{15}-3$
Show Answer

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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