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CAT 2019 Shift 1 Paper

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Question : 76 of 100

Marks: +1, -0

If the population of a town is p in the beginning of any year then it becomes 3+2p inthe beginning of the next year. If the population in the beginning of 2019 is 1000, thenthe population in the beginning of 2034 will be

$(1003)^{15} + 6$
$(997)2^{14} + 3$
$(1003)2^{15} - 3$
$(997)^{15} - 3$

Solution:

We can transform each of the options for 'n' years.

(997) 2^{14}+3 \equiv (p-3) 2^{n-1}+3

(1003) 2^{15}+6 \equiv (p+3) 2^{n}+6

(1003) 2^{15}-3 \equiv (p+3) 2^{n}-3

(997)^{15}-3 \equiv (p-3)^{n}-3

As per the condition, in one year, the population '

p'

becomes

' 3+2 p'

Putting the value of

n=1

in each option, and checking to get

3+2p

, we have

(p-3) 2^{n-1}+3 \equiv 3 \neq 3+2p

(p+3) 2^{n}+6 \equiv (p+3) 2+6 \neq 3+2p

(1003) 2^{15}-3 \equiv (p+3) 2-3=3+2p

(p-3)^{n}-3 \equiv (p-3)-3 \neq p-6

Hence, the right answer is option

3 .

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