If the population of a town is p in the beginning of any year then it becomes 3+2p GMAT Problem Solving

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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

  1. \((997)2^{14}-3\)
  2. \((997)2^{14}+3\)
  3. \((1003)2^{15}−3\)
  4. (997)^15−3
  5. (1003)^15+6

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’ - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide 2022”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

The population at the beginning of 2019 is 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 =
\(2^{15}(1003)−3\)

Hence, we can rearrange and state this as \(2^{15}(1003)−3\)

The correct answer is C.

Correct Answer: C

Model Answer 2
Explanation:

This problem can be solved by analyzing patterns in the progressions.
As per the question, the Population increases as follows:
Let us consider the population is P
Year-0 (2019): P
Year-1 (2020): 2P+3
Year-2 (2021): 4P+9
Year-3 (2022): 8P+21
Year-4 (2023): 16P+45
Year-5 (2024): 32P+93
From the above the progression we see there are two parts (a) a Power of a 2 and (b) multiples of 3

Every subsequent year the (a) increases by 2^(n). That is in year 5 the part (a) is 2^5
So, for Year-15 (2034) the part (a) will 2^15P
We see similar patterns in the part (b),
Year 1: 3 = 3X1 = 3X(2^1 -1)
Year 2: 9 = 3X3 = 3X(2^2 -1)
Year 3: 21 = 3X7 = 3X(2^3 -1)
So, for Year 15 or 2034: 3X(2^15-1)

combining (a)+(b), we get
= 2^15P + 3X(2^15 -1)
= 2^15P + 3*2^15 - 3
= 2^15(P+3) - 3
As per the question, we know P=1000 in 2019,
Hence,
=2^15(1000+3) - 3
=2^15(1003) - 3
The correct answer is C.

Correct Answer: C

Approach Solution 3
Population 2019 --> 10001000
After 1 year, Population = 3+2∗1000=3+2(1003−3)=2(1003)−33+2∗1000=3+2(1003−3)=2(1003)−3
After 2 years, Population = 3+2[2(1003)−3)]=3+22(1003)−6=22(1003)−33+2[2(1003)−3)]=3+22(1003)−6=22(1003)−3
After 3 years, Population = 3+2[22(1003)−3)]=3+23(1003)−6=23(1003)−33+2[22(1003)−3)]=3+23(1003)−6=23(1003)−3
After n years, Population = 2n(1003)−32n(1003)−3
In 2034, i.e, after 15 years, so the Population becomes 215(1003)−3215(1003)−3

The correct answer is C.
Correct Answer
: C

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