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A function p estimates that there were 2{,}000 animals in a population in 1998. Each year from 1998 to 2010,...

GMAT Advanced Math : (Adv_Math) Questions

Source: Practice Test
Advanced Math
Nonlinear functions
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A function \(\mathrm{p}\) estimates that there were \(2{,}000\) animals in a population in \(1998\). Each year from \(1998\) to \(2010\), the function estimates that the number of animals in this population increased by \(3\%\) of the number of animals in the population the previous year. Which equation defines this function, where \(\mathrm{p(x)}\) is the estimated number of animals in the population \(\mathrm{x}\) years after \(1998\)?

A
\(\mathrm{p(x) = 2{,}000(3)^x}\)
B
\(\mathrm{p(x) = 2{,}000(1.97)^x}\)
C
\(\mathrm{p(x) = 2{,}000(1.03)^x}\)
D
\(\mathrm{p(x) = 2{,}000(0.97)^x}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Initial population in 1998: 2,000 animals
    • Each year the population 'increased by 3% of the number of animals in the population the previous year'
    • Need function p(x) for population x years after 1998
  • What 'increased by 3%' means: The population becomes the previous amount PLUS 3% of the previous amount = \(\mathrm{100\% + 3\% = 103\%}\) of the previous amount

2. INFER the mathematical pattern

  • This is exponential growth since each year's population depends on multiplying the previous year by the same factor
  • The growth factor is 1.03 (since \(\mathrm{103\% = 1.03}\) in decimal form)
  • We need the exponential growth formula: \(\mathrm{p(x) = initial\_amount \times (growth\_factor)^x}\)

3. TRANSLATE values into the formula

  • Initial amount (a) = 2,000
  • Growth factor = 1.03
  • Time variable = x years after 1998
  • Therefore: \(\mathrm{p(x) = 2{,}000(1.03)^x}\)

4. Verify by checking the answer choices

Looking at the options, Choice C matches our derived formula exactly.

Answer: C. \(\mathrm{p(x) = 2{,}000(1.03)^x}\)




Why Students Usually Falter on This Problem


Most Common Error Path:

Poor TRANSLATE reasoning: Students misinterpret 'increased by 3%' as meaning the population gets multiplied by 3 each year, rather than understanding it means the population becomes 103% of the previous year's amount.

This leads them to think the growth factor should be 3, causing them to select Choice A (\(\mathrm{p(x) = 2{,}000(3)^x}\)).


Second Most Common Error:

Weak TRANSLATE skill: Students correctly understand that 3% growth means multiplying by something greater than 1, but get confused about whether an increase means adding or subtracting the percentage from 1.

They might think '3% increase' means \(\mathrm{1 - 0.03 = 0.97}\), leading them to select Choice D (\(\mathrm{p(x) = 2{,}000(0.97)^x}\)), which actually represents a 3% decrease.


The Bottom Line:

The key challenge is accurately translating percentage language into mathematical operations. Students must understand that 'increased by r%' means 'becomes (100 + r)% of the original,' not just 'multiply by r.'

Answer Choices Explained
A
\(\mathrm{p(x) = 2{,}000(3)^x}\)
B
\(\mathrm{p(x) = 2{,}000(1.97)^x}\)
C
\(\mathrm{p(x) = 2{,}000(1.03)^x}\)
D
\(\mathrm{p(x) = 2{,}000(0.97)^x}\)
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