A model estimates that at the end of each year from 2015 to 2020, the number of squirrels in a...
GMAT Advanced Math : (Adv_Math) Questions
A model estimates that at the end of each year from 2015 to 2020, the number of squirrels in a population was 150% more than the number of squirrels in the population at the end of the previous year. The model estimates that at the end of 2016, there were 180 squirrels in the population. Which of the following equations represents this model, where \(\mathrm{n}\) is the estimated number of squirrels in the population \(\mathrm{t}\) years after the end of 2015 and \(\mathrm{t \leq 5}\)?
\(\mathrm{n = 72(1.5)^t}\)
\(\mathrm{n = 72(2.5)^t}\)
\(\mathrm{n = 180(1.5)^t}\)
\(\mathrm{n = 180(2.5)^t}\)
1. TRANSLATE the percent increase
- Given information:
- Population is "150% more" each year than the previous year
- At end of 2016: 180 squirrels
- t = years after end of 2015
- What "150% more" means mathematically:
- New population = Old population + 150% of old population
- New population = Old population × (1 + 1.5) = Old population × 2.5
- Growth factor = 2.5
2. INFER the exponential model structure
- Since population grows by a fixed factor each year, this is exponential growth
- General form: \(\mathrm{n = n_0 \times (2.5)^t}\)
- Strategy: Use the given data point to find \(\mathrm{n_0}\) (initial population at end of 2015)
3. SIMPLIFY to find the initial population
- At end of 2016, \(\mathrm{t = 1}\) and \(\mathrm{n = 180}\)
- Substitute: \(\mathrm{180 = n_0 \times (2.5)^1}\)
- Solve: \(\mathrm{n_0 = \frac{180}{2.5} = 72}\)
4. Write the final equation
- \(\mathrm{n = 72(2.5)^t}\)
Answer: B. n = 72(2.5)t
Why Students Usually Falter on This Problem
Most Common Error Path:
Poor TRANSLATE reasoning: Students often interpret "150% more" as meaning "multiply by 1.5" instead of "multiply by 2.5"
They think: "150% more means 150% = 1.5, so growth factor is 1.5"
But "150% more" actually means: Original + 150% of original = 1.0 + 1.5 = 2.5 times the original
This leads them to incorrectly use growth factor 1.5, giving them \(\mathrm{n_0 = \frac{180}{1.5} = 120}\), and they select Choice A (n = 120(1.5)t) - but this isn't even an option! This forces them to guess or get confused.
Second Most Common Error:
Weak INFER skill: Students might correctly get the growth factor as 2.5 but then use the wrong approach to find \(\mathrm{n_0}\)
They might think \(\mathrm{t = 0}\) corresponds to 2016 instead of 2015, leading them to use \(\mathrm{n_0 = 180}\) directly and select Choice D (n = 180(2.5)t)
The Bottom Line:
The trickiest part is correctly interpreting "150% more than" - it means the new amount is 250% of the original (or 2.5 times), not 150% of the original. Once you get the growth factor right, the rest follows standard exponential model setup.
\(\mathrm{n = 72(1.5)^t}\)
\(\mathrm{n = 72(2.5)^t}\)
\(\mathrm{n = 180(1.5)^t}\)
\(\mathrm{n = 180(2.5)^t}\)