IB Mathematics SL 2.5 Exponential growth and decay models AI SL Paper 1- Exam Style Questions- New Syllabus
Question
An entomologist is monitoring the population of a newly established ant colony. At the beginning of the first week, the population is recorded as \( 200 \) ants. The population is observed to grow at a consistent rate, increasing by \( 15\% \) each week.
(a) Calculate the expected number of ants in the colony at the start of the \( 11^{\text{th}} \) week.
The colony’s population first exceeds \( 3000 \) individuals during week \( k \).
(b) Determine the value of \( k \).
Most appropriate topic codes (IB Mathematics: applications and interpretation):
• SL 1.3: Geometric sequences and series; use of the formula for the \( n^{\text{th}} \) term — part (a)
• SL 2.5: Modelling with exponential growth functions; growth and decay — part (b)
• SL 2.5: Modelling with exponential growth functions; growth and decay — part (b)
▶️ Answer/Explanation
(a)
This is a geometric sequence with \( u_1 = 200 \) and \( r = 1.15 \).
At the start of week 11: \( n = 11 \)
\( u_{11} = 200 \times (1.15)^{10} \)
\( = 200 \times 4.045557… \)
\( \approx 809.111… \)
\( \boxed{809 \text{ ants}} \) (accept 810)
(b)
We require \( 200 \times (1.15)^{k-1} > 3000 \).
Divide by 200: \( (1.15)^{k-1} > 15 \)
Take logs: \( (k-1) \ln(1.15) > \ln(15) \)
\( k-1 > \frac{\ln(15)}{\ln(1.15)} \approx \frac{2.70805}{0.13976} \approx 19.376… \)
Thus \( k > 20.376… \) so smallest integer \( k = 21 \).
Check: \( 200 \times (1.15)^{20} \approx 3273.30 \) > 3000
\( \boxed{21} \)