Question

The capacity of a school is 1100 students and the current enrollment is 980 students. If the student population increases at a rate of 5 percent per year, what is the smallest integer $n$ such that the enrollment will exceed the capacity in $n$ years?

Answer 1

Answer:

3 years

Step-by-step explanation:

In this problem, the initial number of students at time t = 0 is

$n_0 = 980$

We know that the number of students increases by 5 % every year. This means that we can write the student's population as

$n(t)=(1.05)^t n_0$

where

t is the time, measured in years

Here we want to find after how many years t the student's population will exceed the maximum capacity of the school, which is

N = 1100

To solve the problem, we just put n = 1100 and we solve for t. We find:

$1100 = (1.05)^t n_0\\t=log_{1.05} (\frac{1100}{980})=2.37 y$

Which means that the student's population reaches the maximum capacity of the school after 2.37 years. Since we want this number to be an integer, this means that the enrollment will exceed the capacity in 3 years.

Answer 2

Answer:

3 years

Step-by-step explanation: