Question

Cory has 15 die-cast cars in his collection. Each year his collection increases by 20%. Roger has 40 cars in his collection. Each year he collects 1 additional car.

Part A: Write functions to represent Cory and Roger's collections throughout the years.
Part B: How many cars does Cory have after 6 years? How many does Roger have after the same number of years?
Part C: After approximately how many years is the number of cars that Cory and Roger have the same? Justify your answer mathematically.

Answer 1

Part A:
For Roger: $y=x+40$
For Cory: $y=15 (1.2)^{x}$
where y is the total  number of cars each boy collects over the year, x as the number of years passed.

Part B:
For Roger: 46
For Cory: 44.79~ 45 cars

Part C:
Here, we are trying to find the number of years (x) where Cory and Roger have the same number of cars.
By equating Roger and Cory's functions, we can solve for x
$x+40=15 (1.2)^{x}$.
Since we cannot solve the value of x directly, we use trial and error to estimate the year.
When x=1
$41 \neq 18$
When x=2
$42 \neq 21.6$
When x=3
$43 \neq 25.92$
When x=4
$44 \neq 31.10$
When x=5
45=/=37.3
When x=6
46=/=44.8
When x=7
47=/=53.7

The years that pass by before  Cory and Roger have nearly the same number of cars is 6.

Answer 2

Answer:

Step-by-step explanation:

Part A:

For Roger:

For Cory:

where y is the total number of cars each boy collects over the year, x as the number of years passed.

Part B:

For Roger: 46

For Cory: 44.79~ 45 cars