At a chess tournament, the number of competitors in each round is 50% of the number of competitors in the previous round. What type of relationship most appropriately models this situation?

A. linear increase

B. exponential decay

C. exponential growth

D. linear decrease

Answer: Option B. exponential decay

Solution:

If initially the number of competitors is n

If the number of round is "r" and the number of competitors after "r" rounds is N (r)

1) After the first round (r=1), the number of competitors is:

N (r) = N (1) = n * 50% = n * 50/100 →N (r) = N (1) = n*0.5=n*0.5^1

2) After the second round (r=2), the number of competitors is:

N (r) = N (2) = N (1) * 50% = (0.5 n) * (50/100) = (0.5n) * (0.5) →N (r) = N (2) = n*0.5^2

3) After the third round (r=3), the number of competitors is:

N (r) = N (3) = N (2) * 50% = (n*0.5^2) * (50/100) = (n*0.5^2) * (0.5) →N (r) = N (3) = n*0.5^3

Then:

For r=1→N (r) = N (1) = n*0.5^1

For r=2→N (r) = N (2) = n*0.5^2

For r=3→N (r) = N (3) = n*0.5^3

In general: N (r) = n*0.5^r

This is an exponential function because the independent variable "r" is in the exponent, and because the number of competitors (funcion N (r)) decrease with the number of rounds the type of relationship most appropiately models this situation is an exponential decay