Question

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 1

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