This is a carbon dating problem. It works of the principle that the radioactive C14 isotope is present in living lifeforms to some extent as they eat and breathe. When an organism dies, no new carbon is coming into the organism so any C14 that decays is not replaced.
I found from wikipedia that the half life of C14 is about 5730 years. Meaning that after 5730 years, 50% of the C14 that was originally in the sample will be left.
We can start on this problem using the half life equation:
N(t) = N0(0.5)^(t/halflife), where N(t) is how much is left and N0 is how much was originally there.
For part 1, they give us t = 505 years, and are asking for what percent of carbon is left. Let's make things easy and have N(0) be 1.
N(t) = 1(0.5)^(505y/5730y)
N(t) = 0.94
We can use the same equation for part 2. They give us N(t) = 0.97, and are asking for year.
0.97 = 1(0.5)(t/5730y)
0.97 = 0.5(t/5730y)
log0.5(0.97) = log0.5(0.5(t/5730y))
log0.5(0.97) = t/5730y
5730y*log0.5(0.97) = t
t = 251.8y
Now a followup practice question! Could the scratch paper really have been used by Columbus in 1492?