Decaying after 75 days how much left 
After $75$ days, a radioactive substance has decayed to $26.7\%$ of its original amount. After an additional $75$ days, what percent of its original amount will it have decayed to?

I got $100-26.7=19.6$, meaning there is $73.3$ from original and multiplied $0.267$ to $73.3$ and got $19.57$. Is this right?
 A: You're close, the 26.7% implies that there is 26.7% of the substance left after the 75 days, meaning a loss of 73.3%. Decaying the substance for another 75 days results in another 73.3% loss (in this case, 73.3% of 26.7), meaning you subtract the result you came up with (19.57%) from the 26.7%, giving you an answer of approximately 7.13
A simple way is simply taking 26.7% of 26.7% (the resulting amount of substance left after 75 additional days) resulting in the same answer of 7.13
A: Here's an answer with a bit of formalism to complete that of @xav MX.
Let $r$ be the daily rate of decay. That means every day, the substance's radioactivity decays by $r$ compared to the day before. So in $75$ days, it decays by $r^{75}$. By hypothesis, $r^{75}=0.267$.
And in after another $75$ days, it will have decayed by $r^{75+75}=\left(r^{75}\right)^2=0.267^2=0.0713$, which gives $7.13%$ as the answer.
The reason for using some formalism is to be able to generalize to a different period of time. For instance, if they asked you what the decay would be after an additional $100$ days instead of $75$, you'd compute $r^{75+100}=\left(r^{75}\right)^{\frac {175}{75}}=0.267^{\frac {175}{75}}=0.046$ and the answer would be 4.6%.
