# The half-life of radon-222 is 3.8 days. What was the original activity if it has an activity of 10 Bq after 7.6 days?

$40 \setminus \text{Bq}$
After $7.6$ days, radon-$222$ has elapsed $\frac{7.6}{3.8} = 2$ half-lives. So, its original activity would be ${2}^{2} = 4$ times the amount of activity it has right now.
If it has $10$ becquerels of activity right now, that must mean that its original activity was $10 \setminus \text{Bq"*4=40 \ "Bq}$.