Question

A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determined using the equation f(t)=100(1/2)^t/10. After three days, the scientist receives a second shipment of 100 milligrams of the same element. The equation used to represent the amount of shipment 2 remaining after t days is f(t)=100(1/2)^t-3/10. After any time, t, the mass of the element remaining in shipment 1 is what percentage of the mass of the element remaining in shipment 2?
A.78.1%
B.81.2%
C.123.1%
D.128.0%

Answer 1

The mass of the first shipment at time t is
$m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}$

The mass of the second shipment at time t is
$m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }$

At time t, the ratio of m₁ to m₂ is
$\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123$

Therefore as a percentage,
$\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%$

Answer: B. 81.2%