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

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?

• 78.1%
• 81.2% <<<CORRECT
• 123.1%
• 128.0%

Explanation:

Edge 2021

Answer 2

The mass of the first shipment at time t is

`m_(1)=100( (1)/(2) )^{ (t)/(10)}`

The mass of the second shipment at time t is

`m_(2)=100( (1)/(2) )^{ (t-3)/(10) }`

At time t, the ratio of m₁ to m₂ is

`(m_(1))/(m_(2)) = (100)/(100). ((1/2)^(t/10))/((1/2)^((t-3)/10)) \\ = ((1/2)^(t/10))/((1/2)^(t/10)). (1)/((1/2)^(-3/10)) \\ = (1/2)^(3/10) \\ = 0.8123`

Therefore as a percentage,

`(m_(1))/(m_(2)) =100*0.8123 = 81.23 \%`

Answer: B. 81.2%