You go to the doctor and he gives you 11 milligrams of radioactive dye. After 20 minutes, 4.25 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived? Give your answer to the nearest minute.

The answer is 27 minutes (to the nearest minute)

Step-by-step explanation:

we are told that at arrival (0 minute), the concentration of the radioactive dye = 11 mg

after 20 mins, level = 4.25 mg

Therefore after 20 minutes, the level of the radioactive dye that disappears;

= 11 - 4.25 = 6.75

hence it takes 20 minutes for 6.75 mg to disappear;

20 minutes = 6.75 mg

∴ 1 minute = 6.75 ÷ 20 = 0.3375 mg

meaning that after every minute 0.3375 mg dis appears

Next we are told that the detector sounds if more than 2 mg of the substance is still in the body, hence, to pass through successfully without setting off the alarm, out of the original 11 mg of the radioactive substance, at least 9 mg must have disappeared out of the system, to be left with just 2 mg, and since we know how many milligrams disappears every minutes, we will find the time it takes for 9 mg to dis appear as follows;

Disappearance rate:

0.3375 mg  = 1 minute

∴ 9 mg = (1 ÷ 0.3375) × 9 = 26.67 minute = 27 minute (to the nearest minute)