## **Core Concept**
The question tests the understanding of first-order elimination kinetics of drugs. In first-order kinetics, the rate of elimination of a drug is directly proportional to its concentration in the body. This is characterized by a constant fraction of the drug being eliminated per unit time, which leads to a logarithmic decrease in drug concentration over time.

## **Why the Correct Answer is Right**
Given that 80 mg of the drug was eliminated in the first 120 minutes, 80 mg remained after 120 minutes because the initial dose was 160 mg. In first-order kinetics, the elimination rate constant (Kel) and half-life (t1/2) are crucial. The half-life is the time required for the plasma concentration of a drug to reduce by half. Since 80 mg was eliminated in the first 2 hours, 80 mg remained. This means that after another 2 hours (total 4 hours), half of 80 mg will remain, which is 40 mg. After another 2 hours (total 6 hours), half of 40 mg will remain, which is 20 mg.

## **Why Each Wrong Option is Incorrect**
- **Option A:** This option suggests a calculation error or incorrect understanding of first-order kinetics.
- **Option B:** This option is incorrect because if 80 mg remained after 2 hours, the subsequent decrease should follow first-order kinetics, leading to 40 mg at 4 hours and 20 mg at 6 hours, not 40 mg at 6 hours.
- **Option D:** This option suggests 60 mg remaining, which does not align with the principles of first-order elimination kinetics given the information provided.

## **Clinical Pearl / High-Yield Fact**
A key point to remember is that for drugs following first-order kinetics, the half-life (t1/2) is constant and independent of the drug concentration. This property allows for predictable drug accumulation and elimination. For any drug administered, knowing its half-life helps in predicting how long it will take to reach steady state (approximately 4-5 half-lives) and how much of the drug will remain at any given time.

## **Correct Answer:** . 20 mg