The normal dosage of a particular medicine is $t$ tablets per day for each patient. A hospital's current supply of these tablets will last $p$ patients for $d$ days. If the recommended dosage increases by 20% and the number of patients decreases by one-third, then for how many days will the hospital's supply last?

### Concept & Constant Total Supply The total quantity of a resource (tablets) remains constant even when consumption rates or consumer numbers change. $$ \text{Total Supply} = \text{Patients} \times \text{Tablets per day} \times \text{Days} $$ ### Step-by-Step Solution 1. **Define initial total supply:** - Initial supply = $p \times t \times d$ 2. **Define the new variables:** - New dosage ($t'$) = Increased by 20% = $t + 0.2t = 1.2t = \frac{6}{5}t$ - New patients ($p'$) = Decreased by one-third = $p - \frac{1}{3}p = \frac{2}{3}p$ - Let the new number of days be $d'$. 3. **Equate the old supply to the new consumption rate:** - $p \times t \times d = p' \times t' \times d'$ - $p \times t \times d = (\frac{2}{3}p) \times (\frac{6}{5}t) \times d'$ 4. **Simplify and solve for $d'$:** - Cancel out $p$ and $t$ from both sides: - $d = \frac{2}{3} \times \frac{6}{5} \times d'$ - $d = \frac{12}{15} \times d'$ - $d = \frac{4}{5} \times d'$ - $d' = \frac{5}{4} \times d = \frac{5d}{4}$ ### Exam Strategy & Shortcut Use proportional multipliers. The days are inversely proportional to both patients and dosage. New Days = Old Days $\times$ (Inverse Ratio of Patients) $\times$ (Inverse Ratio of Dosage). Patient multiplier went from $1$ to $\frac{2}{3}$, so its inverse ratio is $\frac{3}{2}$. Dosage multiplier went from $1$ to $\frac{6}{5}$, so its inverse ratio is $\frac{5}{6}$. New Days = $d \times \frac{3}{2} \times \frac{5}{6} = d \times \frac{15}{12} = \frac{5d}{4}$. ### Common Pitfall Miscalculating the percentage increase or fractional decrease. A 20% increase means multiplying by $1.2$ (or $\frac{6}{5}$), not just $0.2$. A decrease *by* one-third means $\frac{2}{3}$ remains. ### Final Answer Therefore, the correct answer is **$\frac{5d}{4}$**.