Question:medium

Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was

Updated On: Jan 15, 2026
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Solution and Explanation

Let \( x \) represent the number of patients in hospital A, and \( x + 21 \) represent the number of patients in hospital B.

The total recovery days for hospital A patients is 200, and for hospital B patients is 152. The average recovery days for hospital A patients were 3 days longer than for hospital B patients.

Step 1: Formulate the equation

The average recovery days for hospital A is \( \frac{200}{x} \), and for hospital B is \( \frac{152}{x + 21} \). The problem states that: \[ \frac{200}{x} = \frac{152}{x + 21} + 3 \]

Step 2: Simplify the equation

Isolate the term with the unknown by subtracting \( \frac{152}{x + 21} \) from both sides: \[ \frac{200}{x} - \frac{152}{x + 21} = 3 \] To clear the denominators, multiply the entire equation by \( x(x + 21) \): \[ 200(x + 21) - 152x = 3x(x + 21) \] Expand both sides: \[ 200x + 4200 - 152x = 3x^2 + 63x \] Combine like terms: \[ 48x + 4200 = 3x^2 + 63x \] Rearrange into standard quadratic form: \[ 3x^2 + 15x - 4200 = 0 \]

Step 3: Solve the quadratic equation

Divide the equation by 3 to simplify: \[ x^2 + 5x - 1400 = 0 \] Apply the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1400)}}{2(1)} \] \[ x = \frac{-5 \pm \sqrt{25 + 5600}}{2} = \frac{-5 \pm \sqrt{5625}}{2} \] \[ x = \frac{-5 \pm 75}{2} \] This yields two potential solutions: \[ x = \frac{-5 + 75}{2} = 35 \quad \text{or} \quad x = \frac{-5 - 75}{2} = -40 \] Since the number of patients cannot be negative, \( x = 35 \) is the valid solution.

Final Answer:

Hospital A admitted \( \boxed{35} \) patients.

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