Question

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

Answer 1

Let the number of patients admitted in hospital A be denoted as \( x \) and the number of patients in hospital B as \( x + 21 \).

The sum of recovery days for patients in hospital A is 200, and in hospital B is 152. We are also told that the average recovery days for patients in hospital A was 3 more than the average in hospital B.

Step 1: Set up the equation

The average recovery days for patients in hospital A is \( \frac{200}{x} \), and the average for hospital B is \( \frac{152}{x + 21} \). The given condition is: \[ \frac{200}{x} = \frac{152}{x + 21} + 3 \]

Step 2: Solve the equation

To solve the equation, first subtract \( \frac{152}{x + 21} \) from both sides: \[ \frac{200}{x} - \frac{152}{x + 21} = 3 \] Now, let's multiply both sides of the equation by \( x(x + 21) \) to eliminate the denominators: \[ 200(x + 21) - 152x = 3x(x + 21) \] Expanding both sides: \[ 200x + 4200 - 152x = 3x^2 + 63x \] Simplifying: \[ 48x + 4200 = 3x^2 + 63x \] Rearranging: \[ 3x^2 + 15x - 4200 = 0 \]

Step 3: Solve the quadratic equation

Divide through by 3: \[ x^2 + 5x - 1400 = 0 \] Now, solve using the quadratic formula: \[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1400)}}{2(1)} \] \[ x = \frac{-5 \pm \sqrt{25 + 5600}}{2} = \frac{-5 \pm \sqrt{5625}}{2} \] \[ x = \frac{-5 \pm 75}{2} \] So, we have two possible 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, the solution is \( x = 35 \).

Final Answer:

The number of patients admitted to hospital A is \( \boxed{35} \).

Answer 2

Let the number of patients admitted to hospital A be \( x \), and the number of patients admitted to hospital B be \( x + 21 \).

Given:

• Total recovery days for patients in hospital A: 200
• Total recovery days for patients in hospital B: 152
• The average recovery days for patients in hospital A exceeds those in hospital B by 3 days

Formulate the equation from the given data:

The average recovery days for Hospital A is \( \frac{200}{x} \), and for Hospital B is \( \frac{152}{x + 21} \).

Since the average recovery days for A exceeds B by 3, \[ \frac{200}{x} - \frac{152}{x + 21} = 3 \]

Solve the equation:

Combine fractions: \[ \frac{200(x + 21) - 152x}{x(x + 21)} = 3 \] \[ \frac{200x + 4200 - 152x}{x(x + 21)} = 3 \] \[ \frac{48x + 4200}{x(x + 21)} = 3 \]

Multiply both sides by the denominator: \[ 48x + 4200 = 3x(x + 21) \] \[ 48x + 4200 = 3x^2 + 63x \]

Rearrange: \[ 3x^2 + 63x - 48x - 4200 = 0 \] \[ 3x^2 + 15x - 4200 = 0 \] Divide through by 3: \[ x^2 + 5x - 1400 = 0 \]

Factor the quadratic: \[ (x + 40)(x - 35) = 0 \] So, \[ x = -40 \quad \text{or} \quad x = 35 \]

Since \( x \) can't be negative, we take: \[ \boxed{x = 35} \]

Answer:

The number of patients admitted to hospital A is 35.