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

Entered answer:

✅ Correct Answer: 35

Let $n =$ number of patients in hospital B

Then hospital A has $(n - 21)$ patients since A admitted 21 less than B

For hospital B:

Total recovery days $= 152$

Number of patients $= n$

Average recovery days $= \frac{152}{n}$

For hospital A:

Total recovery days $= 200$

Number of patients $= (n - 21)$

Average recovery days $= \frac{200}{n - 21}$

The average recovery days in hospital A is 3 more than hospital B.

$\frac{200}{n-21} = \frac{152}{n} + 3$

$200n = 152(n-21) + 3n(n-21)$

$200n = 152n - 152 \times 21 + 3n(n-21)$

$200n = 152n - 3192 + 3n^2 - 63n$

$200n = 89n - 3192 + 3n^2$

$3n^2 - 111n - 3192 = 0$

$n^2 - 37n - 1064 = 0$

$(n - 56)(n + 19) = 0$

This gives us: $n = 56$ or $n = -19$

Since the number of patients cannot be negative, $n = 56$ .

Hospital B has $56$ patients.

Therefore, hospital A has: $56 - 21 = 35$ patients

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