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
Approach Solution - 1
Consider the number of patients admitted in hospital A and hospital B to be 'x' and ‘x + 21'
Given that, the sum of recovery days for patients in hospitals A and B were 200 and 152, respectively.
The average recovery days for patients admitted in hospital A was 3 more than the average in-hospital B.
\(\frac{200}{x}=\frac{152}{x+21}+3\)
By solving the above equation we get x = 35.
Hence, the number of patients admitted to hospital A is 35.
Approach Solution -2
Let's denote the number of patients admitted to hospital A as 'x' and to hospital B as 'x + 21'. Given that the total recovery days for patients in hospitals A and B were 200 and 152, respectively, and that the average recovery days for patients admitted to hospital A exceeded those in hospital B by 3.
\(\frac{200}{x}-\frac{152}{x+21}=3\)
\(\frac{200x+4200-152x}{x(x+21)}=3\)
\(\frac{48x+4200}{x(x+21)}=3\)
\(\frac{48x+4200}{3}=x(x+21)\)
\(16x+1400=x^2+21x\)
\(x^2+5x-1400\)
\((x+40)(x-35)=0\)
x=-40,35
x can't be negative so that means the value of x is 35
So, the answer is 35
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