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
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
Option: 35
Solution:
Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:
\(\frac{{200}}{x} - \frac{{152}}{{x + 21}} = 3\)
\(\frac{{\left( {200x + 4200 - 152x} \right)}}{{x\left( {x + 21} \right)}} = 3\)
\(\frac{{\left( {48x + 4200} \right)}}{{x\left( {x + 21} \right)}} = 3\)
\(16x + 1400 = x\left( {x + 21} \right)\)
\({x^2} + 5x - 1400 = 0\)
(x+40)(x-35)=0
Hence, x=35.
Solution:
Let the number of Covid patients in Hospitals A and B be x and x+21 respectively. Then, it has been given that:
\(\frac{{200}}{x} - \frac{{152}}{{x + 21}} = 3\)
\(\frac{{\left( {200x + 4200 - 152x} \right)}}{{x\left( {x + 21} \right)}} = 3\)
\(\frac{{\left( {48x + 4200} \right)}}{{x\left( {x + 21} \right)}} = 3\)
\(16x + 1400 = x\left( {x + 21} \right)\)
\({x^2} + 5x - 1400 = 0\)
(x+40)(x-35)=0
Hence, x=35.
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