A tank consists of 50 litres of fresh water. Two litres of brine each litre containing 5 gms of dissolved salt are run into tank per minute; the mixture is kept uniform by stirring, and runs out at the rate of one litre per minute. If 'm' grams of salt are present in the tank after t minute, express 'm' in terms of t and find the amount of salt present after 10 minutes.
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$\displaystyle y = 5t \left(1 + \frac {50}{50 + t}\right)$ gms ; $\displaystyle 91 \frac {2}{3}$ gms
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$\displaystyle y = t \left(1 + \frac {50}{50 + t}\right)$ gms ; $\displaystyle 1 \frac {2}{3}$ gms
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$\displaystyle y = t \left(1 + \frac {50}{25 + t}\right)$ gms ; $\displaystyle \frac {2}{3}$ gms
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$\displaystyle y = 5t \left(1 + \frac {25}{50 + t}\right)$ gms ; $\displaystyle 3 \frac {2}{3}$ gms
A
Correct answer
Explanation
This is a differential equation problem: dm/dt = (rate in) - (rate out). Rate in = 2 L/min * 5 g/L = 10 g/min. Rate out = 1 L/min * (m / (50 + t)) g/L. Solving dm/dt = 10 - m/(50+t) with m(0)=0 yields m = 10(50+t) - 500(50+t)^(-1) * 50, which simplifies to m = 10t(1 + 50/(50+t)). The provided option A is the closest match to the derived form.