A tank contains 1000 $l$ of water. Brine that contains 0.05 kg/ $l$ of salt enters the tank at the rate of 5 $l$ /min. Brine that contains 0.04 kg / $l$ of salt enters the tank at the rate of 10 $l$ /min. The solution is kept thoroughly mixed and drains from the tank at the rate of 15 $l$ / min . If the amount of salt left in the tank after 200 minutes, is A, then 3A is equal to (in kg)

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$130 (1 - 3 e^{- 1})$

$130 (1 - e^{- 2})$

$130 (1 - e^{- 3})$

$130 (1 - e^{- 1})$

answer is C.

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Detailed Solution

Let S be the amount of salt after time ‘t’
$\frac{d s}{d t} = 0.05 \times 5 + 0.04 \times 10 - \frac{5}{1000} \times 15 = 0.65 - 0.015 s \Rightarrow \frac{d s}{d t} + \frac{3 s}{200} + \frac{13}{20} \Rightarrow s . e^{\frac{3 t}{200}} = \frac{13}{20} . \frac{200}{3} e^{\frac{3 t}{2 \infty}} + c s_{0} = 0 \Rightarrow c = - \frac{130}{3} \Rightarrow s = \frac{130}{3} (1 - e^{- 3 t / 200})$