pure water is poured at the rate of 3 gal/min to a tank containing 300 lbs. of salt dissolved in 100 gals. of water and solution kept while stirring, forced out at 2 gals/min. Find the amount of salt at the end of 1 hr. and 30 minutes?

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- dumbcow

Tank loses 2 gal/min while adding 3 gal/min for a net of 1 gal/min
\[Volume = 100 + t\]
Salt content in the tank starts at 300 and loses 2C lbs every min
where C(t) is the concentration of salt at time t
C(t) = Salt content/Volume
\[C(t) = \frac{300-2t*C(t)}{100+t}\]
solving for C(t)
\[\rightarrow C(t) = \frac{300}{100+3t}\]
Let S(t) be salt content at time t
S(t) = C(t)*Volume
\[\rightarrow \ s(t) = \frac{300(100+t)}{100+3t}\]
when t=90
\[s(90) = \frac{300(190)}{100+270} = 203.57\]

- anonymous

thanks for the effort dumb cow, but i already got the answer. . it's 83.10 lbs.

- dumbcow

Can you please explain to me how you got that answer?
Thanks

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## More answers

- anonymous

this problem is actually related to mixture problem related to differential equation application. . have you encountered this one?

- dumbcow

no

- anonymous

here,
let me show you how i got it. .

- anonymous

|dw:1328684363825:dw|
we got 100 +t as denominator for the because the rate of going in minus the rate in going out (3t - 2t) resulted to t. .

- anonymous

to make the equation--\[dQ/dt=3(0) - 2(Q/100+t)\] we're finding the change of concentration

- anonymous

|dw:1328684946025:dw||dw:1328685032254:dw| cancel ln.|dw:1328685126662:dw|
this is now our equation. .

- anonymous

now let's find the value for C or constant, given the initial number of salt (Q) which is 300 at t=0
|dw:1328685267814:dw| now that we have the value for constant, we can now find the remaining salt in lbs at t=90 mins.

- anonymous

|dw:1328685463206:dw|

- anonymous

that's how i got the answer. . kindly check if there are irregularities.

- dumbcow

No it looks fine, thanks for the explanation.
i didn't see where i could solve dQ/dt explicitly like that

- anonymous

aah, yea me either got difficulties in solving the problem. but anyways, thanks for the medal.

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