1540.41 – Brine Tank

A marine research lab is studying the camouflage capabilities of sea creatures, and it’s time to freshen up Knifey the Cuttlefish’s tank, among others. A 700 gallon tank is being prepared for back-up for the whole lab. It’s currently holding 600 gallons of salt water; the amount of salt dissolved in the water is a half-pound per gallon.

An intake valve is opened and another 100 gallons of brine is allowed to flow into the tank; this brine has 2 pounds of salt per gallon. The tank is now full. Its contents are thoroughly stirred, and then an outlet valve is opened and 100 gallons of the mixture are drained out.

This process of adding, stirring, and draining is repeated two more times. At the end, how much salt per gallon is in the 600 gallons that remain in the tank?

Solution

At the beginning: $600 \times \dfrac{1}{2} = 300 \ \lb$ of salt in the tank. With each addition, 200 pounds of salt are added. With each drainage, one-seventh of the total amount of salt is removed, so that six-sevenths remain.

Addition #1: $300 + 200 = 500 \ \lb \ \left(\dfrac{5 \ \lb}{7 \ \gal}\right)$

Drain #1: $500 \times \dfrac{6}{7} = \dfrac{3000}{7} \ \lb$ of salt in $600 \ \gal$ .

Addition #2: $\dfrac{3000}{7} + 200 = \dfrac{4400}{7} \ \lb$ of salt in $700 \ \gal \ \left(\dfrac{44 \ \lb}{49 \ \gal}\right)$

Drain #2: $\dfrac{4400}{7} \times \dfrac{6}{7} = \dfrac{26400}{49} \ \lb$

Addition #3: $\dfrac{26400}{49} + 200 = \dfrac{36200}{49} \ \lb \ \left(\dfrac{362 \ \lb}{343 \ \gal}\right)$

Drain #3: $\dfrac{36200}{49} \times \dfrac{6}{7} = \dfrac{217000}{343} \ \lb$

The final of salt per gallon is $\dfrac{217000}{343} \div 600 \approx \dfrac{1.05 \ \lb}{\gal}$ .

This is more salty, by far, than the salt concentration of sea water, which is about 5 ounces of salt per gallon. The water in the big tank will be diluted in another process before Knifey gets her new water.