Concentration and Geom

[The Dragon]

13A-36. A 55-gal saltwater fish tank is prepared by adding 1/2 cup of salt to each gallon of fresh water. Over time, 3 gallons of pure water (no salt) evaporated from the tank. To replenish the tank, an additional 8 gallons of saltwater was removed. How many cups of salt should be mixed with fresh water to make up the 11 gallons needed to restore the salt concentration of the tank to the proper level? Answer: 4.23 or 4.23*10^0

13A-60 Geometry problem.

[sxk1693]

[sxk1693]

You should get around [unparseable or potentially dangerous latex formula]

[ilovenachos]

I'm lost on the salt thing. If you lose 11 gallons and 4 cups worth of salt. And you regain 11 gallons wouldn't you just need to put back in 4 cups of salt? I go through the process of 55 gallons/27.5 cups of salt. -3 gallons/-0 cups of salt. So now you're at 52 gallons/27.5 cups of salt. The concentration is different. You remove 8 gallons of saltwater which = -8 gallons/-4 cups of salt. So now you're at 44 gallons/23.5 cups of salt. You have to add back 11 gallons meaning your new liquid total is going to be 55 gallons yet again and you have to get back to the 50% concentration. So wouldn't you just add 4 cups of salt back in? I'm lost as to where the ".23" comes from. I know i'm probably overlooking something really simple.

[Zubair_Marediya]

[Zubair_Marediya]

[ilovenachos]

I was just assuming it was half a cup of salt per gallon of saltwater. I didn't even take into account that when you add the 11 gallons you're not just adding 5.5 cups of salt. You're just adding water with an unknown amount of salt. I really make the simplest mistakes on these things xD