Question: An aquarium holds an equal number of clownfish and blowfish. 26 of the blowfish stay in their own tank, and the remaining blowfish swim into a display tank.  An equal number of clownfish join the blowfish in the display tank, but then a third of these clownfish swim back into their own tank.  If the aquarium holds a combined total of 100 fish, how many clownfish are now in the display tank? Is final answer 15?
Answer: There is an equal number of clownfish and blowfish, which means there are 100 fish / 2 = <<100/2=50>>50 blowfish.
Since some stay in their tanks, there are 50 blowfish – 26 blowfish = <<50-26=24>>24 blowfish in the display tank.
There were the same amount of clownfish until 24 clownfish / 3 = <<24/3=8>>8 clownfish swam back to their own tank.
This leaves 24 clownfish – 8 clownfish = <<24-8=16>>16 clownfish in the display tank.
#### 16