There are 80 fish in Tank A at first and 16 fish from Tank A are sold away.

Number of fish in Tank A at the end
= 80 - 16
= 64

There are a total of 160 fish left in Tank A and Tank B at the end.

Number of fish in Tank B at the end
= 160 - 64
= 96

24 fish from Tank B are sold.

Number of fish in Tank B at first
= 96 + 24
= 120

So there are 120 fish in Tank B at first and 24 fish in Tank B are sold.

Fraction of fish sold (as a fraction of the starting number)
= 24/120
= 1/5 (after dividing both numerator and denominator by 24)

Percentage of fish sold in Tank B
= 1/5 * 100%
= 20%

In TankB not A

Kc, you will need to be patient with the solutions as I am using phone to type and I can’t view the question without clicking on “done” which sends in part of my solutions first. I need to refer to the questions several times while typing, hence we as tutors usually post the workings part by part.

Updated in the first post.

So answer for part A is 96?

Yup, there are 96 fish from Tank B in the end.

I do not recommend drawing models for this question as they can get misleading. I use a combination of forward approaches (calculating number of fish in Tank A at the end) and backward approaches (calculating number of fish in Tank B at first) to do this question.