Hello Roland2rule.
Let's take a look at your question.
Let a, b, c and d the number of chocolates that have A, B, C and D respectively.
First, we know that A had 80 more chocolates then B initially, that is to say, $$\left(1\right)\ \ \ \ \ \ a=80+b\ .$$ Now, A gave 1/3rd of the chocolates with her to B, so A at the end had $$\left(2\right)\ \ \ a-\frac{a}{3}=\frac{2a}{3}\ chocolates$$ and B has $$\left(3\right)\ \ b+\frac{a}{3}=\frac{3b+a}{3}\ chocolates.$$ But B gave 1/4th of what she then had to C, this implies that B at the end has $$\left(4\right)\ \ \frac{3b+a}{3}\ -\frac{\frac{3b+a}{3}}{4}=\frac{3b+a}{3}-\frac{3b+a}{12}=\frac{9b+3a}{12}=\frac{3b+a}{4}$$ and C now has $$\left(5\right)\ c+\frac{\frac{3b+a}{3}}{4}=c+\frac{3b+a}{12}.$$ Let's stop here for a moment. We know that all the four girls had an equal number of chocolates at the end.
Hence, we have that (2) and (4) are equal, and using (1) we can get $$\frac{2a}{3}=\frac{3b+a}{4}\ \Leftrightarrow\ \ 8a=9b+3a\ \Leftrightarrow\ 5a=9b\ \Leftrightarrow\ 400+5b=9b$$ $$400=4b\ \Leftrightarrow\ \ b=100\ and\ then\ a=180.$$ Now, we can rewrite some of the previous equations as follows:
B at the beggining had 100 chocolates, then she had $$100+\frac{a}{3}=100+\frac{180}{3}=160$$ and she gave 1/4 to C, so she gave 40 chocolates to C. Now, C has c+40 and C gave 1/5 to D. Finally, C had $$c+40-\frac{c+40}{5}=\frac{5c+200-c-40}{5}=\frac{4c+160}{5}$$ and it has to be equal to 120. Hence, $$\frac{4c+160}{5}=120\ \Leftrightarrow\ 4c+160=600\ \ \Leftrightarrow\ c=110.$$ Therefore, c+40=150 and C gave 1/5 to D, then she gave 30 chocolates to D.
Now, D has d+30 chocolates, and it has to be equal to 120, therefore d=90.
In conclusion, the difference between the number of chocolates that C and D initially had is c-d=110-90=20. So, the correct answer is the option [spoiler]A=20[/spoiler].
I hope this can help you.
I'm available if you'd like a follow-up.
Regards.
