Cups

We have three cups. In the cups, we had fluid and boredom we started to shed.

1 We shed one-third of the fluid from the second glass into the first and third.
2 Then, we shed one-quarter cup of liquid from the first to the second and to the third.
3 Then, we shed one-fifth of the liquid from the third to the first cup and to the second.

Finally, 10, 12, and 9 dl fluids remain in cups. How many dl of liquid was originally in the cups?

Final Answer:

a =  8.5 dl
b =  16.5 dl
c =  6 dl

Step-by-step explanation:

$$\begin{gathered} x=10\text{dl} \\ y=12\text{dl} \\ z=9\text{dl} \\ z=m\cdot 3/5 \\ m=\frac{5}{3}\cdot z=\frac{5}{3}\cdot 9=\frac{5\cdot 9}{3}=\frac{45}{3}=15\text{dl} \\ k=x-m/5=10-15/5=7\text{dl} \\ l=y-m/5=12-15/5=9\text{dl} \\ m=15\text{dl} \\ k=d/2 \\ d=2\cdot k=2\cdot 7=14\text{dl} \\ e=l-d/4=9-14/4=\frac{11}{2}=5.5\text{dl} \\ f=m-d/4=15-14/4=\frac{23}{2}=11.5\text{dl} \\ b=3\cdot e=3\cdot \frac{11}{2}=\frac{3\cdot 11}{2}=\frac{33}{2}=16.5\text{dl} \\ a=d-b/3=14-\frac{33}{2}/3=\frac{17}{2}=8.5\text{dl} \\ c=f-b/3=\frac{23}{2}-\frac{33}{2}/3=6\text{dl} \\ \text{ Verifying Solution: } \\ {V}_1=x+y+z=10+12+9=31\text{dl} \\ {V}_2=a+b+c=\frac{17}{2}+\frac{33}{2}+6=31\text{dl} \\ {V}_1=V_2 \end{gathered}$$

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