Barrel Water Weight Empty

The barrel of water weighed 64 kg. When We poured 28% of the water from it on the first day and a third of the rest on the second day, it weighed 38 kg. Calculate the empty barrel's weight and the water's initial weight.

Final Answer:

s =  14 kg
v =  50 kg

Step-by-step explanation:

$$\begin{gathered} r=100\%-28\%=1-\frac{28}{100}=0.72 \\ s+v=64 \\ s+r\cdot v-r\cdot v/3=38 \\ s+0.72\cdot v-0.72\cdot v/3=38 \\ s+v=64 \\ 3s+1.44v=114 \\ Pivot:Row1\leftrightarrow Row2 \\ Row2-\frac{1}{3}\cdot Row1\to Row2 \\ 0.52v=26 \\ v=\frac{26}{0.52}=50 \\ s=\frac{114-1.44v}{3}=\frac{114-1.44\cdot 50}{3}=14=14\text{kg} \\ s=14 \\ v=50 \end{gathered}$$

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