Feed to hay

How many tons of water do we have to dehydrate from 8 tons of feed if the original moisture is 54 percent to get hay with a moisture content of 15 percent? Round to two decimal places.

Correct answer:

x =  3.67 t

Step-by-step explanation:

$$\begin{gathered} m_1=8\text{t} \\ {p}_1=54\%=\frac{54}{100}=\frac{27}{50}=0.54 \\ {p}_2=15\%=\frac{15}{100}=\frac{3}{20}=0.15 \\ {v}_1=p_1\cdot m_1=0.54\cdot 8=\frac{108}{25}=4.32\text{t} \\ s=m_1-v_1=8-4.32=\frac{92}{25}=3.68\text{t} \\ {p}_2=v_2/(s+v_2) \\ {p}_2\cdot (s+v_2)=v_2 \\ 0.15\cdot (3.68+v_2)=v_2 \\ 0.85v_2=0.552 \\ {v}_2=\frac{0.552}{0.85}=0.64941176 \\ {v}_2=\frac{276}{425}\doteq 0.649412 \\ x=v_1-v_2=4.32-0.6494=\frac{312}{85}\doteq 3.6706\text{t} \\ \text{ Verifying Solution: } \\ {p}_2=\frac{v_1-y}{s+v_1-y} \\ {p}_2\cdot (s+v_1-y)=v_1-y \\ 0.15\cdot (3.68+4.32-y)=4.32-y \\ 0.85y=3.12 \\ y=\frac{3.12}{0.85}=3.67058824 \\ y=\frac{312}{85}\doteq 3.670588 \\ y=x \end{gathered}$$

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