Convert ton / bushel to pound / gallon
Learn how to convert
1
ton / bushel to
pound / gallon
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(\dfrac{ton}{bushel}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{pound}{gallon}\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)\)
\(\text{Left side: 1.0 } \left(\dfrac{ton}{bushel}\right) = {\color{rgb(89,182,91)} \dfrac{10^{5}}{3.523907016688}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(89,182,91)} \dfrac{10^{5}}{3.523907016688}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}\)
\(\text{Right side: 1.0 } \left(\dfrac{pound}{gallon}\right) = {\color{rgb(125,164,120)} \dfrac{5.0}{4.54609 \times 10^{-2}}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(125,164,120)} \dfrac{5.0}{4.54609 \times 10^{-2}}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(\dfrac{ton}{bushel}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{pound}{gallon}\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{10^{5}}{3.523907016688}} \times {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{5.0}{4.54609 \times 10^{-2}}}} \times {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} \dfrac{10^{5}}{3.523907016688}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{5.0}{4.54609 \times 10^{-2}}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{10^{5}}{3.523907016688}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{5.0}{4.54609 \times 10^{-2}}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}}\)
\(\text{Conversion Equation}\)
\(\dfrac{10^{5}}{3.523907016688} = {\color{rgb(20,165,174)} x} \times \dfrac{5.0}{4.54609 \times 10^{-2}}\)
Switch sides
\({\color{rgb(20,165,174)} x} \times \dfrac{5.0}{4.54609 \times 10^{-2}} = \dfrac{10^{5}}{3.523907016688}\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{4.54609 \times 10^{-2}}{5.0}\right)\)
\({\color{rgb(20,165,174)} x} \times \dfrac{5.0}{4.54609 \times 10^{-2}} \times \dfrac{4.54609 \times 10^{-2}}{5.0} = \dfrac{10^{5}}{3.523907016688} \times \dfrac{4.54609 \times 10^{-2}}{5.0}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{5.0}} \times {\color{rgb(99,194,222)} \cancel{4.54609}} \times {\color{rgb(166,218,227)} \cancel{10^{-2}}}}{{\color{rgb(99,194,222)} \cancel{4.54609}} \times {\color{rgb(166,218,227)} \cancel{10^{-2}}} \times {\color{rgb(255,204,153)} \cancel{5.0}}} = \dfrac{{\color{rgb(255,204,153)} \cancelto{10^{3}}{10^{5}}} \times 4.54609 \times {\color{rgb(255,204,153)} \cancel{10^{-2}}}}{3.523907016688 \times 5.0}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{10^{3} \times 4.54609}{3.523907016688 \times 5.0}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx258.01418587\approx2.5801 \times 10^{2}\)
\(\text{Conversion Equation}\)
\(1.0\left(\dfrac{ton}{bushel}\right)\approx{\color{rgb(20,165,174)} 2.5801 \times 10^{2}}\left(\dfrac{pound}{gallon}\right)\)
