# 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 $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx258.01418587\approx2.5801 \times 10^{2}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{ton}{bushel}\right)\approx{\color{rgb(20,165,174)} 2.5801 \times 10^{2}}\left(\dfrac{pound}{gallon}\right)$$