How many loads of gravel will be needed to cover 2.0 mi of road 35 ft wide, to a depth of 3.0 in, if one truckload contains 8.0 cu yards of gravel?

notice, the volume of the area to be gravelled is just length*width*height.

the length is 2miles, there are 63360 inches in a mile so (2*63360) inches then.

the width is 35 feet, there are 12 inches in a foot, so (12 * 35) inches then.

the height is just 3 inches.

so the volume is (2*63360)(12*35)(3), or 159667200 inches.

now     [tex]\bf \begin{array}{ll} yd&in\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ yd&36in\\ (yd)^2&(36in)^2\\ (yd)^3&(36in)^3\\ yd^3&36^3in^3\\ &46656in^3 \end{array}[/tex]

how many yd³ is there in 159667200 in³?

[tex]\bf \begin{array}{ccll} yd^3&in^3\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ 1&46656\\ x&159667200 \end{array}\implies \cfrac{1}{x}=\cfrac{46656}{159667200}\implies \cfrac{159667200}{46656}=x \\\\\\ \cfrac{30800}{9}=x\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{there are }8~yd^3\textit{ in one load, how many loads in }\cfrac{30800}{9}~yd^3? \\\\\\ \cfrac{\frac{30800}{9}}{8}\implies \cfrac{\frac{30800}{9}}{\frac{8}{1}}\implies \cfrac{30800}{9}\cdot \cfrac{1}{8}\implies \cfrac{30800}{72}\implies \cfrac{3850}{9} \\\\\\ \stackrel{loads}{427\frac{7}{9}}[/tex]