How many loads of gravel are needed to cover 4 mi of roadbed

 How many loads of gravel are needed to cover 4 mi of roadbed, 72 ft wide, to a depth of  5 in. if one truckload capacity is  8 yd^3?   

Solution

length of road bed= 4 mile= 4* 5280 ft= 21120 ft
wide= 72 ft
depth= 5 in = 5*0.08333= 0.4166 ft
volume of gravel required= 21120*72*0.4166 = 633574.656 ft³
truckload capacity= 8 yd³= 8*27 ft³ = 216 ft³
no. of truckloads required= 633574.656/ 216 = 2934 trucks