Using a load factor of 50%, how many 10-yard dump truck loads are required to export 6 inches of paving removed from a 120 ft by 4 ft trench?

Using a load factor of 50%, how many 10-yard dump truck loads are required to export 6 inches of paving removed from a 120 ft by 4 ft trench?

• A. 2
• B. 3
• C. 4
• D. 5

Answer: (A)

The idea being tested is converting a measured trench volume into the number of truckloads when only half of each truck’s capacity can be used. Start by finding the volume of the removed paving: multiply the length, width, and depth. The trench is one hundred twenty feet long, four feet wide, and the paving is six inches thick, which is half a foot. So the volume is one hundred twenty times four times half cubic feet, equal to two hundred forty cubic feet. Convert that to cubic yards since truck capacity is given in yards: there are twenty seven cubic feet in a cubic yard, so two hundred forty cubic feet is about nine cubic yards. A ten-yard dump truck would carry ten cubic yards when full. With a fifty percent load factor, the useful load is five cubic yards per trip. Dividing the total volume by per-trip capacity gives about nine divided by five, which is roughly one and a half trips. Since you can’t make a fraction of a trip, you round up to two trips. Two truckloads are required.

The idea being tested is converting a measured trench volume into the number of truckloads when only half of each truck’s capacity can be used. Start by finding the volume of the removed paving: multiply the length, width, and depth. The trench is one hundred twenty feet long, four feet wide, and the paving is six inches thick, which is half a foot. So the volume is one hundred twenty times four times half cubic feet, equal to two hundred forty cubic feet.

Convert that to cubic yards since truck capacity is given in yards: there are twenty seven cubic feet in a cubic yard, so two hundred forty cubic feet is about nine cubic yards.

A ten-yard dump truck would carry ten cubic yards when full. With a fifty percent load factor, the useful load is five cubic yards per trip. Dividing the total volume by per-trip capacity gives about nine divided by five, which is roughly one and a half trips. Since you can’t make a fraction of a trip, you round up to two trips.

Two truckloads are required.