Math in the Workplace: Overview

Math in the Workplace - Geometry/Analysis & Probability

Problem:

A company has a contract to remove 15,000 cubic yards (truck volume) of pit run (unprocessed) volcanic cinders from Horse Butte Cinder Pit. The cinders are being placed on a new road in a nearby subdivision. The contractor's truck drivers are required to place a signed and dated load ticket in a ticket box located at the pit entrance each time a load of cinders is hauled out of the pit. You, the inspector, have measured his bottom-dumping trucks to determine the volume they hold. They all have the following shape and measurements:

When you once visited the pit, you noticed that trucks heaped to overflowing sailed right by the ticket box without leaving tickets on several occasions. (Your presence was not observed). You mention this to the contractor, but still observe some truckers not stopping.

After the contractor has finished hauling cinders, you count all load tickets. There are 479. You are concerned and decide to make measurements of the road in the subdivision to determine the volume of cinders hauled. Cinders were placed on 6.6 miles of road.

These are your measurements taken at several places along the road:

Distance (mi)	Width (ft)	Thickness (in)
1.0	24	8
2.0	25	10
3.0	23	9
4.0	24	6
5.0	23	9
6.0	25	7
The cinders on the road were partially compacted by the haul trucks and traffic from the subdivision.

Was the contractor in violation of the permit (figure 10% compacted)? If so, how much?

Solution:

Calculations for truck volume problem

24 ft. x 6 ft. x 4 ft = 576 cu. ft.

[

(18 ft. x 3 ft. + 24 ft. x 6 ft) 2

]

(2)

= 198 cu. ft.

    27 cu. ft. in 1 cu. yd.

576 + 198 = 774 cu. ft. 27 cu. ft. = 28.6 cu. yd.

479 loads x 774 cu. ft. = 370,746 cu. ft. 13,731 cu. yd.
 

Heaped Truck:

576 + 198(below) [Volume calculated above]

+

198(above) [heaped]

= 972 cu.ft = 36 cu. yd.

Distance = 6.6 miles
Average width = 24 ft.
Average thickness = 8.2 inches

(5280 x 6.6)(24)(8.212) = 571,507 cu. ft. = 21,166 cu. yd.

Material compacted about 10%:

21,166 .9

= 23,518 cu. yd.

The contract volume is 15,000 cu.yds. Therefore, the contractor is in violation (+8,518 cu.yds)

*This problem demonstrates a case of having more information given than is actually needed.

Math in the Workplace: Overview