How many tons of asphalt concrete will be required to overlay a 2.5 mile long 24 ft wide road with 3 in of wearing course asphalt concrete that has a density of 1479 pcf?

2 Answers

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

To determine the weight of the asphalt needed, first determine the area of the surface to be paved.

Since the shape of the road is a rectangle, use the formula of area of rectangle.

A `=` Length `xx` Width

Before plugging-in the values, convert the 2.5 miles to feet in order that the dimension to have same unit.


`2.5 mi xx (5280 ft)/(1mi) = 13200ft`

Now that the length is in feet, plug-in the values of dimension of the road to the formula of area of rectangle.



Hence, the area to be paved is 316 800 `ft^2` .

Next, determine the volume. Use the formula:

V `=` Area `xx` thickness

Before substituting the values, convert the thickness to feet in order to have same units.


`3 i n xx (1ft)/(12 i n)=0.25ft`

Then, substitute the values of the area and thickness to the formula of volume.


Thus, the volume of the asphalt needed is 79200 ft^3.

Next, to determine its weight , multiply that with the given density 1479 pounds per cubic feet (1479 lb/ft^3).


So, the weight is 11 7136 800 lbs.

Then, convert it to tons.

`117136800 lbs xx (1t o n)/(2000lbs) = 58568.4 t o n s`

Therefore, the amount of asphalt needed is 58 568.4 tons.

Zaca's profile pic

Zaca | Student, Undergraduate | (Level 1) Salutatorian

Posted on

Steps: Find volume -> Find Pounds -> Find Tons

Convert everything to feet (3in=.25ft):

`(2.5 mi) * (5280 ft)/(1 mi) = 13200 ft`

Find Volume:

`13200ft * .25ft * 24ft = 79200 ft^3`

Use density to convert from volume to pounds:

`(79200 ft^3) * (1479 lbs)/(1 ft^3)`

`= 117136800 lbs`

Convert from pounds to tons:

`117136800 lbs * (1 tn)/(2000 lbs) = 58568.4 tns`