The volume of the gravel would equal 100 times 60 which is 6000!
However, since one yard’s depth is 6 inches out of 36 inches, we multiply the 6000 by 1/6 which would equal 1000 yd^3!
So, the required gravel volume would be 1000 yd^3 (cubic yards)
The gravel in a full truck carrying 9 cubic yards would cost less than 9 separate trucks!
As the school needs 1000 cubic yards of gravel, we divide this by 9 which would equal 111 full trucks of gravel containing 999 cubic yards of gravel plus 1 separate cubic yard!
We do the math :
$95 (full truck) times 111 trucks + = $10,545 thousand
Plus $12.50 for the one separate one cubic yard of gravel = $ 10,557.50
So, the minimum cost of gravel would be
$10,557.5 thousands
Hope this helps :)