SOLUTIONS TO ADDITIONAL PROBLEMS ON VOLUME
(1) V = L x W x H
_______.8 = 138.88 cubic yd
1 3 8 __
) 3 7 5 0.0 must be removed
V = 1,500 feet x 1 foot x 2.5 feet
27
V = 3,750 cubic feet
2 7
1 0 5
27 cubic feet = 1 yard
8 1
2 4 0
2 1 6
2 4 0
2 1 6
________9 7.__5 4 = 597.45 gal of
5 ___ 4 ___
(2) V = L x W x H
)
V = 42 in x 62 in x 53 in
2 3 1 1 3 8 0 1 2.0 0 0 = bbls of
V = 138012 cubic inches
1 1 5 5 stored
231 cubic inches = 1 gallon
2 2 5 1
2 0 7 9
1 7 2 2
1 6 1 7
1 0 5 0
__9 2 4
1 2 6 0
1 1 5 5
1 0 5 0
2
9
4
(3) V = π r2h
V = 3.14 x (7.5 feet)2 x 6.5 feet
V = 3.14 x 56.25 square feet x 6.5 feet
_______ 0 4.2 8 = 204.28
2__
V = 1,148.06 cubic feet
)
1 bbl = 5.62 cubic feet
5 6 2 1 1 4 8 0 6
bbls of
oil in
tank
(4) V = π r2
V = 3.14 x (.25 foot)2 x 9,500
r = 3 inches or .25 foot
V = 3.14 x .0625 square foot x 9,500 feet
V = 1,864.4 cubic feet
Oil = __55 lbs.__ x 1,864.4 cubic feet = 102,542.0 pounds
cubic feet
(5) V = π r2h
V = 3.14 x (4 feet)2 x 10 feet
V = 3.14 x 16 square feet x 10 feet
V = 502.40 cubic feet
1 cubic foot 7.48 gallons
502.40 cubic feet x 7.48 = 3,757.952 gallons which can be stored
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