I'm 98% sure that I have the right answer but show me your answer and how you got it. it would be greatly appreciated.
QUSETION: How many tons of sand will cover 123,000 sq ft. at 4 in. thick?
FACT: there are 2430 lbs of sand in a cubic yard
Basically, I want to cover a baseball infield (which is 123,000 square feet) with a special type of sand. I'll be laying it at 4 inches thick. How many tons of this sand will i need to order?
10 points to correct answer with best explanation.............
post script - i apologize to readers located in France or any other country who uses the metric system. i need this answer in tonage because i can't call the concrete company and make my order in kilograms or whatever. thanks again............
2006-06-21
06:03:02
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8 answers
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asked by
kris
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Science & Mathematics
➔ Mathematics
POSTED 21 June 2006 6:43 pm-----------
to tom d & jimbob,
please note that when you grade a baseball field, you have to worry about much more than in-between the bases (i.e. 90x90). you have to worry about the space from one dug out to the other, all the way to the warning track. when grading a ballfield, sod is layed last (over sand). I think I confused both of you when I said "infield." I should have just said ballfield as a whole..................thanks though for the concern.
2006-06-21
11:44:49 ·
update #1
123,000 sqft x 1sq yard /9 sqft = 13,666.67 sq yards
4 in x 1 yd / 36 in = 0.111 yards
so the volume used is 0.11111111 x 13,666.6667
which is 1518.518 sq yards
and since there are 2430 lbs of sand in a cubic yard
2430 lbs x 1 ton / 2000 lbs = 1.215 tons
we have 1518.518 x 1.215 = 1845 tons
2006-06-21 06:11:32
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answer #1
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answered by dhaval70 2
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"Mr. Scientist" and "dhavel70" have the correct answer for the values you gave, but the entire infield of a baseball field, up to the grass, doesn't cover an area anywhere near 123,000 sq. ft. That
area represents a square plot of ground approximately 351 ft. by 351 ft. Don,t take my word for it, but I'd guess the area of the infield is somewhere in the ball park (excuse the pun) of 16,500 sq. ft.
As a matter of curiosity, why do you want to cover the infield with four inches of sand?
2006-06-21 08:57:07
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answer #2
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answered by tom d 2
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First of all, are you sure that your area is 123,000 sq ft? That sounds like an awful lot (considering that the "diamond" between the bases is only 90*90 = 8100 sq ft)
Assuming you are correct, you need to calculate volume:
V = b * h
b = 123,000 sq ft = 13666.67 sq yd
h = 4 in = 0.1111 yd
V = 13666.67 * 0.1111 = 1518.519 cubic yd
There are 2430 lbs of sand in a cubic yd, so you need 2430 * 1518.519 = 3690000 lbs of sand
In tons this is 3690000 / 2000 = 1845 tons
2006-06-21 09:50:00
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answer #3
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answered by jimbob 6
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123,000 sq ft. divided by 9 = 13666.66 sq yards
2430 lbs/ yd^3 divided by 9 = 270 lbs / 3ftx3ftx4in.
13666.66 x 270 = 3,689,999.99982 lbs of sand needed
3,690,000.00 divided by 2000 = 1845 tons of sand.
Be sure to get some extra to fill in where needed.
2006-06-21 06:28:23
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answer #4
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answered by Mr. Scientist 4
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123000 square feet * 1/3 feet (4 inches) = 40590 cubic feet
40590 cubic feet/ 27 cubic feet per cubic yard = 1873.7 cubic yards
1873.7 cubic yards * 2430 lbs per yard = 4550391 lbs
4550391 lbs / 2000 lbs per ton = 2276.55 tons
2006-06-21 06:30:18
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answer #5
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answered by Don J 1
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123000 sq.ft. = 123000 * (1/3)(1/3) = 123000 * (1/9) = 123000/9 4/36 = (1/9)
(123000/9)(1/9) = (123000/81) cubic yard
2430 * (123000/81) = 3690000 lbs
(369000/2000) = 369/20 = 1845 tons
ANS : 1845 tons
2006-06-21 09:37:47
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answer #6
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answered by Sherman81 6
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2016-12-08 11:10:59
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answer #7
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answered by dematos 3
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4in. 4/12 ft = 1/3 ft.
So you need
123000 * 1/3 cuft = 41000 cuft
1 cu yard = 1.953125 cu ft
So, 41000 cuft = 41000/1.953125 cu yard = 20992 cu yard
Thus wt. of sand = 2430*20992 lbs.
2006-06-21 06:19:03
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answer #8
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answered by ag_iitkgp 7
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