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Advanced Math/Mixture - Concrete - Solution


Here is another concrete related problem:

A 1:2:4 mixture of concrete contains 1 part of cement, 2 parts of sand, and 4 parts of gravel/rocks. Determine amount of cubic yards of each to make 1 cubic yard of concrete.

33 1/3 % of voids (air space) are allowed in the sand and 45 % of voids are allowed in the gravel/rocks.

Let x = cubic yards of cement.

Let y = cubic yards of sand.

Let z = cubic yards of gravel/rocks.

The z cu. yds. of gravel/rocks count as only .55 z cu. yd. of the finished concrete, and the y cu. yd. of sand count as only 2/3 y of the solid, finished concrete.

From equations:

1) x + 2/3 y + .55 z = 1 cu. yd.

2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1

1) x + 2/3 y + .55 z = 1 cu. yd.

2) y = 2x and z = 4x from the mixture.

Substituting y and z from (2) in (1) :

x + 4x/3 + 2.2x = 1

3x + 4x + 6.6x = 3 (multiplying by 3).

13.6x = 3 (combining like terms).

13.6x/13.6 = 3/13.6 (dividing by 13.6 to get x alone).

x = .22058 = .22 cu. yds. of cement.

y = .44116 = .44 cu. yds. of sand.

z = .88232 = .88 cu. yds. of gravel.

I don't think those are correct. Where did I make the error(s)?  Thanks.

I'm guessing at that the 1:2:4 refers to amounts excluding voids.

1 ydł of concrete contains:
 1/7 ydł cement
 2/7 ydł sand (excluding voids)
 4/7 ydł gravel (excluding voids)

Use 1/7 ydł of cement.
Since the sand is ⅓ voids, use (2/7)×(1/(⅔)) = 3/7 ydł of sand.
Since the gravel is 45% voids, use (4/7)×(1/55%) = 80/77 ydł of gravel.

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Janet Yang


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