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Mixture - Concrete - Solution/Explanation

Sun 19 Apr, 2015 05:49 pm
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) is allowed in the sand and 45 % of voids are allowed in the gravel/rocks.

What is known:

A mixture of concrete is 1:2:4 and contains 1 part of cement, 2 parts sand, and 4 parts of gravel/rocks.

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

Want to know:

Amounts of each to make 1 cu. yd. of cement.

The following is provided with problem:

The percent of voids.

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

My attempt:

3x + 12x + 6.6x = 3 (multiplying by 3)

21.6x = 3

21.6x/21.6 = 3/21.6

x = .138888 cu. yd. of cement.

y = .276666 cu. yd. of sand.

z = .55552 cu. yd. of gravel/rocks.

I don't think those are correct. Where did I make the error(s) ? Also, I don't understand the equations.

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Randy Dandy

1
Mon 20 Apr, 2015 10:58 am

Not a homework question.
0 Replies

Rickoshay75

1
Mon 20 Apr, 2015 11:27 am
@Randy Dandy,
Randy Dandy wrote:

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.

The following is provided with problem:

The percent of voids.

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

My attempt:

3x + 12x + 6.6x = 3 (multiplying by 3)

21.6x = 3

21.6x/21.6 = 3/21.6

x = .138888 cu. yd. of cement.

y = .276666 cu. yd. of sand.

z = .55552 cu. yd. of gravel/rocks.

I don't think those are correct. Where did I make the error(s) ? Also, I don't understand the equations.

Why don't you just buy premixed concrete?
0 Replies

timur

1
Mon 20 Apr, 2015 12:18 pm
Randy Dandy wrote:
x + 4x/3 + 2.2x = 1

My attempt:

3x + 12x + 6.6x = 3 (multiplying by 3)

Failed attempt, Randy.

(4x/3)*3=4x

With this new value and following your train of thought you should get the right result.
Randy Dandy

1
Mon 20 Apr, 2015 12:30 pm
@timur,
I don't understand.
timur

1
Mon 20 Apr, 2015 01:21 pm
@Randy Dandy,
look what you wrote:
x + 4x/3 + 2.2x = 1

3x + 12x + 6.6x = 3 (multiplying by 3)

When you multiply 4x/3 by 3 you don't get 12x, you get 4x.

Randy Dandy

1
Mon 20 Apr, 2015 02:17 pm
@timur,
I understand now.

x + 4x/3 + 2.2x = 1
3x + 4x + 6.6x = 3
13.6x = 3
13.6x/13.6 = 3/13.6
x = .22058 = .22 cu. ft. of cement.
y= .44116 = .44 cu. ft. of sand.
z = .88232 = .88 cu. ft. of gravel/rocks.

Correct ?
timur

2
Tue 21 Apr, 2015 10:21 am
@Randy Dandy,
No, Randy, the total should be x+y+z=1

Look:

x= .22058
y=4x/3= .294
z=2.2x= .485

.221 + .294 + .485 = 1 cu. yard.

Randy Dandy

1
Tue 21 Apr, 2015 12:26 pm
@timur,
A question:

If the mixture is 1:2:4, why is the sand not twice and the gravel/rocks not four times the amount of cement?
0 Replies

usery

1
Tue 21 Apr, 2015 10:09 pm
@Randy Dandy,
The question states:

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

That's why there are equations to solve. Do you understand the equations?

Randy Dandy

1
Tue 21 Apr, 2015 11:32 pm
@usery,
I thought I did. I guess not.
0 Replies

usery

1
Thu 23 Apr, 2015 12:05 am
@Randy Dandy,
The ratio of 1:2:4 refers to the containers that the ingredients come in.

The cement container is size 1 and contains no air.

The sand container is size 2 but 33 1/3 % of the box's contents is just air between the sand particles.

The gravel container is size 4 but 45 % of the box's contents is just air between the rocks.

Knowing how much air there is allows for the equations (that you quoted) to be deduced.

When the mixture of wet cement, sand and gravel is poured as concrete there is no longer any air in the mix.

Do you understand everything about this topic now or do you have further questions?
Randy Dandy

1
Thu 23 Apr, 2015 07:52 am
@usery,
The mixture is 1:2:4. The total is 7 parts. The cement is 1/7, sand is 2/7, and
gravel is 4/7.

1/7 = 0.142 cement.
2/7 = 0.285 sand.
4/7 = 0.571 gravel.

Would these calculations be used in any way ?

timur

1
Thu 23 Apr, 2015 08:06 am
@Randy Dandy,
No, Randy.

Obviously you didn't understand usery's explanation.

Try with this concept:

Randy Dandy

1
Thu 23 Apr, 2015 08:25 am
@timur,
I understand that. I don't understand how to calculate the amounts.
usery

1
Thu 23 Apr, 2015 11:35 pm
@Randy Dandy,
We know that 1/7 + 2/7 + 4/7 equals 7/7 = 1

Quote:
The mixture is 1:2:4. The total is 7 parts. The cement is 1/7, sand is 2/7, and gravel is 4/7.
1/7 = 0.142 cement.
2/7 = 0.285 sand.
4/7 = 0.571 gravel.
Would these calculations be used in any way ?

The calculation of the VOLUME amounts is done in the manner shown by timur WHICH USES THE AIR SPACE AND THE RATIO IN THE FORMULA.

Those calculations allow the builder to determine the volume of concrete which will be poured if a mix of 1:2:4 is used and the ingredients are delivered in the sizes of containers containing the amounts of free space (air) that the question states.

GO BACK AND READ THE QUESTION THAT YOU POSTED ESPECIALLY THE BITS ABOUT X,Y,Z AND THE AIR COMPONENT

YOU WERE GIVEN THE EQUATION TO SOLVE.

Coincidentally, I'm thinking of adding a driveway.
Randy Dandy

1
Fri 24 Apr, 2015 06:44 am
@usery,
"GO BACK AND READ THE QUESTION THAT YOU POSTED ESPECIALLY THE BITS ABOUT X,Y,Z AND THE AIR COMPONENT " I read both.

" YOU WERE GIVEN THE EQUATION TO SOLVE." I still don't understand.
timur

1
Fri 24 Apr, 2015 07:59 am
@Randy Dandy,
Well, one last try: Randy Dandy

1
Fri 24 Apr, 2015 08:23 am
@timur,
I still don't know how to apply to solve. Thanks anyway.
0 Replies

Randy Dandy

1
Fri 24 Apr, 2015 01:46 pm
@usery,
I'll bet you're glad I am not determining amount of concrete for your driveway. I can't calculate mixture for ONE cubic yard, much less for a driveway, sidewalk, posts, etc.
0 Replies

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