Mixture - Concrete - Solution/Explanation

Forums: Algebra, Math, Maths, Lazy Student, Do My Homework
Email this Topic • Print this Page

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.

Topic Stats
Top Replies
Link to this Topic

Type: Question • Score: 1 • Views: 3,855 • Replies: 34
Topic Closed

No top replies

Please respond. Thanks.

Not a homework question.

0 Replies

@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

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.

1 Reply

@timur,

I don't understand.

1 Reply

@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.

1 Reply

@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 ?

1 Reply

@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.

1 Reply

@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

@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?

1 Reply

@usery,

I thought I did. I guess not.

0 Replies

@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?

1 Reply

@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 ?

1 Reply

@Randy Dandy,

No, Randy.

Obviously you didn't understand usery's explanation.

Try with this concept:

1 Reply

@timur,

I understand that. I don't understand how to calculate the amounts.

1 Reply

@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.

2 Replies

@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.

1 Reply

@Randy Dandy,

Well, one last try:

1 Reply

@timur,

I still don't know how to apply to solve. Thanks anyway.

0 Replies

@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

Mixture - Concrete - Solution/Explanation

Related Topics

Quick Links

My Account

able2know