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Mon 29 Sep, 2014 02:01 pm
In one kind of concrete the parts of cement, sand, and gravel are as 1:2:4. In another kind 3 parts are as 1:2:5. Determine how much more concrete in pounds needed in 1 ton of one than the other.
What is known:
One kind of concrete has parts of cement, sand, and gravel in parts 1, 2, 4.
Another kind has parts of 1, 2, 5.
1 ton= 2000 pounds.
Want to know:
How much more concrete ( in pounds) needed in 1 ton of one more than the other.
Let 1st kind = A
Let 2nd kind = B
Let C = Cement
Let S = Sand
Let G = Gravel
A = 7 parts- 1 part cement, 2 parts sand, 4 parts gravel - 1/7 cement, 2/7 sand, 4/7 gravel.
B = 8 parts- 1 part cement, 2 parts sand, 5 parts gravel - 1/8 cement, 2/8 sand, 5/8 gravel.
1 ton = 2000 pounds.
I don't know how to proceed.
@Randy Dandy,
It's called cross-multiplication.
Let's put it this way:
8 parts 2000 pounds
_____ = __________
1 part d pounds
Solve according to the above formula.
@Randy Dandy,
1/7 of 2000lb = how much cement in the first mixture?
1/8 of 2000lb = how much cement in the second mixture?
@engineer,
1/7 of 2000 lbs = 285.714 = 286 lbs.
1/8 of 2000 lbs = 250 lbs.
@Randy Dandy,
Right.
Now just multiply by the number of parts of the other components.
type 1
1/7 = cement ; 1/7 x 2000 = 285.714
2/7 = sand ; 2/7 x 2000 = 571.428
4/7 = gravel ; 4/7 x 2000 = 1142.857
---
7/7 = concrete
type 2
type 2
1/8 = cement ; 1/8 x 2000 = 250
2/8 = sand ; 2/8 x 2000 = 500
5/8 = gravel ; 4/7 x 2000 = 1250
---
8/8 = concrete
type 1 uses 285.714 lbs of cement per ton; type 2 uses 250 lbs;
Answer: type 1 uses 35.714 lbs more.
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