SOLUTION: I am not sure if I am posting this in the right place
A contractor mixes concrete from bags of​ pre-mix. How many bags with 10 % cement should he mix with 4 bags of 12.5%
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: I am not sure if I am posting this in the right place
A contractor mixes concrete from bags of​ pre-mix. How many bags with 10 % cement should he mix with 4 bags of 12.5%
Log On
Question 1083425: I am not sure if I am posting this in the right place
A contractor mixes concrete from bags of pre-mix. How many bags with 10 % cement should he mix with 4 bags of 12.5% cement to produce a mix containing 11% cement?
A single 10% cement bag has 10 pounds of pure cement (0.1*100 = 10)
x of these bags would have 10x pounds of pure cement
In total, we have 100*x = 100x pounds of material (cement+other stuff)
A single 12.5% bag has 12.5 pounds of pure cement (0.125*100 = 12.5)
Four of these bags would have 4*12.5 = 50 pounds of pure cement
In total, we have 4*100 = 400 pounds of material (cement+other stuff)
Mixing the x 10% bags and the four 12.5% bags yields
10x+50 pounds
of pure cement
This would be out of 100x+400 pounds total (of cement plus other material)
Divide the two expressions (10x+50 and 100x+400) and set this ratio equal to the desired percentage we want, which is 11% = 0.11, so
(amount of pure cement)/(amount total) = percentage desired
(10x+50)/(100x+400) = 0.11
Solve for x
(10x+50)/(100x+400) = 0.11
[(10x+50)/(100x+400)]*(100x+400) = 0.11*(100x+400)
10x+50 = 0.11*(100x)+0.11*(400)
10x+50 = 11x+44
10x+50-44 = 11x+44-44
10x+6 = 11x
10x+6-10x = 11x-10x
6 = x
x = 6
(