Question 5727: A contractor mixed two batches of concrete that were 9.3 and 11.3 percent cement to obtain 4500 lb of concrete that was 10.8 percent cement. How many pounds of each type of concrete was used?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! You can start by letting x = the # of lbs of the 9.3%-cement concrete and y = the # lbs of the 11.3%-cement concrete.
Next, you can write the necessary equations you'll need to solve the problem.
x + y = 4500 lbs of 10.8%-cement concrete. Rewrite as: y = 4500 - x
9.3%x + 11.3%y = 10.8%(4500)
You can solve this system of equations by the "substitution" method. In the 2nd equation, you'll substitute y = 4500 - x and solve for x. Change the percents to decimal. 9.3% = 0.093, 11.3% = 0.113, and 10.8% = 0.108
0.093x + 0.113(4500 - x) = 0.108(4500) Simplify and solve for x.
0.093x + 508.5 - 0.113x = 468
-0.02x = -22.5
x = 1125 lbs.
y = 4500 - x
y = 4500 - 1125
y = 3375 lbs.
Finally:
1125 lbs of 9.3%-cement concrete were used.
3375 lbs of 11.3%-cement concrete were used.
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