Harshgmat wrote:
Concrete mixture A is composed of 35 percent cement, 40 pecent sand and 25 percent gravel; concrete mixture B is composed of 15 percent cement, 30 percent sand and 55 percent gravel. If a mixture of A and B contains 25 percent cement, what percent of the weight of the mixture is A?
A 20%
B 33 1/3%
C 45%
D 50%
E 55 1/3%
We need to track only on cement percents, and we need either the ratio of A to B (part-part) or the ratio of A to (A + B) (part-whole) to answer the question.
Let % = percent cement in a mixture
A = volume / weight of A
B = volume / weight of B
(% A)(Vol A) + (% B)(Vol B) = (% A+B)(Vol A+B)
\((.35*A)+(.15*B)=.25(A+B)\)
\(35A+15B=25A+25B\)
\(10A=10B\)\(A=B\) in the final mixture
\(A\) is what percent of final mixture's weight?
Part-whole. Percent A is
\(\frac{A}{A+B}*100\)From above
\(A=B\)Percent A:
\(\frac{A}{A+A}=\frac{1A}{2A}=\frac{1}{2}=(.50*100) = 50\) percent
Answer D
Part-PartFrom above, \(A=B\)
Part + Part = Whole
Half + Half = Whole
A is half = 50%
That is, logically, a mixture containing only A and B and equal parts A and B is 50% A.
Answer D
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