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