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Bunuel
Rice mixture A is 25 percent brown rice and 75 percent wild rice by weight; and rice mixture B is 40 percent brown rice and 60 percent white rice by weight. If a blend of mixture A and mixture B contains 30 percent brown rice by weight, what percent of the weight of the blend is mixture A?

(A) 32 1/2%
(B) 33 1/3%
(C) 50%
(D) 65%
(E) 66 2/3%

let A=weight of mixture A in blend
.25A+.4B=.3(A+B)
A/B=2/1
A/(A+B)=2/3=66 2/3%
D
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alligations and mixtures problem
Draw a number line ,with combined mean in middle and calculate the difference max-mean,mean-min

---->5<--->10<--
25-----30-------40
mix(A) =======mix(B)

A/B=10/5=2:1
Quantity of A =2/3 * 100 => answer (e) 66 2/3
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Bunuel
Rice mixture A is 25 percent brown rice and 75 percent wild rice by weight; and rice mixture B is 40 percent brown rice and 60 percent white rice by weight. If a blend of mixture A and mixture B contains 30 percent brown rice by weight, what percent of the weight of the blend is mixture A?

(A) 32 1/2%
(B) 33 1/3%
(C) 50%
(D) 65%
(E) 66 2/3%


Another way to spot the solution trough


A.................M.............................B
25%......5.....30%.........10%.........40%

The distance is is doubled with mixture being close to A.

\(\frac{A}{B}\)=\(\frac{10}{5}\)=\(\frac{2}{1}\)

\(\frac{A}{A+B}\)=\(\frac{2}{3}\)

It means that A has 66 2/3% in the misxture

Answer: E
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Bunuel
Rice mixture A is 25 percent brown rice and 75 percent wild rice by weight; and rice mixture B is 40 percent brown rice and 60 percent white rice by weight. If a blend of mixture A and mixture B contains 30 percent brown rice by weight, what percent of the weight of the blend is mixture A?

(A) 32 1/2%
(B) 33 1/3%
(C) 50%
(D) 65%
(E) 66 2/3%

We can create the equation:

0.25A + 0.4B = 0.3(A + B)

25A + 40B = 30A + 30B

10B = 5A

2B = A

Thus, A/(A+B) = 2B/(2B + B) = 2/3 = 66 2/3%.

Answer: E
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