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Other Arithmetic problems from 11th edition

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Other Arithmetic problems from 11th edition

by jfranco23 » Mon Feb 23, 2009 4:17 pm
Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixture Y is 25% ryegrass and 75% fescue. If a mixture of X and Y contains 30% ryegrass, what % of the weight of the misture is X?

A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%
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by billzhao » Mon Feb 23, 2009 5:55 pm
We have: (40%*X+25%*Y)/(X+Y)=30%, after simplication, we have Y=2*X

So X/(X+Y)=1/3=33 1/3%

Answer is (B)
Yiliang
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