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Applied problems

by pullagurla » Thu Sep 02, 2010 10:46 pm
Seed mixture X is 40 percent ryegrass and 60 percent
bluegrass by weight; seed mixture Y is 25 percent
ryegrass and 75 percent fescue. If a mixture of X and
Y contains 30 percent ryegrass, what percent of the
weight of the mixture is X ?
(A) 10%
(B) 33 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%

Use backsolving
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by blaster » Thu Sep 02, 2010 10:53 pm
0.4x + 0.25y = 0.3(x + y)

x=5/10 , 1/2 - hence 50%
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by singhsa » Fri Sep 03, 2010 12:14 am
let weight of mixture X be X and let weight of mixture Y be Y.

X(40-30)=Y(30-5)
10X=5Y
2X=Y

Now, we are asked to find (X/X+Y)*100
so the eqn will be (X/X+2X)*100 = X/3X *100= 33.33%
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by diebeatsthegmat » Mon Sep 13, 2010 11:27 am
pullagurla wrote:Seed mixture X is 40 percent ryegrass and 60 percent
bluegrass by weight; seed mixture Y is 25 percent
ryegrass and 75 percent fescue. If a mixture of X and
Y contains 30 percent ryegrass, what percent of the
weight of the mixture is X ?
(A) 10%
(B) 33 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%

Use backsolving
in this PS, just ignore blue and fes stuff... just focus on red one
30(x+y)=40x+25y so 5x=10y or x=y/2
thus x/(x+y)=(y:2)/(y+y/2)=1/3y*100%=331/3%
the correct answer is B
hope it helps
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