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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]
07 Oct 2005, 10:55
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 this mixture is X ?
(B) 33 1/3%
(E) 66 2/3 %
The equation I get is
40% of ryegrass in X + 25% ryegrass in Y = 30% ryegrass in X+Y
i.e. 0.40X + 0.25Y = 0.30(X+Y)
=> 0.40X - 0.30X = 0.30Y - 0.25Y
=> 0.1X = 0.05Y
=> X/Y = 1/2, we can now work with this ratio to get
X/X+Y = 1/1+2 = 1/3 = 33 1/3% or B. _________________
well on this type of question you can also logically guess...
we know that Mix X has 40% Ryegrass
we know the Mix Y has 25% Ryegrass.
now if we both X+Y are combined the resulting ryegrass is 30%
well we logically see that since resulting rye grass is less than what it was in Mix X, then Mix X has to be less than 50% ...however we can also see that the resulting MIX has has 30% Ryegrass which is more than Mix Y, then answer choice 10% can also be easily elminated...we are now left with 2 choices 33% or 40% (you are down to 50-50 chance of getting it right)