SOLUTION: ;A gardener has 10 lb of grass seed which cost $0.80 per pound. How many pounds of a grass seed which cost $1.20 per pound should be mixed with the 10 lb of grass seed to produce a

Question 1096223: ;A gardener has 10 lb of grass seed which cost $0.80 per pound. How many pounds of a grass seed which cost $1.20 per pound should be mixed with the 10 lb of grass seed to produce a mixture which sells for $1.10 per pound?

Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
---misread----misunderstood problem description

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Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

I don't know what problem tutor josgarithmetic sees with the problem. You are mixing grass seed that costs $0.80 per pound with other grass seed that costs $1.20 per pound to get a mixture that sells for $1.10 per pound. As long as the $1.10 is between $0.80 and $1.20, the problem has an answer.

(1) Standard algebraic solution....
You are mixing 10 pounds of grass seed A that costs $0.80 per pound with unknown amount x of grass seed B that costs $1.20 per pound to get a mixture of (10+x) pounds that costs $1.10 per pound:

The gardener needs to mix 30 pounds of grass seed B with the 10 pounds of grass seed A to get a mixture that sells for $1.10 per pound.

And here is a solution by a method that will get you to the answer much faster, if you understand how to use it....

The cost per pound of the mixture, $1.10, is "3 times as close" to $1.20 as it is to $0.80. That is, $1.20-$1.10 = $0.10; $1.10-$0.80 - $0.30.

That means the mixture must contain 3 times as much of grass seed B as it does of grass seed A.

And since there are 10 pounds grass seed A, he needs 30 pounds of grass seed B.
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