document.write( "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?
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Algebra.Com's Answer #710769 by greenestamps(13200) $\"\"$ $\"About$

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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.

\n" ); document.write( "(1) Standard algebraic solution....
\n" ); document.write( "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:
\n" ); document.write( " $\"10%280.80%29+%2B+x%281.20%29+=+%2810%2Bx%29%281.10%29\"$
\n" ); document.write( " $\"8+%2B+1.2x+=+11+%2B+1.1x\"$
\n" ); document.write( " $\"0.1x+=+3\"$
\n" ); document.write( " $\"x+=+30\"$

\n" ); document.write( "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.

\n" ); document.write( "And here is a solution by a method that will get you to the answer much faster, if you understand how to use it....

\n" ); document.write( "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.

\n" ); document.write( "That means the mixture must contain 3 times as much of grass seed B as it does of grass seed A.

\n" ); document.write( "And since there are 10 pounds grass seed A, he needs 30 pounds of grass seed B. \n" ); document.write( "

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