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14 votes
14 votes
You have one type of nut that sells for $4.40 a pound and another type of nut that sells for $6.10 a pound. You would like to have 8 1/2 pounds of a nut mixture that sells for $5.80 a pound. How much of each not will you need to obtain the desired mixture?

User David Murdoch
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1 Answer

20 votes
20 votes

Let x be the number of pounds of the first type ($4.40) of nut, and y be the number of pounds of the other one, then we can set the following system of equations:


\begin{gathered} x+y=8(1)/(2)=8.5, \\ 4.40x+6.10y=5.80\cdot8.50=49.30. \end{gathered}

Solving the first equation for x we get:


x=8.5-y\text{.}

Substituting the above result in the second equation we get:


\begin{gathered} 4.40(8.5-y)+6.10y=49.30, \\ 37.40-4.40y+6.10y=49.30, \\ 1.7y=11.90, \\ y=7. \end{gathered}

Substituting y=7 in the third equation on the board we get:


x=8.5-7=1.5.

Answer: You will need to add 1.5 pounds of the $4.40 nut and 7 pounds of the other one.

User Tula
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