word problem

[LEG7930]

A merchant blends tea that sells for $4.65 a pound with tea that sells for $2.45 a pound to produce 110 lb of a mixture that sells for $3.85 a pound. How many pounds of each type of tea does the merchant use in the blend?

If someone can set it up I could probably get it, but I'm not sure where to start.

 

[galactus]

Let x=the amount of $4.65 tea.

Since we need 110 lbs altogether, the amount of $2.45 tea would be the remaining amount: 110-x.

Therefore, we have $\displaystyle 4.65x+2.45(110-x)=3.85(110)$

Solve for x.

 

[LEG7930]

Thank you so much! I did a similar one before but the money was throwing me off!

 

[BigGlenntheHeavy]

$\displaystyle LEG7930, \ if \ I \ may \ be \ so \ presumptuous, \ when \ doing \ mixture \ problems, \ always \ try \ to \ formulate$

$\displaystyle a \ formula, \ to \ wit:$

$\displaystyle \ lbs. \ of \ tea \ \ \ X \ \ \ price \ per \ lb. \ \ = \ \ Total \ Value$

$\displaystyle High-Priced \ Tea \ \ x \ \ X \ \ 4.65 \ \ = \ \ 4.65x$

$\displaystyle Low-Priced \ Tea \ \ 110-x \ \ X \ \ 2.45 \ = \ 269.50-2.45x$

$\displaystyle Medium-Priced \ Tea \ 110 \ X \ 3.85 \ = \ 423.50$

$\displaystyle Ergo, \ adding \ up \ the \ total \ values, \ we \ get \ 4.65x+269.50-2.45x \ = \ 423.50$

$\displaystyle Solving \ for \ x \ gives \ x \ = \ 70 \ and \ 110-x \ = \ 40$

$\displaystyle Check: \ (70)(4.65)+(40)(2.45) \ = \ 423.50$

 

[soroban]

Hello, LEG7930!

When I was leaning, I was shown a table for setting up the problem.

A merchant blends tea that sells for $4.65 a pound with tea that sells for $2.45 a pound
to produce 110 lb of a mixture that sells for $3.85 a pound.
How many pounds of each type of tea does the merchant use in the blend?

$\displaystyle \text{We have two types of tea: }\;\begin{array}{ccc} \text{type A} & \text{at} & \$4.65\text{ per lb.} \\ \text{type B} & \text{at} & \$2.45\text{ per lb.}\end{array}$

$\displaystyle \text{We want: }110\text{ lbs. of mixture worth \$3.85 per pound.}$
. . . $\displaystyle \text{Its value is: }\:110 \times \$3.85 \:=\:\$423.50$

$\displaystyle \text{Write these facts into our chart:}$

. . $\displaystyle \begin{array}{|c||c|c||c|} \hline & \text{No. of } & \text{Unit} & \text{Total} \\ & \text{pounds} & \text{price} & \text{value} \\ \hline \hline \text{type A } & & \$4.65 & \\ \hline \text{type B} & & \$2.45 & \\ \hline \hline \text{Mixture} & 110 & \$3.85 & \$423.50 \\ \hline \end{array}$


$\displaystyle \text{Let }x\text{ = number of pounds of type A.}$
$\displaystyle \text{Then: }\,110-x\text{ = number of pounds of type B.}$

$\displaystyle \text{Write those quantities in our chart:}$

. . $\displaystyle \begin{array}{|c||c|c||c|} \hline & \text{No. of } & \text{Unit} & \text{Total} \\ & \text{pounds} & \text{price} & \text{value} \\ \hline \hline \text{type A } & x & \$4.65 & \\ \hline \text{type B} & 110-x& \$2.45 & \\ \hline \hline \text{Mixture} & 110 & \$3.85 & \$423.50 \\ \hline \end{array}$


$\displaystyle \text{Read across the rows.}$

\(\displaystyle \text{There is }x\text{ pounds of type A at \$4.65 per pound: }\text{ Total Value}\:=\x)(4.65) \:=\:4.65x\text{ dollars.}\)

\(\displaystyle \text{There is }(110-x)\text{ pounds of type B at \$2.45 per pound: }\text{ Total Value}\:=\2.45)(110-x) \:=\:2.45(110-x)\text{ dollars.}\)


$\displaystyle \text{Write those quantities in our chart:}$

. . $\displaystyle \begin{array}{|c||c|c||c|} \hline & \text{No. of } & \text{Unit} & \text{Total} \\ & \text{pounds} & \text{price} & \text{value} \\ \hline \hline \text{type A } & x & \$4.65 & 4.65x\\ \hline \text{type B} & 110-x& \$2.45 & 2.45(110-x)\\ \hline \hline \text{Mixture} & 110 & \$3.85 & \$423.50 \\ \hline \end{array}$


$\displaystyle \text{The equation is waiting for us in the last column!}$

. . $\displaystyle \text{(Total value of type A) }+\text{ (Total value of type B)} \;=\;\text{(Total value of Mixture)}$


$\displaystyle \text{Therefore, we have: }\;4.65x + 2.45(110-x) \;=\;423.50$


$\displaystyle \text{And }that\text{ is where the equation comes from . . .}$