A coffee merchant wants to make 6 lb of blend @ $5/lb.

[igor_iv837]

A coffee merchant wants to make 6 lb of a blend of coffee costing $5 per pound. The blend is made using a $6 per pound grade and $3 per pound grade of cofee. How many pounds of each of these grades should be used?

the formula is like this Amount x*6
6-x *3
6*5 ...I don't get why is it 6x + 3(6-x) =5.6 I get the 5*6 but where they got 6-x? and times 3?

 

[mmm4444bot]

The merchant doesn't know what kind of mix to mix ...

Hello Igor:

6 - x comes from subtacting the $6-grade coffee from the mix.

I know I've told you this before.

6 - x is the $3 coffee.

Start with 6 pounds of mix and remove x pounds.

6 - x pounds remains.

3(6 - x) stands for the value of this part because each pound is worth $3.

Cheers,

~ Mark

PS: There are two different 6's in the formula. Be careful to understand the difference between 6 lb and $6 when looking at the equation.

 

[soroban]

Re: A coffee merchant wants

Hello, igor_iv837!

A coffee merchant wants to make 6 lb of a blend of coffee costing $5 per pound.
The blend is made using a $6 per pound grade and $3 per pound grade of cofee.
How many pounds of each of these grades should be used?

\(\displaystyle \text{The formula is like this:}\;\;\begin{array}{c}\text{Amount} \\ x\cdot6 \\ 6-x\cdot3 \\ 6\cdot5 \end{array}\)

I don't get why is it: 6x + 3(6 - x) = 5*6
I get the 5*6, but where they got (6-x) and times 3 ?

\(\displaystyle \text{He wants a total of 6 lbs of a blended coffee.}\)

\(\displaystyle \text{If he uses }x\text{ lbs of one brand, then he uses }(6-x)\text{ lbs of the other brand }\hdots\quad See?\)


\(\displaystyle \text{He will use }x\text{ lbs of the \$6 grade.}\)
. . \(\displaystyle \text{This has a value of: }\:x\cdot6 \,=\,6x\text{ dollars.}\)

\(\displaystyle \text{He will use }6-x\text{ lbs of the \$3 grade.}\)
. . \(\displaystyle \text{This has a value of: }\6-x)\cdot3 \:=\:3(6-x)\text{ dollars.}\)

\(\displaystyle \text{So the total value of the mixture is: }\:\boxed{6x + 3(6-x)}\text{ dollars.}\)


\(\displaystyle \text{We are told that the mixture will be 6 lbs worth \$5 per lb.}\)
. . \(\displaystyle \text{The total value of the mixture is: }\:6\cdot5 \:=\:\boxed{30}\text{ dollars.}\)


\(\displaystyle \text{And }there\text{ is our equation! }\quad 6x+3(6-x) \;=\;30\)


Got it?