RATL 1 | Lesson 4 | Explore (Multiple Strategies Solutions)

Here are some different ways to think about the solution to the problem:

How many pounds of coffee worth $7 per pound must be mixed with 12 pounds of coffee worth $4 per pound to make a mixture worth $6 per pound?

1) Creating an Equation:

Let \(x\) represent the number of pounds of the $7 per pound coffee. There are 12 pounds of the $4 per pound coffee and x pounds of $7 per pound mixture means that the total is \(x+12\)pounds.

Knowing the cost for each type of coffee gives us this equation to solve:

\[4(12)+7(x)=6(x+12)\]\[48+7x=6x+72\]\[x=24\]

, \(24\) pounds of the \($7\) per pound coffee is needed.


2) Creating a Table:

Number of Pounds Cost Per Pound Total Cost
Cheaper Coffee \(12\) \($4\) \($48\)
Expensive Coffee \(x\) \($7\) \($7x\)
Combined Coffee \(12+x\) \($6\) \($6(12+x)\)

Use the Total Cost Column to create this equation to solve:

\[48+7(x)=69x+12)\]\[48+7x=6x+72\]\[x=24\]

Notice the solution is \(24\).


3) Using a Balance :

http://www.gardenista.com/products/teeter-totter

\(12\)               \(12+x\)                \(x\)         weight

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\($4\)                \($6\)                  \($7\)       cost per lb

Set up a proportion using the weight in pounds and difference between price per pound.

\[($6 – $4 =2)  \ and \  ($7 – $6 = 1)\]

\[\frac{12}{1}=\frac{x}{2}\]

Solve the proportion:

\(x=24\)


Go back to Explore to complete the problem set.