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:


, \(24\) 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:


Notice the solution is \(24\).

3. Using a Balance :


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


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

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

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


Solve the proportion:


Return to Explore to complete the problem set.

Continue to Practice (Mixture Problems)