How many pounds of type A coffee did she use?

Question

Nicole's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Nicole $4.45

 per pound, and type B coffee costs $5.90

 per pound. This month, Nicole made 144

 pounds of the blend, for a total cost of $752.45

. How many pounds of type A coffee did she use? 

Answer 1

For questions like this, start by just writing what the problem is asking into algebraic notation. First, let's look at the fact that 144 pounds of blend were made. Let "a" be the pounds of coffee A and "b" be pounds of coffee B. Since the blend is made up of both types,

a + b = 144

Next, let's look at prices. The total price is $752.45. This price is made up from the prices of the two types, which is just the price per pound times the number of pounds of coffee. We already have "a" and "b" to symbolize the number of pounds, so we get:

$4.45a + 5.90b = $752.45

When ever you see two equations with two variables, you should immediately think about if you can use substitution. The first equation looks easier to deal with, so let's use that for the substitution. We can solve for either b or a. The question only asks for how many pounds of type A, so let's solve for b, so we can get rid of that variable when we do the substitution.

b = 144 - a

Now you can substitute that into the second equation:

$4.45a + $5.90(144 - a) = $752.45

Now just multiply out and simplify:

$4.45a + $849.6 - $5.90a = $752.45

-$1.45a + $849.6 = $752.45

subtract $849.6 from both sides

-$1.45a = -$97.15

divide by -$1.45: notice that the dollar unit and the negative sign go away

a = 67

Keep in mind, that I could have solved for a first and substituted that into the second equation to find b, then use that value in the first equation to find a. I could also have solved for a or b in the second equation and substituted that into the first, etc. There is no one "right" way to solve it, and if you are not sure which is the best way, you should just try. You start to notice patterns and efficient strategies as you get practice.

So for questions like this, just remember to first write down everything the problem is telling you as algebraic expressions and equations. Then solve for one variable to substitute into the other equation. Then just simplify and manipulate the equation to isolate the variable you want to solve. Don't be afraid to just keep trying valid algebraic manipulations; have faith that you will get to the answer eventually even if it looks messy. With some practice, you'll begin to see the easier strategies right away.

Answer 2

This is a classic systems of equations problem. One equation will involve amounts of coffee and the other equation will involve the cost of the coffee.

Let A = the amount of Type A coffee, in pounds

Let B = the amount of Type B coffee, in pounds

To represent the total amount of coffee, we would add the amount of type A and the amount of type B, which has to equal a total of 144 pounds. In algebraic terms:

A + B = 144

To represent the cost of the coffee: Nicole spent $4.45 per pound for however many pounds of type A coffee and $5.90 per pound for however many pounds of type B coffee and spent a total amount of $752.45. In algebraic terms:

4.45A + 5.9B = 752.45

Now put the two equations together:

A + B = 144

4.45A + 5.9B = 752.45

We only want to solve for A. We can use substitution or elimination to do so. Substitution will be easier since the coefficients in the first equation are both one. Solve the first equation for B:

A + B = 144

B = 144 - A

Substitute 144-A for B in the second equation:

4.45A + 5.9B = 752.45

4.45A + 5.9 (144 - A) = 752.45

Solve for A:

4.45A + 849.6 - 5.9A = 752.45

-1.45A + 849.6 = 752.45

-1.45A = -97.15

A = 67 pounds of type A coffee