Q. 143 - Intermediate Algebra

Question

Twelve-year old Melody wants to sell bags of mixed candy at her lemonade stand. She will mix M&M’s that cost \(4.89 per bag and Reese’s Pieces that cost \)3.79 per bag to get a total of twenty- five bags of mixed candy. Melody wants the bags of mixed candy to cost her $4.23 a bag to make. How many bags of M&M’s and how many bags of Reese’s Pieces should she use?

Step-by-Step Solution

Answer

Bags of M&M’s =10

Bags of Reese’s Pieces =15

Step 1. Given Information

M&M’s  costs $4.89 per bag.

Reese’s Pieces  costs $3.79 per bag.

Total Bags=25.

Mixed candy costs $4.23 per bag.

Step 2. Formation of two linear Equations

Let number of bags of M&M’s  = $x$.

Let number of bags of Reese’s Pieces   = $y$.

As we want   25 bags in total so,

$x+y=25$

And if we will balance the total amount of money then,

$$\begin{gathered} 4.89x+3.79y=25\times 4.23 \\ 4.89x+3.79y=105.75 \end{gathered}$$

Step 3. Solving the equations

From the first equation,

$x=25-y$

Now we will substitute this value in the second equation.

$$\begin{gathered} 4.89\left(25-y\right)+3.79y=105.75 \\ 122.5-4.89y+3.79y=105.75 \\ -1.1y=-16.5 \\ 1.1y=16.5 \\ 11y=165 \\ y=15 \end{gathered}$$

Therefore the bags of Reese’s Pieces  = 15 .

Now put this value in the equation 

$$\begin{gathered} x=25-y \\ x=25-15 \\ x=10 \end{gathered}$$

Therefore the bags of  M&M’s   =10.

Step 4. Check the solution

Substitute $10$ for $x$ and $15$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 25\\ 10+15& =& 25\\ 25& =& 25\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}4.89x+3.79y& =& 105.75\\ 4.89·10+3.79·15& =& 105.75\\ 48.9+56.85& =& 105.75\\ 105.75& =& 105.75\end{array}$

This is also a true statement.

So the point satisfies both the equations.