High School

A store sells yogurt in small, medium, and large sizes. A small yogurt costs [tex] \$2 [/tex], a medium costs [tex] \$3 [/tex], and a large costs [tex] \$5 [/tex]. In one hour, the shop sold 27 yogurts for [tex] \$98 [/tex]. There were five more large yogurts than small yogurts. The matrix below represents the system of equations in this situation.

[tex]
\begin{bmatrix}
1 & 1 & 1 & | & 27 \\
2 & 3 & 5 & | & 98 \\
1 & 0 & -1 & | & -5
\end{bmatrix}
[/tex]

How many of each size yogurt did the store sell during that hour?

A. 7 small yogurts, 8 medium yogurts, 12 large yogurts
B. 12 small yogurts, 8 medium yogurts, 7 large yogurts
C. 27 small yogurts, 98 medium yogurts, 5 large yogurts
D. 98 small yogurts, 27 medium yogurts, 5 large yogurts

Answer :

To solve the problem, we need to determine the number of small, medium, and large yogurts sold given the conditions. Let's break it down step by step:

1. Define Variables:
- Let [tex]\( x \)[/tex] be the number of small yogurts sold.
- Let [tex]\( y \)[/tex] be the number of medium yogurts sold.
- Let [tex]\( z \)[/tex] be the number of large yogurts sold.

2. Set Up Equations:
The problem gives us the following information:
- The total number of yogurts sold is 27:
[tex]\[
x + y + z = 27 \quad \text{(Equation 1)}
\][/tex]
- The total cost of the yogurts is \[tex]$98. With the costs being \$[/tex]2 for small, \[tex]$3 for medium, and \$[/tex]5 for large, this gives us:
[tex]\[
2x + 3y + 5z = 98 \quad \text{(Equation 2)}
\][/tex]
- There were five more large yogurts than small yogurts:
[tex]\[
z = x + 5 \quad \text{(Equation 3)}
\][/tex]

3. Substitute and Solve:
- Substitute Equation 3 into Equation 1:
[tex]\[
x + y + (x + 5) = 27 \quad \Rightarrow \quad 2x + y + 5 = 27 \quad \Rightarrow \quad 2x + y = 22 \quad \text{(Equation 4)}
\][/tex]

- Now solve the system of Equations 2 and 4:
[tex]\[
2x + 3y + 5z = 98
\][/tex]
Substituting [tex]\( z = x + 5 \)[/tex] from Equation 3:
[tex]\[
2x + 3y + 5(x + 5) = 98 \quad \Rightarrow \quad 2x + 3y + 5x + 25 = 98
\][/tex]
Simplifying this:
[tex]\[
7x + 3y = 73 \quad \text{(Equation 5)}
\][/tex]

- We now have:
[tex]\[
2x + y = 22 \quad \text{(from Equation 4)}
\][/tex]
[tex]\[
7x + 3y = 73 \quad \text{(from Equation 5)}
\][/tex]

- Solve these linear equations:
- Multiply Equation 4 by 3 to eliminate [tex]\( y \)[/tex]:
[tex]\[
6x + 3y = 66
\][/tex]
- Subtract this from Equation 5:
[tex]\[
7x + 3y - (6x + 3y) = 73 - 66 \quad \Rightarrow \quad x = 7
\][/tex]

4. Find the Remaining Variables:
- Substitute [tex]\( x = 7 \)[/tex] back into Equation 4:
[tex]\[
2(7) + y = 22 \quad \Rightarrow \quad 14 + y = 22 \quad \Rightarrow \quad y = 8
\][/tex]
- Use [tex]\( x \)[/tex] in Equation 3 to find [tex]\( z \)[/tex]:
[tex]\[
z = x + 5 \quad \Rightarrow \quad z = 7 + 5 = 12
\][/tex]

Therefore, the store sold 7 small yogurts, 8 medium yogurts, and 12 large yogurts.