College

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

[tex]\[

\left[\begin{array}{ccc|c}

1 & 1 & 1 & 27 \\

2 & 3 & 5 & 98 \\

1 & 0 & -1 & -5

\end{array}\right]

\][/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 of finding out how many small, medium, and large yogurts the store sold during that hour, we can use the given system of equations derived from the matrix.

Let's denote the number of small yogurts as [tex]\( s \)[/tex], medium yogurts as [tex]\( m \)[/tex], and large yogurts as [tex]\( l \)[/tex]. Based on the problem description:

1. The total number of yogurts sold is 27, so:
[tex]\[
s + m + l = 27
\][/tex]

2. The total sales amount to $98, so the equation for the total cost is:
[tex]\[
2s + 3m + 5l = 98
\][/tex]

3. The problem states that there were five more large yogurts sold than small yogurts, so:
[tex]\[
s - l = -5 \quad \text{(or equivalently, \( l = s + 5 \))}
\][/tex]

Now, we have a system of three equations:
1. [tex]\( s + m + l = 27 \)[/tex]
2. [tex]\( 2s + 3m + 5l = 98 \)[/tex]
3. [tex]\( s - l = -5 \)[/tex]

To solve this system:

- From equation 3, [tex]\( s - l = -5 \)[/tex], we can express [tex]\( l \)[/tex] in terms of [tex]\( s \)[/tex]:
[tex]\[
l = s + 5
\][/tex]

- Substitute [tex]\( l = s + 5 \)[/tex] into the first equation:
[tex]\[
s + m + (s + 5) = 27
\][/tex]
Simplifying, we get:
[tex]\[
2s + m + 5 = 27
\][/tex]
[tex]\[
2s + m = 22 \quad \text{(Equation 4)}
\][/tex]

- Now substitute [tex]\( l = s + 5 \)[/tex] into the second equation:
[tex]\[
2s + 3m + 5(s + 5) = 98
\][/tex]
Simplifying, we get:
[tex]\[
2s + 3m + 5s + 25 = 98
\][/tex]
[tex]\[
7s + 3m = 73 \quad \text{(Equation 5)}
\][/tex]

Now we have a simpler system with equations 4 and 5:
1. [tex]\( 2s + m = 22 \)[/tex]
2. [tex]\( 7s + 3m = 73 \)[/tex]

To solve these equations, we can use substitution or elimination. Here, we'll use elimination:

- Multiply equation 4 by 3:
[tex]\[
3(2s + m) = 3 \times 22
\][/tex]
[tex]\[
6s + 3m = 66 \quad \text{(Equation 6)}
\][/tex]

Subtract equation 6 from equation 5:
[tex]\[
(7s + 3m) - (6s + 3m) = 73 - 66
\][/tex]
[tex]\[
s = 7
\][/tex]

Now that we know [tex]\( s = 7 \)[/tex], substitute back into equation 4:
[tex]\[
2(7) + m = 22
\][/tex]
[tex]\[
14 + m = 22
\][/tex]
[tex]\[
m = 8
\][/tex]

Finally, use the value of [tex]\( s \)[/tex] to find [tex]\( l \)[/tex] using [tex]\( l = s + 5 \)[/tex]:
[tex]\[
l = 7 + 5 = 12
\][/tex]

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