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(Real-World Proportional Problems MC)

Three students are participating in a walkathon. The data for each student is shown.

Student A: The equation [tex]y=13x[/tex] represents the amount of money raised ([tex]y[/tex]) for walking ([tex]x[/tex]) miles.

Student B: [tex]\[tex]$ 36[/tex] is raised after walking 3 miles.

Student C:
\[
\begin{array}{|c|c|c|}
\hline
Miles \, Walked & 4 & 5 \\
\hline
Money \, Raised & \$[/tex] 52 & \$ 65 \\
\hline
\end{array}
\]

Determine which student raised the least amount of money after walking 7 miles.

A. Student A
B. Student B
C. Student C
D. All three students raised the same amount of money after walking 7 miles.


Sagot :

To determine which student raised the least amount of money after walking 7 miles, let’s evaluate the information provided for each student:

### Student A
The equation given for Student A is [tex]\( y = 13x \)[/tex].
- [tex]\( x \)[/tex] represents the miles walked.
- [tex]\( y \)[/tex] represents the amount of money raised.

When [tex]\( x = 7 \)[/tex]:
[tex]\[ y = 13 \times 7 = 91 \][/tex]

### Student B
Student B raises [tex]$36 for walking 3 miles. First, find the rate per mile: \[ \text{Rate per mile} = \frac{36}{3} = 12 \] Then calculate the money raised after walking 7 miles: \[ \text{Money raised} = 12 \times 7 = 84 \] ### Student C Student C raises $[/tex]52 for walking 4 miles and [tex]$65 for walking 5 miles. First, find the rate per mile: \[ \text{Rate per mile} = \frac{65}{5} = 13 \] Then calculate the money raised after walking 7 miles: \[ \text{Money raised} = 13 \times 7 = 91 \] ### Comparison From the calculations: - Student A raised \$[/tex]91.
- Student B raised \[tex]$84. - Student C raised \$[/tex]91.

Thus, the student who raised the least amount of money after walking 7 miles is Student B, who raised \$84.

[tex]\[ \boxed{\text{Student B}} \][/tex]