Answered

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Each of two bike rental shops charges a set fee plus an hourly rate for each rental. The cost of renting a bike from two shops is shown in the table.

\begin{tabular}{|c|c|c|}
\hline
Hours Rented & Shop A Cost (\[tex]$) & Shop B Cost (\$[/tex]) \\
\hline
1 & 9 & 7 \\
\hline
5 & 25 & 27 \\
\hline
\end{tabular}

Mike wants to rent a bike for a certain number of hours and realizes that it would cost the same at both shops. For how many hours does Mike want to rent a bike? Do not include units in your answer.

[tex]\[\square\][/tex]

Sagot :

GinnyAnswer
To solve this problem, we need to determine the set fee and hourly rate for each shop, and then determine the number of hours at which the costs from both shops are equal.

First, let's define the cost function for each shop:
- For Shop A: [tex]\( \text{Cost}_A = \text{set\_fee}_A + \text{hourly\_rate}_A \times \text{hours} \)[/tex]
- For Shop B: [tex]\( \text{Cost}_B = \text{set\_fee}_B + \text{hourly\_rate}_B \times \text{hours} \)[/tex]

We know from the table:
- For Shop A, 1 hour costs [tex]$9 and 5 hours costs $[/tex]25.
- For Shop B, 1 hour costs [tex]$7 and 5 hours costs $[/tex]27.

From these points, we can set up two equations for each shop:

Shop A:
1. [tex]\( 9 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 1 \)[/tex]
2. [tex]\( 25 = \text{set\_fee}_A + \text{hourly\_rate}_A \times 5 \)[/tex]

Subtract the first equation from the second:
[tex]\[ 25 - 9 = (\text{set\_fee}_A + 5 \times \text{hourly\_rate}_A) - (\text{set\_fee}_A + 1 \times \text{hourly\_rate}_A) \][/tex]
[tex]\[ 16 = 4 \times \text{hourly\_rate}_A \][/tex]
[tex]\[ \text{hourly\_rate}_A = 4 \][/tex]

Substitute back into the first equation:
[tex]\[ 9 = \text{set\_fee}_A + 4 \][/tex]
[tex]\[ \text{set\_fee}_A = 5 \][/tex]

Shop B:
1. [tex]\( 7 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 1 \)[/tex]
2. [tex]\( 27 = \text{set\_fee}_B + \text{hourly\_rate}_B \times 5 \)[/tex]

Subtract the first equation from the second:
[tex]\[ 27 - 7 = (\text{set\_fee}_B + 5 \times \text{hourly\_rate}_B) - (\text{set\_fee}_B + 1 \times \text{hourly\_rate}_B) \][/tex]
[tex]\[ 20 = 4 \times \text{hourly\_rate}_B \][/tex]
[tex]\[ \text{hourly\_rate}_B = 5 \][/tex]

Substitute back into the first equation:
[tex]\[ 7 = \text{set\_fee}_B + 5 \][/tex]
[tex]\[ \text{set\_fee}_B = 2 \][/tex]

Now, we have:
- [tex]\( \text{Cost}_A = 5 + 4 \times \text{hours} \)[/tex]
- [tex]\( \text{Cost}_B = 2 + 5 \times \text{hours} \)[/tex]

We need to find the number of hours where the costs are equal:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]

Solving for hours:
[tex]\[ 5 + 4 \times \text{hours} = 2 + 5 \times \text{hours} \][/tex]
[tex]\[ 5 - 2 = 5 \times \text{hours} - 4 \times \text{hours} \][/tex]
[tex]\[ 3 = \text{hours} \][/tex]

Hence, Mike wants to rent a bike for [tex]\( \boxed{3} \)[/tex] hours.
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