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Jorden drives to the store at 30 miles per hour. On her way home, she averages only 20 miles per hour. If the total driving time takes half an hour, how far does she live from the store?

1. Fill out the Rate column in the table below.
[tex]\[
\begin{array}{l}
a = 30 \\
b = 20
\end{array}
\][/tex]

2. Fill out the Distance column.
[tex]\[
c = \square \\
d = \square
\][/tex]

3. Complete the following table:

[tex]\[
\begin{array}{|l|c|c|c|}
\hline
\text{Direction} & \text{Distance} & \text{Rate} & \text{Time} \\
\hline
\text{To store} & c & 30 & \\
\hline
\text{Return home} & d & 20 & \\
\hline
\end{array}
\][/tex]



Answer :

GinnyAnswer
Let's solve the problem step-by-step to find out the distance from Jorden's home to the store and fill out the table accordingly.

1. Identify the Given Data:
- Rate to store: [tex]\( 30 \)[/tex] miles per hour.
- Rate returning home: [tex]\( 20 \)[/tex] miles per hour.
- Total driving time: [tex]\( 0.5 \)[/tex] hours (30 minutes).

2. Setting Up the Problem:
- Let [tex]\( d \)[/tex] be the distance to the store (in miles).

3. Time Calculation:
- Time to store = [tex]\( \frac{d}{30} \)[/tex] hours.
- Time returning home = [tex]\( \frac{d}{20} \)[/tex] hours.
- Total time = Time to store + Time returning home = [tex]\( 0.5 \)[/tex] hours.

4. Formulating the Equation:
- [tex]\( \frac{d}{30} + \frac{d}{20} = 0.5 \)[/tex]

5. Find the Distance:
- Solving the equation for [tex]\( d \)[/tex]:
[tex]\[ \frac{d}{30} + \frac{d}{20} = 0.5 \][/tex]
To solve for [tex]\( d \)[/tex], find a common denominator (60) and solve:
[tex]\[ \frac{2d}{60} + \frac{3d}{60} = 0.5 \][/tex]
[tex]\[ \frac{5d}{60} = 0.5 \][/tex]
[tex]\[ \frac{d}{12} = 0.5 \][/tex]
[tex]\[ d = 0.5 \times 12 \][/tex]
[tex]\[ d = 6 \][/tex]

Thus, the distance from Jorden’s home to the store is 6 miles.

6. Fill Out the Table:
- The rate heading to the store is [tex]\( 30 \)[/tex] mph.
- The rate returning home is [tex]\( 20 \)[/tex] mph.
- Distance to the store (c) is [tex]\( 6 \)[/tex] miles.
- Distance returning home (d) is [tex]\( 6 \)[/tex] miles.
- Time heading to the store = [tex]\( \frac{6}{30} = 0.2 \)[/tex] hours.
- Time returning home = [tex]\( \frac{6}{20} = 0.3 \)[/tex] hours.

[tex]\[ \begin{array}{|c|c|c|c|} \hline \multicolumn{2}{|c|}{\text{Distance}} & \text{Rate} & \text{Time} \\ \hline \text{To store} & 6 & 30 & 0.2 \\ \hline \text{Return} \text{home} & 6 & 20 & 0.3 \\ \hline \end{array} \][/tex]

This completes the solution and fills out the table correctly.

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