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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?

Fill out the Rate column in the table below.
[tex]\[
\begin{array}{|c|c|c|c|}
\hline \multicolumn{2}{|c|}{ Distance } & Rate & Time \\
\hline To store & & 30 & \\
\hline \begin{tabular}{c}
Return \\
home
\end{tabular} & & 20 & \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnswer
Sure, let's solve the problem step-by-step:

1. Given Information:
- Jorden drives to the store at 30 miles per hour.
- On her way home, she averages only 20 miles per hour.
- The total driving time is half an hour (0.5 hours).

2. Unknown:
- Distance to the store (and back) which we'll denote as [tex]\( d \)[/tex].

3. Relationships and Formulas:
- Time to store [tex]\( t_1 \)[/tex] and time back home [tex]\( t_2 \)[/tex].
- Using the relationship: [tex]\[ \text{Time} = \frac{\text{Distance}}{\text{Rate}}. \][/tex]

4. Step-by-Step Solution:
- Let the distance to the store be [tex]\( d \)[/tex] miles.
- Time to drive to the store: [tex]\( t_1 = \frac{d}{30} \)[/tex] hours.
- Time to drive back home: [tex]\( t_2 = \frac{d}{20} \)[/tex] hours.
- Total driving time is given as 0.5 hours.

5. Set up the equation:
- The total time to and from the store: [tex]\[ t_1 + t_2 = 0.5. \][/tex]
- Substituting the expressions for [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] we get: [tex]\[ \frac{d}{30} + \frac{d}{20} = 0.5. \][/tex]

6. Solve the equation:
- To solve the equation, find a common denominator for the fractions. The common denominator of 30 and 20 is 60.
- Rewrite the equation: [tex]\[ \frac{2d}{60} + \frac{3d}{60} = 0.5. \][/tex]
- Combine the fractions: [tex]\[ \frac{2d + 3d}{60} = 0.5. \][/tex]
- Simplify: [tex]\[ \frac{5d}{60} = 0.5. \][/tex]
- Multiply both sides by 60 to clear the fraction: [tex]\[ 5d = 0.5 \times 60. \][/tex]
- Simplify: [tex]\[ 5d = 30. \][/tex]
- Solve for [tex]\( d \)[/tex]: [tex]\[ d = \frac{30}{5}. \][/tex]
- Hence, [tex]\( d = 6 \)[/tex] miles.

7. Conclusion:
- Jorden lives 6 miles from the store.

8. Filling the Table:

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

Thus, the table is filled with the calculated distances, rates, and times.
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