Let's denote the distance to Maria's appointment as [tex]\( d \)[/tex] miles.
1. Time taken to travel to the appointment in the morning:
Maria drives at a speed of 30 mph. Therefore, the time taken to travel to the appointment in the morning can be calculated using the formula:
[tex]\[
t_{\text{morning}} = \frac{d}{\text{speed}_{\text{morning}}}
= \frac{d}{30} \, \text{hours}
\][/tex]
2. Time taken to return from the appointment in the afternoon:
On the return trip, Maria's speed was 20 mph. Thus, the time taken for the return trip is:
[tex]\[
t_{\text{afternoon}} = \frac{d}{\text{speed}_{\text{afternoon}}}
= \frac{d}{20} \, \text{hours}
\][/tex]
3. Relationship between the morning and afternoon travel times:
It is given that the return trip in the afternoon took [tex]\( \frac{1}{2} \)[/tex] hour longer than the morning trip. So, we have:
[tex]\[
t_{\text{afternoon}} = t_{\text{morning}} + 0.5
\][/tex]
4. Substituting the expressions for [tex]\( t_{\text{morning}} \)[/tex] and [tex]\( t_{\text{afternoon}} \)[/tex] into the equation:
Substituting the expressions [tex]\( t_{\text{morning}} = \frac{d}{30} \)[/tex] and [tex]\( t_{\text{afternoon}} = \frac{d}{20} \)[/tex] into the equation from step 3, we get:
[tex]\[
\frac{d}{20} = \frac{d}{30} + 0.5
\][/tex]
5. Solving the equation for [tex]\( d \)[/tex]:
To eliminate the fractions, we can multiply every term by 60 (the least common multiple of 20 and 30):
[tex]\[
60 \left( \frac{d}{20} \right) = 60 \left( \frac{d}{30} \right) + 60 \left( 0.5 \right)
\][/tex]
Simplifying each term:
[tex]\[
3d = 2d + 30
\][/tex]
Subtract [tex]\(2d\)[/tex] from both sides to isolate [tex]\(d\)[/tex]:
[tex]\[
d = 30
\][/tex]
Therefore, the distance to Maria's business appointment is [tex]\( 30 \)[/tex] miles.
