To find the distance to the destination, let's define the distance as [tex]\( D \)[/tex] miles. We know the person drives to the destination at 30 mph and returns at 50 mph. The total travel time returns one hour less than the time going.
Let's denote:
- Speed going: [tex]\( 30 \)[/tex] mph
- Speed returning: [tex]\( 50 \)[/tex] mph
- Time difference: [tex]\( 1 \)[/tex] hour
### Step-by-step Solution:
1. Define the time taken for each part of the journey:
The time taken to go to the destination can be expressed as:
[tex]\[
\text{Time going} = \frac{D}{30}
\][/tex]
The time taken to return from the destination can be expressed as:
[tex]\[
\text{Time returning} = \frac{D}{50}
\][/tex]
2. Set up the time difference equation:
According to the problem, the time returning is one hour less than the time going:
[tex]\[
\frac{D}{30} - \frac{D}{50} = 1
\][/tex]
3. Find a common denominator to combine the terms:
The common denominator for 30 and 50 is 150. Rewrite the fractions:
[tex]\[
\frac{D}{30} = \frac{5D}{150}
\][/tex]
[tex]\[
\frac{D}{50} = \frac{3D}{150}
\][/tex]
4. Substitute these into the equation:
[tex]\[
\frac{5D}{150} - \frac{3D}{150} = 1
\][/tex]
5. Combine and simplify:
[tex]\[
\frac{5D - 3D}{150} = 1
\][/tex]
[tex]\[
\frac{2D}{150} = 1
\][/tex]
6. Solve for [tex]\( D \)[/tex]:
To solve for [tex]\( D \)[/tex], multiply both sides of the equation by 150:
[tex]\[
2D = 150
\][/tex]
Divide both sides by 2:
[tex]\[
D = \frac{150}{2}
\][/tex]
[tex]\[
D = 75
\][/tex]
### Conclusion:
The distance to the destination is [tex]\( \boxed{75} \)[/tex] miles.
