3.32 A person drives to a destination at a rate of 30 mph and returns over the same route at 50 mph. How far is the destination if the time returning is one hour less than the time going?



Answer :

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.