Question

Aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours jogging and walking?

• A. \(\frac{xt}{y}\)
• B. \(\frac{x+t}{xy}\)
• C. \(\frac{xyt}{x+y}\)
• D. \(\frac{x+y+t}{xy}\)
• E. \(\frac{y+t}{x} - \frac{t}{y}\)

Hint:

This formula for average speed problems is very common. The total distance is \(2d\), total time is \(t\). The average speed is \(\frac{2xy}{x+y}\). The distance from home, \(d\), can be found by (Average Speed \(\times\) Total Time) / 2. This gives \(\frac{1}{2} \times \frac{2xyt}{x+y} = \frac{xyt}{x+y}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a classic distance, rate, and time problem. The key relationship is Distance = Rate \(\times\) Time, which can be rearranged as Time = Distance / Rate.
Step 2: Key Formula or Approach:
Let d be the distance from home that Aaron jogs (in miles). This is the value we need to find. The total time spent is the sum of the time jogging out and the time walking back. \[ t_{total} = t_{jog} + t_{walk} \] Step 3: Detailed Explanation:
Let's express the time for each part of the journey in terms of d, x, and y.

Distance jogging out = d
Speed jogging out = x mph
Time jogging out (\(t_{jog}\)) = \(\frac{\text{Distance}}{\text{Rate}} = \frac{d}{x}\)


Distance walking back = d (same route)
Speed walking back = y mph
Time walking back (\(t_{walk}\)) = \(\frac{\text{Distance}}{\text{Rate}} = \frac{d}{y}\)
We are given that the total time is t. So we can set up the equation: \[ t = \frac{d}{x} + \frac{d}{y} \] Now, we need to solve this equation for d. Factor out d from the right side: \[ t = d \left(\frac{1}{x} + \frac{1}{y}\right) \] To combine the fractions in the parenthesis, find a common denominator (xy): \[ \frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x+y}{xy} \] Substitute this back into the equation: \[ t = d \left(\frac{x+y}{xy}\right) \] To isolate d, multiply both sides by the reciprocal of the fraction: \[ d = t \times \left(\frac{xy}{x+y}\right) \] \[ d = \frac{xyt}{x+y} \] Step 4: Final Answer:
The distance from home Aaron can jog is \(\frac{xyt}{x+y}\) miles.