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Zorluk: OrtaAlgebraic Word Problems and Equation Modeling

A courier service dispatches a delivery driver from Facility A to Facility B along a straight 120120-mile route at a constant speed of rr miles per hour. On the return journey from Facility B to Facility A along the same route, traffic congestion reduces the driver's constant speed by 2020 miles per hour. If the total driving time for the entire round trip is 55 hours, what was the driver's speed, in miles per hour, on the return trip?

  1. A
    38
  2. 40Cevap
  3. C
    48
  4. D
    60
  5. E
    8

Cevap

The driver's speed on the return trip was 40 miles per hour.
Setting the total time equation to 120/r + 120/(r - 20) = 5 leads to the quadratic equation r^2 - 68r + 480 = 0. Factoring gives roots r = 60 and r = 8. Since r = 8 yields a negative return speed, r must be 60. Subtracting 20 gives the correct return speed of 40 mph.

Adım Adım Çözüm

1
Set up expressions for outbound and return travel times in terms of r.
Outbound time = \frac{120}{r} hours; Return time = \frac{120}{r - 20} hours.
Time equals distance divided by speed.
2
Formulate the equation for total round-trip time.
\frac{120}{r} + \frac{120}{r - 20} = 5
The sum of the travel times for both legs equals 5 hours.
3
Simplify the equation and convert it into standard quadratic form.
r^2 - 68r + 480 = 0
Dividing by 5 gives \frac{24}{r} + \frac{24}{r - 20} = 1, and multiplying by r(r - 20) yields 24(r - 20) + 24r = r(r - 20).
4
Solve the quadratic equation for r and eliminate extraneous solutions.
r = 60 mph (since r = 8 gives a negative return speed).
Factoring gives (r - 60)(r - 8) = 0. The root r = 8 is physically invalid because r - 20 must be positive.
5
Calculate the return speed r - 20.
60 - 20 = 40 mph
The return speed is 20 mph slower than the outbound speed.

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