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Difficulty: Very hardRatios, Rates, and Proportions

An express delivery truck travels from Warehouse A to Warehouse B at a constant speed, and then returns from Warehouse B to Warehouse A along the same route at a different constant speed. On the first trip, the ratio of the outbound travel time to the inbound travel time is 33 to 55. On a second trip along the same route, the truck's outbound speed is 50%50\% greater than its outbound speed on the first trip, and its inbound speed is 25%25\% less than its inbound speed on the first trip. If the total round-trip travel time for the second trip is 1313 hours, what is the total round-trip travel time, in hours, for the first trip?

Answer: 12 hours

Answer

The total round-trip travel time for the first trip is 12 hours.
The correct answer is 1212. By setting the outbound and inbound travel times of the first trip as t1t_1 and t2t_2 with t1=0.6t2t_1 = 0.6t_2, we apply the inverse relationships of speed and time to express the second trip's times as t1′=23t1t_1' = \frac{2}{3}t_1 and t2′=43t2t_2' = \frac{4}{3}t_2. Substituting these into the second trip's total time equation 23t1+43t2=13\frac{2}{3}t_1 + \frac{4}{3}t_2 = 13 yields t2=7.5t_2 = 7.5 and t1=4.5t_1 = 4.5, summing to 1212 hours.

Step-by-Step Solution

1
Set up the ratio of outbound to inbound travel times for the first trip.
t1=0.6t2t_1 = 0.6t_2 (or t1=35t2t_1 = \frac{3}{5}t_2)
The ratio of the outbound travel time (t1t_1) to the inbound travel time (t2t_2) is given as 33 to 55.
2
Express the new travel times for the second trip (t1′t_1' and t2′t_2') in terms of the original travel times.
t1′=23t1t_1' = \frac{2}{3}t_1 and t2′=43t2t_2' = \frac{4}{3}t_2
For a constant distance, travel time is inversely proportional to speed. A 50%50\% speed increase multiplies time by 11.5=23\frac{1}{1.5} = \frac{2}{3}, and a 25%25\% speed decrease multiplies time by 10.75=43\frac{1}{0.75} = \frac{4}{3}.
3
Create an equation for the total travel time of the second trip.
23t1+43t2=13\frac{2}{3}t_1 + \frac{4}{3}t_2 = 13
The sum of the outbound and inbound travel times on the second trip is given as 1313 hours.
4
Substitute the first trip's relationship into the equation and solve for t2t_2.
t2=7.5t_2 = 7.5
Substituting t1=35t2t_1 = \frac{3}{5}t_2 gives 23(35t2)+43t2=13  ⟹  2615t2=13  ⟹  t2=13×1526=7.5\frac{2}{3}\left(\frac{3}{5}t_2\right) + \frac{4}{3}t_2 = 13 \implies \frac{26}{15}t_2 = 13 \implies t_2 = 13 \times \frac{15}{26} = 7.5.
5
Determine t1t_1 and calculate the total travel time for the first trip.
t1+t2=12t_1 + t_2 = 12
Since t2=7.5t_2 = 7.5, then t1=0.6(7.5)=4.5t_1 = 0.6(7.5) = 4.5. The total round-trip time is 4.5+7.5=124.5 + 7.5 = 12.

Key Concept

Inverse proportionality of speed and time, and solving linear systems involving ratios.
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Topics

Ratios, Rates, and Proportions

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