Bunuel wrote:
A bus travelling from city A to city B breaks down midway. After 30 minutes, the passengers manage to catch another bus and reach city B on time. Had the breakdown occurred 30 km. earlier, the passengers would have had 37.5 minutes to catch the other bus. If the speed of the second bus is 20 km/hr more than that of the first bus, what is the speed of the first bus, in km/hr ?
A. 82
B. 80
C. 72
D. 60
E. 48
The hypothetical breakdown is 30km earlier than the actual breakdown.
HYPOTHETICAL BREAKDOWN <-----30 km-----> ACTUAL BREAKDOWN
For the case with the HYPOTHETICAL breakdown, the distance above is traveled by the FASTER BUS.
For the case with the ACTUAL breakdown, the distance above is traveled by the SLOWER BUS.
When the faster bus travels the distance above, the rest time between buses = 37.5 minutes.
When the slower bus travels the distance above, the rest time between buses = 30 minutes.
The slower bus reduces the rest time by 1/8 hour.
Implication:
The slower bus requires 1/8 LONGER than the faster bus to travel the 30 km above.
We can PLUG IN THE ANSWERS, which represent the speed of the slower bus.
When the correct answer is plugged in, the slower bus will take 1/8 hour longer than the faster bus to travel 30 km.
B: 80 km per hour for the slower bus, implying 100 km for the faster bus
Time for the faster bus to travel 30 km \(= \frac{distance}{rate} = \frac{30}{100} = \frac{3}{10}\) hour
Time for the slower bus to travel 30 km \(= \frac{distance}{rate} = \frac{30}{80} = \frac{3}{8}\) hour
Time difference \(= \frac{3}{8} - \frac{3}{10} = \frac{15}{40} - \frac{12}{40} = \frac{3}{40}\) hour
Eliminate B.
D: 60 km per hour for the slower bus, implying 80 km for the faster bus
Time for the faster bus to travel 30 km \(= \frac{distance}{rate} = \frac{30}{80} = \frac{3}{8}\) hour
Time for the slower bus to travel 30 km \(= \frac{distance}{rate} = \frac{30}{60} = \frac{1}{2}\) hour
Time difference \(= \frac{1}{2} - \frac{3}{8} = \frac{4}{8} - \frac{3}{8} = \frac{1}{8}\) hour
Success!
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