After travelling a certain distance, a car met with an accident; its speed then became \(\frac{5}{6}\) of its usual speed, and it reached its destination 32 minutes late. If the accident had happened 36 km further on, the car would have reached its destination only 24 minutes late. If the total distance travelled is 194 km, after how many kilometres did the accident happen?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 50 km
At \(\frac{5}{6}\) of the usual speed, a stretch takes \(\frac{6}{5}\) of the usual time, so every stretch run slowly adds \(\frac{1}{5}\) of its usual time. The 36 km that are run at full speed in the second case save 8 minutes, which fixes the usual speed.
Let the usual speed be \(v\) km per minute; a stretch of \(s\) km run slowly causes a delay of \(\displaystyle \frac{s}{5v}\) minutes
The two delays differ by \(32-24=8\) minutes because of 36 km: \(\displaystyle \frac{36}{5v}=8\), so \(v=0.9\) km per minute (54 km/hr)
Slow stretch in the first case: \(\displaystyle \frac{s}{5\times0.9}=32\), so \(s=32\times4.5=144\) km
Accident after \(194-144=50\) km
The accident happened after 50 km.