A man walked for two days. On the second day, he walked 2 hrs longer and at an average speed of 1 Km per hour faster than he walked on the first day. If during the two days he walked a total of 64 Km and spent a total of 18 hrs walking, what was his average speed on the first day?
Answer with explanationAnswer: Option B
Let he walked for t hour on first day, so he walked for (t + 2) hour on second day.
Let s km/h be the speed of the man on first day. Then (s + 1) km/h be the speed of the man on second day.
So, according to the question,
t + t + 2 = 18
=> 2t = 16
=> t = 8 h
And st + (s + 1)(t + 2) = 64
=> 8s + 10s + 10 = 64
=> 18s = 54
=> s = 3 km/h