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A father wants to take his two sons to visit their grandmother, who
lives 33 kilometers away. His motorcycle will cover 25 kilometers per
hour if he rides alone, but the speed drops to 20 kph if he carries one
passenger, and he cannot carry two. Each brother walks at 5 kph.
Can the
three of them reach grandmother’s house in 3 hours?
Do you think it's impossible? Click here!
What was the challenge in the journey?
Yes, all three can reach at grandmother's home within 3 hours. Here is how.
Let M be the speed of motorcycle when father is alone, D be the speed of motorcycle when father is with son and S is speed of sons. Let A and B are name of the sons.
As per data, M = 25 kph, D = 20 kph, S = 5 kph.
1. Father leaves with his first son A while asking second son B to walk. Father and A drives for 24 km in 24/20 = 6/5 hours. Meanwhile, son B walks (6/5) x 5 = 6 km.
2. Now father leaves down son A for walking and drives back to son B. The distance between them is 24 -6 = 18 km.
3. To get back to son B, father needs 18/(M+S) = 18/(25+5) = 18/30 = 3/5 hours & in that time son B walks for another (3/5) x 5 = 3 km. Now, son B is 6 + 3 = 9 km away from source where he meets his father. While son A walks another (3/5) x 5 = 3 km towards grandmother's home.
4. Right now father and B are 24 km while A is 6 km away from grandmother's home. So in another 24/20 = 6/5 hours father and B drive to grandmother's home. And son B walks further (6/5) x 5 = 6 km reaching grandmother's home at the same time as father & brother B.
In this way, all three reach at grandmother's home in (6/5) + (3/5) + (6/5) = 3 hours.
In this journey, both sons walks for 9 km spending 9/5 hours and drives 24 km with father taking (6/5) hours. Whereas, father drives forward for 48 km (24 km + 24 km) in (6/5) + (6/5) hours and 15 km backward in 3/5 hours.
