A Visit To Grandmother's Home!

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!

Planning Journey Towards Grandmother's Home

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. 
  

The Plane Landing Ahead of Schedule

A motorcyclist was sent by the post office to meet a plane at the airport.

The plane landed ahead of schedule, and its mail was taken toward the post office by horse. After half an hour the horseman met the motorcyclist on the road and gave him the mail.

The motorcyclist returned to the post office 20 minutes earlier than he was expected.

How many minutes early did the plane land?

Here is calculation of scheduled arrival time. 

Finding Scheduled Arrival Time of The Plane

What is the data given for calculation?

Since, motorcyclist returned to the post office 20 minutes earlier than he was expected, it mean, the horse had saved his 20 minutes of journey. That is, after meeting with horse at some point, the motorcyclist would have needed 20 minutes to go to & come back from airport to the same point. That's how the horse managed to save 20 minutes of motorcyclist. 

Let 'T' be the time at which horse met with motorcyclist.

It also means that, the motorcyclist would have taken another 10 minutes to reach at the airport exactly when plane was scheduled for landing. So the scheduled time of arrival of plane is T+10.

However, plane arrived at time T-30 where horse left airport with mail & met motorcyclist exactly half hour later at time T.

In short, plane landed at time T-30 instead of scheduled T+10 shows that plane landed 40 minutes ahead of schedule.