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 Coordinating arrival (Posted on 2015-09-15)
Alex wants to take his two sons to visit their grandmother,
living 33 kilometers away.
His motorcycle can run 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.
The walking speed of each of the boys is 5 kph.

How can the three of them reach grandmotherâ€™s house in 3 hours?

Source : Russian Mathematical Olympiad (1999)

 See The Solution Submitted by Ady TZIDON No Rating

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 solution | Comment 1 of 2
Let x be the number of km short of the destination that Alex lets his first son off to walk the rest of the way (that is x km) while he goes back to get the second son.  We want to time it so he gets back to his first son just as he's arriving at grandmother's house.

The first phase of the trip takes (33-x)/20 hours.

The second phase, while Alex travels back home, takes (33-x)/25 hours.

The third phase, all the way from home to grandmother's with the second son, takes 33/20 hours.

During the second and third phases, the first son is going x km at 5 km/h, taking x/5 hours, which should match the total time of phases two and three.

x/5 = (33-x)/25 + 33/20

Multiplying by 100:

20x =  132 - 4x +  165

24x =  297

x =  12.375 km before grandmother's house.

Let's see if this is fast enough:

The first phase takes (33 - x)/20 hours =  1.03125 hours.

The second and third combined take x/5 hours =  2.475 hours.

This adds up to  3.50625 hours and the goal has not been met.

Aha! The second son must start walking toward grandmother's house when Alex and the first son first start out.  That way, Alex need not go all the way home to get the second son.

Starting over:

The first phase of the trip takes (33-x)/20 hours, same as before.

By now his second son has traveled 5*(33-x)/20 km.

The second phase, while Alex travels back toward home, takes (33-x - 5*(33-x)/20)/30 hours, the 30 being Alex's speed back plus the second son's speed toward grandmother's house.

This second phase puts Alex and the second son x + 25*(33-x - 5*(33-x)/20)/30 km from grandmother's house.

The third phase, from that point to grandmother's with the second son, takes (x + 25*(33-x - 5*(33-x)/20)/30)/20 hours.

So now, (33-x - 5*(33-x)/20)/30 + (x + 25*(33-x - 5*(33-x)/20)/30)/20 = x/5.

Wolfram Alpha solves this as x = 9 km before grandmother's house.

Phase 1 takes 6/5 hours.

Phases 2 and 3 together take 9/5 hour.

Total: 3 hours.

 Posted by Charlie on 2015-09-15 10:57:22

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