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The more well explained the answer, the better.

Two cities, Kensington MD and Reston VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington. It flies at the speed of 10 km/hour and passes the hiker from its city. When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker. It keeps doing so until the hikers meet. At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston. How many kilometers has the fly flown?

This one is really tricky. I've found 45 but my method is slow.

I've done it using iterations.
di=distance travelled by the fly during iteration i
Li=distance between the hikers after iteration i
Di=cumulative distance travelled by the fly after iteration i

- First iteration:
The fly's speed is 2 times the speed of the hikers so the fly will travel twice the distance travelled by the Reston hiker when it reaches him.
So d1=D1=20km and distances travelled by both hikers are 10km so L1=10km

- 2nd iteration:
The fly will travel twice the distance travelled by the Kensington hiker when it reaches him i.e. d2=2/3*L1=20/3 and distances travelled by both hikers is L1/3=10/3 so L2=10/3
The cumulative distance is D2=20+20/3
...
- nth iteration
From above we can infere that:
Ln=30/(3^n)
dn=2/3*Ln-1=2/3*30/(3^n)=20/(3^n)
Dn=20*(1+1/3+....+1/(3^n))

When n tends to infinite 1+....+1/(3^n)=3/2
So total cumulative distance is D=20*3/2=30

Finally I add the last 15km travelled on the shoulder of the kensington hiker and I get 45.

I used a rather intuitive approach ( this problem is similiar to that of the bunny and the turtle):

The distance between the two men is 30 km's. Each hiker travels at 5km/h; the fly at 10 km/h

Hiker 1 = the one who comes from Kensington
Hiker 2 = the one that comes from Reston

- The fly meets 1 after 20 km's (2 hours of flying, 1 walked 10 km)
- In the meantime 2 has also went 10 km's
- the fly and hiker 2 are 10 km apart from each other
- the fly and the hiker together have a speed of 15 so they meet each other in 2/3 hour, so the fly flies 2/3 hours at 10 km/h, or 20/3 km - in the meantime hiker 1 has made 10/3 km's and the distance between fly and hiker 1 is now 3 1/3 km's apart from 1
.
.
.
the pattern goes on and on and the time taken as well as the distance flown (of course) becomes marginal. The pattern continues till the two hikers meet each other three hours after they went off.

Since we shall say how many kilometers the fly has flown I say it's 30.
Can't wait to look for the answer. If my calculation is true I could try to make it more "accessible".

{B} This is a great question. One should not calculate how much the fly covered in distance, just in time, as we know its speed. The calculation's pretty easy, now: The fly flies 3 hrs (30 km / [5+5] km/h) at 10 km/h making 30 km

Dont worry, its the learning process that matters. Its much better to feel frustrated and actually learn something from 10 difficult problems that solving 100 easy problems. Thats how we designed our challenge questions.

Both hikers are travelling at 5km/h towards each other. This means for each hour, they would have closed off 10km from the 30km journey. Thus, the two hikers would need 30/10 = 3 hours to meet.

In this time, the fly would have flown 10*3 = 30 km.

Both hikers are travelling at 5km/h towards each other. This means for each hour, they would have closed off 10km from the 30km journey. Thus, the two hikers would need 30/10 = 3 hours to meet.

In this time, the fly would have flown 10*3 = 30 km.

neat, brother!!!! _________________

Success is my only option, failure is not -- Eminem

Both hikers are travelling at 5km/h towards each other. This means for each hour, they would have closed off 10km from the 30km journey. Thus, the two hikers would need 30/10 = 3 hours to meet.

In this time, the fly would have flown 10*3 = 30 km.