Question

Frequent flyer. Two cyclists, Kaitlin and Josh, simultaneously started toward each other from two towns 40 km apart. Josh rode at 23 km/h, and Kaitlin rode at 17 km/h. Before departure, a fly landed on Josh's nose. At the moment of departure, it started to fly toward Kaitlin at 40 km/h. When it reached Kaitlin, it immediately turned back and flew in the opposite direction at 30 km/h (the wind blew toward Kaitlin). As soon as the fly reached Josh, it turned back again, and so on. Find the total distance flown by the fly until the cyclists met (the speed of the fly was constant in each direction

Answer 1

The total time the fly flows is one hour. If we have the same speed in both ways, say 40 km/h, then the distance will just be 40 km. However the speed differs in each direction, the fly will be moving in 40 km/s by a ratio 23:17 compared to the other direction. Therefore the average speed will be \(\frac{23}{40}(40)+\frac{17}{40}(30)=32 km/h\), which implies that distance flown is \(32 \text{ km/h } \times 1 h=32 \text{ km }\).

Answer 2

Sorry, calculation mistake! It should be 35.75, NOT 32.

Answer 3

Attached is the solution

Answer 4

Wow, that's a nice job dumbcow :D

but the solution as it's posted is quite a bit simpler...

The fly was airborne for 1 hour (the time that passed until the bikers met). The fly flew with the wind for 23 km more than it flew into the wind (this is the distance traveled by Josh). If we let t1 and t2 be the times during which the fly flew with the wind and into the wind, respectively, we can set up a system of equations:

t1 + t2 = 1,
40t1 - 30t2 = 23.

We find t1 = 53/70 and t2 = 17/70. Therefore, the distance traveled by the fly is

40 × 53/70 + 30 × 17/70 = 37 4/7 km.

Answer 5

yeah I figured there would be a simpler solution..oh well
good question

Answer 6

for more like it check out meta-math if you haven't already.
I posted this there as well once, I forget how people answered it.