M05-14

Bunuel, that is an amazing shortcut. For those who didn't see the shortcut (like me), this is also the sum of a geometric series, with the first term a = 20, and the common ratio r= 1/3. 20*(1/(1-1/3)) = 30.

I think this is a high-quality question and I agree with explanation. OMG!! I spent 5+ minutes trying to figure out the Damn GP and still got it wrong!! And I feel the most stupid person alive after seeing the solution!! Haha!

Same here... lol

same here. too

but i tried to solve it as below in stupid way:)
It would need formula of infinite ratio : sum of total fractions dividing by r each time = r/1-r

Fly speed = 10km/h
each hikers speed = 5km/h

So whatever time they cover, relative distance would be ==> fly: hiker = 2:1
each time distance would be reduced


1st meet distance ==> fly: meet = 20:10 ( total distance to cover 30 km)
2nd meet distance==> fly: meet = 6.66:3.34 ( total distance to cover 10 kms)
3rd distance meet ==> fly: meet = 2.22:1.11 ( total distance to cover 3.33 kms)
from 4th onwards: distance would be less than 1 km each time
Hence equation,
20+ 20/3 +20/9+ 20/27 + .... = 20 ( 1/{1-1/3}) = 20 *3/2= 30 kms ( infinite formula= r/1-r; here r = 1/3

I think this is a high-quality question and I agree with explanation. The question is very very simple if you understand it and is only made "complicated" through the story of the fly. Its pretty straightforward otherwise. I liked how they tried to trick us using the wordy question.

I think this is a high-quality question and I can see it in the official GMAT exam.

an even shorter cut is to realize that the two hikers cover 10km together in one hour, which matches with what the fly covers in one hour.
realizing that alone let us quickly solve this question without the need for calculations.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Please help me understand where I went wrong.
Two people moving at same speed towards each other would meet at the center. So they meet at point which is 15km away from both ends (where fly lands on shoulder).

Fly is flying at 10km/h and Robert hiking at 5km/h. So they would meet at distance covered by fly (say fd = 10t), Robert rd = 5t
here t is same.
fd/10=rd/5 -- i

fd+rd = 30
rd= 30-fd

in i
fd/10= (30-fd)/5
fd/2=30-fd
fd=60-2fd
3fd= 60
fd=20

so fly covered 20km before it met robert, from here it starts to fly back to the point where robert and Kensington met (which we calculated was center)

By this logic now fly has to cover 5km.
Total distance = 25km.

one fault in my logic is, when the fly flies back from here, it would reach the center before Robert. But I couldn't really understand how to accomodate that, so I just let the fly to the midpoint because thats where the hikers meet. Please help me with this.

The combined speed of the fly and the hiker from Reston is 10 + 5 = 15 km/h. Thus, they will meet after 30/15 = 2 hours.

In 2 hours, the fly would cover 2*10 = 20 km, and each hiker would cover 2*5 = 10 km:

K ----------(10 km)---------- (Fly meets Hiker 1) ----------(10 km)---------- (Hiker 2) ----------(10 km)---------- R

Therefore, the distance between the fly and the hiker from Reston would be 10 km, which they would cover in 10/15 = 2/3 of an hour.

In 2/3 of an hour, the fly would cover an additional 2/3*10 = 20/3 km, and each hiker would cover 2/3*5 = 10/3 km:

K -----------(40/3 km)--------------- (Hiker 1) ----(10/3 km)---- (Fly meets Hiker 2) -----------(40/3 km)--------------- R

...

Continuing this pattern, each time the fly covers a third of the distance it did previously, we must sum up the infinite sequence: 20 + 20/3 + 20/9 + ... For an infinite geometric sequence with a common ratio |r| < 1, the sum can be calculated as sum = b/(1 - r), where b is the first term. Therefore, the sum is 20/(1 - 1/2) = 30.

That being said, solving this question using the method shown in the official solution is much quicker. Thus, I suggest understanding that one.

You are unbelievable Bunuel!! I should be punished for making it so cumbersome, calculating the fly to and fro