It is currently 27 Jun 2017, 10:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Question from Challenge 5

Author Message
CEO
Joined: 15 Aug 2003
Posts: 3454

### Show Tags

20 Feb 2006, 11:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

(A) 25
(B) 30
(C) 37.5
(D) 45
(E) 60
Intern
Joined: 16 Aug 2005
Posts: 37

### Show Tags

20 Feb 2006, 12:13
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.

Any faster method ?
Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany

### Show Tags

20 Feb 2006, 13:29
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".
CEO
Joined: 15 Aug 2003
Posts: 3454

### Show Tags

20 Feb 2006, 14:49
ok guys, torture's over

{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.
Manager
Joined: 20 Feb 2006
Posts: 213

### Show Tags

20 Feb 2006, 19:28
I think the answer is 45 = 30 (the distance the fly travels till the two hikers meet) + 15 (It travels with one of the hikers till the end).

Distance travelled by the fly till the hikers meet can easily be calculated as below: (it is based on time as Pretorarian) has said.

Time taken by the hikers to meet = 3 hours.
=> Distance travelled by fly = 3*10 = 30 kms.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

### Show Tags

20 Feb 2006, 20:10
Here's my method, which took less than a minute.

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.
Director
Joined: 24 Oct 2005
Posts: 572
Location: NYC

### Show Tags

20 Feb 2006, 23:07
ywilfred wrote:
Here's my method, which took less than a minute.

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

Senior Manager
Joined: 11 Jan 2006
Posts: 269
Location: Chennai,India

### Show Tags

21 Feb 2006, 05:16
I am jumping same approach as allabout but got the right answer.. 30Km
_________________

vazlkaiye porkalam vazltuthan parkanum.... porkalam maralam porkalthan maruma

Director
Joined: 17 Dec 2005
Posts: 547
Location: Germany

### Show Tags

21 Feb 2006, 05:36
ywilfred wrote:
Here's my method, which took less than a minute.

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.

SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

21 Feb 2006, 07:38
http://www.gmatclub.com/phpbb/viewtopic.php?t=22496
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

21 Feb 2006, 07:38
Display posts from previous: Sort by