Time speed 1 : PS Archive
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Time speed 1

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Manager
Joined: 19 Aug 2009
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21 Sep 2009, 11:18
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Question Stats:

50% (02:15) correct 50% (02:16) wrong based on 2 sessions

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Q.
A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 kms.The speed of 2 cars is 60 kmph each.The bird starts flying from 1st car and moves towards 2nd car, reaches the 2nd car and comes back to 1st car and so on. If the speed at which bird flies is 120 kmph then answer the following questions.Assume the 2 cars have a crash.
1) The total distance travelled by the bird before the crash is
a. 6km
b. 12km
c. 18km

2)The total distance travelled by the bird before it reaches the second car for the 2nd time
a. 10.55 km
b. 11.55km
c. 12.33 km
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22 Sep 2009, 12:59
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virtualanimosity wrote:
Q.
A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 kms.The speed of 2 cars is 60 kmph each.The bird starts flying from 1st car and moves towards 2nd car, reaches the 2nd car and comes back to 1st car and so on. If the speed at which bird flies is 120 kmph then answer the following questions.Assume the 2 cars have a crash.
1) The total distance travelled by the bird before the crash is
a. 6km
b. 12km
c. 18km

2)The total distance travelled by the bird before it reaches the second car for the 2nd time
a. 10.55 km
b. 11.55km
c. 12.33 km

1) There is a very quick solution.
Cars A and B are 12 km farther between each other. They get this distance in 6 minutes. (60minutes*12km/(60kmph+60kmph)=6 minutes). It means after 6 minutes they will crash. Bird is flying at 120 kmph for this 6 minutes. So the bird flyes 120*6/60=12 km. B.

2)This is a scheduling problem.
When bird flyes there is 12 km from bird to Car B. Bird flyes at 120 kmph and car B is approaching at 60 kmph. So they are approaching each other at 180 kmph. But car A and B are approaching each other at only 120 kmph. So bird reach Car B in 12*60/180=4 minutes. In 4 minutes distance between Car A and Car B becomes 4 km (12-120*4/60=4). In 4 minutes bird flyes 8 km's and Car B approaches 4 km. After that the same logic works. Bird and Car A approaches each other at 180 kmph total. 4 km's are covered in 1 minutes 20 seconds. In that time bird covers 2,66 kms and the distance between car A and car B becomes 1,33 km's. In third round bird flyes 2,66/3 km's. In fourth round bird flyes 2,66/(3*3) km's. So all sum to 8 + 2,66 + 2,66/3 + 2,66/9 = 11,8 km's. I think the choice is B.
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24 Sep 2009, 02:58
That is the right answer...thanks for the explanation

maliyeci wrote:
virtualanimosity wrote:
Q.
A naughty bird is sitting on top of a car.It sees another car approaching it at a distance of 12 kms.The speed of 2 cars is 60 kmph each.The bird starts flying from 1st car and moves towards 2nd car, reaches the 2nd car and comes back to 1st car and so on. If the speed at which bird flies is 120 kmph then answer the following questions.Assume the 2 cars have a crash.
1) The total distance travelled by the bird before the crash is
a. 6km
b. 12km
c. 18km

2)The total distance travelled by the bird before it reaches the second car for the 2nd time
a. 10.55 km
b. 11.55km
c. 12.33 km

1) There is a very quick solution.
Cars A and B are 12 km farther between each other. They get this distance in 6 minutes. (60minutes*12km/(60kmph+60kmph)=6 minutes). It means after 6 minutes they will crash. Bird is flying at 120 kmph for this 6 minutes. So the bird flyes 120*6/60=12 km. B.

2)This is a scheduling problem.
When bird flyes there is 12 km from bird to Car B. Bird flyes at 120 kmph and car B is approaching at 60 kmph. So they are approaching each other at 180 kmph. But car A and B are approaching each other at only 120 kmph. So bird reach Car B in 12*60/180=4 minutes. In 4 minutes distance between Car A and Car B becomes 4 km (12-120*4/60=4). In 4 minutes bird flyes 8 km's and Car B approaches 4 km. After that the same logic works. Bird and Car A approaches each other at 180 kmph total. 4 km's are covered in 1 minutes 20 seconds. In that time bird covers 2,66 kms and the distance between car A and car B becomes 1,33 km's. In third round bird flyes 2,66/3 km's. In fourth round bird flyes 2,66/(3*3) km's. So all sum to 8 + 2,66 + 2,66/3 + 2,66/9 = 11,8 km's. I think the choice is B.
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29 Apr 2010, 10:34
Sorry, I didn't follow. How did you get (60minutes*12km/(60kmph+60kmph)=6 minutes)
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15 Feb 2016, 04:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Time speed 1   [#permalink] 15 Feb 2016, 04:05
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