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# There are 7 equally spaced spaced balls on the left side of

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Manager
Joined: 25 Jun 2003
Posts: 93
There are 7 equally spaced spaced balls on the left side of [#permalink]

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15 Jul 2003, 16:24
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There are 7 equally spaced spaced balls on the left side of a
horizontal line, each moving to the right at constant speed 'S'. There
are 9 equally spaced balls on the right side of the same line,
each moving to the left at the same constant speed of 'S'. The balls are all of the same size and mass and the spacing between the balls on the left is
the same as the spacing between the balls on the right. When two balls
collide they will reverse directions, but each will maintain the same
speed S.

How many collisions will occur ?

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Brainless

Director
Joined: 03 Jul 2003
Posts: 652

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15 Jul 2003, 20:22
Is it a GMAT problem? It looks more like a riddle to me

Director
Joined: 03 Jul 2003
Posts: 652

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15 Jul 2003, 20:22
Is it a GMAT problem? It looks more like a riddle to me

SVP
Joined: 03 Feb 2003
Posts: 1604

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15 Jul 2003, 22:36
I vote for 13
the most left and the most right balls do not collide
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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15 Jul 2003, 23:42
I believe there will be a total of 63 collisions!

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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager
Joined: 08 Apr 2003
Posts: 150

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16 Jul 2003, 02:53
Absolutely 63!!!

I solved it by using a different number of balls and generalised the case.

Take 2 balls on left and 3 on right.

It becomes more simpler to see that the number of collisions is 6.

Going by that the number of collisions between 7 on left and 9 on right will be 63.

Intern
Joined: 06 Jul 2003
Posts: 3

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16 Jul 2003, 03:27
Imagine 7 left balls and 7 right balls. The first left ball will collide with the next one 1 time. The second - 2 times, etc.
The total amount of collisions 1+2+3+4+5+6+7+6+5+4+3+2+1=49
But we have two more balls from one side. Those two balls will collide 7 times each and go futher without any obstacle.
Manager
Joined: 25 Jun 2003
Posts: 93

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16 Jul 2003, 08:25
63 is CORRECT !!

Given the scenario , each left ball collides with each of the right ball .
so total collisions will be 7 x 9 = 63
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Brainless

Manager
Joined: 08 Apr 2003
Posts: 150

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16 Jul 2003, 08:33
Brainless wrote:
63 is CORRECT !!

Given the scenario , each left ball collides with each of the right ball .
so total collisions will be 7 x 9 = 63

The answer is fine but explanation is not upto mark.

You cannot say that each ball on the left side collides with each ball on the right hand side. How can the left hand most side ball collide with right hand most side ball.
16 Jul 2003, 08:33
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