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# A boat M leaves Shore A and at the same time boat B leaves

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Director
Joined: 17 Sep 2005
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A boat M leaves Shore A and at the same time boat B leaves [#permalink]  30 Jun 2006, 08:34
A boat M leaves Shore A and at the same time boat B leaves shore B. They move across the river .they met at 500 yards away from A and after that they met 300 yards from shore b without halting at shores. Find the distance between the shore A and B.

Regards,
Brajesh
Senior Manager
Joined: 07 Jul 2005
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Location: Sunnyvale, CA
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1200.

Slightly tricky wordings.

The boats first met after 500yards from A.
Let the total distance between A and B be x.

then,
500/a = x-500/b,
where, a = speed of A,
b =speed of B.

Assuming both boats are coming back to their original start place without halting at the shores,
Distance coverred by A = x + 300
Distance covered by B = 2x - 300

x+300/a = 2x-300/b

From above,
500/x-500 = x+300/2x-300

Solve, for x = 1200
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 121 [0], given: 0

I think the question is ambiguous.
There are three solutions:

1. When both A and B touches the other shore atleast once.
2. When A touches the other shore and B did not even reach the other shore
3. When B touches the other shore and A did not even reach the other shore

1. 500/a = x-500/b i.e a/b = 500/(x-500)

(2x-300)/b = (x+300)/a i.e a/b =

Solving these two we get x = 1200

2. 500/a = x-500/b i.e a/b = 500/(x-500)

(x+300)/a = 300/b i.e a/b = (x+300)/300

Solving these two we get x = 100(1+SQRT(31))

3. 500/a = x-500/b i.e a/b = 500/(x-500)

(2x-300)/b = (x-300)/a i.e a/b = (x-300)(2x-300)

Solving these two we get x = 100(7+SQRT(19))
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 28 Dec 2005
Posts: 758
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There is no ambiguity.

A boat M leaves Shore A and at the same time boat B leaves shore B. They move across the river .they met at 500 yards away from A and after that they met 300 yards from shore b without halting at shores. Find the distance between the shore A and B.

I think sgrover has the right approach.
CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 121 [0], given: 0

Futuristic wrote:
There is no ambiguity.

A boat M leaves Shore A and at the same time boat B leaves shore B. They move across the river .they met at 500 yards away from A and after that they met 300 yards from shore b without halting at shores. Find the distance between the shore A and B.

I think sgrover has the right approach.

Please see the attched figure. There are three cases....
Attachments

Cases.JPG [ 9.89 KiB | Viewed 1074 times ]

_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Director
Joined: 17 Sep 2005
Posts: 930
Followers: 3

Kudos [?]: 38 [0], given: 0

sgrover wrote:
1200.

Slightly tricky wordings.

The boats first met after 500yards from A.
Let the total distance between A and B be x.

then,
500/a = x-500/b,
where, a = speed of A,
b =speed of B.

Assuming both boats are coming back to their original start place without halting at the shores,
Distance coverred by A = x + 300
Distance covered by B = 2x - 300

x+300/a = 2x-300/b

From above,
500/x-500 = x+300/2x-300

Solve, for x = 1200

One thing I don't understand.
Why did you equate the time taken by boat A and time taken by boat B?
[500/a = x-500/b]

Regards,
Brajesh
Manager
Joined: 27 Jan 2006
Posts: 156
Location: Europe
Followers: 1

Kudos [?]: 2 [0], given: 0

not clear question..
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Stay hungry, Stay foolish

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