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shaji Senior Consultant
Joined: 25 Apr 2005 Posts: 34
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Posted: Tue Apr 26, 2005 1:40 pm Post subject: Error in Latest Quiz |
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Two objects leave points A and B simultaneously and travel towards points B and A respectively on the same path at constant but different speeds. After meeting at a point C between A and B, they proceed towards their destinations. The object from A reaches B 4 hours after the two objects met at C. The object from B reaches A 6 hours after the two objects met at C. How long did the slower of the two objects take to reach its destination from its origin? (Assume that they spent '0' minutes at point C)
24 hours
36 hours
10 hours
12 hours
14 hours
The correct answer is 10.89 hours in my opinion. Please explain how its 12 hours as this considred as correct.
Thanks.
Shaji |
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ksb GMAT Tutor
Joined: 28 Dec 2004 Posts: 54 Location: Chennai, India
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Posted: Fri Apr 29, 2005 12:31 pm Post subject: |
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The numbers had an error. The question should read that B reaches A 9 hours after they meet.
Here is the solution after revising the numbers
Let the speeds of the object leaving A be 'x' kmph and that leaving B be 'y' kmph
Let 't' hours be the time taken by the objects before they met.
The distance covered by the object leaving A before the meeting = xt kms
This is the distance, the object from B has to cover after meeting the object from A = 9y kms
Similarly, the distance covered by the object leaving B before the meeting = yt kms
This is the distance that the object from A has to cover after meeting the object from B = 4x kms
Equating the relation in the first pair of statements, xt = 9y or x = 9y/t --- (1)
Equating the relation in the second pair of statements, yt = 4x --- (2)
Substitute the value of x as 9y/t derived in equation (1) in equation (2)
We get yt = 4(9y/t)
or yt = 36y/t
or t^2 = 36
or t = 6 hours.
Therefore, the time taken by object from B to reach its destination will be 6 + 9 = 15 hours.
KSB |
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shaji Senior Consultant
Joined: 25 Apr 2005 Posts: 34
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Posted: Fri Apr 29, 2005 9:01 pm Post subject: Error in Latest Quiz |
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The solution is correct with the revised numbers . 15 hours is not figuring in the answer chioces. Therfefore the awswer choices need modification or the question needs yet another correction.
Thanks;
Shaji
| ksb wrote: | The numbers had an error. The question should read that B reaches A 9 hours after they meet.
Here is the solution after revising the numbers
Let the speeds of the object leaving A be 'x' kmph and that leaving B be 'y' kmph
Let 't' hours be the time taken by the objects before they met.
The distance covered by the object leaving A before the meeting = xt kms
This is the distance, the object from B has to cover after meeting the object from A = 9y kms
Similarly, the distance covered by the object leaving B before the meeting = yt kms
This is the distance that the object from A has to cover after meeting the object from B = 4x kms
Equating the relation in the first pair of statements, xt = 9y or x = 9y/t --- (1)
Equating the relation in the second pair of statements, yt = 4x --- (2)
Substitute the value of x as 9y/t derived in equation (1) in equation (2)
We get yt = 4(9y/t)
or yt = 36y/t
or t^2 = 36
or t = 6 hours.
Therefore, the time taken by object from B to reach its destination will be 6 + 9 = 15 hours.
KSB |
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