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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

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

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