A starts from X at 9:00 am and reaches Y at 1:00 pm. B starts from Y

9am-1pm = 4hr
9am-3pm = 6hr

d/(d/4+d/6) = 1/(1/4+1/6) = 24/10 = 2.4 = 2hr 2/5hr = 2hr 24min --->9+2:24 = 11:24am

Let the total distance is d and the time taken to meet t
d/4×t + d/6×t = d
1/4×t + 1/6×t = 1
T = 2.4 hrs

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This question sounds more like a Time & work question, especially if you solve it like the way I wish to solve it.

The distance between X and Y is constant. A takes 4 hours to cover this distance whereas B takes 6 hours to cover the same distance.

This way, we can assume the distance between X and Y as the LCM of 4 and 6 i.e. 12 miles. Therefore,
Speed of A = \(\frac{12 }{ 4}\) = 3 miles per hour and Speed of B = \(\frac{12 }{ 6}\) = 2 miles per hour.

Since they start at the same time, Relative Speed = 5 miles per hour.

Time taken to meet = \(\frac{Distance between A and B }{ Relative Speed\\
}\)

i.e. Time = \(\frac{12 }{ 5}\) hours = \(\frac{12 }{ 5}\) * 60 = 144 minutes = 2 hours 24 minutes from 9 AM which is nothing but 11:24 AM.

The correct answer option is C.

Hope that helps!

The rate for A is D/4, and the rate for B is D/6, and they both travel for T hours. Thus:

DT/4 + DT/6 = D

D(T/4 + T/6) = D

T/4 + T/6 = 1

Multiplying by 12, we have:

3T + 2T = 12

5T = 12

T = 12/5 = 2.4 hours = 2 hours 24 minutes. So, A and B met at 9 am + 2 hour 24 minutes = 11:24 am.

Answer: C

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