A can walk from P to Q and back in 20 mins, whereas B can walk from Q to P and back in 30 mins.If A and B start towards each other from P and Q respectively at 8.00 am, when will they meet.
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8:08
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8:09
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8:06
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8:05
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8:07
A's speed is 2PQ/20 = PQ/10 per min. B's speed is 2PQ/30 = PQ/15 per min. When moving toward each other, relative speed is PQ/10 + PQ/15 = PQ/6 per min. Time to meet = Distance / Relative Speed = PQ / (PQ/6) = 6 mins. 8:00 + 6 mins = 8:06.
To solve this problem, we can set up a system of equations to represent the distances traveled by A and B.
Let's assume that the distance between P and Q is x.
Since A can walk from P to Q and back in 20 minutes, their walking speed is 2x/20 = x/10 per minute.
Similarly, B can walk from Q to P and back in 30 minutes, so their walking speed is 2x/30 = x/15 per minute.
Now, let's consider the time it takes for A and B to meet after they start walking towards each other.
Let t represent the time it takes for them to meet.
A would have traveled a distance of (x/10) * t, and B would have traveled a distance of (x/15) * t.
Since they start at 8:00 am and walk towards each other, the total distance traveled by both A and B should be equal to the distance between P and Q, which is x.
Therefore, we can write the equation: (x/10) * t + (x/15) * t = x
Simplifying the equation, we get: 3x * t/30 = x t/10 = 1 t = 10
So, A and B will meet after 10 minutes.
Since they start at 8:00 am, they will meet at 8:06 am.
Therefore, the correct answer is option C) 8:06.