Two persons A and B are at P and Q,respectively. A walks at ‘V’ km/h and B is 2 km/h faster than A, starting simultaneously from where they stand.If they walk towards each other,they meet in 72 minutes.If they walk in the same direction the faster over takes the slower in 6 hours .What are their respective speeds in km/h?
Solution
Speed of A = V
Speed of B = V + 2
Let the distance between points P and Q = D
Now,
$$\frac{D}{V + (V + 2)} = \frac{72}{60} = 1.2$$
D = 1.2(2V + 2)
D = 2.4V + 2.4 .......... (1)
$$\frac{D}{(V + 2) - V} = 6$$
D = 12 ......... (2)
From (1) and (2):
12 = 2.4V + 2.4
V = 4
Speed of A = V = 4 km/h
Speed of B = V + 2 = 6 km/h
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