A man jogs at 6mph over a certain journey and walks over the same route at 4mph what is his

2*6*4/(4+6)=4.8
avg speed=4.8mph
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TCS giving this type of question? o.O
bdw answer will be 2xy/x+y
i.e (2*6*4)/10= 4.8 mph
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average speed of the same distance will be calculated as = (2*s1*s2)/(s1+s2)
thus the answer will be 4.8mph
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avg speed during hole journey =(2xy/(x+y))
=(2*6*4)/(6+4)
=4.8mph (rqd. speed)
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its should be like this :

2*ab/a+b
because its relative speed

i.e 2(*6*4)/6+4 = 48/10 = 4.8 Ans
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x=6
y=4
average speed =(2*x*y)/(x+y)
=(2*6*4)/(6+4)
=4.8
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avg spd=2x*y/x+y
2*24/10
4.8
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A man jogs at 6mph over a certain journey and walks over the same route at 4mph.what is his speed for the journey