<PRE><B>Question 954012</B>
Suppose t is the normal time to travel from A to B. 

And normal speed is 750 km/hr.

Speed on return trip is 900 km/hr.

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Time taken is 20 min less than the original ( which is t here). 

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I.e.    {{{t-(20/60) = t-(1/3)}}} hr

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In both the trip distance is same that is...

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{{{750*t = 900*(t-(1/3))}}}
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5t = 6t - 2
t = 2 hr.
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Returning trip take 20 min less than the original, hence time is--  

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  2 hr - 20 min = 1 hr 40 min.

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