A man with3of his usual speed reaches 5 the destination

3/5 of usual speed will take 5/3 of usual time.
[∵ time & speed are inversely proportional]

∴	5	of usual time
	3

= usual time +	5
	2

⇒	2	of usual time =	5	⇒ usual time
	3		2

=	5	×	3	=	15	= 3	3	hours.
	2		2		4		4

Second Method :

Here, A = 3, B = 5, t = 2	1
	2

Usual time =	A	× time
	Diff. of A and B

=	3	× 2	1
	5 − 3		2

=	3	×	5
	2		2

=	15	= 3	3	hours
	4		4