Question

A man travelling at a speed of 20 km/hr, reached his office 10 minutes late. Next day he travelled at a speed of 30 km/hr and he reached his office 10 minutes earlier. The distance between his office and home is:

Answer 1

The time which is need to reach office from his home is t s.
Here,
10 minutes = (10×60) s =600 s.
In the first day,
the speed of the man was, V= 20 km/hr =5•56 m/s
and he needs (t +600) s to reach office.
so, T= (t+600) s
We know,
S=VT [here, S is the distance between his office
and home]
=>S=5•56×(t +600) m.......................(1)
Again in the next day,
the speed of the man was,V'=30 km/hr =8•33 m/s
and he needs (t - 600) s to reach office.
So, T'=(t-600) s
We know,
S= V'T' [ here, S is the distance between his office
and home]
=>S=8•33×(t-600) m..........................(2)

From equation (1) &(2) ,
5•56×(t+600)=8•33×(t-600）
=>5•56t+3336=8•33t- 4998
=>3336+4998=8•33t-5•56t
=>8334=2•77t
=> t =8334÷2•77
=>t =3008•66 s

Now, we get from equation (1) ,
S =5•56×(t+600)
=>S= 5•56×(3008•66+600)
=>S= 20064•15 m
So, the distance between his office and home is 20064•15 m