Question

A man started 20 min late and by travelling at a speed of 1 1/2 times of his usual speed , reaches his office in time.the time taken by the manto reach his office at his usual speed is

Answer 1

Let the distance between the office and starting point be 'd'
Let the usual speed to reach the office be 's'
Let the usual time taken to reach the office be 't'
According to the formula,
Distance = Speed * Time
=> d = s * t
=> d = ts --------------------(1)
The units of speed is kilometer per hour or meter per second
Let us consider meter per second
20 min = 20 * 60 = 1200 seconds
The man started 20 min late so we have to subtract from usual time
The man travelled at 1.5 times speedier so we have multiply it to the usual speed
Distance = Speed * Time
d = ( 1.5s) * (t - 20)
d = 1.5ts - 1800s ----------(2)
Even though the speed and time varies, distance doesn't change so (1) and (2) are equal
1.5ts - 1800s = ts
1.5ts - ts = 1800s
0.5ts = 1800s
0.5t = 1800
t = 1800/0.5
t = 3600 seconds = 60 minutes = 1 hour
The usual time taken to reach the office is 1 hour

Answer 2

Let's assume the distance from the man's house to the office is 1 km and the normal speed is Xkm/h.

The new speed on the day he leaves 20 minutes late will be:

1.5 times the normal speed which is

1.5X

The normal time taken is:

Time =Distance/speed

1/X hrs

The time taken when speed is 1.5X is:

1/1.5X since the distance traveled is the same.

The relative time is 20 minutes which is 20/60= 1/3hrs.

Therefore :

1/X - 1/1.5X=1/3hrs

Lcm is 3X

(3-2)/3X=1/3

1/3X=1/3

3X=3

X=1km/h

Time =Distance/speed

1/1=1hr

Time is 1 hr