[aryan007]

Originally Posted by adnanalam202003

This is a basic but a good problem,
A train is travelling from point A to point B. If it travels at a speed 10 km/hr faster it reaches B 1 hour early. If it travels at a speed 6 km/hr less it reaches B, 1 hour late. What is the distance between the two points.
Solve with steps.

time taken are in arithmetic so velocities are in harmonic
v+10 v-6
t-1 t+1
v=2(v+10)(v-6)/(2v+4)
v=30

v+10/v-6=t+1/t-1
t=4
distance = 4*30=120

[adnanalam202003]

Correct Aryan....
How can you say that the time is in artihmetic? can you please explain..

[aryan007]

Originally Posted by adnanalam202003

Correct Aryan....
How can you say that the time is in artihmetic? can you please explain..

when distance is constant , time is inversely proportional to velocity
question says
If it travels at a speed 10 km/hr faster it reaches B 1 hour early. If it travels at a speed 6 km/hr less it reaches B, 1 hour late

so times are t-1,t and t+1 , where t is time with velocity v
t-1,t,t+1 is a arithmetic progression, hence velocity will be in harmonic progression , as it is inversely proportional to time.