what is answer??

I'm assuming that the train travels at its normal rate for the first 120 km and then reduces its rate by 4km/hr for the rest of the trip.......

There is no specific answer for this.......to see why...let the normal rate be = R, the normal time = T, and the distance = D...and we have....

RT = D  →  T = D/R

So the time for the first part of the trip = D/R  = 120/R

And the fime for the second part of the trip = [D - 120] / (R - 4]

And the sum of these  equal to the normal time plus 1 hour.....so we have.....

120/R  + [D - 120] / [R - 4]  = T + 1 

120/R  + [D - 120] / [R - 4] =  D/R + 1       simplify

120[R - 4] + R[D - 120]  = (R^2 - 4R] [D/R + 1]

120R - 480  +  RD - 120R  = RD + R^2 - 4D - 4R     rearrange

R^2 - 4R - [4D - 480] = 0  ......   now let D = 480km     and we have

R^2 - 4R - [4(480) - 480] = 0

R^2 - 4R  - 1440  = 0    factor

(R + 36) (R - 40)  = 0     .......  and R = 40km/hr

So...when D = 480km, R = 40km/hr and the total time = 480/40 + 1 = 13 hrs   

Now,  let D = 960km   and we have

R^2 - 4R - [4(960) - 480] = 0

R^2 - 4R - 3360 = 0      factor

(R -60)(R + 56) = 0      and R = 60km/hr 

So, when D = 960km, R = 60km/hr, and the total time = 960/60 + 1 = 17 hrs

Thus......there are infinite possibilities........

The wording is a bit confusing.

If I understand correctly, you mean to say the train takes 1 hour to travel 120 kilometres. It starts at X speed and slows down by 4_kmh^-1.

However, do you mean the speed reduction happened before it left the first station, or are we suggesting it gradually slowed down by this amount uniformly across the 120_km?

Average speed:

^{120 km}/_{1 hour} = 120_kmh^-1

If we suggest it slowed down uniformly across the 120_km, then 120_kmh^-1 = ^{X + (X-4)} / ₂

120_kmh^-1 = X-2.

X = 118_kmh^-1

If we suggest is slowed down before it even started the 120kmh^-1 stretch, then 120_kmh^-1 = X-4

124_kmh^-1 = X

Thanks Sir-Emo-Chappington and CPhill,

This question looks really interesting :))