Choo-Choo!

Case1:Let 360 km distance is travelled at speed ' s ' km/h in time ' t ' hour.

Case 2 : As per given statement , in this case the same distance 360 km is travelled at speed ' s+5 ' km/h in time ' t-5 '.

Distance = speed x time

From case1, t x s =360.........eqn(a).

From case2, $(t-1) (s+5) = 360$ $ts+5t-s-5= 360$ ts=360 $5(t-1) -s = 0$ $s = 5(t-1)$ Substituting in equation (a), $5t^{2} -5t-360=0$

t = 9 OR t = -8

Time can't be negative -----> t=9, Again from eqn(a) $t\times s =360$ $s = \frac {360}{t} = \frac {360}{9}$ $s=40$

Let $t$ be the time taken by the train to reach its destination, and $v$ its velocity. From what we're given by the problem, we can write the following:

$\frac{360}{t} = v$ (I) and

$\frac{360}{t - 1} = v + 5$ (II)

These equations can also be written as:

$\frac{t}{360} = \frac{1}{v}$ (III) and

$\frac{t - 1}{360} = \frac{1}{v+5}$ (IV)

We can substitute (III) in (IV):

$\frac{1}{v} - \frac{1}{360} = \frac{1}{v + 5}$ (V)

Then, we can write:

$\frac{1}{v} - \frac{1}{v + 5} = \frac{1}{360}$

$\frac{5}{v^{2}+5v} = \frac{1}{360}$

Thus: $v^{2} + 5v - 1800 = 0$ , and from here we derive two solutions: $v_{1} = 40$ and $v_{2} = -45$ . Since velocity must be always a positive number, we find that it must equal $40$ .

d1=360, s1=x since t=d/s then t1=360/x; d2=360, s2=x+5, t2=360/x+5

since t1 - t2 = 1 hour,

solve for x:

360/x - 360/x+5 = 1

Rember that $time (t) = \frac{distance (d)}{speed (s)}$

If the train is 5 km/hr faster than the time is 1 hour less so $t-1 = \frac{d}{s+5}$

Replace t with $\frac{d}{s}$

We obtain $\frac{d}{s+5} + 1 - \frac{d}{s} = 0$

After few steps we have $v^{2}+5v-5d = 0$

-> v = 40 km/hr

The distance travelled is 360 km. Let time taken be x hrs and speed be y km/h. BTP, 360/x=y........(1). Also, 360/x+5=y-1.....(2). Subtracting (2) from(1), we get 360/x - 360/x+5 =1.Taking LCM and solving, we get the quadratic equation x²+5x-1800=0. By Quadratic formula, x=40, -45. Since speed cannot be negative, it is 40 km/hr.