A train travelled a certain distance at a uniform speed. Had the speed been $6$ km per hour more, it would have needed $4$ hours less. Had the speed been $6$ km per hour less, it would have needed $6$ hours more. The distance, in km, travelled by the train is

800

720

780

640

✅ Correct Option: 2

We know that Distance $=$ Speed $\times$ Time.

In our two scenarios, the distance always remains the same.

Scenario	Speed	Time	Distance
Original	$s$	$t$	$st$
Case 1 (6km/hr faster, 4 hours less)	$s + 6$	$t - 4$	$(s + 6)(t - 4)$
Case 2 (6km/hr slower, 6 hours more)	$s - 6$	$t + 6$	$(s - 6)(t + 6)$

Since the distance is always the same. We can equate them:

Distance in Case 1 $=$ Distance of Original Train

$\Rightarrow (s + 6)(t - 4) = st$

$\Rightarrow st - 4s + 6t - 24 = st$

$\Rightarrow -4s + 6t - 24 = 0$

$\Rightarrow -4s + 6t = 24$ ... (equation i)

Distance in Case 2 $=$ Distance of Original Train

$\Rightarrow (s - 6)(t + 6) = st$

$\Rightarrow st + 6s - 6t - 36 = st$

$\Rightarrow 6s - 6t - 36 = 0$

$\Rightarrow 6s - 6t = 36$ ... (equation ii)

Solving equation (i) and equation (ii) by adding:

$\begin{aligned} -4s + 6t &= 24 \\ (+) \quad 6s - 6t &= 36 \\ \hline 2s &= 60 \\ s &= 30 \text{ km/hr} \end{aligned}$

Substituting $s = 30$ in equation (i):

$\begin{aligned} -4s + 6t &= 24 \\ -4(30) + 6t &= 24 \\ -120 + 6t &= 24 \\ 6t &= 144 \\ t &= 24 \text{ hours} \end{aligned}$

Original Distance $=st = (30)(24) = 720$ kilometers.

In speed-time-distance problems where distance is constant, setting up equations using the fact that different speed-time combinations yield the same distance is the most efficient approach.

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A train travelled a certain distance at a uniform speed. Had the speed been $6$ km per hour more, it would have needed $4$ hours less. Had the speed been $6$ km per hour less, it would have needed $6$ hours more. The distance, in km, travelled by the train is