Time, Speed and Distance Question 1 - CAT 2024
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:
Step-by-Step Explanation
Let the original uniform speed of the train be v km/h and the original time taken be t hours. The total distance covered is D = vt.
Scenario 1 (Speed increased by 6 km/h):
The new speed is (v + 6) and the new time is (t − 4).
(v + 6)(t − 4) = vt
vt − 4v + 6t − 24 = vt
6t − 4v = 24
3t − 2v = 12 --- (Equation 1)
Scenario 2 (Speed decreased by 6 km/h):
The new speed is (v − 6) and the new time is (t + 6).
(v − 6)(t + 6) = vt
vt + 6v − 6t − 36 = vt
6v − 6t = 36
v − t = 6 ⟹ v = t + 6 --- (Equation 2)
Solve for t and v:
Substitute Equation 2 into Equation 1:
3t − 2(t + 6) = 12
3t − 2t − 12 = 12 ⟹ t = 24 hours
Now compute the speed v:
v = 24 + 6 = 30 km/h
Calculate the total distance:
D = v × t = 30 × 24 = 720 km
Correct Option: A
💡 FundaMakers Tip:
For constant distance problems involving two symmetric speed changes (±Δv), you can use the speed variation constant framework:
D = [v(v + Δv₁)/Δv₁] × Δt₁ = [v(v − Δv₂)/Δv₂] × Δt₂
Substituting the values directly bypasses the time variables:
[v(v + 6)/6] × 4 = [v(v − 6)/6] × 6
4(v + 6) = 6(v − 6)
4v + 24 = 6v − 36
2v = 60 ⟹ v = 30 km/h
Knowing the speed is 30 km/h, the first condition tells us the updated speed is 36 km/h. The ratio of speeds is 30 : 36 = 5 : 6, meaning the time ratio is 6 : 5. The drop of 1 unit equals 4 hours, making the original time 6 × 4 = 24 hours. Total distance = 30 × 24 = 720 km. This approach completely avoids simultaneous equations.