Time, Speed and Distance Question 3 - CAT 2025
Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was:
Step-by-Step Explanation
Define the baseline metrics:
Normal travel window = 5 pm to 11 pm = 6 hours. Let his initial speed be x km/h.
Total Distance (D) = 6x
Let y be the distance covered before the stoppage occurs. The remaining distance to cover is (6x − y).
Scenario 1 (20-minute stop, speed +3 km/h):
Stoppage = 20 minutes = 1/3 hour. The available running time is:
6 − 1/3 = 17/3 hours
y/x + (6x − y)/(x + 3) = 17/3 --- (Equation 1)
Scenario 2 (30-minute stop, speed +5 km/h):
Stoppage = 30 minutes = 1/2 hour. The available running time is:
6 − 1/2 = 11/2 hours
y/x + (6x − y)/(x + 5) = 11/2 --- (Equation 2)
Solve by reverse-engineering the choices:
Solving this system analytically leads to complex quadratics. Let us test the multiple-choice options for x:
Test Option B (x = 15 km/h):
Total Distance = 6 × 15 = 90 km.
From Equation 1:
60/15 + (90 − 60)/18 = 17/3
4 + 30/18 = 17/3
This satisfies the equation with y = 60 km.
Check if y = 60 satisfies Equation 2:
60/15 + (90 − 60)/(15 + 5)
= 4 + 30/20
= 4 + 1.5
= 5.5 hours
= 11/2 hours
This matches the problem parameters perfectly. The initial speed was 15 km/h.
Correct Option: B
💡 FundaMakers Tip:
When algebraic setups for Time-Speed-Distance create systems of fractional fractions, switch to testing the options immediately. Choose options that form clean integers with the numbers in the question text. Testing 15 provides smooth denominators (15, 18, 20) that resolve quickly.