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NTPC Graduate Tier 1 2025 Shift-1 📅 09 Jun, 2025

Babita travels a certain distance at a speed of 11 km/hr and double the earlier distance at 44 km/hr. She then returns to the starting point through the same route. If her average speed for the entire journey is 34 km/hr, then what is her speed (in km/hr) for the return journey?

A
70
B
79
C
74.8
D
76.8
Result Summary
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APEDIA

NTPC Graduate Tier 1
2025 • 09 Jun, 2025 • Shift-1
Babita travels a certain distance at a speed of 11 km/hr and double the earlier distance at 44 km/hr. She then returns to the starting point through the same route. If her average speed for the entire journey is 34 km/hr, then what is her speed (in km/hr) for the return journey?
Correct Answer
74.8
Distance Allocation: Let the initial conceptual distance be mapped as 'D'. The next leg is double, meaning '2D'. The return trip covers the exact entire outgoin......
💡 Analysis & Explanation
Distance Allocation ▼
Let the initial conceptual distance be mapped as 'D'. The next leg is double, meaning '2D'. The return trip covers the exact entire outgoing distance, so the return distance is D + 2D = '3D'. The absolute total distance covered is 6D.
Time Formulation ▼
Time is purely Distance / Speed. Time for leg 1 = D/11. Time for leg 2 = 2D/44 = D/22. Let the unknown return speed be 'V'. Time for return trip = 3D/V. The total time structurally equals D/11 + D/22 + 3D/V = 3D/22 + 3D/V.
Applying Average Speed Equation ▼
Average Speed is Total Distance / Total Time. We are given the average speed is 34. Thus, 6D / (3D/22 + 3D/V) = 34. Canceling the common D variable simplifies this to 6 / (3/22 + 3/V) = 34. Dividing numerator by 3 gives 2 / (1/22 + 1/V) = 34.
Solving for Variable V ▼
This implies 1/22 + 1/V = 2/34 = 1/17. Rearranging to isolate V gives 1/V = 1/17 - 1/22. Finding a common denominator (374), we get 1/V = (22 - 17) / 374 = 5 / 374. Inverting the fraction yields V = 374 / 5 = 74.8.
Conclusion ▼
Her required average speed for the final return leg is exactly 74.8 km/hr.