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19 Feb 2026, 00:17
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E
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Difficulty:
85%
(hard)
Question Stats:
37% (02:02) correct
63%
(01:44)
wrong
based on 49
sessions
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A train always travels at one of two speeds: 160 km/hr in rural areas and 40 km/hr in urban areas. Is its average speed from A to B greater than 100 km/hr?
(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance between A and B is 150 kilometers.
(1) More than 2/3 of the distance from A to B is through rural areas.
(2) The distance between A and B is 150 kilometers.
19 Feb 2026, 02:10
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Key concept: Weighted average speed (harmonic mean)
The instinct here is to think "if more than 2/3 of the distance is rural, the average speed must be closer to 160 than to 40, so it's probably above 100." That's the trap — average speed is NOT a simple weighted average of the two speeds. Because time = distance/speed, the faster leg takes less time, so it contributes less to the total travel time than you'd expect. You have to use the harmonic mean framework.
Setting up the math:
Let D = total distance, r = fraction of distance through rural areas.
Total time = rD/160 + (1−r)D/40
Average speed = D ÷ [rD/160 + (1−r)D/40] = 1/[r/160 + (1−r)/40]
For average speed > 100:
1/[r/160 + (1−r)/40] > 100
→ r/160 + (1−r)/40 < 1/100
→ [r + 4(1−r)]/160 < 1/100
→ 4 − 3r < 1.6
→ 3r > 2.4
→ r > 0.8 (i.e., more than 80% must be rural)
Statement (1): More than 2/3 of the distance is rural (r > 0.667)
We need r > 0.8. If r = 0.7, average speed < 100. If r = 0.9, average speed > 100. Two different answers → INSUFFICIENT
Statement (2): Distance = 150 km
Tells us nothing about the rural/urban split. We still don't know r. → INSUFFICIENT
Both together: We know r > 2/3 and D = 150 km, but r could still be 0.7 or 0.9. The total distance adds no constraint on the split. → STILL INSUFFICIENT
Answer: E
Takeaway: Whenever a DS question involves average speed over two-speed journeys, set up the harmonic mean inequality first — find the exact threshold before evaluating each statement.
The instinct here is to think "if more than 2/3 of the distance is rural, the average speed must be closer to 160 than to 40, so it's probably above 100." That's the trap — average speed is NOT a simple weighted average of the two speeds. Because time = distance/speed, the faster leg takes less time, so it contributes less to the total travel time than you'd expect. You have to use the harmonic mean framework.
Setting up the math:
Let D = total distance, r = fraction of distance through rural areas.
Total time = rD/160 + (1−r)D/40
Average speed = D ÷ [rD/160 + (1−r)D/40] = 1/[r/160 + (1−r)/40]
For average speed > 100:
1/[r/160 + (1−r)/40] > 100
→ r/160 + (1−r)/40 < 1/100
→ [r + 4(1−r)]/160 < 1/100
→ 4 − 3r < 1.6
→ 3r > 2.4
→ r > 0.8 (i.e., more than 80% must be rural)
Statement (1): More than 2/3 of the distance is rural (r > 0.667)
We need r > 0.8. If r = 0.7, average speed < 100. If r = 0.9, average speed > 100. Two different answers → INSUFFICIENT
Statement (2): Distance = 150 km
Tells us nothing about the rural/urban split. We still don't know r. → INSUFFICIENT
Both together: We know r > 2/3 and D = 150 km, but r could still be 0.7 or 0.9. The total distance adds no constraint on the split. → STILL INSUFFICIENT
Answer: E
Takeaway: Whenever a DS question involves average speed over two-speed journeys, set up the harmonic mean inequality first — find the exact threshold before evaluating each statement.
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