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18 May 2026, 19:00
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B
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65% (02:02) correct
35%
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wrong
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A field technician left a service depot at 1:20 PM for a maintenance visit at Harbor Clinic scheduled for 4:05 PM the same day. If Harbor Clinic was at least 150 kilometers from the service depot and the technician drove at an average (arithmetic mean) speed of at least 60 kilometers per hour, did the technician arrive late for the maintenance visit?
(1) The technician drove at an average (arithmetic mean) speed of no more than 75 kilometers per hour.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
(1) The technician drove at an average (arithmetic mean) speed of no more than 75 kilometers per hour.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
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M47-159
09 Jun 2026, 01:00
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GMAT Club Official Solution:
A field technician left a service depot at 1:20 PM for a maintenance visit at Harbor Clinic scheduled for 4:05 PM the same day. If Harbor Clinic was at least 150 kilometers from the service depot and the technician drove at an average (arithmetic mean) speed of at least 60 kilometers per hour, did the technician arrive late for the maintenance visit?
The technician had from 1:20 PM to 4:05 PM, so the available time was 2 hours 45 minutes.
Let d be the distance from the service depot to Harbor Clinic, and let s be the technician’s average speed. Given that d >= 150 and s >= 60, the question asks:
Is d/s > 2 hours 45 minutes?
(1) The technician drove at an average speed of no more than 75 kilometers per hour.
So:
60 <= s <= 75
But the distance is still not fixed.
For example, if d = 150 and s = 75, then d/s = 150/75 = 2 hours. The technician was not late.
But if d = 240 and s = 60, then d/s = 240/60 = 4 hours. The technician was late.
Not sufficient.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
So:
d < 160
Since s >= 60, the travel time, d/s, was less than:
160/60 = 8/3 hours = 2 2/3 hours = 2 hours 40 minutes.
Therefore, the technician was not late.
Sufficient.
Answer: B.
A field technician left a service depot at 1:20 PM for a maintenance visit at Harbor Clinic scheduled for 4:05 PM the same day. If Harbor Clinic was at least 150 kilometers from the service depot and the technician drove at an average (arithmetic mean) speed of at least 60 kilometers per hour, did the technician arrive late for the maintenance visit?
The technician had from 1:20 PM to 4:05 PM, so the available time was 2 hours 45 minutes.
Let d be the distance from the service depot to Harbor Clinic, and let s be the technician’s average speed. Given that d >= 150 and s >= 60, the question asks:
Is d/s > 2 hours 45 minutes?
(1) The technician drove at an average speed of no more than 75 kilometers per hour.
So:
60 <= s <= 75
But the distance is still not fixed.
For example, if d = 150 and s = 75, then d/s = 150/75 = 2 hours. The technician was not late.
But if d = 240 and s = 60, then d/s = 240/60 = 4 hours. The technician was late.
Not sufficient.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
So:
d < 160
Since s >= 60, the travel time, d/s, was less than:
160/60 = 8/3 hours = 2 2/3 hours = 2 hours 40 minutes.
Therefore, the technician was not late.
Sufficient.
Answer: B.
General Discussion
ankushsambare
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18 May 2026, 20:46
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Time available from 1:20 PM to 4:05 PM = 2 hours 45 minutes = 2.75 hours
Required distance ≥ 150 km
Speed ≥ 60 km/h
We check whether arrival is late (i.e., actual travel time > 2.75 hours).
Basic bounds:
Minimum possible time = distance / maximum speed
Maximum possible time = distance / minimum speed
So:
Fastest case: 150 / 75 = 2 hours
Slowest case: 150 / 60 = 2.5 hours
So even in worst case, time = 2.5 hours < 2.75 hours
This already suggests he could arrive on time or early, but not necessarily late.
Now check statements.
Statement (1): speed ≤ 75 km/h
So speed is between 60 and 75.
For fixed minimum distance 150 km:
Time range:
150/75 = 2 hours to 150/60 = 2.5 hours
Still always ≤ 2.75 hours
So still cannot conclude lateness or not (always on time or early)
Not sufficient.
Statement (2): distance < 160 km
So distance is between 150 and 159.
Now combine with base speed ≥ 60:
Maximum time = 159/60 ≈ 2.65 hours < 2.75 hours
Even slowest case still arrives early
So still cannot determine “late or not” (he is always not late)
So we can conclude definitively he is NOT late.
Thus statement (2) alone is sufficient.
Final Answer: B
Required distance ≥ 150 km
Speed ≥ 60 km/h
We check whether arrival is late (i.e., actual travel time > 2.75 hours).
Basic bounds:
Minimum possible time = distance / maximum speed
Maximum possible time = distance / minimum speed
So:
Fastest case: 150 / 75 = 2 hours
Slowest case: 150 / 60 = 2.5 hours
So even in worst case, time = 2.5 hours < 2.75 hours
This already suggests he could arrive on time or early, but not necessarily late.
Now check statements.
Statement (1): speed ≤ 75 km/h
So speed is between 60 and 75.
For fixed minimum distance 150 km:
Time range:
150/75 = 2 hours to 150/60 = 2.5 hours
Still always ≤ 2.75 hours
So still cannot conclude lateness or not (always on time or early)
Not sufficient.
Statement (2): distance < 160 km
So distance is between 150 and 159.
Now combine with base speed ≥ 60:
Maximum time = 159/60 ≈ 2.65 hours < 2.75 hours
Even slowest case still arrives early
So still cannot determine “late or not” (he is always not late)
So we can conclude definitively he is NOT late.
Thus statement (2) alone is sufficient.
Final Answer: B
