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05 Aug 2026, 23:13
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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.
05 Aug 2026, 23:13
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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 \(d \geq 150\) and \(s \geq 60\), the question asks:
Is \(\frac{d}{s} > 2\) hours 45 minutes?
(1) The technician drove at an average speed of no more than 75 kilometers per hour.
So:
\(60 \leq s \leq 75\)
But the distance is still not fixed.
For example, if \(d = 150\) and \(s = 75\), then \(\frac{d}{s} = \frac{150}{75} = 2\) hours. The technician was not late.
But if \(d = 240\) and \(s = 60\), then \(\frac{d}{s} = \frac{240}{60} = 4\) hours. The technician was late.
Not sufficient.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
So:
\(d < 160\)
Since \(s \geq 60\), the travel time, \(\frac{d}{s}\), was less than:
\(\frac{160}{60} = \frac{8}{3}\) hours \(= 2\frac{2}{3}\) 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 \(d \geq 150\) and \(s \geq 60\), the question asks:
Is \(\frac{d}{s} > 2\) hours 45 minutes?
(1) The technician drove at an average speed of no more than 75 kilometers per hour.
So:
\(60 \leq s \leq 75\)
But the distance is still not fixed.
For example, if \(d = 150\) and \(s = 75\), then \(\frac{d}{s} = \frac{150}{75} = 2\) hours. The technician was not late.
But if \(d = 240\) and \(s = 60\), then \(\frac{d}{s} = \frac{240}{60} = 4\) hours. The technician was late.
Not sufficient.
(2) Harbor Clinic was less than 160 kilometers from the service depot.
So:
\(d < 160\)
Since \(s \geq 60\), the travel time, \(\frac{d}{s}\), was less than:
\(\frac{160}{60} = \frac{8}{3}\) hours \(= 2\frac{2}{3}\) hours \(= 2\) hours 40 minutes.
Therefore, the technician was not late.
Sufficient.
Answer: B
