A field technician left a service depot at 1:20 PM for a maintenance

Question

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.

ankushsambare · Answer

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

BongBideshini · Answer

1:20 PM to 4:05 PM = 2 hours 45 minutes = 2.75 hours
Given constraints:
 
 Distance ≥ 150 km 
 Speed ≥ 60 km/h  
 Statement (1):  Speed ≤ 75 km/h
Combined with the given information:
60 ≤ speed ≤ 75 km/h
distance ≥ 150 km
 Case 1: 
Distance = 150 km, speed = 60 km/h
time = 150/60 = 2.50 hours
Technician arrives on time.
 Case 2: 
Distance = 200 km, speed = 60 km/h
time = 200/60 = 3.33 hours
Technician arrives late..
 Therefore, Statement (1) alone is INSUFFICIENT.

Statement (2):  Distance < 160 km
distance < 160 km
speed ≥ 60 km/h
t =  \frac{<160}{≥60} 
t < 2.66 hours
Since:
2.66 hours < 2.75 hours available,

Therefore, Statement (2) alone is SUFFICIENT.

Answer: B

SS990 · Answer

Given: D>=150km, S=60km/hr, for which travel time t=2.5hrs

Start time is 1.20pm, ETA at destination is 4.05pm i.e time for travel to reach on time = 2hr 45min or 2(3/4)hrs

To find: did technician reach on time?

1) S<=75. This is insufficient since we dont know the upper limit of D

2) D<=160km. We know 60<= S<=75 lets assume S=60. Then 160/60 is 2(2/3)hrs less than available time of 2(3/4) hrs given to reach destination

Hence B is sufficient.

A field technician left a service depot at 1:20 PM for a maintenance

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A field technician left a service depot at 1:20 PM for a maintenance

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