A person left a city P to go to city Q at 6.30 a.m. He travelled at a speed of 90 km/hr for 2 hours and 30 minutes. After that, the speed was reduced to 50 km/hr. If the distance between the cities is 345 km, at what time did he reach the city Q?
(1) 10.25 a.m
(2) 11.15 a.m
(3) 11.24 a.m
(4) 11.50 a.m
Answers
number of km left =345-225=120
with speed of 50kmph time taken =120/50= 2.4 i.e 2hrs and 24mins
so total time of travel= 4hours and 54mins
the time he will reach q will be 11.24am.. option C
Answer:
The correct answer is option (3).11.24
Step-by-step explanation:
Formula:-
Speed = distance/time
It is given that,
A person left a city P to go to city Q at 6.30 a.m.
He traveled at a speed of 90 km/hr for 2 hours and 30 minutes.
Total distance = 345km
To find the distance covered in 1st journey
Speed = distance/time
distance = speed*time = 90*2.5 = 125
To find the remaining distance
Remaining distance = total distance - 125 = 345 -125 = 120 km
To find the time in the 2nd journey
Speed = distance/time
time = distance/speed = 120/50 = 2.4 hours = 2 hours 24 minutes
To find total time
total time = 2 hours and 30 minutes + 2 hours 24 minutes = 4 hours 54 minutes
To find the time to reach
time at which the person reach = 6.30 + 4.54 = 11.24
The correct answer is option (3).11.24