. A, B and C can walk at the rates of 3, 4 and 5 km per hour respectively. They starts from Pune at 1, 2, 3 o’clock respectively. When B catches A, B sends him back with a message to C. When will C get the message?
Answers
Answered by
55
hey there ,,,, here is the answer of your question !!!
Distance already covered by A when B
starts moving at 2o’clock = 3km/hr × (2 - 1) hr. = 3 kms
Now,
time taken by B to catch A
=(3 km )/((4-3) km/hr)= 3 hrs
B will catch A at (2’o clock + 3 hrs) i.e. at 5 o’clock
Distance where B will catch A = 4 km/hr × (5 - 2) hr. = 12 km.
Also at 5o, clock, C will be at 5 km/hr × (5 - 3) hrs = 10 kms
i.e. at 5’o clock distance between (A or B) and C = (12 - 10) = km = 2km
Now
, time to meet A & C
(2 km )/((5+3) km/hr)
=1/(4 hr)=15 min
Time of meeting of of A & C
= 5 : 15 o’clock
hope that helps
# himanshu jha
✌✌
Distance already covered by A when B
starts moving at 2o’clock = 3km/hr × (2 - 1) hr. = 3 kms
Now,
time taken by B to catch A
=(3 km )/((4-3) km/hr)= 3 hrs
B will catch A at (2’o clock + 3 hrs) i.e. at 5 o’clock
Distance where B will catch A = 4 km/hr × (5 - 2) hr. = 12 km.
Also at 5o, clock, C will be at 5 km/hr × (5 - 3) hrs = 10 kms
i.e. at 5’o clock distance between (A or B) and C = (12 - 10) = km = 2km
Now
, time to meet A & C
(2 km )/((5+3) km/hr)
=1/(4 hr)=15 min
Time of meeting of of A & C
= 5 : 15 o’clock
hope that helps
# himanshu jha
✌✌
Anonymous:
ok
Answered by
28
Distance covered by A when B starts moving at 2o'clock
=3km/hr×(2-1)hr
=3hrs
Now,
Time taken by B to catch A
=3km/(4-3)km/hr
=3 hrs.
B will catch A at (2o'clock+3hrs) i.e. at 5o'clock.
Distance where B will catch A=4km/hr×(5-2)hr=12 km.
Also, at 5o'clock C will be at 5km/hr×(5-3)hrs=10 kms.
That means at 5o'clock distance between (A or B) and C is (12-10)=2 kms.
Now,
time to meet A&C
=2km/(5+3)km/hr
=1/4hr
=(1/4×60) min
=15 minutes.
So, time of meeting of A&C
=5:15 o'clock.
______________________________
=3km/hr×(2-1)hr
=3hrs
Now,
Time taken by B to catch A
=3km/(4-3)km/hr
=3 hrs.
B will catch A at (2o'clock+3hrs) i.e. at 5o'clock.
Distance where B will catch A=4km/hr×(5-2)hr=12 km.
Also, at 5o'clock C will be at 5km/hr×(5-3)hrs=10 kms.
That means at 5o'clock distance between (A or B) and C is (12-10)=2 kms.
Now,
time to meet A&C
=2km/(5+3)km/hr
=1/4hr
=(1/4×60) min
=15 minutes.
So, time of meeting of A&C
=5:15 o'clock.
______________________________
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