Question

. 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?

Answer 1

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
✌✌

Answer 2

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.

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