A starts from a place at 11.00 am and travels at a speed of
4 kmph , B starts at 1.00 pm and travels with speeds of 1
kmph for 1 hr , 2 kmph for the next hr , 3 kmph for the
next hr and so on. At wht time will B catch up with A ?
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Answer / jhilam bera
At 1:00 pm the dif'n between A & B = 8 km
after 2:00 pm ................... = 11 km (as B's speed
is 1 and A's 4km, then eqv speed=(4-1)=3km)
After 3:00....................... = 13km (as B's speed 2km)
After 4:00....................... = 14km
after 5:00....................... = 14km (A's speed= B's
speed)
after 6:00....................... = 13km
after 7:00....................... = 11km
after 8:00....................... = 8km
after 9:00....................... = 4km
and now the eqv speed is= (9-4) =5km/hr;
and the remaning distance is 4km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48min=9:48 pm;
| Is This Answer Correct ? | 44 Yes | 5 No |
Answer / vivek kumar 2k6 bit mesra
lets say it takes time = T hr + t(fraction)hour
then for for T hours distance traveled by A = 4(T+2)
and that by B = T(T+1)/2
also Dist travelled by A > dist by B .......in T hours as in T+t hours it B reaches to A
4(T+2)>T(T+1)/2
-1.815 < T < +8.815
so integral value of T= 8 hours
So in 8 hr A travelled 4(8+2)=40
B travelled 8(8+1)/2=36
so this lag in distances will be covered up in t minutes
now speed of B will be 9 in the 9th hour
so, 9t-4t=4
t=4/5hr
=48 minutes
So the total time taken is T + t
= 8hr + 48 min
So answer will be 1pm+8:48 = 9:48pm............Ans
| Is This Answer Correct ? | 14 Yes | 1 No |
Answer / satish raj
A covers 8km till 01.00:pm and B starts at 01.00pm.hence at this point distance among them is 8km.
hence after 1hour at 02.00pm distance among them will be
11 km(i.e with a relative speed of 4-1=3km/h A will cover increase 3km gap).
At 03.00 pm,distance among them will be 13km(i.e. with a relative speed 4-2=2km/h and hence increase 2km gap)
hence,At 04.00- distance among them will be 14km.
at 05.00-distance among them will be 14km.
now 05:00pm-06:00pm slot relative speed will be countable and going with it- we will get 13hr-11hr-8hr-4hr and this 4 gap of will be completed in 4/5 hour=48min(because at that slot relative speed of B will be 5km/h).
ans:-09.48:pm
| Is This Answer Correct ? | 2 Yes | 1 No |
Answer / vikram mandal
A goes by (11am-1pm)2h 8km.
A covers 3hours=B covers 1hour.
so,A covers the distance of 10hours=40km
where B covers the distance of 8hours=36km
now,speed of B's is 9km/h.
where speed of A's is 4km/h.
within next 1hour A will cover 44km.
so,(44-36)km=8km.to cover this distance B needs time=(8/9)h=.88h
So,B catch up to A same day to 08.54p.m.
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / avinash
They will meet after 1 pm.A has unique speed 4 km/hr.so,A cross 8 km at 1 pm.But,B increase speed like(1+2+3+4+5+6+7+8+9 = 45).For A 9*4=36 + 8=44. so before 9 pm and after 8:30 pm they should meet.So, Answer is : 8:45 PM.
| Is This Answer Correct ? | 1 Yes | 3 No |
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