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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 Posted / 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 |
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