Ques:- 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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Ques:- P, Q, R, S, T and U are six cities which are in a row in the same order. A bus, b1 travels from P to U and another bus b2 travel from U to P. The bus b1 reaches S at 10:40 and bus b2 reaches Q at 10:35 if the travel time between any two adjacent cities is 40 minutes and the stoppage time at each city is 15 minutes. Which city do the two buses meet?
Recent Answer : :
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
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the answer is At 9:48 PM
At 1:00 pm the difference between A & B = 8 km
after 2:00 pm ………………. = 11 km (as B’s speed
is 1 and A’s 4 km, then eqv speed=(4-1)=3 km)
After 3:00………………….. = 13 km (as B’s speed 2 km)
After 4:00………………….. = 14 km
after 5:00………………….. = 14 km (A’s speed= B’s
speed)
after 6:00………………….. = 13 km
after 7:00………………….. = 11 km
after 8:00………………….. = 8 km
after 9:00………………….. = 4 km
and now the eqv speed is= (9-4) =5 km/hr;
and the renaming distance is 4 km;
then, time=(60*4)/5=48 min;
then the meeting time is=9:00+48 min=9:48 pm;