Time speed distance

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Three cars started simultaneously from Ajmer to Banaras along the same highway. The second car travelled with a speed that was 10km/h higher than the first car's speed and arrived at Banaras 1 hour earlier than the first car. The third car arrived...
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Solution to above problem

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Can someone post the solution

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@pimpalekaran  ·  68 karma

Of....?

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@vishiac  ·  5,092 karma

@harsimran29fe 15% more means 115%= 120.. 100% = 120*100/115= 104.34

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Prabhjeet made a trip from hyderabad to Bengaluru to attend an  interview. he had been tprabhjeet tog at a destination h have ensured that he reaches the interview on time. He had flat tyre on the way and it took 100 min to mend it.he arrived 15 min late.distance between two cities 260km.

1.if the speed after flat tyre was 20% more than original speed.fimd the normal time by prabhjeet to reach destination from site of accident.

@vishiac  ·  5,092 karma

@sayakaiyth diff in 85 min, and speed ratio is 5:6, so time ratio is 6:5, so actual time is 6*85min= 8.5hrs

@vishiac  ·  5,092 karma

@sayakaiyth if u haven't checked... Question was posted in 2014...☺

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A pedestrian left point A for a walk, going with the speed of 5 km/h. When the pedestrian was at distance of 6 km from A, a cyclist followed him from A and cycling at the speed 9 km/h higher than that of the pedestrian. When the cyclist overtook the pedestrian, they turned back and returned  to A together, at the speed of 4 km/h. At what speed will the time spent by the pedestrian on his journey from A to A be least?

5 km/h
6km/h
6.1 km/h
5.5 km/h

Jaideep travels from Alaska, which is on highway, to Burgeon, which is 16 km from the highway. The distance between Alaska and Burgeon along a straight line is 34 km. At what point should Jaideep turn from the highway to reach Burgeon in the shortest possible time, if his speed along the highway is 10 km/h and 6 km/h otherwise.

30 km away from A
20 km away from A
18 km away from A
15 km away from A

Three sprinters A,B,C had to sprint from points P to Q and back again (starting in that order). The time interval between the starting times of the three sprinters was 5 seconds each. Thus C started 10 seconds after A, while B started 5 seconds after A. The three sprinters passed a certain point R, which is somewhere between P and Q, simultaneously (none of them having reached point Q yet). Having reached Q and reversed the direction, the third sprinter met the second 9 m short of Q and met the first sprinter 15 m short of Q. Find the speed of the first sprinter if PQ=55 km.

4 m/s
3 m/s
2 m/s
1 m/s