TCS
Company
Numerical Ability
Time Distance and Speed
A 300m race is held between two players running 3m/s and 5m/s in swimming pool of 50m long. how many times the two player meet in opposite direction till the faster one completes the race.
Read Solution (Total 17)
-
- 4 times is right answer.
- 9 years agoHelpfull: Yes(34) No(15)
- time fast= 60s
distance slower in 60 s=3*60=180m
first time same direction
so crossing=3
- 9 years agoHelpfull: Yes(9) No(4)
- The answer is 3 times. I ont know how to explain this. But I got the exact answer by calculating it in my mind and drawing it in the paper.
Anyway I ll try to narrate it. Get it if u understand:
Ok let me assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap.
So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th minute both will complete a lap A's 5th and B's 3rd. So they will start next lap(A's 6th and B's 4th) in the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other. So totally they met 3 times in opposite direction. - 8 years agoHelpfull: Yes(7) No(8)
- 2 damn sure
- 9 years agoHelpfull: Yes(6) No(10)
- Exact answer is "4"times
Bcz Every 10sec of time 1st will cover 30m
2nd will cover 50m,To complete total race by faster one i,e 2nd will take 50sec,
In that duration there are 4 crossings had been taken place,that's ans is 4 ...write ...I hopE U uNderstOOd. - 9 years agoHelpfull: Yes(5) No(6)
- 4 is correct solution.
for the first 10 sec both swim in same direction. then faster one will change direction and suppose at time t they meet in opposite direction then at this instant 5t+3t=50 ,this gives t=6.25
But first meet will occur after initial 10 sec so the first meet occurs at 16.25 sec
Similarly next will occur at 16.25+6.25=22.5 sec.
Now at t=30 sec faster one will be at the second end of pool while the slower has not crossed the first end till 30 sec . so now they will meet at 22.5+6.25+6.25=35.00 sec . And at 50 sec both will be at second end of the pool.
as we see the faster one will take 300/5=60 sec to win the race .
so now they will not meet after the 50th sec
so they meet 4 times i.e. at 16.25 , 22.5 , 35.0 , 50.0
so answer is 4 times
- 8 years agoHelpfull: Yes(3) No(1)
- i think they meet 2 times in opp direction.. as they move at the same tym for the first 50m.. they are not opp to eachother..the 2nd runner is faster than the first one..the 2nd runner takes 60sec to complete..so he runs 6 tyms through the swimming pool..the 1st one runs 100sec to complete.in between they meet 2 tyms in opp direction
- 9 years agoHelpfull: Yes(2) No(3)
- the Answer should be 3...Deepak can u explain this pls
- 8 years agoHelpfull: Yes(2) No(2)
- 3 times in opposite direction
- 8 years agoHelpfull: Yes(2) No(0)
- Answer is 2,
In 6 rounds by the faster one, only during 3rd and 4th round they will meet in opposite direction. - 8 years agoHelpfull: Yes(2) No(1)
- IT IS THREE TIMES....
- 8 years agoHelpfull: Yes(1) No(0)
- 3 is fine
- 8 years agoHelpfull: Yes(1) No(0)
- options are 2,3,4,5
- 9 years agoHelpfull: Yes(0) No(2)
- 4 is correct because, at the last lab, 1st player starts at 50 sec & 2nd player ends 3rd lab at 49.8sec.hence another collision occurs
- 8 years agoHelpfull: Yes(0) No(2)
- 2 times will be the answer. the total time taken by faster one will be 60s hence we have to calculate within 60s time period. the first meet will be at 12.5s and second meet will be at 37.5s at the 50s they will again meet bt it will be in same direction.now 250 meter has been completed by the faster one and slower has completed 150 meter so their faces are in the same direction as the race is only of 300m hence there will no meet again. so answer is two times.
- 8 years agoHelpfull: Yes(0) No(2)
- Answer is 4 times.
time 2nd= 60s
distance for 1st in 60 s=3*60=180m
they meet from the beginning since in the start when the faster one covers 50 m the slower one covers only 30 mtrs.
so crossing=4 - 7 years agoHelpfull: Yes(0) No(1)
- 3 times is right answer
- 5 years agoHelpfull: Yes(0) No(0)
TCS Other Question