Two swimmers start swimming in a swimming pool from opposite ends. they met first at a distance of 50m from east and return back, they met again for second time at 20m from west then find the length of pool. Assume their speeds are constant.

• ans:70 let a swimmer start at the east(e) and another at west(w).now,during the opposite direction of swimming,e will cover 50m(as e and w meet 50m from east)and w will cover x meters.now while returning back,e will cover 50m and w will cover the same x meters.now,during the second time,w will cover 20m(as speeds of w and e are constant).as long as e and w don't increase their speed,e and w will meet 50m from the east and 20m from west.therefore, total length of the pool=50+20=70.click yes if u understood r else click no.
• 10 years agoHelpfull: Yes(33) No(13)
• Let thr be 2 swimmer E n W swimming from east and west direction respectively and thr speed be Se and Sw.
  CASE 1 :
  Time taken by E = 50/Se
  Time taken by W = d-50/Sw
  Clearly, 50/Se=d-50/Sw => Sw/Se=d-50/50 .... eq(1)
  
  CASE 2:
  Dist. covered by E = 50+50+(d-50-20)
  = d+30 Since, he return back n thn meet at 20m frm west
  Dist. covered by W = d-50+20
  = d-30
  Time taken by E = d+30/Se
  Time taken by W = d-30/Sw
  Therefore, Sw/Se= d-30/d+30
  From eq(1),
  d-30/d+30=d-50/50
  => d=70 (ans)
  
• 10 years agoHelpfull: Yes(21) No(4)
• options are:
  a-70
  b-100
  c-130
  d-not remember.
  answer ll be (d)130 ...try to solve using concept v directl proportional to distance...it is easy...
• 10 years agoHelpfull: Yes(14) No(9)
• in 1st case from east, distance =50 and rest is d-50(d is total length), now as speed is constant so t1*s1=50----eqn1 and t2*s2=d-50-----eqn2 as time taken is equal so we get s1/s2=50/d-50 now in second case from west distance is 20 and rest is d-20 so now s2*t2=20 and s1*t1=d-20 time taken same so we get s1/s2=d-20/20 after solving we get d=0 and d=70 , so d=0 not possible so we get 70 as answer
• 10 years agoHelpfull: Yes(7) No(4)
• ans=130
  ratio of speeds of 2 persons is same for the 2 meets.
  50/(d-50)=(d-30)/d+30)
  solving this we get d= 130
• 10 years agoHelpfull: Yes(5) No(2)
• As the speed is constant and they return back , so they will cover same distance both of the times.
  first time, the distance was given from the east side ie 50m.
  second time, the distance was given from the west side ie 20m.
  so total distance will be 70m.
  i think this is the simplest explanation.
• 10 years agoHelpfull: Yes(3) No(1)
• answer will be 130.
  ms vedika got confused with east side and west side so she made the eq opposite
  the eq will come as (d-30)/(d+30)=50/(d-50)
• 8 years agoHelpfull: Yes(2) No(1)
• as it is given speed is constant
  1st time
  so in time t swimmer 'a' covers 'x=50' from east and
  swimmer 'b' covers 'y' distance so the total distance = 50+y
  similarly
  2nd time
  swimmer 'a' covers 'x' distance from east
  swimmer 'b' covers 'y=2o' from west
  as there speed are constant and nothing is given about time
  let us assume that time is also constant
  so in constant speed and time both of them will cover same distance both the time
  thus total distance = X+Y= 50+20=70
• 10 years agoHelpfull: Yes(1) No(1)
• W−−−−−−−−> ∙ ∙ ====> swimmer.
  
  In the first leg she swam 50m50m, so in the 2nd leg she must have swum 100m100m
  In the 2nd leg, she swam L−50+20,L−50+20, so equating
  
  L−50+20=100L−50+20=100, which yields L=130m
• 8 years agoHelpfull: Yes(1) No(0)
• Ans will be 130 u have perhaps u have mistaken to solving the equation. and m not solving just explaining many of them solve already whose ans is 130.
  i wll tell u why 70 wont be ans.
  east denote with E and west denotes with W. accord to ques 1st meeting is 50 from east so if u are saying 70 is length then W travell 20 meter so its means E is faster than W cs when W cover 20 E cover 50.
  2nd meeting 20 meter from west so its means E travell 40 meter and W travel 100 meter its not possible. E is faster than W.
  now assume 130 when E cover 50 w cover 80 in 1st meeting
  2nd meeting E cover 80+20=100
  W cover 50+110=160
  because W is faster i hope u ll understand.
• 7 years agoHelpfull: Yes(1) No(0)
• answer is 70 since they both have taken same time
  let total distance between them is x
  so d1/s1=d2/s2
  (x-50)/u=50/u .........i)
  
  20/u=(x-20)/u .........ii)
  solve above two equations
  we get 70 as a ans
  
• 10 years agoHelpfull: Yes(0) No(0)
• Mmmmmmmmnmmm
• 9 years agoHelpfull: Yes(0) No(0)
• let the two swimmers be at east as e and at west be w..
  initially they started swimming in opposite direction to each other..
  the total length of pool be x..
  they meet at 50m from e and x-50 m from w..
  after meeting they return in opposite direction and again meet at 20m from west..
  distance traveled by e is 50+50+(x-50-20)..
  distance traveled by w is x-50-20..
  for first meet
  50/Ve=(x-50)/Vw............1
  for second meet
  (50+50+(x-50-20))/Ve=(x-50-20)/Vw.........2
  solving 1 and 2 we get x=70
• 7 years agoHelpfull: Yes(0) No(1)
• W−−−−−−−−>∙∙ swimmer.
  
  In the first leg she swam 50m, so in the 2nd leg she must have swum 100m
  In the 2nd leg, she swam L−50+20, so equating
  
  L−50+20=100, which yields L=130m
• 6 years agoHelpfull: Yes(0) No(0)
• W−−−−−−−−>∙∙ swimmer.
  
  In the first leg she swam 50m, so in the 2nd leg she must have swum 100m
  In the 2nd leg, she swam L−50+20, so equating
  
  L−50+20=100, which yields L=130m
• 5 years agoHelpfull: Yes(0) No(0)