Re: 1 of toughest TSD question ever | |

Good work Mohit |

Re: 1 of toughest TSD question ever | |

Hi siddharth, You need to grasp this logic. Otherwise you can verify it by assuming some values. Time taken will be same. |

Re: 1 of toughest TSD question ever | |

Sir , I am not getting it. Please explain ! |

Re: 1 of toughest TSD question ever | |

just equate the total time of man and cork... 1/vr=1/2+(1+1/2(vm-vr))/(vr+vm) after solving u'll get vr=1 vr=speed of cork (or river) vm-speed of man |

Re: 1 of toughest TSD question ever | |

1 KM/hour let 'x' be the start position, 'y' be the meeting position ( 1 KM from x) and 'z' be the position until which the man swims Let speed of stream= 'a' kmph and speed of man(constant) ='b' kmph xy= 1 km yz = (a-b)*1/2 the time taken by the cork to reach the bridge after meeting with the man = 1/b so from the passage we can say that, 1/b= ½ + (time taken by the man to travel ‘zy’) + (time taken by the man to travel ‘yx’) 1/b= ½+ (a-b)/(2*(a+b)) + 1/(a+b) 1/b= (a+b)/ 2*(a+b) + (a-b)/(2*(a+b)) + 2/ 2*(a+b) 1/b = (a+1)/(a+b) b=1 and a can hold any value. |

Re: 1 of toughest TSD question ever | |

Consider the speed of the stream to be zero. In that scenario, the swimmer would have reached the hat in 1 hour (half hour to go forward +half hour to move back to the position where he first saw the cork). But, because the hat was 1 km behind that point (reverting to the original case), it means that the hat covered 1km in 1 hour i.e. the speed of water= 1km/hr |

Re: 1 of toughest TSD question ever | |

let speed of swimmer = x, water = y time taken by the swimmer to meet the cork after turning back = t so yt = 1 if you check the distance, (1/2)(x-y) + y((1/2)+t) = (x+y)t solve and get y = 1 First question of IE Irodov..Brings back memories |