### Discussion :: Estimating and Costing - Section 1 (Q.No.35)

Saikumar said: (Sep 17, 2015) | |

How it is calculated? |

Prabhakar said: (Sep 27, 2015) | |

How it is taken? |

Gance said: (Dec 25, 2015) | |

1 in 120. So for 30 m there is rise of 0.25 m. R.L of formation level at 6 chain = 100+0.25 = 100.25. Now apply the similar triangles as 1/x = 1.5/(30-X). Where, 1 = 100.25-99.25. x = 12. So the answer B. |

Ashik said: (Aug 11, 2016) | |

Can you explain it? Please. |

Kamruz said: (Jan 6, 2017) | |

I think it is A. |

Asmaul Husna said: (May 9, 2017) | |

@Gance. Where is 1.5 from? |

Karthick said: (Jun 24, 2017) | |

6+ 12 chain how? |

Vishnu said: (Jul 11, 2017) | |

Can anyone explain clearly? |

Chandra said: (Nov 28, 2017) | |

How this will occurs? |

Maya said: (Dec 20, 2017) | |

@6 difference b/w level is (100.25-99.25)=1. @7 RL=100.25+0.25=100.5. So difference= 102-100.5=1.5, let x be distance at no filling or cutting, distance b/w 7 and 6 is 30, By similar triangle, 1 -> x, 1.5 ->30-x. |

Liam Payne said: (Jan 10, 2018) | |

The number of neutrons will be 6 so the total number becomes 6 + 6 is equal to 12. |

Mainak said: (Jun 6, 2018) | |

99.25+(2.75/30)*x = 100+(x+30)/120. By solving it x=12m from chainage 6 towards ch. 7, so answer will be (6+12). |

Abhishek said: (Dec 13, 2018) | |

It is (6+8) chains. |

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