Civil Engineering - Estimating and Costing - Discussion

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

35. 

The reduced levels of points, 30 metres apart along the longitudinal section of a road portion between chainages 5 and 9 are shown in the given figure. If there is a uniform up-gradient of the road 120 in 1, the chainage of the point with no filling or cutting is

[A]. (6 + 15) chains
[B]. (6 + 12) chains
[C]. (6 + 18) chains
[D]. None of these.

Answer: Option B

Explanation:

No answer description available for this question.

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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