Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 23 (Q.No. 40)
40.
For a sleeper density of (n + 5), the number of sleepers required for constructing a broad gauge (BG) railway track of length 650 m is
Discussion:
3 comments Page 1 of 1.
Priyabrata Nayak said:
1 year ago
Simple trick:
Sleeper density, n+5=13+5=18
(n=standard length of rail=13m appx)
13m ---> 18 sleeper
1m ---> 18/13
650m --->18/13 * 650 = 900 sleeper (ans).
Sleeper density, n+5=13+5=18
(n=standard length of rail=13m appx)
13m ---> 18 sleeper
1m ---> 18/13
650m --->18/13 * 650 = 900 sleeper (ans).
Harkesh meena said:
6 years ago
Sleeper density =12.8 + 5 =17.8 almost 18.
No of rail's = 650 ÷ 12.8 = 50.78 almost 51.
So no of sleepers = 18 * 51 = 918.
No of rail's = 650 ÷ 12.8 = 50.78 almost 51.
So no of sleepers = 18 * 51 = 918.
Sadashiva said:
8 years ago
Sleeper density = (n+5).
Length of one rail for broad gauge track 12.8 say 13(I,e n),
Sleeper density=13+5=18,
The Number of sleepers required for 650m lenth=(18/13)* 650.
=899.999 say 900.
Length of one rail for broad gauge track 12.8 say 13(I,e n),
Sleeper density=13+5=18,
The Number of sleepers required for 650m lenth=(18/13)* 650.
=899.999 say 900.
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