Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 14 (Q.No. 22)
22.
The standard time meridian in India is 82°30'E. If the standard time at any instant is 20 hours 10 minutes, the local mean time for the place at a longitude of 20°E would be
4h PM
4h 10m PM
lh 20m PM
0h 20m AM
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Sofiya Fayaz said:   3 years ago
82°30'-20°= 62°30' now converting...
(62°30'/360)°-24= 4°10'= 4h 10'
Now, 20h 10'- 4h 10'= 16h.
And 16h is 4pm.
(4)

Tawseef Nabi said:   2 years ago
82°30' - 20° = 62° 30'.

62*4 = 248 min
And for 30' = 0.5° *4 = 2.0 min
Total time we get = 248 + 2 = 250 min
250/60 = 4 hours and 10 min
Given standard time 20h 10 min
Therefore:- 20h 10 min - 4h 10 min = 16hours, Which is at 4 pm as per local mean time.
(4)

Kaushik said:   3 years ago
But 4 pm is standard time at 20 degrees E.

We have asked for local time at 20 degrees E.
(3)

Pitambar Jena said:   3 years ago
@Phani.

How 62*4=240? It's 248.
(1)

Phani said:   7 years ago
82 30 - 20 = 62 30.

62*4 = 240,
30*4 = 120,
So add 250 which is equal to 4h 10m,
Now 20h 10m - 4h 10m = 16h,
Regarding to question it is 24hrs format.
So 16h - 12h = 4h which is A.

SONAM CHERING said:   4 years ago
@Phani.

How, 250 is equal to 4h 10m? Explain.

Mir said:   3 years ago
62° x 4 min = 248min.
Half degree). 5x4 =2 min.
Add 248+2 = 250 min.
250 min = 4 hr 10 min (4x60=240 and 10 min).

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