Room 206 JPD Building 1955 CM Recto Avenue, Manila
Telephone Number: (02) 516 7559 E-Mail: megareview_2008 @yahoo.com
REFRESHER MODULE 1 - SURVEYING
PROBLEM 1
A 0.65kg, 50m tape was standardized and supported throughout its whole
length and found to be 0.00205m longer at an observed temperature of
31.8°C and a pull of 10KN. This tape was used to measure a 4% grade line
which was found to be 662.702m. During measurement, the temperature is
15°C and the tape is suspended under a pull of 20KN. E=200GPa, crosssectional area of tape is 3mm2 and the coefficient of linear expansion is
0.0000116m/°C.
1. Compute the standard temperature.
a. 28.27°C
b. 30.24°C
c. 26.27°C
d. 32.24°C
2. Compute the total correction per tape length.
a. -0.2378
b. +0.2378
c. -0.2187
d. +0.2187
3. Compute the true horizontal distance.
a. 659.46m
b. 659.27m
c. 659.20m
d. 659.35m
PROBLEM 2
A line measures 7800m at elevation 900m. The average radius of the
curvature in the area is 6400km.
1. Compute the sea level distance.
a. 7768.90m
b. 7778.90m
c. 7788.90m
d. 7798.90m
2. Compute the reduction factor.
a. 0.99886
b. 0.99986
c. 0.99686
d. 0.99786
PROBLEM 3
A line 125 m. long was paced by a surveyor for four times with the
following data 161,165, 159 and 158. Then another line was paced for five
times with the following results, 520, 525, 524, 522 and 518.
1. Determine the pace factor.
a. 0.7756
b. 0.7776
c. 0.7796
d. 0.7806
2. Determine the number of paces for the new line.
a. 521.8
b. 519.8
c. 523.8
d. 524.8
3. Determine the distance of the new line.
a. 405.75m
b. 402.75m
c. 407.75m
d. 408.75m
PROBLEM 4
From the measured values of distance AB, the following data were
recorded.
DATA
DISTANCE
No. OF MEASUREMENTS
1
120.68
1
2
120.84
4
3
120.76
6
4
120.64
8
1. Find the probable weight of data 2.
a. 4
b. 6
2. Find the probable error of the mean.
a. +0.0106
b. +0.0126
3. Find the standard deviation.
a. +0.0856
b. +0.0836
4. Find the standard error.
a. +0.0187
b. +0.0167
5. Find the probable distance of AB.
a. 120.72
b. 120.92
c. 8
d. 1
c. +0.0146
d. +0.0166
c. +0.0816
d. +0.0796
c. +0.0147
d. +0.0137
c. 120.52
d. 120.32
PROBLEM 5
The following interior angles of a triangle traverse were measured with
the same precision.
ANGLE
VALUE (DEGREES)
No. OF MEASUREMENTS
A
39
5
B
65
4
C
75
3
1. Determine the most probable value of angle A.
a. 39° 15.32’
b. 39° 19.15’
c. 39° 25.53’
2. Determine the most probable value of angle B.
a. 65° 13.35’
b. 65° 15.32’
c. 65° 19.15’
3. Determine the most probable value of angle C.
a. 75° 19.15’
b. 75° 13.35’
c. 75° 25.53’
d. 39° 13.35’
d. 65° 25.53’
d. 75° 19.15’
PROBLEM 6
From the given data of a different leveling as shown in the tabulation:
STA.
1
2
3
4
5
6
7
B.S.
5.87
7.03
3.48
7.25
10.19
9.29
F.S.
ELEV.
392.25
6.29
6.25
7.08
5.57
4.45
4.94
1. Find the diff. in elevation of station 7 and station 5.
a. 10.09m
b. 8.09m
c. 12.09m
2. Find the height of instrument at station 4.
a. 400.26m
b. 396.26m
c. 392.26m
d. 14.09m
d. 386.26m
PROBLEM 7
The table shows the values of backsight, foresight & intermediate
foresight reading taken from BM1 to BM2. Elevation of BM1 = 328.70m.
STA.
BM1
1
2
3
4
TP1
5
6
7
TP2
8
9
10
11
BM2
B.S.
2.32
F.S.
I.F.S.
ELEVATION
328.70
1.7
2.2
1.2
0.9
2.77
3.43
2.2
3.7
1.6
2.22
3.06
2.8
3.6
2.0
1.1
2.45
1. Find the difference in elevation between stations 5 & 9.
a. 2.64m
b. 2.44m
c. 2.24m
2. Find the elevation of TP2.
a. 325.30m
b. 327.30m
c. 329.30m
d. 2.74m
d. 323.30m
PROBLEM 8
In a two peg – test of a dumpy level, the following observations were
taken.
Rod reading on A
Rod reading on B
Instrument at C
0.296
0.910
Instrument at D
1.563
2.140
Point C is equidistant from A and B. D is 2.5m from A and 72.5m from B.
1. What is the difference in elevation between A and B?
a. 0.614
b. 0.624
c. 0.634
d. 0.644
2. Determine the error in the rod reading at B with the instrument at D.
a. 0.0393
b. 0.0383
c. 0.0373
d. 0.0363
3. What is the corresponding rod reading on A for a horizontal line of sight
with instrument still at D.
a. 1.574
b. 1.564
c. 1.554
d. 1.544
PROBLEM 9
Considering the effects of curvature and refraction, the difference in
elevation of points B & C is found out to be 111.356m. From point A, the
angle of elevation of B is 18⁰30’ & that of C is 8⁰15’. A is between B & C.
1. If C is 2000m from A, how far is B from C?
a. 1.6km
b. 1.4km
c. 1.2km
d. 1.0km
2. Find the elevation of A if elevation of B is 450m.
a. 48.79m
b. 48.59m
c. 48.19m
d. 48.39m
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Room 206 JPD Building 1955 CM Recto Avenue, Manila
Telephone Number: (02) 516 7559 E-Mail: megareview_2008 @yahoo.com
REFRESHER MODULE 1 - SURVEYING
PROBLEM 10
Two hills A & C have elevations of 600m & 800m respectively. In between
A & C, is hill B which has an elevation of 705m. B is located 12km from A
& 10km from C. Determine the clearance or obstruction of the line of sight
if an observer is @ A so that C will be visible from A. Ans. 3.95m
a. 3.95m (obstruction)
b. 3.95m (clearance)
c. 3.75m (obstruction)
d. 3.75m (clearance)
PROBLEM 11
A field is in the form of a regular pentagon. The directions of the bounding
sides were surveyed with an assumed meridian 5⁰ to the right of the true
north & south meridian. As surveyed, the bearing of one side AB is N
33⁰20’ W.
1. Compute the true bearing of Line CD.
a. S65⁰20’E
b. S66⁰20’E
c. S64⁰20’E
d. S63⁰20’E
2. Compute the true azimuth of Line BC.
a. 220⁰40’
b. 221⁰40’
c. 222⁰40’
d. 223⁰40’
PROBLEM 12
The bearing of a line from A to B was measured as S16°30’W. It was found
that there was a local attraction at both A & B and therefore a forward and
a backward bearing were taken between A & a point C at which there was
no local attraction. If the bearing of AC was S30°10’E & that of CA was
N28°20’W, what is the corrected bearing of AB?
a. S18°20’W
b. S16°20’W
c. S17°20’W
d. S19°20’W
PROBLEM 13
From the given closed traversed shown.
LINES
BEARING
DISTANCES
A-B
44.37 m.
S. 35 30’ W
B-C
137.84 m.
N. 57 15’ W
C-D
12.83 m.
N. 1 45’ E
D-E
?
64.86 m.
E-A
?
106.72
1. Find the Bearing of line D-E.
a. N 66.17°E
b. N 68.17°E
c. N 70.17°E
d. N 72.17°E
2. Find the Bearing of line E-A.
a. S54.2°E
b. S52.2°E
c. S50.2°E
d. S48.2°E
PROBLEM 14
From the data below:
LINE
AB
BC
CD
DE
EA
LATITUDE
-36.13
+74.56
+12.82
+19.90
-68.40
1. Solve for area by Transit Rule.
a. 6753.29m2
b. 6726.62m2
2. Solve for area by Compass Rule.
a. 6753.29m2
b. 6726.62m2
DEPARTURE
-25.77
-115.93
+0.39
+61.74
+69.57
c. 6726.29m2
d. 6753.62m2
c. 6726.29m2
d. 6753.62m2
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