CE Board Problems in Surveying

MATHEMATICS SURVEYING CE LICENSURE EXAMINATION PROBLEMS SURVEYING

CORRECTIONS IN TAPING 5.

Using a 50-m tape that is 0.02 m too long, the measured distance from A to B is 160.42 m. What is the correct distance from A to B? (N94 M 25) a. 160.484 m c. 160.4 m b. 160.356 m d. 160.44 m

6.

The sides of a square lot having an area of 2.25 hectares were measured using a 100-m tape that was 0.04 m too short. Compute the error in the area in sq. m. (M95 M 25) a. 18.0036 c. 11.9984 b. 12.0016 d. 17.9964

7.

The distance from D to E, as measured, is 165.2 m. If the 50 m tape used is 0.01 m too short, what is the correct distance in meters? (M96 M 19) a. 165.299 c. 165.167 b. 165.365 d. 165.233

8.

A steel tape is 100 m long at a standard pull of 65 N. Compute the pull correction in mm if during measurement the applied pull is 40 N. The tape has a cross sectional area of 3.18 mm2 and a modulus of elasticity E = 200 GPa. (N96 M 29) a. -7 c. -6 b. -4 d. -5

9.

The correct distance between points E and F is 213.5 m. If a 100-m tape that is 0.025 m too long is used to measure EF, what will be the measured distance in meters? (N97 M 21) a. 212.765 c. 213.447 b. 212.659 d. 213.553

10.

When the temperature was 3 C, the distance from E to F was measured using a steel tape that has a standard length at 20 C with a coefficient of thermal expansion of 0.0000116 / C. If the correct distance from E to F is 836.5 m, what is the measured distance in meters? (M98 M 23) a. 835.675 c. 836.005 b. 836.335 d. 836.665

PACING 1.

2.

A surveying student had recorded the following after repeated pacing: First distance = 100 m No. of paces: 142, 145, 143, 146.5 Second distance = ? No. of paces: 893.5, 896, 891.5, 897 Find the second distance in meters. (N98 M 4) a. 620 c. 630 b. 650 d. 640 A student recorded the following number of paces after walking a distance of 50 m repeatedly as 71.5, 72.0, 70.0, and 69.5. He wanted to measure the distance between two points C and D. He recorded the following number of paces from C to B or back as 465, 468, 463, and 460. What is the distance from C to D? (N02 M 28) a. 328 m c. 462 m b. 378 m d. 421 m

CHAINING 3.

A line was measured to have 5 tallies, 6 marking pins, and 63.5 links. How long is the line? (M94 M 38) a. 1205.50 cm c. 5663.50 ft b. 1887.50 m d. 119.50 in

4.

A line was measured with a 50-m tape. There were 2 tallies and 3 pins, and the distance from the last pin and the end of the line was 2.25 m. Find the length of the line in meters. (M97 M 20) a. 1152.25 c. 117.25 b. 1017.25 d. 517.25

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MATHEMATICS SURVEYING 11.

A line was measured to be 412 m using a 30-m tape, which is of standard length at a temperature of 20 C. During measurement, the temperature was 52 C. If the coefficient of thermal expansion of the tape material is 0.0000116 m/m C, calculate the correct length of the line. (M99 M 22) a. 411.847 m c. 412.153 m b. 413.215 m d. 409.845 m

12.

Using a 25-m tape, a square lot was measured and found to have an area of 1 hectare. If the total error in area is 4.004 square meter short, what is the error in each tape length? (N01 M 9) a. 0.005 m too short c. 0.008 m too long b. 0.008 m too short d. 0.005 m too long

13.

A baseline measures 25 km at elevation 520 m. If the average radius of curvature is 6400 km, compute the sea-level distance. (M02 M 6) a. 24996.24 m c. 24997.97 m b. 24998.63 m d. 24995.24 m

16.

With the transit at point A and line of sight horizontal, the stadia intercept at B is 0.6 m. If the stadia interval factor is 99.96 and the stadia constant is 0.3, find the distance AB. (M01 M 18) a. 54.21 m c. 78.32 m b. 43.87 m d. 60.28 m

17.

The length intercepted on the stadia rod is 1.8 m and the line of sight makes an angle of 430’ with the horizontal. Find the horizontal distance, in meters, from the center of the instrument to the rod, if the stadia constant is 0.3 m and the stadia interval factor is 100. (M02 M 25) a. 172.43 m c. 187.36 m b. 179.74 m d. 196.87 m

DIFFERENTIAL LEVELING 18.

DISTANCE BY TACHYMETRY (STADIA METHOD) 14.

With the transit at point A and the line of sight horizontal, the stadia intercept at B was found to be 1.94 m. If the stadia constant is 0.3 and AB = 194.2 m, find the stadia interval factor. (M98 M 25) a. 99.958 c. 99.968 b. 99.938 d. 99.948

15.

The length intercepted on the stadia rod is 3.6 m and the line of sight makes an angle of 315’ with the horizontal. Find the distance, in meters, from the center of the instrument to the rod, if the stadia constant is 0.3 m and the stadia interval factor is 100. (N98 M 17) a. 363.33 c. 357.62 b. 361.28 d. 359.14

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Using the following 22) Station BM12 1 2 BM13 3 4 5 6 BM14 a. 255.45 b. 245.02

notes, what is the elevation of BM14? (N95 M BS 4.64 5.80 2.25 6.02 8.96 8.06 9.45 12.32

FS

Elev. 209.65

5.06 5.02 5.85 4.94 3.22 3.71 2.02 1.98 c. d.

225.05 235.35

MATHEMATICS SURVEYING 19.

20.

21.

The following notes were taken during a differential leveling. What is the difference in elevation between BM1 and BM2? (M96 M 24) Station BS FS Elev. BM1 7.11 751.05 1 8.83 1.24 2 11.72 1.11 BM2 10.21 a. 17.7 c. 15.1 b. 18.2 d. 16.4 Based on the following leveling notes, find the elevation of Sta 5 in meters. (M97 M 8) Station BS FS Elev. 1 8.26 458.45 m 2 9.98 2.39 3 12.87 2.26 4 8.65 11.36 5 5.32 a. 476.88 c. 478.82 b. 479.26 d. 477.28 Determine the difference between the elevations of Sta 6 and Sta 5, in meters, using the following notes: (N97 M 15) Station BS FS Elev. 1 4.90 463.8 m 2 6.06 5.32 3 2.51 5.28 4 6.28 6.11 5 9.22 4.60 6 3.48 a. 5.74 c. 5.53 b. 5.47 d. 5.66

LEVELING ADJUSTMENTS 22.

To make a peg adjustment, the following notes were taken: Rod reading at P Rod reading at Q

Wye level at 1 0.75 2.766

Wye level at 2 1.906 3.798

Point 1 is on the line PQ and midway between P and Q. Point 2 is on the same line as P and Q but not between them. Point 2 is 25 m from P and 230 m from Q. With the wye level at point 2, what is the rod reading at P for a level sight? (N94 M 26) a. 1.921 c. 1.98 b. 1.962 d. 1.954 SENSITIVITY OF A BUBBLE 23.

With the use of an engineer’s level, the reading on a rod 80 m away was found to be 2.82 m. The bubble was leveled through 5 spaces on the level tube and the rod reading increased to 2.884 m. What is the radius of curvature of the level tube if one space on the tube is 0.6 mm long? (N02 M 24) a. 3.75 m c. 3.35 m b. 4m d. 3.50 m

EARTH CURVATURE AND REFRACTION 24.

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Point A is between points B and C. The distances of B and C from point A are 1000 m and 2000 m, respectively. Measured from point A, the angle of elevation of point B is 1830’, while that of point C is . The difference in the elevations of B and C is 44.4 m, with C being lower than B. Considering the effects of curvature and refraction, the value of  is nearest to: (M95 M 26) a. 730’ c. 815’ b. 645’ d. 520’

MATHEMATICS SURVEYING 25.

26.

The top of a tower signal at B 2000 m from A away was sighted through a transit with recorded vertical angle of 230’. The height of the mast is 12 m and the HI of the transit above the point where it is set is 1.10 m. The elevation of the point under the transit A is 133.3 m. Compute the elevation of the base of the signal B. (N96 M 26) a. 225 c. 215 b. 220 d. 210 Point A is between points B and C. The distances of B and C from point A are 1000 m and 2000 m, respectively. Measured from point A, the angle of elevation of point B is 1830’, while that of point C is 815’. Find the difference in the elevations of B and C. Consider the effects of curvature and refraction. (N00 M 2) a. 44.4 m c. 48.7 m b. 32.6 m d. 52.1 m

29.

DECLINATION 30.

If the polar distance of a star is 230’, what is its declination? (M94 M 26) a. 17730’ c. 9230’ b. 8730’ d. 6230’

31.

A line has a magnetic bearing of S4130’E when the declination was 130’E. What is the true bearing of the line if a local attraction is 330’ to the east of the vicinity? (M94 M 39) a. S 3530’ E c. S 45 E b. S 45 E d. S 4330’ E

AREA BY TRAPEZOIDAL RULE 27.

Find the area of a piece of land with an irregular boundary as follows: Station Offset Distance (m) 0 + 000 5.59 0 + 010 3.38 0 + 020 2.30 0 + 030 3.96 0 + 040 4.80 The stations are on straight-line boundary. Find the area of the land in m2 by Trapezoidal Rule. (M96 M 2) a. 138.4 c. 118.5 b. 128.5 d. 148.4

AREA BY TRIANGULATION 32.

ANGLES AND DIRECTIONS 28.

b. 322 d. 218 Lot ABCDEFA is a closed traverse in the form of a regular hexagon with each side equal to 100 m. The bearing of AB is N25E. What is the bearing of CD? (M03 M 26) a. S 35 E c. S 30 E b. S 45 E d. S 40 E

The forward azimuth of a line is known to be 52. What is its back azimuth? (N99 M 26) a. 148 c. 232 ------- 4 -------

To determine the area of a triangular lot ABC, a surveyor set-up a transit at point P inside the lot and recorded the following bearing and distances of each corners of the lot from P. Corner Bearing from P Distance from P A N 4732’ W 36.25 m B N 6852’ E 48.32 m C Due South 65.25 m Determine the area of the lot in square meter. (N99 M 20) a. 3256 c. 4253 b. 3127 d. 2586

MATHEMATICS SURVEYING

CLOSED TRAVERSE (MISSING DATA) 33.

34.

35.

A closed traverse has the following data: Course Bearing Distance (m) 1-2 N 9.27 E 58.70 2-3 S 88.43 E 27.30 3-4 4-5 S 5.30 E 35.00 5-1 S 72.07 W 78.96 What is the length of course 3-4? (N94 M 27) a. 39.3 m c. 38.65 m b. 37.5 m d. 35.2 m A closed traverse has the following data: Course Bearing Distance (m) 1-2 N 9.27 E 58.70 2-3 S 88.43 E 27.30 3-4 N 86.78 E 35.20 4-5 S 5.30 E 35.00 5-1 What is the bearing of line 5-1? (M95 M 27) a. S 78.31 W c. S 85.16 W b. S 76.05 W d. S 72.07 W

36.

From the given data of a closed traverse, compute the bearing of line 3-4. (N96 M 8) Line Bearing Distance (m) 1-2 N 58 E 80 2-3 Due N 50 3-4 4-1 S 36.74 E 89.8 a. N 60 E c. N 64.3 E b. N 55.60 E d. S 80.4 W

37.

A closed traverse has the following data: Line Distance (m) Bearing AB 64.86 N 7210’ E BC 107.72 S 4813’ E CD 44.37 S 3530’ W DE 137.84 EA 12.83 Find the bearing of line DE. (N97 M 26) a. N 5715’ W c. N 5655’ W b. N 5944’ W d. N 5825’ W

38.

A closed traverse has the following data: Line Bearing Distance (m) AB 44.47 BC 137.84 CD N 145’ E 12.83 DE N 7210’ E 64.86 EA S 4813’ E 107.72 Find the bearing of line AB. (M98 M 17) a. S 3718’ W c. S 3834’ W b. S 3530’ W d. S 3646’ W

A closed traverse has the following data: Course Bearing Distance (m) AB S 1536’ W 24.22 BC S 6911’ E 15.92 CD N 5758’ E DA S 8043’ W Find distance DA in meters? (N95 M 24) a. 77 c. 75 b. 79 d. 73

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MATHEMATICS SURVEYING

39.

40.

A closed traverse has the following data: Line Bearing Distance (m) AB 60.00 BC 72.69 CD S 1720’ E 44.83 DE S 7036’ W 56.45 EA N 7430’ W 50.00 Find the bearing of line BC. (M00 M 24) a. S 4552’ E c. S 8223’ E b. N 1811’ E d. N 1530’ E

TRANSIT RULE

A closed traverse has the following data: Line Bearing Distance (m) AB N 86.78 E 35.20 BC S 5.30 E 35.00 CD S 72.07 W 78.95 DE N 9.27 E 58.70 EA Find the length of side EA. (M01 M 28) a. 27.3 m c. 54.9 m b. 29.7 m d. 31.6 m

42.

A closed traverse has the following data: Line Distance Bearing AB 895 S 7029’ E BC 315 S 2628’ E CD 875 S 6533’ W DE 410 N 4531’ W EA 650 N 1000’ E Determine the correct bearing of line EA using the transit rule. (M97 M 19) a. N 944’34” E c. N 950’28” E b. N 947’21” E d. N 953’1” E

43.

A closed traverse has the following data: Line Distance Bearing AB 895 S 7029’ E BC 315 S 2628’ E CD 875 S 6533’ W DE 410 N 4531’ W EA 650 N 1000’ E Find the corrected bearing of line CD by transit rule. (N98 M 12) a. S 6544’12” W c. S 6540’18” W b. S 6542’33” W d. S 6548’29” W

44.

A closed traverse has the following data: Line Latitude 1-2 +9.15 2-3 -8.41 3-4 -24.15 4-5 +6.21 5-1 +17.10 Using the transit rule, determine the corrected latitude of line 3-4. (N99 M 27) a. -24.113 c. -24.214

SUBDIVISION OF LOTS 41.

A closed triangular traverse has the following data: Line Bearing Distance (m) AB N 60 E 1000 BC Due South CA N 60 W An area of 280,000 square meter is cut-off starting from corner A to point F on line BC. What is the length of line AF? (N03 M 29) a. 878.35 m c. 863.14 m b. 893.25 m d. 914.75 m

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MATHEMATICS SURVEYING b.

-24.187

d.

-24.044 a. b.

C 40.35 40.16

63

2 c. d.

40.57 40.77

PROBABLE VALUES ERROR IN TRANSIT READING 45.

The observed interior angles of a triangle and their corresponding number of observations are as follows: Corner Angle No. of observations 1 39 3 2 65 4 3 75 2 Determine the corrected angle at corner 1. (M99 M 29) a. 3828’12.2” c. 3918’27.7” b. 4032’14” d. 3936’32”

46.

The observed interior angles of a triangular piece of land ABC are as follows: A = 3514’37”, B = 9630’09”, and C = 4815’05”. The most probable value of angle B is nearest to: (M00 M 23) a. 9630’15” c. 9630’06” b. 9630’03” d. 9630’12”

47.

The observed interior angles of a triangle and their corresponding number of observations are as follows: Corner Angle No. of observations A 41 5 B 65 6 C 75 2 Determine the most probable value of angle C. (N01 M 16) a. 7534’21” c. 7448’56” b. 7354’32” d. 7425’23”

48.

49.

The horizontal axis of a transit was inclined at 4’ with the horizontal due to non-adjustment. The first sight had a vertical angle of 50, the next has -30. Determine the error in the measured horizontal angle. (M03 M 29) a. 7’4.6” c. 8’4.6” b. 7’9.2” d. 8’9.2”

HYDROGRAPHIC SURVEY 50.

In a hydrographic survey using a current meter with meter constant a = 0.232 and b = 0.022, it is required to determine the velocity of the water at that point if the time of observation was recorded to last for 50 seconds and the recorded number of revolution is taken as 10. (M94 M 42) a. 0.0575 m/s c. 0.0247 m/s b. 0.0346 m/s d. 0.0684 m/s

51.

The area bounded by the waterline of a reservoir and the contours at an interval of 2 m are as follows: A1 = 10,250 m2, A2 = 8,350 m2, A3 = 7,750 m2, A4 = 6,900 m2, and A5 = 5,250 m2. Calculate the volume of the reservoir by end-area method. (M99 M 24) a. 54,250 m3 c. 61,500 m3 b. 72,450 m3 d. 84,550 m3

52.

In a hydrographic survey, a staff gage reading of 8.15 m was observed at the instant the depth of the sounding was 17.6 m. The zero mark of the staff gage is at elevation 148.2 m. Find the elevation of the point where the sounding was made. (M01 M 21) a. 139.25 m c. 136.25 m b. 138.75 m d. 137.75 m

The following interior angles of a triangular traverse were measured with the same precision. What is the most probable value of angle A, in degrees? (M03 M 27) Angle Value (degrees) No. of observations A 41 5 B 77 6 ------- 7 -------

MATHEMATICS SURVEYING 58.

53.

Given the following areas bounded by the waterline of a lake and the contours 1, 2, 3, 4, and 5. Each contour is at 2 m interval. A1 = 6,150 m2 A4 = 4,140 m2 2 A2 = 5,010 m A5 = 3,150 m2 2 A3 = 4,650 m Determine the volume of water in the lake. (N02 M 26) a. 35,500 m3 c. 37,800 m3 b. 36,800 m3 d. 38,900 m3

59.

60.

Find the radius of a simple curve having a degree of curve of 5 using chord basis. (N96 M 18) a. 229.26 m c. 114.74 m b. 142.3 m d. 201.5 m

61.

A circular curve has the following data: Azimuth of back tangent = 205 Azimuth of forward tangent = 262 Middle ordinate, M = 5.8 m Find the length of the tangent, in meters. (M97 M 28) a. 23.77 c. 25.99 b. 24.88 d. 22.83

62.

From point A on a simple curve, the perpendicular distance to the tangent, at point Q, is 64 m. The tangent passes through the PC. The distance from Q to PC is 260 m. Find the radius of the curve, in meters. (M98 M 4) a. 580 c. 560 b. 540 d. 520

63.

From the PC, the deflection angles of two intermediate points A and B of a simple curve are 315’ and 815’, respectively. The chord distance between A and B is 40 m long. Find the length of the curve from the PC to B, in meters. (N98 M 1) a. 74.3 c. 70.5 b. 66.1 d. 62.4

64.

A 4-degree simple curve has an angle of intersection of 24. Find the length of the long chord. Use arc basis. (M99 M 20)

IDENTITIES 54.

If coversed sin  is 0.134, find the value of . (M94 M 10) a. 60 c. 45 b. 20 d. 30

55.

If versed sin  is 0.148, what is the value of ? (N03 M 4) a. 31.57 c. 24.78 b. 58.43 d. 11.24

SIMPLE CURVES 56.

57.

The tangents of a simple curve have bearings of N7512’E and S7836’E, respectively. What is the central angle of the curve? (N94 M 28) a. 38.2 c. 29.2 b. 26.2 d. 27.2 A 3-degree curve has an angle of intersection of 24. What is the length of the long chord in meters? Use chord basis. (M95 M 7) a. 158.85 c. 171.28 b. 183.42 d. 162.46

What is the angle of intersection of the two tangents of a simple curve if their bearings are N7512’E and S7836’E, respectively? (N95 M 23) a. 2612’ c. 324’ b. 1842’ d. 2238’ A 3-degree curve has an external distance of 8.53 m. What is the central angle? Use chord basis. (M96 M 18) a. 23 c. 25 b. 24 d. 22

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MATHEMATICS SURVEYING a. b.

65.

66.

67.

68.

69.

119.13 m 125.10 m

c. d.

112.21 m 132.56 m

70.

From point A on a simple curve, the perpendicular distance to the tangent, at point Q, is 64 m. The tangent passes through the PC. The distance from Q to PC is 260 m. Find the length of the curve from PC to A, in meters. (M00 M 20) a. 270.37 c. 298.56 b. 352.47 d. 254.12 From a point A on a simple curve, the perpendicular distance to the tangent at point Q is x. The tangent passes through PC. Point A is at station 20+250 and PC is at station 20+150. If the radius of the curve is 800 m, find x. (N00 M 18) a. 6.24 m c. 7.89 m b. 25.47 m d. 62.32 m The offset distance from PC to PT of a simple curve is 8 m. If the angle of intersection of the curve is 20, what is the radius of the curve? (M01 M 13) a. 123.87 m c. 153.98 m b. 132.65 m d. 113.62 m Determine the central angle of a 350-m simple curve if the nearest distance from the curve to the point of intersection of the tangents is 18 m. (N01 M 1) a. 32 c. 36 b. 39 d. 56 The tangent of a simple curve from PI to PC has a bearing of N65E and the other tangent from PI to PT has a bearing of N55W. At a point 150 m from PC along the tangent through PC, the right angle offset to point F on the curve is 6.2 m. Find the radius of the curve. (N02 M 23) a. 1657.5 m c. 1752.3 m b. 1547.2 m d. 1817.6 m

What is the central angle in degrees of the 200 m and the distance of the midpoint of 14.20 m? (M03 M 28) a. 48 c. b. 46 d.

curve whose radius is the curve to the PI is 42 44

COMPOUND CURVES 71.

Station PT of a compound curve is at 15+480.14, I1 = 30, I2 = 36, D1 = 4 and D2 = 5. What is the stationing of PCC? Use arc basis. (N94 M 29) a. 15+319.63 c. 15+325.28 b. 15+336.14 d. 15+342.5

72.

A compound curve has the following data: I1 = 28, I2 = 31, D1 = 3, D2 = 4, and Sta PI = 30+120.5. Find the stationing of PCC. Use Sta PI = 30+120.5. (M96 M 1) a. 30+110.73 c. 30+114.88 b. 30+118.46 d. 30+106.97

73.

The elements of a compound curve are as follows: I1 = 18, I2 = 23.5, D1 = 2, D2 = 4, and Sta PI = 42+89.6. Find the stationing of PC. Use arc basis. (N97 M 5) a. 41+908.38 c. 41+917.23 b. 41+934.92 d. 41+925.19

74.

A compound curve has the following properties: I1 = 32 Length of long chord from PC to PCC, L1 = 235.98 m I2 = 24 Length of long chord from PCC to PT, L2 = 178.23 m Find the length of chord from PC to PT. (M01 M 19) a. 402.15 m c. 476.45 b. 376.54 m d. 234.76

75.

The length of the common tangent of a compound curve is 321 m. D1 = 2.5, I1 = 36, I2 = 62. Find the degree of the second curve. Use arc basis. (M02 M 14) a. 3 c. 3.5

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MATHEMATICS SURVEYING b.

76.

2

d.

Stationing of PC = 67+345.23 Determine the stationing of PRC (point of reversed curvature). Use arc basis. (N01 M 2) a. 67+505.23 c. 67+467.23 b. 67+549.23 d. 67+689.23

4

A compound curve has a common tangent 520 m long. The first curve passing through the PC is a 3-degree curve with a central angle of 50. Find the radius of the second curve if its central angle is 35. (M03 M 22) a. 1.084.3 c. 1,143.7 b. 1,265.2 d. 1,304.3

80.

A reversed curve of equal radii connects two parallel tangents 12 m apart. The length of chord from PC to PT is 140 m. Determine the total length of the reversed curve. (N02 M 27) a. 142.85 m c. 140.17 m b. 145.1 m d. 146.7 m

REVERSED CURVES

SPIRAL CURVES

77.

81.

78.

79.

The common tangent BC of a reversed curve is 280.5 m and has a bearing of S4731’E. AB is the tangent of the first curve whose bearing is N7245’E. CD is a tangent of the second curve whose bearing is N3813’E. A is at the PC while D is at the PT. The radius of the first curve is 180 m. The PI is at Sta 12+523.37. Find the stationing of the PT. (M95 M 8) a. 12+883.65 c. 12+889.54 b. 12+893.24 d. 12+878.16 The common tangent of a reversed curve is 280.5 m and has an azimuth of 31229’. BC is a tangent of the first curve whose azimuth is 25245’. DE is a tangent of the second curve whose azimuth is 21813’. The radius of the first curve is 180 m. PI is at Sta 16+523.37. B is at PI. What is the stationing of PI2? (N95 M 20) a. 16+754.8 c. 16+774.8 b. 16+764.8 d. 16+784.8

A spiral easement curve has a length of 100 m with a central curve having a radius of 300 m. Determine the offset distance from the tangent to the second-quarter point of the spiral. (M02 M 26) a. 0.69 m c. 6.77 m b. 1.99 m d. 3.58 m

VERTICAL CURVES 82.

A parabolic curve AB, 400 m long is connected by tangents having an upgrade of +6.5% and a downgrade of -3% intersecting at Sta 20+800 at elevation 102.5 m. Find the distance from B to the highest point of the curve. (N96 M 17) a. 125.6 c. 126.3 b. 123.5 d. 124.8

83.

A grade of 6.5% meets a grade of -3% at Sta 10+800 whose elevation is at 1560 m. A parabolic curve AB, 400 m long connects the gradelines with A on the back tangent. Find the stationing of A. (M97 M 11) a. 10+300 c. 10+500 b. 10+600 d. 10+400

A reversed curve has the following properties: Degree of curve, D1 = 3 Central angle, I1 = 24 Degree of curve, D2 = 4 Central angle, I2 = 38 ------- 10 -------

MATHEMATICS SURVEYING 84.

A vertical sag curve has tangent grades of -3.5% and +4.6% meeting at point A whose elevation is 67 m. If the length of the curve is 440 m, find the elevation of the PC. (N99 M 18) a. 75.2 m c. 59.3 m b. 74.7 m d. 73.2 m

85.

A vertical parabolic sag curve has tangent grades of -1.2% and +0.6%. If the grade changes uniformly at 0.18% per 20 m, find the length of the curve. (M00 M 22) a. 200 m c. 300 m b. 150 m d. 250 m

86.

A parabolic curve AB 400 m long connects two tangent grades of +6.5% and -3%. If the elevation of the summit is 123.256 m, what is the elevation of point B? (N00 M 21) a. 122.111 m c. 121.361 m b. 119.625 m d. 120.542 m

a. b.

1994.26 2134.65

c. d.

2113.32 1973.62

89.

The area in cut of two irregular sections 35 m apart are 34 m2 and 56 m2, respectively. The base width is 10 m and the side slope is 1:1. Find the corrected volume of cut in m3, using the prismoidal correction formula. (N95 M 21) a. 1564.84 c. 1458.45 b. 1386.84 d. 1249.26

90.

Find the area of the given cross-section if the width of roadway is 10 m. (N96 M 21) Left Center Right 9.8 0 11.2 3.2 2.80 4.21 a. 47.925 c. 65.35 b. 59.421 d. 53.625

91.

Two irregular sections 80 m apart have areas in cut of 26 m2 and 84 m2. The base width is 8 m and the side slope is 1:1. Find the corrected volume of cut in m3 between the two stations using the prismoidal correction formula. (N97 M 12) a. 4,252.2 c. 4,241.5 b. 4,234.8 d. 4,263.4

92.

The areas in cut of two irregular sections 70 m apart are 26 m2 and 84 m2, respectively. Base width = 8 m. Side slope is 1:1. Using the prismoidal correction formula, find the corrected volume of cut, in m3, between the two stations. (M98 M 9) a. 3,715.2 c. 3,710.5 b. 3,705.4 d. 3,720.8

93.

Find the corrected volume of cut in m3 between two stations 60 m apart if the areas of the irregular sections in cut at the stations are 32 m2 and 68 m2, respectively. Base width = 8 m. Side slope is 1:1. Use the prismoidal correction formula. (N98 M 7) a. 2,937.6 c. 2,948.8 b. 2,919.4 d. 2,936.8

EARTHWORKS 87.

88.

The cross-section notes shown below are for a ground excavation for a 10-m wide roadway. Sta 16+100 8.2 L 0 13.65 R +2.15 +3.5 +5.8 What is the cross-sectional area, in m2, at Sta 16+100? (N94 M 30) a. 64.1125 c. 58.1125 b. 46.1125 d. 52.1125 The cross-section notes shown below are for a ground excavation for a 10-m wide roadway. Sta 25+100 7.85 L 0 8.45 R +1.90 +3.20 +2.30 Sta 25+150 9.35 L 0 10.37 R +2.90 +2.60 +3.80 What is the volume of excavation, in m3, between the two stations? Use the prismoidal formula. (M95 M 9)

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MATHEMATICS SURVEYING

94.

The longitudinal ground profile and the gradeline shows that the length of cut is 560 m while the length of fill is 740 m. The width of the roadbed is 10 m for both cut and fill. The profile areas between the groundline and the gradeline, which are parallel, are 4,800 m2 for cut and 5,829 m2 for fill. Find the difference between the volume of fill and the volume of cut in m3 if the side slopes are 1.5:1 for cut and 2:1 for fill. (M01 M 29) a. 43,768.2 m3 c. 40,413.7 m3 3 b. 35,783.2 m d. 56,298.2 m3

95.

Given the following cross-section notes of an earthwork on an 8-m wide roadway: 6.70 0 4.75 +1.80 +1.40 +1.50 Determine the area of the cross-section. (N02 M 25) a. 12.208 m2 c. 11.428 m2 2 b. 11.837 m d. 12.615 m2

96.

Given a side slope of 2:1, a road width of 10 m and a crosssectional area of 31.7 m2, find the value of x in the following crosssection notes: (M03 M 24) 9.8 0 7.4 +2.4 x +1.2 a. +2.93 m c. +2.51 m b. +2.71 m d. +2.64 m

97.

The cross-sectional area of a road with width of 10 m is 33.1 m2. The cross-sectional area is as follows: 9.8 0 7.4 2.4 x 1.2 Determine the value of x. (N03 M 28) a. 3.2 m c. 2.4 m b. 2.8 m d. 1.6 m

GRADE OF FINISHED ROADWAY 98.

The ground makes a uniform slope of 4.8% from Sta 12+180 to Sta 12+240. At Sta 12+180, the center height of the roadway is 1.2 m fill. At the other station, the center height is 2.5 m cut. Find the grade of the finished road. (M97 M 4) a. -4.15 % c. -1.37 % b. -2.48 % d. -3.26 %

99.

From Station A with center height of 1.4 m in fill, the ground makes a uniform slope of 5% to Station B whose center height is 2.8 m in cut. Assuming both sections to be level sections having a width of roadway of 10 m and side slope of 2:1 for both cut and fill, compute the cross-sectional area of fill 8 m from Station A. Distance from Station A to Station B is 60 m. (M99 M 23) a. 12.54 m2 c. 15.81 m2 2 b. 8.41 m d. 9.81 m2

100.

The ground makes a uniform slope of 4.8% from Sta 12+180 to Sta 12+240. At Sta 12+180, the center height of the roadway is 1.2 m fill. At the other station, the center height is 2.5 m cut. Find the length of cut in meters. (M00 M 21) a. 30.85 c. 46.32 b. 40.54 d. 50.28

101.

The ground makes a uniform slope of 4.8% from Sta 12+180 to Sta 12+250. At Sta 12+180, the center height of the roadway is 1.2 m fill. At the other station, the center height is 3.11 m cut. Find the grade of the finished road. (N00 M 19) a. -4.214 % c. -1.427 % b. -2.149 % d. -1.357 %

102.

The ground makes a uniform slope of 5% from Sta 12+180 to Sta 12+240. At Sta 12+180, the center height of the roadway is 1.4 m fill. At the other station, the center height is 2.8 m cut. Find the grade of the finished road. (N01 M 15)

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MATHEMATICS SURVEYING a. b. 103.

-2 % -1.6 %

c. d.

-2.7 % -3 %

From Sta 12+180 with center height of 1.4 m in fill, the ground makes a uniform slope of 5% to Sta 12+240 whose center height is 2.8 m in cut. How far is the point of intersection of the ground and the road from Sta 12+180? (M02 M 13) a. 25 m c. 15 m b. 20 m d. 30 m

BORROW PITS 104.

The volume of a borrow pit 90 m by 90 m is to be determined by unit area method. The area is divided by equal square areas 30 m by 30 m. The height of earth at each unit area are as follows: Row 1 3.2 m 2.8 m 3.6 m 4.2 m Row 2 3.8 2.5 3.2 3.8 Row 3 3.1 2.9 3.3 3.0 Row 4 3.0 2.5 2.8 2.6 Find the volume of the borrow pit in cubic meters. (N99 M 17) a. 28,045 c. 24,515 b. 22,045 d. 25,065

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