Free Sample Questions

1. A flight plan for an area 10 miles wide & 15 miles long is required. The average terrain in the area is 1500ft above the mean sea level. The camera has a 6 inch focal length with 9×9inch format. End lap & side lap are to be 50% & 30% respectively. The required photo scale is 1: 12000. What is the distance between exposures?
A. 4000ft/photo
B. 4800ft/photo
C. 4500ft/photo
D. 4200ft/photo
ANS: C
EXP: Distance between exposures = (1-endlap) ×photo scale ×photo width.

2. Kinematic GPS survey technique requires at least two receivers. One receiver is placed on the known point and remains stationary, while the rover is moved from point to point. Where kinematic GPS survey technique is applicable?
A. Topographic surveys
B. Monographic survey
C. Chronographic surveys
D. Polytrophic surveys
ANS: A
EXP: Kinematic GPS survey technique is applicable for topographic surveys.

3. As a part of contract the distance (for example 1,000feet) is established as a free haul distance. This means the material hauled 1,000 feet or less is hauled free. What is meant by the term overhaul?
A. Total haul plus free haul.
B. Total haul minus free haul.
C. Total haul multiplied to free haul.
D. Total haul divided by free haul.
ANS: B
EXP: overhaul is the total haul minus the free haul.

4. Since the mass curve is plotted from excavation and embankment quantities, the cut and fill between points at which any horizontal line cuts off a loop of the mass curve will exactly balance, called balancing line. What is indicated by the loop below the balancing line?
A. Hauling from right to left.
B. Hauling from left to right.
C. Hauling from top to bottom.
D. Hauling from bottom to top.
ANS: A
EXP: The loop below the balancing line indicates hauling from right to left.

5. Consider an earthwork data having shrinkage factor is estimated to be about 25%. Determine the adjusted excavation (CY) corresponding to the station 5+00. Consider the excavation(CY) = 2000 corresponding to the station 5+00
A. 1250
B. 1500
C. 2000
D. 2500
ANS: B
EXP: Adjusted excavation = 2000 X .75 = 1500

6. Among different types of errors the random errors are the accidental errors which are very small and their algebraic signs are matters of chance. This random errors are also can be expressed by another term, what is that term?
A. Compensating errors.
B. Cumulative errors.
C. Instrumental errors.
D. Natural errors.
ANS: A
EXP: Random errors are also known as compensating errors.

7. Before packing of the machined products it is necessary to check the desired allowance and clearance of the product and to calculate the product error. If the rectangular lot size of the machined products is as like - 55.00’ ± 0.01’ x 110.00’ ± 0.02’ how much will be the error of the area if the area is represented in ft2?

A. ± 1.156 ft2
B. ± 1.56 ft2
C. ± 1.12 ft2
D. ± 0.1236 ft2
ANS: B
EXP: Eprod = ± √ (A2Eb2 +B2Ea2) = ±√ {(55)2 (0.02)2 + (110)2 (0.01)2} = ± 1.56 ft2.

8. Product error of a rectangular lot can be determined from the lot dimension and the desired dimension. If it becomes necessary to calculate the product error of the series Eseries for n number of measurements?
A. ± E √ (E n4)
B. ± E √ (n)
C. ± E √ (n2En)
D. ± E √ (n-2)
ANS: B
EXP: Error of a series E series = ± E √ (n).

9. Rhombic prism is a parallelopiped without any angles 90o and the volume of rhombic prism can be represented as Volume = Bh. How does “h” can be represented for the expression of volume of rhombic prism?
A. Horizontal distance between faces
B. Perpendicular distance between opposite faces
C. Central distance between two prisms
D. Tapered distance of the prism
ANS: B
EXP: h= Perpendicular distance between opposite faces.

10. A FRUSTRUM with top and base parallel have top face area T and base face area B. If h be the perpendicular distance of the top and base face of the FRUSTRUM how does its volume can be expressed?

A. Volume = 1/2 h [TB + hB + √(TB2)]
B. Volume = 1/3 h [T + B + √(TB)]
C. Volume = 4/3 h [T + B + √(T2)]
D. Volume = 1/4 h [T + B + √(TB)]

ANS: B
EXP: Volume of FRUSTRUMS with top and base parallel; Volume = 1/3 h [T + B + √ (TB)].

11. The length of each arm of a triangle BC, CA and AB are respectively a, b, and c and the included angles are A, B, C. If the value of the angle A and B and the length of the arm BC=a is given. How to determine the angle C and the length AB=c?
A. C= 180-(A+B); c = a (sin C)/(sin B)
B. C= 180-(A+B); c = a (sin B)/(sin A)
C. C= 180-(A+B); c = a (sin C)/(sin A)
D. C= 180-(A+B); c = b (sin C)/(sin A)

ANS: C
EXP: C= 180-(A+B); c = a (sin C)/(sin A).

12. The maximum length of line between connection may be increased to 100km double run for second order, class ll, double run third order is used where the first-order control has not been fully established. What is the maximum length can be increase for double run third order?
A. 75
B. 50
C. 100
D. 175
ANS: B
EXP: The maximum length of line between connection may be increased to 100km double run for second order, class ll, 50 km for double run third order in those areas where the first-order control has not been fully established.

13. General specification for vertical control of U.S. government recommended spacing of lines for national network. Which one of below is correct answer?
A. Net A: 100 to 300 for 1st Order
B. Net B: 75 to 100 for 1st Order
C. 10 to 25 for 3RD Order
D. 20 to 50 for 2ND Order Class 2
ANS: A
EXP: General specification for vertical control of U.S. government recommended spacing of lines for national network. Net A: 100 to 300 for 1st Order. Net B: 50 to 100 for 1st Order. 20 to 50 for 2ND Order Class 1 . 10 to 25 for 2ND Order Class 2.As needed for third Order.

14. Middle ordinate in vertical line curve expressed by “m” ; B.V.C and E.V.C. shows two end point of curve. Select the formula to measure the middle ordinate of vertical line curve.
A. m= ((EI B.V.C. + EI.E.V.C.)/2- EI.V)
B. m= ((EI B.V.C. + EI.E.V.C.)- EI.V)/2
C. m= ((EI B.V.C. + EI.E.V.C.)/2- EI.V)/2
D. m= ((EI B.V.C. -EI.E.V.C.)/2- EI.V)/2

ANS: C
EXP: Middle ordinate in vertical line curve measured from B.V.C and E.V.C. by this formula.

15. A rod reading at the top of a pipe is 13.20’, BS =4.73’, BM =195.35’ pipe = 42” RCP(OD), and with a thickness of 3”. There must be a 1.5’ minimum cover. What is the flow line elevation?
A. 181.38’
B. 183.63’
C. 179.06’
D. 179.84’
ANS: B
EXP: A rod reading at the top of a pipe is 13.20’, BS =4.73’, BM =195.35’ pipe = 42” RCP(OD), and with a thickness of 3”. There must be a 1.5’ minimum cover. The flow line elevation is 183.63’

16. You are given a profile sheet with a beginning station of 1+45, elevation 350.00 ft, and the following: (1)slope-0.30% to station 4+50 (2) slope+1.35% to station 10+45 (3) slope-10.8% to station 12+50. What is the elevation at station 12+50?
A. 335.98’
B. 336.98’
C. 334.98’
D. 335.00’
ANS: C
EXP: You are given a profile sheet with a beginning station of 1+45, elevation 350.00 ft, and the following: (1)slope-0.30% to station 4+50 (2) slope+1.35% to station 10+45 (3) slope-10.8% to station 12+50. The elevation at station 12+50 is 334.98’

17. A utility company will place 4 pipes of 8”, 12”, 16” and 24”(OD) with a 0.50” of wall thickness in a trench. The specifications call for a minimum clearance of 8” in between each pipe plus 3-feet of structural backfill compacted to 95% relative compaction over the top pipe. What is the flow line elevation of the bottom pipe, if the ground elevation at top of the trench is 255.00 feet?
A. 248.25’
B. 245.05’
C. 242.80’
D. 250.25’
ANS: B
EXP: A utility company will place 4 pipes of 8”, 12”, 16” and 24”(OD) with a 0.50” of wall thickness in a trench. The specifications call for a minimum clearance of 8” in between each pipe plus 3-feet of structural backfill compacted to 95% relative compaction over the top pipe. The flow line elevation of the bottom pipe is 245.05’, if the ground elevation at top of the trench is 255.00 feet

18. A transit (hi=5.47ft) at station K (elevation=871.20 ft) is slighted at station M with the following results: stadia interval=2.65ft, crosshair rod reading=9.33ft with a vertical angle of -4029’.The elevation of station of M is-
A. 844.50’
B. 848.75’
C. 846.69’
D. 842.80’
ANS: C
EXP: A transit (hi=5.47ft) at station K (elevation=871.20 ft) is slighted at station M with the following results: stadia interval=2.65ft, crosshair rod reading=9.33ft with a vertical angle of -4029’.The elevation of station of M is-846.69’

19. A reading of 5.86’ is obtained on station 3+00 that has an elevation of 125.05’. The instrument is now turned to TBM “R” and the following readings are obtained: vertical angle of -10o30’, horizontal distance of 85.00’, and a rod reading of 3.57. Determine the elevation of TBM “R”?
A. 110.00’
B. 100.00’
C. 120.00’
D. 111.60’
ANS: D
EXP: A reading of 5.86’ is obtained on station 3+00 that has an elevation of 125.05’. The instrument is now turned to TBM “R” and the following readings are obtained: vertical angle of -10o30’, horizontal distance of 85.00’, and a rod reading of 3.57. The elevation of TBM “R” is 111.60’

20. The magnetic bearing of an old survey line originally recorded as N7o25’ W is now N2o30’ E. What has been the change in magnetic declination.
A. 9o55’ W
B. 9o55’ E
C. 4o55’ W
D. N9o55’ W
ANS: A
EXP: The magnetic bearing of an old survey line originally recorded as N7o25’ W is now N2o30’ E. The change in magnetic declination is 9o55’ W

21. If the coordinates of point A are 1035 N, 1585 W and those of point B are 435 N, 900 W, the bearing of course AB is-
A. N41o12’50”W
B. S48o47’04”E
C. S41o12’50”E
D. S41o12’50”W
ANS: B
EXP: If the coordinates of point A are 1035 N, 1585 W and those of point B are 435 N, 900 W, the bearing of course AB is S48o47’04”E

22. A back sight with inverted telescope is sighted at a point that bears N 48o E from the transit. The telescope is plunged and a right deflection angle of 68o is turned. The azimuth of the new line measured from north is-
A. 285o
B. 286o
C. N74o W
D. S74o E
ANS: B
EXP: A back sight with inverted telescope is sighted at a point that bears N 48o E from the transit. The telescope is plunged and a right deflection angle of 68o is turned. The azimuth of the new line measured from north is 286o

23.  Based on the mass diagram shown below, determine the avg. overhaul. The distance involved in moving this material will range from 600´ to 1200´. 100 feet is equivalent to 1 station.

A.      300 feet or 3 stations.

B.      600 feet or 3 stations.

C.      600 feet or 6 stations.

D.      300 feet or 6 stations.

 ANS: A

EXP: Avg. haul distance = (1200+600)/2 = 900feet. Avg. overhaul = 900-600 = 300 feet or 3 stations

24. Rectangular plates are produced in an industry and to calculate the error of the product a lot is taken. If the symbol E_a and E_b  represents the typical error symbol how the error of the rectangular lot can be calculated?

A.     E_prod =  ± √( A²E_b³ +B²E_a³)

B.     E_prod =  ± √( AE_b +B²E_a)

C.     E_prod =  ± √( A²E_b² +B²E_a²)

D.     E_prod =  ± √( A²E_a² +B²E_b²)

ANS: C

EXP: Error of product for a rectangular lot is E_prod = ± √ (A²E_b² +B²E_a²).

25.                      This is the picture of a sphere. Where it is divided in to 7 portions. Each portion has its own area which can be determined by using some known equation. Determine the area of portion 6 where L₁+L₂=h=6?

A. 18²unit

B. 3  unit

C. 27 unit

D. 18 unit

ANS: D

EXP: A=0.5xh (L₁+L₂)

26.          The existing elevations of the front and back sides of the lot are shown below. During a 100 year flood, the river raises to an elevation of 178.00 feet. Determine the flooded area in square feet?

A.      20000

B.      18050

C.      16670

D.      20830

  ANS: D

  EXP: The flooded area in square feet is 20830

27.          Using a line perpendicular to AB through x, cut off one-third area to include corner B, and determine length xy in feet.

A.      175

B.      165

C.      185

D.      200

  ANS: A

  EXP: The length of xy in feet is 175

28.          The parcel below contains 18000 square feet. If cut-off line HG is parallel to EF, determine the length of HG in feet.

A.      265

B.      250

C.      280

D.      220

  ANS: A

  EXP: the length of HG in feet is 265