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 = (1endlap) ×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 √ (n2)
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 firstorder 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 firstorder 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)slope0.30% to station 4+50
(2) slope+1.35% to station 10+45 (3) slope10.8% 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)slope0.30% to station 4+50
(2) slope+1.35% to station 10+45 (3) slope10.8% 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 3feet 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 3feet 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 is846.69’
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 = 900600 = 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
}=
± √( A^{2}E_{b}^{3
}+B^{2}E_{a}^{3})
B.
E_{prod
}=
± √( AE_{b}^{
}+B^{2}E_{a})
C.
E_{prod
}=
± √( A^{2}E_{b}^{2
}+B^{2}E_{a}^{2})
D.
E_{prod
}=
± √( A^{2}E_{a}^{2
}+B^{2}E_{b}^{2})
ANS: C
EXP: Error of product
for a rectangular lot is E_{prod
}= ± √ (A^{2}E_{b}^{2
}+B^{2}E_{a}^{2}).
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 L_{1}+L_{2}=h=6?
A. 18^{2}unit
B. 3
unit
C. 27 unit
D. 18 unit
ANS: D
EXP: A=0.5xh (L_{1}+L_{2})
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 onethird 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 cutoff 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
