300+ REAL TIME Surveying Objective Questions & Answers 2023

250+ TOP MCQs on Flow Irrigation Survey and Answers

Surveying Multiple Choice Questions on “Flow Irrigation Survey”.

1. Which of the following is a classification of irrigation survey?
a) Direct method
b) Weir method
c) Notch method
d) River method
Answer: a
Clarification: Irrigation survey project can be classified into direct and indirect methods. Each of these classifications can be explained under certain sections. Both are applied in case of surface water schemes, sanitary water scheme etc.

2. The area that can be irrigated by a canal can be determined as ___________
a) Canal area
b) Paddy area
c) Irrigated area
d) Commanded area
Answer: d
Clarification: Commanded area is the area that can be irrigated by a canal system which can be either gross command area or culturable command area. The application of both of these depends upon the area that has to be irrigated.

3. Percentage of irrigation proposed to be irrigated annually yields ________
a) Density of irrigation
b) Intensity of irrigation
c) S.G of irrigation
d) Extent of irrigation
Answer: b
Clarification: Intensity of irrigation deals with the percentage of irrigation to be irrigated annually. In general, the area which has to be irrigated needs to be expressed in percentage of CCA which is able to represent the intensity of irrigation for the crop season.

4. Storage irrigation method is also known as __________
a) Weir method
b) Direct method
c) Indirect method
d) Notch method
Answer: c
Clarification: Indirect method or the storage irrigation method involves in the irrigation of excess water of a river during monsoon season. This is done to flow the excess water into the river which will be later filled in the reservoir or the tank.

5. Which of the following methods is adopted in case of the flow of excess water?
a) Weir method
b) Traverse irrigation method
c) Direct irrigation method
d) Storage irrigation method
Answer: d
Clarification: The usage of storage irrigation method is done in case of excess water scheme. This helps in storing the excess water in the upstream side of the dam which is constructed across the river, which is later adopted when flow of river is excess.

6. Which of the following indicates a storage structure?
a) Dam
b) Weir
c) Notch
d) Culvert
Answer: a
Clarification: A structure can be determined as a storage structure if it is having the ability to store excess water in case of floods. In general dams, spill ways and sluices are considered as the storage structures and also as appurtenant works.

7. Which of the following can be used as a diversion structure?
a) Dam
b) Barrage
c) Reservoir
d) Culvert
Answer: b
Clarification: A diversion structure involves the change in the direction of flow of stream or river, which can decrease the load or flow of water in a particular direction. Diversion structures generally include a barrage, a canal head and a river training head.

8. Which of the following can be indicated as the classification of the surface irrigation method?
a) Gradient method
b) Irrigation method
c) Flooding method
d) Drowning method
Answer: c
Clarification: Surface irrigation method generally involves flooding method in which water is allowed to cover the surface of the land in a continuous sheet of water which is sufficient to allow the field to absorb the water for increasing the field capacity.

9. A small basin is adopted in case of ____________
a) Sandy soil
b) Clay soil
c) Rocky soil
d) Metamorphic soil
Answer: a
Clarification: The usage of the basin not only depends upon slope but also on the type of soil it is placed. A small basin is adopted in case of steeper slope of the land, sandy soil, small stream size and small irrigation depth.

10. Flat land can adopt large basins.
a) False
b) True
Answer: b
Clarification: The usage of the large basin is done when the slope of the land is flat, presence of clay soil, large stream size, large depth of irrigation and in the presence of a mechanized field.

250+ TOP MCQs on Surveying Definition and Answers

Surveying Multiple Choice Questions on “Definition”.

1. Determining the relative positions of points on above or beneath the surface of the earth by means of direct or indirect measurements of distance and direction and elevation is called as _________
a) Surveying
b) Levelling
c) Measuring
d) Contouring
Answer: a
Clarification: Surveying is defined as determining the relative positions of points above or beneath the surface of the earth by means of direct or indirect measurements of distance and direction and elevation.

2. Finding the elevations of a point with respect to a given or assumed and establish points given elevation or at different elevations with respect to given or assumed dactum is ________
a) Surveying
b) Levelling
c) Bearing
d) Contouring
Answer: b
Clarification: Finding the elevations of a point with respect to a given or assumed and establish points given elevation or at different elevations with respect to given or assumed dactum is levelling.

3. Type of surveying in which the mean surface of the earth is considered as a plane and the spheroidal shape is neglected called as ________
a) Topographic Surveying
b) Hydrographic Surveying
c) Geodetic Surveying
d) Plane Surveying
Answer: d
Clarification: Type of surveying in which the mean surface of the earth is considered as a plane and the spheroidal shape is neglected is plane surveying.

4. Type of surveying in which the shape of the earth taken into account is __________
a) Topographic Surveying
b) Hydrographic Surveying
c) Geodetic Surveying
d) Plane Surveying
Answer: c
Clarification: Type of surveying in which the shape of the earth taken into account is geodetic surveying. The survey which deals with bodies of water for purpose of navigation, water supply, harbor works or for the determination of mean sea level is hydrographic surveying.

5. Horizontal projection of an area and shows only horizontal distances of the points is __________
a) Contour lines
b) Levelling
c) Surveying
d) Plan
Answer: d
Clarification: Horizontal projection of an area and shows only horizontal distances of the points is plan or map. Finding the elevations of a point with respect to a given or assumed and establish points given elevation or at different elevations with respect to given or assumed dactum is levelling.

6. What type of surveys needs to fix the boundaries of municipalities and of state and federal jurisdictions?
a) Topographic Surveying
b) Hydrographic Surveying
c) Cadastral Surveying
d) City Surveying
Answer: c
Clarification: The surveys need to fix the boundaries of municipalities and of state and federal jurisdictions are cadastral surveying. Survey which deals with bodies of water for purpose of navigation, water supply, harbor works or for the determination of mean sea level is hydrographic surveying.

7. Determining the absolute location of any point or the absolute location and direction of any line on the surface of the earth is called _______
a) Topographic Surveying
b) Astronomical Surveying
c) Cadastral Surveying
d) Hydrographic Surveying
Answer: b
Clarification: Determining the absolute location of any point or the absolute location and direction of any line on the surface of the earth is astronomical surveying.

8. Determining different strata in the earth’s crust is called as_______
a) Mine Survey
b) Geological Survey
c) Geodetic Survey
d) Archaeological Survey
Answer: b
Clarification: Determining different strata in the earth’s crust is the Geological Survey. Type of surveying in which the shape of the earth taken into account is geodetic surveying.

9. Determining unearthing relics of antiquity is called as_______
a) Mine Survey
b) Geological Survey
c) Geodetic Survey
d) Archaeological Survey
Answer: d
Clarification: Determining the unearthing relics of antiquity is an archaeological survey. Determining different strata in the earth’s crust is the geological survey.

10. In which surveying, shape of earth is taken into consideration?
a) Plane surveying
b) Geodic surveying
c) Topographic surveying
d) Geological surveying
Answer: b
Clarification: In Geodic surveying, shape of the earth is considered for carrying out high precise work. In case of Plane surveying, spheroid shape is neglected and entire area is considered in the form of triangles. Topographic surveying consists of vertical and horizontal locations of points whereas Geological surveying determines earth’s strata.

11. Representing large scale on the surface of the earth is____________
a) Plan
b) Map
c) Scale
d) Area
Answer: a
Clarification: For any representation, if it consists large scale then it represents plan and for small scale it represents map.

12. Which of the following units measurement system is generally employed?
a) Centesimal system
b) Hours system
c) Minutes system
d) Sexagesimal system
Answer: d
Clarification: Since most surveying instruments are graduated according to this system, Sexagesimal system is widely used in India. Centesimal system is having a great approach in Europe for its adaptability in interpolation. Hours system is having its use in navigation.

13. The ratio of map distance to corresponding ground distance is called as__________
a) Representative factor
b) Representation factor
c) Reciprocating factor
d) Recurring factor
Answer: a
Clarification: This factor can be used for determining the ratio of map distance to ground distance which would be helpful for further calculations.

14. Which among the following scales is used to determine the original scale when the plan on the drawing sheet shrinks due to atmospheric conditions?
a) Vernier scale
b) Plane scale
c) Shrunk scale
d) Diagonal scale
Answer: c
Clarification: Shrunk scale is used to determine the original scale when any plan shrinks due to atmospheric conditions, which can be determined by a formula. By using the original scale further calculations can be done.

15. Which among the following methods is used for determining the precise position on the earth surface?
a) Geological surveying
b) Geodic surveying
c) Land surveying
d) Plane Surveying
Answer: b
Clarification: In Geodic surveying, spheroid surface of the earth is considered which might be possible for determining the precise position by avoiding any further assumptions.

16. Which among the following is one of the principles of surveying?
a) Taking measurements
b) Covering entire area
c) Determining the elevation differences
d) Working from whole to part
Answer: d
Clarification: By working from whole to part, it is possible to eliminate the errors and to localise the errors. Otherwise, it might expand in magnitude.

17. Design a vernier for a theodolite circle divided into degrees and one fourth degrees to read to 20ꞌꞌ.
a) 55
b) 45
c) 65
d) 35
Answer: b
Clarification: W.K.T, L.C = s/n
S = (1/4)˚= 15ꞌ and L.C=20ꞌꞌ=20/60 min.
So, 20/60 = 15/n
n = 45.

18. Horizontal angle measured clockwise from geographic meridian to the direction of progress of a line is known as _______
a) Horizontal meridian
b) Vertical meridian
c) Azimuth
d) Horizontal bearing
Answer: c
Clarification: Azimuth is the angle measured from geographic meridian which is quite different from bearing as it is measured w.r.t north direction.

19. The formula for shrunk scale can be given as___________
a) Original scale*shrinking factor
b) Shrunk scale*shrinking factor
c) Vernier scale* shrinking factor
d) Diagonal scale* shrinking factor
Answer: a
Clarification: The shrinkage factor obtained by the ratio of shrunk length to actual length multiplied by the original scale will give the shrunk scale formula.

250+ TOP MCQs on Chain Surveying – Obstacles in Chaining and Answers

Surveying Questions and Answers for Aptitude test on “Chain Surveying – Obstacles in Chaining”.

1. How many kinds of obstacles of chaining are there?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Obstacles of chaining are of three kinds. They are obstacles to ranging, obstacles to chaining, obstacles to both chaining and ranging.

2. Which of the following is not one among the three major kinds of obstacles of chaining?
a) obstacles to ranging
b) obstacles to chaining
c) obstacles to levelling
d) obstacles to ranging and chaining
Answer: c
Clarification: Obstacles to levelling is not a kind of obstacles to chaining. Obstacles of chaining are of three kinds. They are obstacles to ranging, obstacles to chaining, obstacles to both chaining and ranging.

3. Both ends of the lines may be visible from intermediate points on the line. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining
Answer: a
Clarification: Obstacles to ranging but not ranging is a type of obstacle, in which the ends are not Intervisible, is quite expected in a flat country. There may be two cases of this obstacle, both ends of the lines may be visible from intermediate points on the line and both ends of the line may not be visible from intermediate points on the line.

4. When it is possible to chain round the obstacle, i.e a pond, hedge etc. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining
Answer: b
Clarification: There may be two cases of this obstacle i.e obstacle to chaining but not ranging, when it is possible to chain round the obstacle, i.e a pond, hedge etc and when it is not possible to chain round the obstacle e.g. a river.

5. Both ends of the line may not be visible from intermediate points on the line. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining
Answer: a
Clarification: There may be two cases of this obstacle, both ends of the lines may be visible from intermediate points on the line and both ends of the line may not be visible from intermediate points on the line.

6. When it is not possible to chain round the obstacle e.g. a river. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining
Answer: b
Clarification: There may be two cases of this obstacle i.e obstacle to chaining but not ranging, when it is possible to chain round the obstacle, i.e a pond, hedge etc and when it is not possible to chain round the obstacle e.g. a river.

7. To continue a survey line AB past an obstacle, a line BC 100 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the lengths which must be chained off along CD in order that ED may be in AB produced?
a) 100 m
b) 200 m
c) 300 m
d) 400 m
Answer: b
Clarification: Here angle ABC is 90°. From, ∆ BCD, CD = BC sec 60° = 100 × 2 = 200m.

8. To continue a survey line AB past an obstacle, a line BC 100 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the lengths which must be chained off along CE in order that ED may be in AB produced?
a) 141.42 m
b) 282.84 m
c) 140.14 m
d) 267.7 m
Answer: a
Clarification: Angle ABC is 90°. From ∆BCE, and CE = BC sec 45° = 100 × 1.4142 = 141.42 m.

9. To continue a survey line AB past an obstacle, a line BC 300 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the obstructed length BE?
a) 250 m
b) 600 m
c) 452.28 m
d) 300 m
Answer: d
Clarification: Here angle ABC is 90°. BE = BC tan 45° = 300 × 1 = 300 m.

10. If we select two points A and B on either side of the obstacle and equal perpendiculars AC and BD are set out. Then AB is equal to?
a) AC
b) CD
c) DA
d) BD
Answer: b
Clarification: Since AC and BD are two perpendiculars set either side of the obstacle of equal length. Therefore CD is parallel and equal to AB. Therefore, AB = CD.

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250+ TOP MCQs on Levelling – Curvature and Refraction and Answers

Surveying Multiple Choice Questions on “Levelling – Curvature and Refraction”.

1. Horizontal line departs from a level surface because of _____________
a) Refraction
b) Radius of earth
c) Curvature of earth
d) Parallelism
Answer: c
Clarification: From the definition of level surface and a horizontal line it is evident that a horizontal line departs from a level surface because of the curvature of the earth.

2. In the long sights, the horizontal line of sight doesn’t remain straight but it slightly bends downwards having concavity towards earth due to ____________
a) Refraction
b) Radius of earth
c) Curvature of earth
d) Parallelism
Answer: a
Clarification: Due to refraction, in the long sights, the horizontal line of sight doesn’t remain straight but it slightly bends downwards having concavity towards earth.

3. Find the correction for curvature for a distance 1200 m?
a) 0.113 m
b) 0.131 m
c) 0.133 m
d) 0. 313 m
Answer: a
Clarification: Correction for curvature is 0.07849 d². Therefore, here 0.07849*(1.2)² = 0.113 m.

4. Find correction for refraction for a distance of 1200 m?
a) 0.0106 m
b) 0.0160 m
c) 0.0016 m
d) 0.0116 m
Answer: b
Clarification: Correction for refraction is correction for curvature/7.
Therefore, here (0.07849 * (1.2)²)/7 = 0.016 m.

5. Find the correction for curvature for a distance 2.48 km?
a) 0.483 m
b) 0.434 m
c) 0.443 m
d) 0. 403 m
Answer: a
Clarification: Correction for curvature is 0.07849 d². Therefore, here 0.07849*(2.48)² = 0.483 m.

6. Find correction for refraction for a distance of 2.48 km?
a) 0.0066 m
b) 0.0160 m
c) 0.069 m
d) 0.096 m
Answer: c
Clarification: Correction for refraction is correction for curvature/7. Therefore, here (0.07849 * (2.48)²)/7 = 0.069 m.

7. Find combined correction for curvature and refraction for 3400 m?
a) 0.078 m
b) 0.778 m
c) 0.709 m
d) 0.786 m
Answer: b
Clarification: Combined correction for curvature and refraction is given by 0.06728 d². Therefore, here 0.06728 (3.4)² = 0.778 m.

8. Find combined correction for curvature and refraction for 1.29 km?
a) 0.112 m
b) 0.128 m
c) 0.212m
d) 0.221 m
Answer: a
Clarification: Combined correction for curvature and refraction is given by 0.06728 d². Therefore, here, 0.06728(1.29)² = 0.112 m.

9. In order to find the difference in elevation between two points P and Q, a level was set upon the line PQ, 30 m from P and 1280 m from Q. The reading obtained on staff kept at P and Q were respectively 0.545 m and 3.920 m. Find the true difference in elevation between P and Q?
a) 3.226 m
b) 3.343 m
c) 3.265 m
d) 3.345 m
Answer: c
Clarification: Since the distance of P from an instrument is small, the correction for curvature etc is negligible. Combined correction for Q is 0.06728 (1.28)² = 0.110 m. Correct staff reading at Q = 3.920 – 0.110 = 3.810 m. Difference in elevation between P and Q = 3.810 – 0.545 = 3.265 m.

10. A light house is visible just above the horizon at a certain station at the sea level. The distance between the station and light house is 10 km. Find the height of the light house?
a) 0 6728 m
b) 0.06728 m
c) 67.280 m
d) 6.728 m
Answer: d
Clarification: Combined correction for curvature and refraction is given by 0.06728 d². Therefore, here 0.06728 (10)² = 6.728 m.

250+ TOP MCQs on Theodolite Traversing – Errors and Answers

Surveying Multiple Choice Questions on “Theodolite Traversing – Errors”.

1. Which of the following errors can be eliminated by a method of repetition?
i) errors due to eccentricity.
ii) errors due to in adjustments of line of collimation.
iii) error due to inaccurate graduations.
iv) error due to inaccurate bisection of the object.
a) i only
b) i and ii
c) i, ii and 3
d) i, ii, iii and iv
Answer: d
Clarification: By a method of repetition the following errors can be eliminated. They are errors due to eccentricity, errors due to in adjustments of line of collimation, error due to inaccurate graduations, error due to inaccurate bisection of the object etc.

2. Sources of errors in transit work are broadly classified into ______ types.
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Sources of error in theodolite work are 3 types. They are instrumental, personal and natural.

3. Error due to imperfect adjustment of plate levels comes under ________ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: c
Clarification: Instrumental errors are due to the imperfect adjustment of an instrument, structural defects in the instrument and imperfections due to wear.

4. Error due to structural defects in the instrument comes under ________ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: c
Clarification: Instrumental errors are due to imperfect adjustment of an instrument, structural defects in the instrument and imperfections due to wear.

5. Error due to imperfections due to wear comes under ________ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: c
Clarification: Instrumental errors are due to imperfect adjustment of an instrument, structural defects in the instrument and imperfections due to wear.

6. Error due to the line of collimation not being perpendicular to the horizontal axis comes under ________ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: c
Clarification: Instrumental errors are due to imperfect adjustment of an instrument, structural defects in the instrument and imperfections due to wear. Error due to the line of collimation not being perpendicular to the horizontal axis also comes under instrumental error.

7. Error due to in accurate centering comes under ________ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Inaccurate centering comes under errors in manipulation.

8. Inaccurate levelling comes under ______ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Inaccurate levelling comes under errors in manipulation.

9. Slip comes under _____ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Slip comes under errors in manipulation.

10. Manipulating wrong tangent screw comes under _____ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Manipulating wrong tangent screw comes under errors in manipulation.

11. Parallax comes under _____ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Parallax comes under errors due to sighting and reading.

12. Inaccurate bisection of points observed comes under _____ error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: a
Clarification: The personal errors may be due to errors in manipulation, errors in sighting and reading. Inaccurate bisection of points observed comes under errors due to sighting and reading.

13. Un equal atmospheric refraction due to high temperature comes under which sources of errors?
a) Personal
b) Natural
c) Instrumental
d) Personal and natural
Answer: b
Clarification: Un equal atmospheric refraction due to high temperature comes under natural errors. Unequal settlement of tripod, wind vibrations etc., also comes under the same category.

14. Unequal settlement of tripod comes under ____ source of error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: b
Clarification: Un equal atmospheric refraction due to high temperature comes under natural errors. Unequal settlement of tripod, wind vibrations etc., also comes under the same category.

15. Unequal expansion of parts of telescope comes under _____ source of error.
a) personal
b) natural
c) instrumental
d) personal and natural
Answer: b
Clarification: Un equal atmospheric refraction due to high temperature comes under natural errors. Unequal expansion of parts of telescope also comes under a natural source of error.

250+ TOP MCQs on Area Calculation – Area by Co-ordinates and Answers

Surveying Multiple Choice Questions on “Area Calculation – Area by Co-ordinates”.

1. The method of tracing is involved in which of the following procedures?
a) Sub-division into squares
b) Sub-division into area figures
c) D.M.D method
d) Division into trapezoidal figures
Answer: d
Clarification: Area by map measurements involves calculating the area by the map details provided. It has certain classifications that use tracing paper in its procedure. Generally, the division of trapezoidal figure involves usage of tracing paper.

2. Which of the following is not a method involved in area by co-ordinate?
a) Sub-division into area figures
b) Co-ordinate method
c) Meridian method
d) D.M.D method
Answer: a
Clarification: Area by co-ordinate method contains ample classified techniques which can have its own importance. Those include co-ordinate method, meridian method, D.M.D method, and departure and total latitude method.

3. Which of the following is the main thing in the process of calculation of area by co-ordinate method?
a) M.D
b) Departure
c) Latitude
d) Parallels
Answer: c
Clarification: For the calculation of area by the methods of co-ordinates require the calculation of the length of latitude. If they are given it is quite simple. But if they aren’t it would be a tedious procedure.

4. Which of the following shapes are generally considered when the area is computed by map measurements?
a) Square
b) Triangle
c) Rectangle
d) Pyramidal
Answer: b
Clarification: For calculation purpose, triangles are the most preferred shapes as they can be solved with ease. But in reality most of them are in their own complex shape, which may resemble a trapezoid. So, mostly trapezoid and triangles will be used.

5. The figure indicates which of the following processes?

a) Sub-division into squares
b) Sub-division into area figures
c) Division into trapezoidal figures
d) D.M.D method
Answer: a
Clarification: The area calculations by the map measurements involve three classifications, among them the sub-division into squares classification involves the creation of grid lines. It is done so as to improve the accuracy.

6. Which of the following is not a classification in area by map measurement?
a) Division into trapezoidal figures
b) Sub-division into area figures
c) Sub-division into squares
d) D.M.D method
Answer: d
Clarification: Area by map measurement method involves three classifications. They are division into trapezoidal figures, sub-division into squares and sub-division into area figures. Any of the methods can be applied.

7. If the values of latitudes are 223.5 m and 65.31 m and meridians are 16.8 m, 24.67m. Find the area using D.M.D method.
a) 2268.99 sq. m
b) 2862.99 sq. m
c) 2682.99 sq. m
d) 28865.99 sq. m
Answer: c
Clarification: Area by D.M.D method can be given as, A = ¹⁄₂(∑(m*L))
A = ¹⁄₂ (223.5*16.8 + 65.31*24.67)
A = 2682.99 sq. m.

8. Find the area by co-ordinate method if the independent co-ordinates are (400, 400), (423, 456), (478, 498), (400, 400)
a) 1510 cu. m
b) 1150 cu. m
c) 5110 cu. m
d) 115 cu. m
Answer: a
Clarification: The formula in area by co-ordinate method can be given by,
A = ¹⁄₂ (y1(x2-x4) + y3(x3-x4) + y3(x4-x2) + y4(x1-x3)). On substitution of the co-ordinates in the area we get,
A = ¹⁄₂ (100(423-400) + 498(478-400) + 498(400-423) + 400(400-478))
A = -1510 cu. m, negative sign has no significance. So, A = 1510cu. M.

9. The value of total latitudes and its adjoining departures were given. Calculate area by departure and total latitudes method. Total Latitudes = 110, 25,0, 0 and adjoining departures = 245, 245, -245, -245.
a) 15636 sq. m
b) 16536 sq. m
c) 16563 sq. m
d) 15663 sq. m
Answer: b
Clarification: The formula for calculation of area by departure and latitude can be given as,
A = ¹⁄₂ ∑(total latitude * algebraic sum of adjoining departures). On substituting the given values, we get,
A = ¹⁄₂ (110*245 + 25*245 + 0*-245 + 0*-245) = 16536 sq. m.

10. Calculate the area by M.D method, if the value of m1, m2, m3 are given as 233.4 m, 12.78 m, 99.98 m respectively and latitudes are 110 m, -15 m, 89 m.
a) 43372.51 sq. m
b) 34732.15 sq. m
c) 34537.15 sq. m
d) 34372.51 sq. m
Answer: d
Clarification: The formula of meridian and distance method can be given as,
A = ∑ (m*L). On substitution, we get
A = 233.4*110 + 12.78*-15 + 99.98*89
A = 34372.51 sq. m.