Question 1 |
A delivery agent is at a location R. To deliver the order, she is instructed to travel to location P along straight-line paths of \mathrm{RC}, \mathrm{CA}, \mathrm{AB} and \mathrm{BP} of 5 \mathrm{~km} each. The direction of each path is given in the table below as whole circle bearings. Assume that the latitude (L) and departure (D) of R is (0,0) \mathrm{km}. What is the latitude and departure of P (in km, rounded off to one decimal place)?
\begin{array}{|c|c|c|c|c|} \hline Paths & RC & CA & AB & BP \\ \hline \begin{array}{c}\text {Directions} \\ \text {(in degrees)}\end{array} & 120 & 0 & 90 & 240 \\ \hline \end{array}
\begin{array}{|c|c|c|c|c|} \hline Paths & RC & CA & AB & BP \\ \hline \begin{array}{c}\text {Directions} \\ \text {(in degrees)}\end{array} & 120 & 0 & 90 & 240 \\ \hline \end{array}
L=2.5 ; D=5.0 | |
L=0.0 ; D=5.0 | |
L=5.0 ; D=2.5 | |
L=0.0 ; D=0.0 |
Question 1 Explanation:

Latitude of ' C ' =-5 \sin 30^{\circ}=-2.5 \mathrm{Km}
Departure of ' C ' =5 \cos 30^{\circ}=+4.33 \mathrm{Km}
Latitude of ' \mathrm{B} ' = Latitude of ' \mathrm{C} ' +5 \mathrm{Km} =-2.5+5=+2.5 \mathrm{Km}
Departure of 'B' = Departure of ' C ' +5 \mathrm{Km} =+4.33+5=+9.33 \mathrm{Km}
Latitude of ' P ' = Latitude of ' B ' -5 \sin 30^{\circ} =+2.5-2.5=0
Departure of ' P ' = Departure of ' B ' -5 \sin 30^{\circ} =9.33-4.33=+5 \mathrm{Km}
\therefore \quad \mathrm{L}=0.0, \quad \mathrm{D}=5.0
Question 2 |
If the size of the ground area is 6 \mathrm{~km} \times 3 \mathrm{~km} and the corresponding photo size in the aerial photograph is 30 \mathrm{~cm} \times 15 \mathrm{~cm}, then the scale of the photograph is 1 : ____ (in integer).
15000 | |
20000 | |
25000 | |
30000 |
Question 2 Explanation:
Scale of Photograph (\mathrm{S})=\frac{\text { Photo distance }}{\text { Ground distance }}
For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
\Rightarrow \quad \mathrm{S}=\frac{1}{20000}
For Breadth Scale :
\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
S=\frac{1}{20000}
\therefore Scale of photograph, S=1: 20000
For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
\Rightarrow \quad \mathrm{S}=\frac{1}{20000}
For Breadth Scale :
\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
S=\frac{1}{20000}
\therefore Scale of photograph, S=1: 20000
Question 3 |
The magnetic bearing of the sun for a location at noon is 183^{\circ} 30^{\prime}. If the sun is exactly on the geographic meridian at noon, the magnetic declination of the location is ___
3^{\circ} 30^{\prime} \mathrm{W} | |
3^{\circ} 30^{\prime} \mathrm{E} | |
93^{\circ} 30^{\prime} \mathrm{W} | |
93^{\circ} 30^{\prime} \mathrm{E} |
Question 3 Explanation:
At any place 12 O'clock sun will be exactly over the true meridian of that place.

Declination =183^{\circ} 30^{\prime}-180^{\circ}=3^{\circ} 30^{\prime} \mathrm{W}
Magnetic declination will be 3^{\circ} 30 \mathrm{~W}.

Declination =183^{\circ} 30^{\prime}-180^{\circ}=3^{\circ} 30^{\prime} \mathrm{W}
Magnetic declination will be 3^{\circ} 30 \mathrm{~W}.
Question 4 |
Trigonometric levelling was carried out from two stations P and Q to find the reduced level (R.L.) of the top of hillock, as shown in the table. The distance between stations P and Q is 55 \mathrm{~m}. Assume stations P and Q, and the hillock are in the same vertical plane. The R. L. of the top of the hillock (in\mathrm{m} ) is _____ (round off to three decimal places)
\begin{array}{|l|l|l|l|} \hline Station &\begin{array}{c}\text{Vertical} \\ \text{angle of} \\ \text {the top of} \\ \text{hillock }\end{array} &\begin{array}{c}\text {Staff} \\ \text{reading} \\ \text {on} \\ \text{benchmark}\end{array} &\begin{array}{c}\text {R.L. of} \\ \text {benchmark}\end{array} \\ \hline P &18^{\circ} 45^{\prime} &2.340 \mathrm{~m} &100.000 \mathrm{~m} \\ \hline Q &12^{\circ} 45^{\prime} &1.660 \mathrm{~m}& \\ \hline \end{array}
\begin{array}{|l|l|l|l|} \hline Station &\begin{array}{c}\text{Vertical} \\ \text{angle of} \\ \text {the top of} \\ \text{hillock }\end{array} &\begin{array}{c}\text {Staff} \\ \text{reading} \\ \text {on} \\ \text{benchmark}\end{array} &\begin{array}{c}\text {R.L. of} \\ \text {benchmark}\end{array} \\ \hline P &18^{\circ} 45^{\prime} &2.340 \mathrm{~m} &100.000 \mathrm{~m} \\ \hline Q &12^{\circ} 45^{\prime} &1.660 \mathrm{~m}& \\ \hline \end{array}
124.365 | |
137.682 | |
253.142 | |
254.325 |
Question 4 Explanation:

While taking reading from P,
RL of hillock =H I_P+x \tan 18^{\circ} 45^{\prime}
=100+2.34+x \tan 18^{\circ} 45^{\prime}
While taking reading from Q.
\mathrm{RL} of hillock =\mathrm{HI}_Q+(\mathrm{x}+55) \tan 12^{\circ} 45^{\prime}
=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}
Equating both,
100+2.34+x \tan 18^{\circ} 45^{\prime}=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}
2.34+0.339 x=1.66+12.445+0.226 x
\therefore x=104.115
\quad \therefore \quad RL of hillock
=100+2.34+104.115 \tan 18^{\circ} 45^{\prime}
=137.682
Question 5 |
A student is scanning his 10 inch \times 10 inch certificate at 600 dots per inch (dpi) to convert it to raster. What is the percentage reduction in number of pixels if the same certificate is scanned at 300 dpi?
62 | |
88 | |
75 | |
50 |
Question 5 Explanation:
DPI = Dots Per Inch, it should be noted that DPI is not dots per square inch.
\% reduction in number of pixels
=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100
=\frac{600^{2}-300^{2}}{600^{2}} \times 100
=75 \%
\% reduction in number of pixels
=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100
=\frac{600^{2}-300^{2}}{600^{2}} \times 100
=75 \%
There are 5 questions to complete.
Thankyou very much for providing Pyq s with neat explanation…
Que 9 galt h 68° aayega wha 62° nhi ..ans bhi galt diya h 196° aayega ans
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