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# Mensuration Questions for SSC-CPO Set-3 PDF

Download SSC CPO Mensuration Questions with answers set-3 PDF based on previous papers very useful for SSC CPO exams. Very important Mensuration Questions for SSC exams.

Question 1: The volume of a right circular cone is 2464 $cm^3$. If the height of cone is 12 cm, then what will be the radius of its base?

a) 12 cm

b) 8 cm

c) 14 cm

d) 10 cm

Question 2: A hollow iron pipe is 10 cm long and its external diameter is 18 cm. If the thickness of the pipe is 2 cm and iron weighs 8.5 g/cm$^3$, then the weight of the pipe from the following is closest to:

a) 8.54 kg

b) 9.54 kg

c) 7.54 kg

d) 5.54 kg

Question 3: The ratio between the perimeter and the breadth of a rectangle is 3 : 1. If the area of the rectangle is 310 sq. cm, the length of the rectangle is nearly:

a) 11.45 cm

b) 10.45 cm

c) 12.45 cm

d) 13.45 cm

Question 4: A circle, with radius 8 cm, which has the area equal to the area of a triangle with base 8 cm. Then the length of the corresponding altitude of triangle is:

a) $38 \pi$ cm

b) $16 \pi$ cm

c) $18 \pi$ cm

d) $8 \pi$ cm

Question 5: A copper wire is bent in the form of an equilateral triangle of area $4 \sqrt{3} cm^2$.If the same wire is bent into the form of a square, the area of the square will be:

a) 16 sq.cm

b) 9 sq. cm

c) 144 sq. cm

d) 64 sq.cm

Question 6: The length, breadth and height of a room are in the ratio of 4 : 3 : 2. The cost of carpeting the floor at ₹ 10 per square meter is ₹ 480. The height of the room is:

a) 10 m

b) 2 m

c) 6 m

d) 4 m

Question 7: Two concentric circles with radii p cm and (p+ 2) cm are drawn on a paper. The difference between their areas is 44 sq. cm What is the value of p?(Take $\pi=\frac{22}{7}$)

a) 1.5

b) 5

c) 6

d) 2.5

Question 8: A rectangular farm has to be fenced on one long side, one short side and one diagonal. If the cost of fencing is ₹ 10 per meter, the area of the farm is 4800 m$^2$ and the short side is 60 m long, the cost of doing the job will be:

a) ₹2100

b) ₹1800

c) ₹2400

d) ₹3600

Question 9: A wall of 12 m $\times$ 8 m has a door of 3 m $\times$ 1.5 m and two windows each of 1.5 m $\times$ 1.5 m. Find the area of wall that can be painted(free from doors and windows).

a) 80 sq.cm

b) 87 sq.cm

c) 96 sq.cm

d) 65 sq.cm

Question 10: What will be the cost of fencing a circular garden of radius 35 m at the rate of ₹16 per meter?(Take $\pi = \frac{22}{7}$)

a) ₹3520

b) ₹3240

c) ₹2850

d) ₹3160

Question 11: A hall is 15 m long and 12 m broad.If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume(in $m^3$) of the  hall is:

a) 1600

b) 900

c) 1200

d) 720

Question 12: A circle, with radius 8 cm, which has the area equal to the area of a triangle with base 8 cm. Then the length of the corresponding altitude of triangle is:

a) $38\ \pi\ cm$

b) $16\ \pi\ cm$

c) $8\ \pi\ cm$

d) $18\ \pi\ cm$

Question 13: The curved surface area of a right circular cylinder of height 28 cm is 176 $cm^2$ . The volume(in $cm^3$) of cylinder is (Take $\pi = \frac{22}{7}$)

a) 66

b) 110

c) 88

d) 176

Question 14: The area of an equilateral triangular park is equal to $5\sqrt{3}$ times the area of a triangular field with sides 18 m, 80 m and 82 m. What is the side of the triangular park?

a) 125 m

b) 120 m

c) 140 m

d) 100 m

Question 15: The diagonal of a square is 14 cm. What will be the length of the diagonal of the square whose area is double of the area of first square?

a) $28\surd2 cm$

b) $14\surd2 cm$

c) $28 cm$

d) $21\surd2 cm$

Volume of right circular cone = $\dfrac{1}{3} \pi r^2 h$
Given, h = 12 cm and Volume = $2464 cm^3$
Then, $\dfrac{1}{3} \times \dfrac{22}{7} \times r^2 \times 12 = 2464$

=> $r^2 = 196$
=> r = 14 cm
Therefore, Radius of the cone = 14 cm

External radius of the pipe = 9 cm
Internal radius of the pipe = 9-2 = 7cm
Volume of the pipe = $\pi \times (9^2 – 7^2) \times 10 = 1005.71 cm^3$
Weight of the pipe per $cm^3 = 8.5 gm$
Then, Total weight = 1005.71*8.5 = 8548 gm = 8.54 kg

Let the length and breath of the rectangle be l cm and b cm
Given, $\dfrac{2(l+b)}{b} = \dfrac{3}{1}$
=> 2(l+b) = 3b
=> 2l+2b = 3b
=> b = 2l
Given, lb = 310
Substituting b = 2l in above equation
$l\times 2l = 310$
$2l^2 = 310$
=> $l^2 = 155$
We know that $12^2 = 144$ and $13^2 = 169$.
Hence, The value of l lies in between 12 and 13.
Therefore, From the options, l = 12.45 cm

Area of a circle=$\pi*r*r$
Area of a triangle=(1/2)*b*h
h=64$\pi$*2*b/8
h=16$\pi$

Area of a equilateral triangle=$\sqrt{3}a^{2}/4$
$\sqrt{3}a^{2}/4$=$4\sqrt{3}$
a=4 cm
Total length =3*4
=12 cm
Side of each square=12/4
=3 cm
Area=9 sqcm

Let length,breath and height be 4x,3x and 2x respectively
Area of the floor=12$x^{2}$
Total cost of flooring=12$x^{2}$*10
12$x^{2}$*10=480
x=2
l=8 cm,b=6 cm and h=4 cm

Given $\pi R^{2}-\pi r^{2}$=44
$R^{2}-r^{2}$=14
R=p+2,r=p
R-r=2 R+r=2p+2
(R+r)(R-r)=14
2p+2=7
p=2.5

Area of the farm=4800
l*b=4800
given b=60 m
l*60=4800
l=80 m
In a rectangle diagonal=$\sqrt{(l^{2}+b^{2})}$
=$\sqrt{(80^{2}+60^{2})}$
=$\sqrt{(10000)}$
=100 m
So total fencing required=100+80+60=240 m
Cost of fencing each meter=Rs 10
For 240m it is 240*10=Rs 2400

Area of the wall=96 sq cm
Area of door=3*1.5=4.5 sq cm
Area of two windows=2*1.5*1.5
=4.5 cm
Therefore required area=96-(4.5+4.5)
=96-9
=87 sq cm

Circumference of the circular garden =$2\pi r$
Area=2*(22/7)*35
=220 sq mt
Cost for each each mt=Rs 16
Cost for whole garden=220*16
=Rs 3520

Area of the floor and ceiling will be same and it is equal
i.e l*b=15*12=180 sq m
Area of four walls will be 2*l*h+2*b*h=2h(l+b)
2h(l+b)=360
h(12+15)=180
h=180/27
h=20/3
Volume=l*b*h=12*15*20/3
=1200 cubic meters

Area of a circle=$\pi r^{2}$
Area of a triangle=(1/2)bh
Given both are equal
Therefore $\pi r^{2}$=(1/2)bh
$\pi 8^{2}$=(1/2)8h
h=$16\pi$

Lateral surface area of a cylinder=$2\pi r h$
Therefore 2*(22/7)*r*28=176
r=8/8
r=1 cm
Volume of a cylinder=$(\pi) r^{2}h$
=22*1*1*28/7
=88 sq cm

Given sides of the triangle are 18,80 and 82 cm
these are the sides of a right angled triangle as 9,40,41 is a Pythagorean triplet
So area of a right angled triangle=(1/2)bh
=(1/2)*80*18
=720 sq cm
Given area a equilateral triangle=$5\sqrt{3}$*720
$\sqrt{3}a^{2}/4$=$5\sqrt{3}$*720
$a^{2}$=36*100*4
a=6*10*2
a=120 cm

$\sqrt{2}a = 14 => a = \dfrac{14}{\sqrt{2}}$ where a = side of first square
Area of first square $= ( \dfrac{14}{\sqrt{2}})^2 = \dfrac{196}{2} = 98 cm^2$
Area of second square $= 2\times98 = 196 cm^2$
Side of second square $= \sqrt{196} = 14 cm$
Diagonal of second square $=14\times\sqrt{2}$ = $14\sqrt{2} cm$.