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101. Find the perimeter of a regular pentagon whose each side measures 6 metres. Also, find the sum of each interior and exterior angles of the regular pentagon.
  A.  Perimeter : 35 m; sum of each interior and exterior angle: 2π
  B.  Perimeter : 30 m; sum of each interior and exterior angle: π
  C.  Perimeter : 20 m; sum of each interior and exterior angle: 3π
  D.  Perimeter : 25 m; sum of each interior and exterior angle: 5π
     
   
View Answer

Shortcut:
There is a regular polygon of 'n' sides. If length of each side is 'z' metre, then the sum of the exterior angles is given by 2π and the value of each exterior angle is
2π / n

Note: There is a regular polygon of 'n' sides and the length of each side is 'z' metres. Then the sum of each exterior and interior angle is given by π and the perimeter of the regular polygon is given by 'nz' metres.
Here, n = 5, z = 6
Using these values in the theorem given in Note, we get:
Perimeter of the regular polygon = nz = 5 x 6 = 30 m
Sum of each exterior and interior angle = π

102. A room 8 m long, 6 m broad and 3 metres high has two windows 1
1 / 2
m x 1 m and door 2 m x 1
1 / 2
m. Find the cost of papering the walls with paper 50 cm wide at 25P per metre.
  A.  Rs 30
  B.  Rs 33
  C.  Rs 39
  D.  Rs 42
     
   
View Answer

Shortcut:
If a room 'l' metres long, 'b' metres broad and 'c' metres high has
N windows (a1m x b1m, a2m x b2m, ......, anm x bnm)
M doors (x1m x y1m, x2m x y2m, ......, xmm x ymm), then the cost of papering the walls with paper 'd' metre wide at Rs Z per metre is given by
Rs
Z / d
[2(l + b)h − N(a1b1 +a2b2+....+anbn) − M(x1y1 + x2y2 + .....+ xmym)]
or,
Cost of paper per metre / Width of the paper
x Net area of the four walls
Here, Z = 25P = Rs 1/4, d = 50 cm = 1/2 m, n = 2, m = 1, l = 8, b = 6, h = 3, N = 2, M = 1, a1 = 3/2, b1 = 1, x1 = 2, y1 = 3/2
Using these values in the shortcut, we get:
Required cost =
1/4 / 1/2
[2(8 + 6)3 − 2(3/2 x 1) − 1(2 x 3/2)]
=
1 / 2
[14 x 6 − 3 − 3]
=
1 / 2
(84 − 6)
=
1 / 2
x 78 = 39
Hence, the cost of papering the walls is Rs 39

103. The radius of a circular wheel is 1
3 / 4
m. How many revolutions will it make in travelling 11 km?
  A.  1300
  B.  1100
  C.  1600
  D.  1000
     
   
View Answer

Shortcut:
The radius of a circular wheel is 'r' metre. The number of revolutions it will make in travelling 'd' km is given by
d / 2πr

or, No. or revolutions =
Distance / 2πr

Here, r = 7/4, d = 11 km = 11000 m
Using these values in the shortcut, we get:
No. or revolutions =
11000 / 2 x 22/7 x 7/4

=
11000 x 4 / 2 x 22 x 1

=
1000 x 4 / 2 x 2 x 1
= 1000
Hence, the number of revolution are 1000.

104. The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?
  A.  250
  B.  260
  C.  225
  D.  200
     
   
View Answer

Shortcut:
The radius of a circular wheel is 'r' metre. The number of revolutions it will make in travelling 'd' km is given by
d / 2πr

or, No. or revolutions =
Distance / 2πr

Here, Distance covered by wheel in 1 minute =
66 x 1000 x 100 / 60
= 110000 cm
Circumference of wheel = 2πr = 2 x
22 / 7
x 70 = 440 cm
r = 70, d = 66 km = 110000 cm
Using these values in the shortcut, we get:
No. or revolutions =
110000 / 440
= 250
Hence, total number of revolution are 250.

105. The diameter of a wheel is 2 cm. It rolls forward covering 10 revolutions. The distance travelled by it is:
  A.  58.4 cm
  B.  65.8 cm
  C.  62.8 cm
  D.  60 cm
     
   
View Answer

Shortcut:
The radius of a circular wheel is 'r' metre. The number of revolutions it will make in travelling 'd' km is given by
d / 2πr

or, No. or revolutions =
Distance / 2πr

Here, r = 1, revolutions = 10, d = ? cm
Using these values in the shortcut, we get:
No. or revolutions =
d / 2 x 22/7 x 1

10 =
7d / 2 x 22

10 x 44 = 7d ⇒ d =
440 / 7
= 62.8
Hence, the distance is 62.8 cm

106. If the wheel of the engine of a train 4
2 / 7
metres in circumference makes seven revolutions in 4 seconds, the speed of the train is:
  A.  30 km/hr
  B.  27 km/hr
  C.  21 km/hr
  D.  25 km/hr
     
   
View Answer

Shortcut:
The radius of a circular wheel is 'r' metre. The number of revolutions it will make in travelling 'd' km is given by
d / 2πr

or, No. or revolutions =
Distance / 2πr

Here, Distance covered in 4 sec =
30 / 7
x 7 = 30 m
Circumference = 2πr = 2 x
22 / 7
x r
or,
30 / 7
= 2 x
22 / 7
x r
or, 15 = 22 r r =
15 / 22

Speed of train =
30 / 4
=
30 / 4
x
18 / 5
= 27 km/hr

107. The circumference of a circular garden is 1012 metres. Inside the garden, a road of 3.5 m width runs round it. Calculate the area of this road.
  A.  3700.5 sq m
  B.  3605.8 sq m
  C.  3473.7 sq m
  D.  3503.5 sq m
     
   
View Answer

Shortcut:
The circumference of a circular garden is 'c' metres. Inside the garden a road of 'd' metres width runs round it. The area of the ring-shaped road is given by d(c − πd) or πd(2r − d) [∵ c = 2πr]
where r = radius of the circle.
Area of ring-shaped road = width of ring (circumderence of th circl − π x width of ring). Or
π x width of ring(2 x radius of the circle − width of ring)

OAC is a circle of radius = r, there is pathway, inside the circle of width = d.
Here, c = 1012, d = 3.5
Area of ring-shaped road = 3.5(1012 −
22 / 7
x 3.5)
= 3.5(1012 − 22 x 0.5)
= 3.5(1012 − 11) = 3.5(1001) = 3503.5
Hence, the area is 3503.5 sq m.

108. The circumference of a circular garden is 1012m. Find the area. Outside the garden, a road of 3.5 m width runs round it. Calculate the area of this road and find the cost of gravelling it at the rate of 32 paise per sq m.
  A.  Area: 3585.5 sq m; Cost: Rs 1155.76
  B.  Area: 3610.5 sq m; Cost: Rs 1215.26
  C.  Area: 3580.5 sq m; Cost: Rs 1145.76
  D.  Area: 3560.5 sq m; Cost: Rs 1115.42
     
   
View Answer

Shortcut:
The circumference of a circular garden is 'c' metres. Outside the garden, a road of 'd metres width runs around it. The area of the ring-shaped road is given by d(c + πd) or πd(2r + d) [∵ c = 2πr]
where r = radius of the circle.
Area of ring-shaped road = width of ring (circumderence of th circle + π x width of ring). Or

OAC is a circle of radius = r, there is pathway, inside the circle of width = d.
Here, c = 1012, d = 3.5
Area of ring-shaped road = 3.5(1012 +
22 / 7
x 3.5)
= 3.5(1012 + 22 x 0.5)
= 3.5(1012 + 11) = 3.5(1023) = 3580.5
∴ the area is 3580.5 sq m.
Cost of gravelling = 0.32 x 1023 = Rs 1145.76

109. A circular grassy plot of land 70 m in diameter has a path 7 m wide running round it on the outside. How many stones 25 cm by 11 cm are needed to pave the path?
  A.  61600
  B.  60100
  C.  62500
  D.  60800
     
   
View Answer

Shortcut:
The circumference of a circular garden is 'c' metres. Outside the garden, a road of 'd metres width runs around it. The area of the ring-shaped road is given by d(c + πd) or πd(2r + d) [∵ c = 2πr]
where r = radius of the circle.
Area of ring-shaped road = width of ring (circumderence of th circle + π x width of ring). Or
OAC is a circle of radius = r, there is pathway, inside the circle of width = d.
Here, r = 35, d = 7
Area of ring-shaped road = 7(
22 / 7
x 70 +
22 / 7
x 7)
= 7(22 x 10 + 22)
= 7(220 + 22) = 7 x 242 = 1694
∴ the area of path is 1694 sq m.
Converting metres into centimeters.
= 1694 x 100 x 100 sq cm.
Area of one stone = 25 x 11 = 275 sq cm. No. of stones =
1694 x 100 x 100 / 275
= 61,600
Hence, 61,600 stones are needed to pave the path.

110. A circular park of radius 25 m has a path of width 3.5 m all round it. Find the area of the path, if the garden has path both on its outside as well as inside.
  A.  1400 sq metres
  B.  1100 sq metres
  C.  1200 sq metres
  D.  1000 sq metres
     
   
View Answer

Shortcut:
A circular garden has ring-shaped road around it both on its inside and outside, eahof width 'd' units If 'r' is the radius of the garden, then the total area of the path is 4πdr sq units or 2cd [∵ c(perimeter) = 2πr]
Here, r = 25, d = 3.5
Total area path = 4 x
22 / 7
x 3.5 x 25
= 4 x 22 x 0.5 x 25 = 1100
Hence, the area is 1100 sq m.

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