Home ≫ Arithmetic Aptitute ≫ Elementary Mensuration-I ≫ Page 8
71. There are two concentric circles of radii 8 cm and 3 cm respectively. If larger circle makes 120 revolutions to cover a certain distance, then find the number of revolutions made by smaller circle to cover the same distance.
  A.  320 revolutions
  B.  290 revolutions
  C.  300 revolutions
  D.  340 revolutions
     
   
View Answer

Shortcut:
There are two concentric circles of radii R and r respectively. Now consider the following cases.
Case I: If larger circle makes 'n' revolutions to cover a certain distance, then the smaller circle makes
R / r
n revolutions to cover the same distance.
Case II: If smaller circle makes n revolutions to cover a certain distance, then the larger circle makes
r / R
x n revolutions to cover the same distance.
Here, R = 8, r = 3, n = 120
Revolution of smaller circle =
8 / 3
x 120 = 8 x 40 = 320
Hence, the smaller circle will make 320 revolutions to cover the same distance.

72. There are two concentric circles. Radius of the larger circle is 28 cm. Area of the smaller circle is
1 / 3
rd of the area between the two circles. Fidn the area and perimeter of the smaller circle.
  A.  620 sq cm
  B.  616 sq cm
  C.  604 sq cm
  D.  590 sq cm
     
   
View Answer

Shortcut:
There are two concentric circles. Radius of the larger circle is R. If the area of the smaller circle is
x / y
of the area of the region (shaded-portion) between two circles, then
(i) the radius of the smaller circle =
√[
1 / (1 + y/x)
]


(ii) the perimeter of the smaller circle = 2x πR x
√[
1 / (1 + y/x)]
]

(iii) the area of the smaller circle = πR2 x
1 / (1 + y/x)

(iv) the area of the shaded circle = πR2 x
1 / (1 + x/y)

Where, R = radius of the larger circle.
Here, R = 28, x = 1, y = 3
The perimeter of the smaller circle = 2 x π x 28 x
√[
1 / (1 + 3/1)
]

= 2 x
22 / 7
x 28 x
√[
1 / 4
]

= 2 x 22 x 4 x
1 / 2
= 44 x 2 = 88 cm
The area of the smaller circle = πR2 x
1 / (1 + y/x)

=
22 / 7
x (28)2 x
1 / (1 + 3/1)

=
22 / 7
x 28 x 28 x
1 / 4

= 22 x 4 x 28 x
1 / 4
= 22 x 28 = 616 sq cm.

73. How many metres of a carpet 75 cm wide will be required to cover the floor of a room which is 20 metres long and 12 metres broad?
  A.  310 m
  B.  335 m
  C.  320 m
  D.  305 m
     
   
View Answer

Shortcut:
Length of a carpet 'd' metre wide, required to cover the floor of a room which is 'y' metre long and 'z' metre broad, is given by
yz / d
metre.
or Length required =
Length of room x breadth of room / Width of carpet

Here, d = 75, y = 20, z = 12
Length required =
20 x 12 / 0.75
= 320 metre

74. A 75 cm wide carpet is used to cover the floor of a room which is 20 m long and 12 m broad. What amount needs to be spent in carpeting the floor if the carpet is available at Rs 20 per metre?
  A.  Rs 6200
  B.  Rs 5800
  C.  Rs 6400
  D.  Rs 6500
     
   
View Answer

Shortcut:
A 'd' metre wide carpet is used to cover the floor of a room which is 'y' meter long and 'z' metre broad. If the carpet is available at Rs 'M' per metre, then the total amount required to cover the floor of the room is given by Rs M x
yz / d
metre.
or Amount required = Rs Rate x
Length of room x breadth of room / Width of carpet

Note: Length of carpet =
yz / d

Here, d = 75, y = 20, z = 12, M = 20 per metre
Amount required = 20 x
20 x 12 / 0.75
= 6400
Hence, the total amount required is Rs 6400.

75. How many paving stones each measuring 2.5 m x 2m are required to pave a rectangular courtyard 30 m long and 16.5 m wide?
  A.  99
  B.  85
  C.  90
  D.  102
     
   
View Answer

Shortcut:
Number of tiles, each measuring d1 metre x d2 metre, required to pave a rectangular courtyard y m long and z meter wide are given by
yz / d1 x d2
metre.
or Number of tiles required =
Length x breadth of courtyard / length x breadth of each tile

Here, d1 = 2.5, d2 = 2, y = 30, z = 16.5
Tiles required =
30 x 16.5 / 2.5 x 2
= 15 x 66 = 99
Hence, the total number of paving stones required is 99.

76. Certain number of paving stones each measuring 2.5m x 2 m are required to pave a rectangular courtyard 30 m long and 16.5 m wide. What amount needs to be spent if the tiles of the aforesaid dimentsion are available at Re 1 per piece?
  A.  Rs 95
  B.  Rs 81
  C.  Rs 85
  D.  Rs 99
     
   
View Answer

Shortcut:
Certain number of tiles, each measuring d1 metre x d2 metre, are required to pave a rectangular courtyard y m long and z metre wide. If the tiles are available at Rs M per piece, then the amount needs to be spen is given by Rs M x
yz / d1 x d2

or Amount required = Rate x
Length x breadth of courtyard / length x breadth of each tile

Here, d1 = 2.5, d2 = 2, y = 30, z = 16.5, M = Re 1 per piece
Amount required = 1 x
30 x 16.5 / 2.5 x 2
= 15 x 66 = Rs 99

77. A hall-room 39m 10 cm long and 35m 70 cm broad is to be paved with equal square tiles. Find the largest tile so that the tles exactly fit and also find the number of tiles required.
  A.  483
  B.  475
  C.  489
  D.  480
     
   
View Answer

Shortcut:
A room y m long and z m broad is to be paved with square tiles of equal sizes. The largest possible tile so that the tiles exactly fit is given by "HCF" of length and breadth of the room" and the number of tiles required are
yz / (HCF of y and z)2

Here, y = 39.10, z = 35.70
Side of largest possible square tile = HCF of length and breadth of the room
= HCF of (39.10 and 35.70) = 1.70 m
No. of tiles required =
39.10 x 35.70 / (1.70)2

=
39.10 x 35.70 / 1.70 x 1.70
= 483

78. The length and breadth of a room are 10m 75cm and 8m 25 cm respectively. The floor is to be paved with square tiles of the largest possible size. The size of the tiles is:
  A.  25 cm x 25 cm
  B.  20 cm x 25 cm
  C.  35 cm x 25 cm
  D.  15 cm x 25 cm
     
   
View Answer

Shortcut:
A room y m long and z m broad is to be paved with square tiles of equal sizes. The largest possible tile so that the tiles exactly fit is given by "HCF" of length and breadth of the room" and the number of tiles required are
yz / (HCF of y and z)2


Here, y = 10.75, z = 8.25
Largest possible square tile = HCF of (1075 and 825) = 25 cm
∴ The size of the tiles is 25 cm x 25 cm

79. A square field, 10 m long, is surrounded by a path 2 m wide. Find the area of the path.
  A.  98 sq m
  B.  96 sq m
  C.  100 sq m
  D.  88 sq m
     
   
View Answer

Shortcut:
If a square hall y metres long is surrounded by a verandah (on the outside of the square hall) z metres wide, then the area of the verandah is given by 4z(y + z) sq metres.
Here, y = 10, z = 2
Area of path = 4z(y + z) = 4 x 2(10 + 2) = 4 x 2 x 12 = 96 sq m.

80. A square grass plot is 100 m long. It has a gravel path 2.5 m wide all round it on the inside. Find the area of the path.
  A.  950 sq m
  B.  965 sq m
  C.  980 sq m
  D.  975 sq m
     
   
View Answer

Shortcut:
If a square plot is y metres long. It has a gravel path z metre wide all round it on the inside, then the area of the path is given by 4z(y − z) sq metres.
Here, y = 100, z = 2.5
Area of path = 4z(y − z) = 4 x 2.5(100 − 2.5) = 4 x 2.5 x 97.5 = 975 sq m.

Copyright © 2020-2022. All rights reserved. Designed, Developed and content provided by Anjula Graphics & Web Desigining .