Problems on Areas : Aptitude Test

1.

The length of a room is 6 m and width is 4.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 900 per sq. metre. 

   A.) Rs. 25660
   B.) Rs. 25560
   C.) Rs. 25650
   D.) Rs. 26550

Answer: Option 'C'

Area = 6 × 4.75 sq. metre.
Cost for 1 sq. metre. = Rs. 900
Hence total cost = 6 × 4.75 × 900 = 6 × 4275 = Rs. 25650

2.

The ratio of the areas of two squares, one having double its diagonal then the other is:

   A.) 1 : 3
   B.) 3 : 1
   C.) 1 : 4
   D.) 4 : 1

Answer: Option 'D'

Lenth of the diagonals be 2x and x units. 
areas are 1/2 × (2x)2 and (1/2 × x2
Required ratio = 1/2 × 4 x2 : 1/2 x2 = 4 : 1

3.

The sides of a rectangle are in the ratio of 6 : 5 and its area is 1331 sq.m. Find the perimeter of rectangle.

   A.) 120 m
   B.) 121 cm
   C.) 121 m
   D.) None of these

Answer: Option 'C'

Let 6x and 5x be sides of the rectangle 
6x × 5x = 1331 
11x2 = 1331 
x2 = 1331/11 = 121 => x = 11 
Length = 6x = 6 × 11 = 66 
Breadth = 5x => 5 × 11 = 55 
Perimeter = 2 (l+b) => 2 (66+55) = 121 m

4.

A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? 

   A.) 98
   B.) 88
   C.) 99
   D.) 89

Answer: Option 'B'

Given that area of the field = 680 sq. feet
=> lb = 680 sq. feet
Length(l) = 20 feet
=> 20 × b = 680
=> b = 680/20 = 34 feet
Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

5.

A rectangular room 14 m long, 12 m broad is surrounded by a varandah, 3 m wide. Find the area of the varandah? 

   A.) 190 m2
   B.) 170 m2
   C.) 180 m3
   D.) 180 m2

Answer: Option 'D'

Area of varandah = (l+b+2p)2p 
= (14 + 12 + 6 ) 6 
= 180 m2

6.

A rectangular mat has an area of 120 sq.metres and perimeter of 46 m. The length of its diagonal is: 

   A.) 17 cm
   B.) 17 m
   C.) 17 m2
   D.) 16 m

Answer: Option 'B'

rectangular area = l × b = 120 and perimeter = 2(l+b) = 46 
l+b = 23 
(l-b)2 - 4lb = (23)2 - 4 × 120 = (529-480) = 49 = l-b = 7
l+b = 23, l-b = 7, we get l = 15, b = 8 
Diagonal = √152+82 
√225+64 => √289 = 17 m

7.

Area of four walls of a room is 99 m2. The length and breadth of the room are 7.5 m and 3.5m respectively. The height of the room is:

   A.) 4.5 Cm
   B.) 5.5 m
   C.) 4.5 m
   D.) 5.5 Cm

Answer: Option 'C'

2(7.5+3.5)×h = 99 
2(11)h = 99 
22h = 99 
h = 99/22 = 9/2 = 4.5 m

8.

The perimeter of a square circumscribed about a ciecle of radius r is: 

   A.) 12 r
   B.) 16 r
   C.) 64 r
   D.) 8 r

Answer: Option 'D'

Each side of the square = 2r 
Perimeter of the square = (4×2r) = 8r

9.

If the radius of a circle decreased by 50% its area is decreased by:

   A.) 75%
   B.) 50%
   C.) 85%
   D.) 65%

Answer: Option 'A'

Original area = πr2 
New area = π(r/2)2 = (πr2)/4 
Reduction in area = [(πr2 - (πr2)/4]2 = (3πr2)/4 
Reduction percent = ((3πr2)/4 × 1/(πr2 × 100) % = 75%

10.

The inner circumference of a circle race track 18 m wide is 880 m. Find the radius of the outer circle.

   A.) 140 cm
   B.) 150 m
   C.) 140 m
   D.) None of these

Answer: Option 'C'

Let inner radius be r metres 
Then 2πr = 640 
2 × 22/7 r = 880 
44/7 r = 880 
r = 880 × 7/44 = 140 m

11.

A rectangle measures 8 Cm on length and its diagonal measures 17 Cm. What is the perimeter of the rectangle? 

   A.) 46 cm
   B.) 48 cm
   C.) 46 m
   D.) None of these

Answer: Option 'C'

Second side = √172-82 
= √289-64 
= 15 cm 
Perimeter = 2 (l+b) = 2(8+5) Cm = 2(23) = 46 Cm

12.

A rectangular courtyard 3.78 m lang and 5.25 m broad is to be paved exactly with square tiles, all of the same size. The minimum number of such tiles is: 

   A.) 350
   B.) 450
   C.) 460
   D.) 470

Answer: Option 'B'

l = 378 Cm and b = 525 Cm 
Maximum length of a square tile 
= HCF of (378,525) = 21 Cm 
Number of tiles = (378×525)/(21×21) = (18×25) = 450

13.

The area of the largest triangle that can be inscribed in a semicircle pf rasius r Cm, is:

   A.) r3 Cm2
   B.) r2 Cm3
   C.) r2 Cm2
   D.) r3 Cm3

Answer: Option 'C'

Area of the largest triangle = (1/2 × 2r × r) cm2 = r2 cm2

14.

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? a

   A.) 126 sq. ft.
   B.) 128 sq. ft.
   C.) 136 sq. ft.
   D.) 116 sq. ft.

Answer: Option 'A'

Let l = 9 ft.
Then l + 2b = 37
=> 2b = 37 - l = 37 - 9 = 28
=> b = 28/2 = 14 ft
Area = lb = 9 × 14 = 126 sq. ft.

15.

Of the two square fields, the area of the one is 1 hectare, while anothe one is broader by 1%. There differences in area is:

   A.) 200 m
   B.) 201 m
   C.) 201 m2
   D.) None of these

Answer: Option 'C'

Area of one square field = 10000 m2 
10000 × 1 = 10000 
Side of this field = √10000 m = 100 m 
Side of another square = 101 m 
Difference of areas = [ 1012 - 1002 ] m2 
[101+100][101-100] m2
(201)(1) m2 = 201 m2

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