# RRB NTPC - Problems on Areas : Aptitude Test

1.

A large field of 900 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

A.) 395
B.) 385
C.) 415
D.) 405

Let the areas of the parts be x hectares and (900 - x) hectares.
Difference of the areas of the two parts = x - (900 - x) = 2x - 900
one-fifth of the average of the two areas = 1/5[x+(900−x)]/2
= 1/5 × (900/2) = 450/5 = 90
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x - 900 = 90
=> 2x = 990
⇒ x = 990/2 = 495
Hence, area of smaller part = (900 - x) = (900 – 495) = 405 hectares.

2.

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

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

3.

The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?

A.) 20 metres
B.) 25 metres
C.) 27 metres
D.) 18 metres

lb = 460 m2...(Equation 1)
Let the breadth = b
Then length, l = b × (100+15)/100 = 115b/100 ...(Equation 2)
From Equation 1 and Equation 2,
115b/100 × b = 460
b2 = 46000/115 = 400
⇒ b = √400 = 20 m

4.

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

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

5.

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

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

6.

The area of a square field is 4802 m2 the length of its diagonal is:

A.) 94 m
B.) 96 m
C.) 98 m
D.) None of these

Let the diagonal be d metres,
Then 1/2 d2 = 4802
d2 = 9604
d = √9604
d = 98 m

7.

Find the cost of carpenting a room 13 m long and 9 m broad with a carpet 100 Cm broad at the rate of Rs.20 per meter.

A.) 117 m
B.) 107 m
C.) 116 m
D.) 117 m2

Area of the carper = Area of the room
= 13 × 9 => 117 m2
Breadth of the carpet = 100 cm = 1m
Length of carpet = Area/Breadth => 117/1 = 177 m.

8.

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.

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.

9.

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

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

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

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

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

12.

Find the area of the sector whose are is making an angle of 90° at the center of a circle of radius 3.2 Cm.

A.) 46.32 Sq.cm
B.) 54.32 Sq.cm
C.) 56.22 Sq.cm
D.) 56.32 Sq.Cm

Area of the sector = 90/360 × πr2
= 90/360 × 22/7 × 3.2 × 3.2
= (11 × 10.24)/2 = 112.64/2 = 56.32 Sq.Cm

13.

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

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

14.

A girl walking at the rate of 9 Km per hour crosses a square field diagonally in 12 seconds. The area of the field is:

A.) 460 sq.m
B.) 600 sq.m
C.) 510 sq.m
D.) 450 sq.m

Distance covered in (9×1000)/(3600) × 12 = 30 m
Diagonal of squarre field = 30 m.
area of square field = 302/2
= 900/2 = 450 Sq.m

• ## Important Mensuration Formulas

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

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

16.

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

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

17.

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

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

• ## Important Mensuration Formulas

18.

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

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

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%

• ## Important Mensuration Formulas

19.

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

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

20.

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

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