# Problems on Areas : Aptitude Test

• ## Important Mensuration Formulas

#### Basic Computer Knowledge Test Questions and Answers

1.

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

#### Basic Computer Knowledge Test Questions and Answers

2.

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

#### Basic Computer Knowledge Test Questions and Answers

3.

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

#### Basic Computer Knowledge Test Questions and Answers

4.

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

#### Basic Computer Knowledge Test Questions and Answers

5.

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

#### Basic Computer Knowledge Test Questions and Answers

6.

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

#### Basic Computer Knowledge Test Questions and Answers

7.

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

#### Basic Computer Knowledge Test Questions and Answers

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.

#### Basic Computer Knowledge Test Questions and Answers

9.

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

#### Basic Computer Knowledge Test Questions and Answers

10.

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.