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?
Answer: Option 'C'
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.
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
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?
Answer: Option 'A'
Answer: Option 'A'
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:
Answer: Option 'C'
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:
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
6.
The area of a square field is 4802 m2 the length of its diagonal is:
Answer: Option 'C'
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.
Answer: Option 'A'
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
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.
9.
The perimeter of a square circumscribed about a ciecle of radius r is:
Answer: Option 'D'
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.
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?
Answer: Option 'C'
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.
Answer: Option 'D'
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?
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
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:
Answer: Option 'D'
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
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:
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
16.
The ratio of the areas of two squares, one having double its diagonal then the other is:
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
17.
The area of the largest triangle that can be inscribed in a semicircle pf rasius r Cm, is:
Answer: Option 'C'
Area of the largest triangle = (1/2 × 2r × r) cm2 = r2 cm2
18.
If the radius of a circle decreased by 50% its area is decreased by:
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%
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.
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
20.
A rectangular mat has an area of 120 sq.metres and perimeter of 46 m. The length of its diagonal is:
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