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.

**Answer: Option 'C'**

**Let 6x and 5x be sides of the rectangle
6x × 5x = 1331
11x**

2.

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 = √17 ^{2}-8^{2} **

= √289-64

= 15 cm

Perimeter = 2 (l+b) = 2(8+5) Cm = 2(23) = 46 Cm

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:

**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**

4.

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 × x^{2})
Required ratio = 1/2 × 4 x^{2} : 1/2 x^{2} = 4 : 1**

5.

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)**

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.

**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**

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?

**Answer: Option 'A'**

**Answer: Option 'A'
lb = 460 m ^{2}...(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**

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.

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**

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?

**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.**

11.

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 m ^{2}
10000 × 1 = 10000
Side of this field = √10000 m = 100 m
Side of another square = 101 m
Difference of areas = [ 101^{2} - 100^{2} ] m^{2}
[101+100][101-100] m^{2}
(201)(1) m^{2} = 201 m^{2}**

12.

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 m ^{2}
Breadth of the carpet = 100 cm = 1m
Length of carpet = Area/Breadth => 117/1 = 177 m.**

13.

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**

14.

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**

15.

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**

16.

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 d ^{2} = 4802
d^{2} = 9604
d = √9604
d = 98 m**

17.

The diagonal of a rhombus are 65 m and 60 m. Its area is:

**Answer: Option ''**

**Area of the rhombus = 1/2 d1d2 = ( 1/2 × 65 × 60 ) cm ^{2}
= 65 × 30 = 1950 cm^{2}**

18.

The diagonal of a square is 40 m. The area of the square is:

**Answer: Option 'A'**

**Arear = 1/2 × ( diagonal ) ^{2}
= ( 1/2 × 40 × 40 )m^{2}
= 1600/2 = 800 m^{2}**

19.

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) cm ^{2} = r^{2} cm^{2}**

20.

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 × πr ^{2}
= 90/360 × 22/7 × 3.2 × 3.2
= (11 × 10.24)/2 = 112.64/2 = 56.32 Sq.Cm**

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