MCQ: Area - 1 - SSC CGL MCQ

Area of square = 256 cm2 , Side of square = 16 cm
Area of rectangle = l x b
Length = 16 x 1.5 = 24 cm
Breadth = 16 x 1.2 = 19.2 cm
Area of rectangle = 460.8 cm²

Hence, option D is correct.

Let the radius of inner circle = R
Circumference of Inner circle = 2πR
Radius of larger circle = πR
Circumference of larger circle = 2π (πR)
As both of them complete 1 round in same time, the ratio of their speeds will be equal to the ratio of their distance travelled.

When both of them run on the larger track the difference in their distance travelled will be (22 – 7) =
15 units.
15 units = 75m
1unit = 5m
P completes one round. So 1 round = 22units
Total distance = 22 × 5 = 110m
Hence, option C is correct.

Let the radius of the circle = k

Area of rectangle = 126 = L × B = 14 × B → B = 9 cm
Perimeter = 2 × (L + B) = 2 × (14 + 9) = 46 cm
Hence, option C is correct.

Let the length and breadth of park be ‘15x’ cm and ‘8x’ cm respectively,
So, (15x)² + (8x)² = 68²
225x2 + 64x2 = 4624
289x2 = 4624

X = 4
So, the length and breadth of the park is 60 m and 32 m, respectively
Cost of fencing = 12 × 2 × (60 + 32) = Rs.2208
Cost of sodding = 2 × 60 × 32 = Rs.3840
Required difference = 3840 – 2208 = Rs.1632
Hence, option D is correct.

Let the length and breadth of the rectangle be 7a cm and 3a cm respectively.
Perimeter of rectangle = 2 (7a + 3a)
X = 20a

According to question
⇒ 2πr – 20a = 8

⇒ 22a – 20a = 8
⇒ 2a = 8
⇒ a = 4
Length of the rectangle = 7 x 4 = 28cm
Hence, option A is correct.

Area of square = 9 sq. cm so side of the square = root of 9 sq. cm = 3 cm = Radius of the metallic right
circular cone

From the question, height of the cone will become 100% more than 1.5 = 3 cm
Let x number of cones are melted to form a solid sphere

By solving this, x = 32
Hence, option D is correct.

Let the width of the rectangle = x cm
Then, 300% of x = 51

Perimeter of the rectangle = 2(length + width) = 2 × (51 + 17) = 136 cm
Perimeter of the square = (136 + 8) = 144 cm
The sides of the square = 144/4 = 36 cm
The area of the square field without street = (36)² sq. cm = 1296 sq. cm
The area of the square field with street = (36 + 20)² sq. cm = 3136 sq. cm
The area of the street = 3136 – 1296 = 1840 sq. cm
The total cost of constructing the street = 1840 x 25 = Rs. 46,000
Hence, option C is correct.

The area of rectangle = l × b = 14 × 12 = 168 sq. cm
The area of square = 168 + 28 = 196 sq. cm
The side of square = square root of 196 = 14 cm
Radius of incircle of a square 

The reqd. area

Hence, option B is correct.

The number of square tiles required

The price of one tile = Rs. 15
Therefore, the price of 300 tiles = 15 × 300 = Rs. 4500
Hence, option A is correct.

Let the radius of the circle = r cm the length of the rectangle = 2r cm
According to the question, area of the circle = A = πr²= 3 x 2r x 11

r = 21 cm
The perimeter of the rectangle = 2(l + b) = 2 (21 × 2 + 11) = 106 cm
Hence, option D is correct.

Sides of the square = 120 cm and 180 cm respectively [ As the perimeter is given we get it by dividing
the perimeter by 4]
The difference between the area of squares = (180 +120) × (180 – 120) = 300 × 60
= area of the rectangular field
The area of a rectangular field = length × breadth = 60 × length = 300 × 60
Length of the rectangle = 300 cm
The perimeter of the rectangle = 2(length + breadth) = 2(300 + 60) = 720 cm
The total cost of putting a fence around it = 720 × 5 = Rs. 3600
Hence, option C is correct.

The volume of the pizza = πr² × thickness

The volume of small slice of pizza

The volume of the remaining part = 1386 – 115.5 = 1270.5 cm³
Hence, option B is correct.

The length of longest pole that can be placed on the floor of room is nothing but the diagonal of the
rectangle
Therefore, by Pythagorean theorem,
(3x)² + (4x)² = 25²
Where 3x is the length of the room and 5x is the breadth of the room
9x2 + 16x2 = 25x2 = 625
By solving, x = 5 cm
The area of the floor of the room = 3x x 4x = 12 x 5x 5 = 300 sq. cm
The total cost of cementing the floor of the room = 5 x 300 = Rs. 1500
Hence, option B is correct.

Let, breadth and length of rectangle = 4x and 5x
Area of Square = 24x
Area of Rectangle = 20X2
Side of Square = √24X
Perimeter of Square = 4 × √24X
Ratio of Area of Rectangle & Perimeter of Square is 15 : 1

Squaring both sides,

Length of Rectangle = 5X = 30
Hence, option A is correct.

Let the length and the breadth of the rectangle be l and b respectively

⇒ 50 x l x b = 24(l²+b²)
⇒ 12l² – 25 x l x b + 12b² = 0

But l can't be less than b