Answer: Option C

Explanation:
Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7               ----(1)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy 3x - 4y = 12        ----(2)
Solving 1 & 2 , we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Answer: Option C

Explanation:
Let breadth = x m, length = 3x m, height = H metres.
Area of the floor = (Total cost of carpeting)/(Rate/m²)=(270/5)m²=54m².
x× (3x/2) = 54 =>x² = (54×2/3) = 36 x = 6.
So, breadth = 6 m and length =(3/2)×6 = 9 m.
Now, papered area = (1420/10)m² = 142 m².
Area of 1 door and 2 windows = 8 m².
Total area of 4 walls = (142 + 8) m² = 150 m²
2×(9+ 6)× H = 150 H = 150/30 = 5 m.

Answer: Option D

Explanation:
Area of square/area of triangle = a²/(√3/4)a²
                                              = 4/√3 i.e 4:√3

Answer: Option A

Explanation:
area of circle  = πb²/4
                     = 22/7 × 3.5²/4
                     = 9.625sq.m
 

Answer: Option D

Explanation: Let other diagonal = 2x cm.
Since diagonals of a rhombus bisect each other at right angles, we have:
(20)² = (12)² + (x)² =>x =√(20)² – (12)²= √256= 16 cm.                              _I
So, other diagonal = 32 cm.
Area of rhombus = (1/2) x (Product of diagonals) =(1/2× 24 x 32) cm² = 384 cm²

Answer: Option A

Explanation:
Let the two parallel sides of the trapezium be a cm and b cm.
Then, a-b = 4 ------ (1)
And, (1/2) x (a+b) x 10 = 475 =>(a+b) =((250 x 2)/10) => a + b = 50------- (2)
Solving 1 and 2, we get: a=27,b=23
So, the two parallel sides are 27 cm and 23 cm.

Answer: Option B

Explanation:
Clearly, the cow will graze a circular field of area 2244 sq. metres and radius      
equal to the length of the rope.
Let the length of the rope be R metres.
Then, Π(R)²= 2244 => R² = (2244 X (7/22)) = 714=> R = 26.72
Length of the rope = 26.72 m.

Answer: Option A

Explanation:
Area = (17.56 x 10000) m²= 175600 m².
ΠR2 = 175600  ⇔ (R)² = (175600 x (7/22)) ⇔ R = 236.37 m.
Circumference = 2ΠR = (2 x (22/7) x 236.37) m =1485.78 m.
Cost of fencing = Rs. (1485.78   x 3) = Rs. 4457.

Answer: Option D

Explanation:
Distance to be covered in 1 min. = (60 X 1000)/(60) m = 1000 m.
Circumference of the wheel = (2 x (22/7) x 0.60) m = 3.77 m.
Number of revolutions per min. =(1000/3.77) = 265

Answer: Option D

Explanation: Distance covered in one revolution =((60X 1200)/1000)= 72m.
 
2πR = 72 => 2 x (22/7) x R = 72 =>  R = 72 x (7/44) = 11.45 m.