Area of the carpet = Area of the room

Correct Option: D

Area of the carpet = Area of the room
= 8 x 6 = 48 sq m

Width of the carpet = 75/100 = 3/4 m

Length of the carpet = 48 x (4/3)
= 16 x 4 = 64 m

∴ Cost of carpeting = 64 x 20 = ₹ 1280

l₁ = 20 cm, A₁ = 200 sq cm
∴ b₁ = 200/20 = 10 cm
Now, A₂ = 200 x 6/5 = 240 sq cm
b₂ = 10 cm
∴ l₂ = 240/10 = 24 cm

∴ Perimeter of new rectangle = 2(l₂ + b₂)

Correct Option: D

l₁ = 20 cm, A₁ = 200 sq cm
∴ b₁ = 200/20 = 10 cm
Now, A₂ = 200 x 6/5 = 240 sq cm
b₂ = 10 cm
∴ l₂ = 240/10 = 24 cm

∴ Perimeter of new rectangle = 2(l₂ + b₂)
= 2(24 + 10) = 2 x 34 = 68 cm

Area to the rectangular field = 12375/15 = 825 sq m
According to the question,
(L x B) = 825 [L = length and B = breadth]

Correct Option: C

Area to the rectangular field = 12375/15 = 825 sq m
According to the question,
(L x B) = 825 [L = length and B = breadth]
⇒ L x 25 = 825
∴ L = 825/25 = 33 m

According to the question,
4a = 68 [ where a = side]
∴a = 68/4 = 17 cm
∴ Required area = a²

Correct Option: C

According to the question,
4a = 68 [ where a = side]
∴a = 68/4 = 17 cm

∴ Required area = a²
= (17)² = 289 sq cm

Let the radius of the park be r, then
πr + 2r = 288

Area of the park = (1/2)πr²

Correct Option: A

Let the radius of the park be r, then
πr + 2r = 288
(π + 2)r = 288
⇒ [(22/7) + 2]r = 288
⇒ r = (288 x 7)/36 = 56

∴ Area of the park = (1/2)πr²
= (1/2) x (22/7) x 56 x 56
= 4928