Let original radius be r.
Then, according to the questions,
π (r + 1)² - πr² = 22

Correct Option: C

Let original radius be r.
Then, according to the questions,
π (r + 1)² - πr² = 22
⇒ π x [(r + 1)² - r²] = 22
⇒ (22/7) x (r + 1 + r ) x (r + 1 - r) = 22
⇒ 2r + 1 = 7
⇒ 2r = 6
∴ r = 6/2 = 3 cm

Increase in circumference of circle = 5%
∴ Increase in radius is also 5%.
Now, increase in area of circle = 2a + (a²/100) %

Correct Option: B

Increase in circumference of circle = 5%
∴ Increase in radius is also 5%.
Now, increase in area of circle = 2a + (a²/100) %
Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%

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

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 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