To find the class mark (xᵢ) for each interval, the folowing relation is used.
xᵢ = (Upper limit + Lower limit)/2
Given that the mean nocket allawance is ₹18 Taking 18 as.assured mean (a), dᵢ and fᵢdᵢ are calculated as follows:

From the table, we obtain
∑fᵢ = 44 + f, ∑fᵢdᵢ = 2f – 40 and a = 18
mean(X̄) = a + (∑fᵢdᵢ)/∑fᵢ ⇒ 18 = 18 + ((2F – 40)/(44 + F)) ⇒ 0 = (2F – 40)/(44 + f) ⇒ 2f – 40 = 0 ⇒ f = 20
Hence, the value of missing frequency f is 20.

See this video explanation 😮✌👇

(i) Let the chord AB subtends right angle at the centre O.
Area of segment with an angle 90°
= 90°/360° × πr² = 1/4 × π(10)² = 1/4 × 3.14 × 10 × 10 = 78.5 cm²
Area of triangle OAB
= 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm²
Area of minor segment = Area of minor sector Area of triangle OAB
= 78.5 cm² – 50 cm² = 28.5 cm²
(ii) Area of major segment = Area of cirlce Area of minor segment
= πr² – 1/4πr² = (1 – 1/4)πr² = 3/4πr²
= 3/4 × π(10)² = 3/4 × 3.14 × 10 × 10 = 235.5 cm²

Radius of circle 15 cm
Area of sector OPRQ
= 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm²
Area of minor segment
= Area of minor sector OPRQ – Area of △0PQ
= 117.75 cm² – √3/4(15) cm² [As triangle OPQ is an equilateral triangle]
= 117.75 cm² – 225√3/4 cm²
= 117.75 cm² – 56.25 x 1.73 cm²
= 231 cm² – 97.3125 cm²
= 20.4375 cm²
Area of major segment = Area of circle – Area of minor segment
= πr² – 20.4375 cm²
= π(15)² – 20.4375] cm²
= [3.14 x 15 x 15 – 20.4375] cm²
= [706.5 – 20.4375] cm²
= 686.0625 cm²

For better understanding to the above question see here😎👇

The shape of grass field, where the horse can graze is a sector with central angle 90°.
(i) The area of the field, where the horse can graze
= Area of sector OABO with radius 5
= 90°/360° × πr² = 1/4 × (5)²
= 1/4 × 3.14 × 25 = 19.625 m²
(ii) If the rope were 10 m long instead of 5 m, the area of field where the horse can graze = Area of sector 0ABO with radius 10
= 90°/360° × πr² = 1/4 × (10)²
= 1/4 × 3.14 × 100 = 78.50 m²
The increase in grazing area = (78.50 – 19.625) m² = 58.875 m²