Section D
Section D consists of 4 questions of 5 marks each.
Q.32. A motor boat whose speed is 20 km/h in still water takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.
A train covers a distance of 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 48 minutes less for the journey. Find the original speed of the train.
📥 View Answer for Part (A)
Solution for Part (A):
Step 1: Let speed of stream = x km/h
Speed in still water = 20 km/h
Speed upstream = (20 – x) km/h
Speed downstream = (20 + x) km/h
Step 2: Distance = 48 km each way
Time upstream = \(\frac{48}{20 – x}\) hours
Time downstream = \(\frac{48}{20 + x}\) hours
Step 3: According to problem:
Time upstream = Time downstream + 1 hour
\(\frac{48}{20 – x} = \frac{48}{20 + x} + 1\)
Step 4: Multiply by (20-x)(20+x): \(48(20 + x) = 48(20 – x) + (20 – x)(20 + x)\) \(960 + 48x = 960 – 48x + 400 – x^2\)
Step 5: Simplify: \(48x = -48x + 400 – x^2\) \(x^2 + 96x – 400 = 0\)
Step 6: Solve quadratic equation: \(x^2 + 100x – 4x – 400 = 0\) \(x(x + 100) – 4(x + 100) = 0\) \((x – 4)(x + 100) = 0\)
Step 7: x = 4 or x = -100 Speed cannot be negative, so x = 4 km/h
📥 View Answer for Part (B)
Solution for Part (B):
Step 1: Let original speed = x km/h
Distance = 360 km
Original time = \(\frac{360}{x}\) hours
Step 2: New speed = (x + 5) km/h
New time = \(\frac{360}{x + 5}\) hours
Step 3: Time difference = 48 minutes = \(\frac{48}{60} = \frac{4}{5}\) hours \(\frac{360}{x} – \frac{360}{x + 5} = \frac{4}{5}\)
Step 4: Multiply by 5x(x+5): \(5 \times 360(x + 5) – 5 \times 360x = 4x(x + 5)\) \(1800(x + 5) – 1800x = 4x^2 + 20x\)
Step 5: Simplify: \(1800x + 9000 – 1800x = 4x^2 + 20x\) \(9000 = 4x^2 + 20x\) \(4x^2 + 20x – 9000 = 0\)
Step 6: Divide by 4: \(x^2 + 5x – 2250 = 0\)
Step 7: Solve quadratic: \(x^2 + 50x – 45x – 2250 = 0\) \(x(x + 50) – 45(x + 50) = 0\) \((x – 45)(x + 50) = 0\)
Step 8: x = 45 or x = -50 Speed cannot be negative, so x = 45 km/h
Q.33. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds, the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000√3 m, find the speed of the aeroplane.
📥 View Answer for Part (A)
Solution for Part (A):
Step 1: From first point (ground level): tan 60° = \(\frac{h}{x}\) √3 = \(\frac{h}{x}\) h = x√3 …(1)
Step 2: From second point (10 m above ground): Height from this point = (h – 10) m tan 45° = \(\frac{h – 10}{x}\) 1 = \(\frac{h – 10}{x}\) x = h – 10 …(2)
Step 3: Substitute (2) in (1): h = (h – 10)√3 h = h√3 – 10√3
Step 4: Rearrange: h√3 – h = 10√3 h(√3 – 1) = 10√3 h = \(\frac{10√3}{√3 – 1}\
Step 5: Rationalize: h = \(\frac{10√3}{√3 – 1} \times \frac{√3 + 1}{√3 + 1}\) h = \(\frac{10√3(√3 + 1)}{3 – 1}\) h = \(\frac{10√3(√3 + 1)}{2}\) h = 5√3(√3 + 1) h = 5(3 + √3) h = 15 + 5√3 meters
📥 View Answer for Part (B)
Solution for Part (B):
Step 1: Let initial position be point A Angle of elevation = 60° tan 60° = \(\frac{3000√3}{AB}\) √3 = \(\frac{3000√3}{AB}\) AB = 3000 m
Step 2: After 30 seconds at point C Angle of elevation = 30° tan 30° = \(\frac{3000√3}{CB}\) \(\frac{1}{√3} = \frac{3000√3}{CB}\) CB = 3000√3 × √3 = 3000 × 3 = 9000 m
Step 3: Distance travelled AC = CB – AB AC = 9000 – 3000 = 6000 m
Step 4: Speed = Distance ÷ Time Speed = \(\frac{6000}{30} = 200\) m/s
Step 5: Convert to km/h: 1 m/s = 3.6 km/h 200 m/s = 200 × 3.6 = 720 km/h
Q.34. A cylindrical bucket, 32 cm high and with base radius 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of the cone is 3.5 cm, find the volume of the solid. Also find the cost of painting its outer surface at the rate of ₹ 4 per cm².
📥 View Answer for Part (A)
Solution for Part (A):
Step 1: Volume of sand = Volume of cylinder Volume = πr²h = π × (18)² × 32 = π × 324 × 32 = 10368π cm³
Step 2: Volume of conical heap = \(\frac{1}{3}πR^2H\) Where R = radius of cone, H = 24 cm \(\frac{1}{3}πR^2 × 24 = 10368π\)
Step 3: Simplify: \(8πR^2 = 10368π\) \(R^2 = \frac{10368}{8} = 1296\) \(R = √1296 = 36\) cm
Step 4: Find slant height (l): \(l = √(R^2 + H^2) = √(36^2 + 24^2)\) = √(1296 + 576) = √1872 = √(144 × 13) = 12√13 cm
📥 View Answer for Part (B)
Solution for Part (B):
Step 1: Volume of solid = Volume of hemisphere + Volume of cone Step 2: Volume of hemisphere = \(\frac{2}{3}πr^3\) = \(\frac{2}{3} × \frac{22}{7} × 7^3\) = \(\frac{2}{3} × \frac{22}{7} × 343\) = \(\frac{2 × 22 × 49}{3} = \frac{2156}{3}\) cm³
Step 3: Volume of cone = \(\frac{1}{3}πr^2h\) = \(\frac{1}{3} × \frac{22}{7} × 7^2 × 3.5\) = \(\frac{1}{3} × \frac{22}{7} × 49 × 3.5\) = \(\frac{1}{3} × 22 × 7 × 3.5 = \frac{1}{3} × 22 × 24.5\) = \(\frac{539}{3}\) cm³
Step 4: Total volume = \(\frac{2156}{3} + \frac{539}{3} = \frac{2695}{3} = 898.33\) cm³
Step 5: Surface area for painting: = Curved surface area of hemisphere + Curved surface area of cone CSA of hemisphere = \(2πr^2 = 2 × \frac{22}{7} × 49 = 308\) cm²
Step 6: Slant height of cone: \(l = √(r^2 + h^2) = √(7^2 + 3.5^2)\) = √(49 + 12.25) = √61.25 = 7.826 cm
Step 7: CSA of cone = \(πrl = \frac{22}{7} × 7 × 7.826\) = 22 × 7.826 = 172.17 cm²
Step 8: Total surface area = 308 + 172.17 = 480.17 cm²
Step 9: Cost of painting = 480.17 × 4 = ₹1920.68
Q.35. Find the mean, median and mode of the following data:
| Class Interval | Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | 10 |
| 20-30 | 18 |
| 30-40 | 30 |
| 40-50 | 20 |
| 50-60 | 12 |
| 60-70 | 5 |
OR
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f:
| Daily Pocket Allowance (₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
|---|---|---|---|---|---|---|---|
| Number of Children | 7 | 6 | 9 | 13 | f | 5 | 4 |
📥 View Answer for Part (A) – Mean, Median, Mode
| Class Interval | Frequency (f) | Mid-point (x) | f × x | Cumulative Frequency (cf) |
|---|---|---|---|---|
| 0-10 | 5 | 5 | 25 | 5 |
| 10-20 | 10 | 15 | 150 | 15 |
| 20-30 | 18 | 25 | 450 | 33 |
| 30-40 | 30 | 35 | 1050 | 63 |
| 40-50 | 20 | 45 | 900 | 83 |
| 50-60 | 12 | 55 | 660 | 95 |
| 60-70 | 5 | 65 | 325 | 100 |
| Total | Σf = 100 | Σfx = 3560 |
Mean Calculation
Formula: Mean = \(\frac{Σfx}{Σf}\)
Step 1: Σf = 100, Σfx = 3560
Step 2: Mean = \(\frac{3560}{100}\)
Step 3: Mean = 35.6
Mean = 35.6
Median Calculation
Step 1: N = 100, N/2 = 50
Step 2: Median class = 30-40 (cf ≥ 50)
Step 3: l = 30, f = 30, cf = 33, h = 10
Step 4: Formula: Median = \(l + \frac{\frac{N}{2} – cf}{f} × h\)
= \(30 + \frac{50 – 33}{30} × 10\)
= \(30 + \frac{17}{30} × 10 = 30 + 5.67\)
Median = 35.67
Mode Calculation
Step 1: Modal class = 30-40 (highest f = 30)
Step 2: l = 30, f₁ = 30, f₀ = 18, f₂ = 20, h = 10
Step 3: Formula: Mode = \(l + \frac{f₁ – f₀}{2f₁ – f₀ – f₂} × h\)
= \(30 + \frac{30 – 18}{2×30 – 18 – 20} × 10\)
= \(30 + \frac{12}{60 – 38} × 10 = 30 + \frac{120}{22}\)
= \(30 + 5.45 = 35.45\)
Mode = 35.45
Final Answers for Part A:
Mean = 35.6 | Median = 35.67 | Mode = 35.45
📥 View Answer for Part (B) – Missing Frequency
| Class Interval | Number of Children (f) | Mid-point (x) | f × x |
|---|---|---|---|
| 11-13 | 7 | 12 | 84 |
| 13-15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17-19 | 13 | 18 | 234 |
| 19-21 | f | 20 | 20f |
| 21-23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
| Total | Σf = 44 + f | Σfx = 752 + 20f |
Step-by-Step Solution:
Step 1: Calculate total frequency
Σf = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f
Step 2: Calculate sum of f×x
Σfx = (7×12) + (6×14) + (9×16) + (13×18) + (f×20) + (5×22) + (4×24)
= 84 + 84 + 144 + 234 + 20f + 110 + 96
= 752 + 20f
Step 3: Apply mean formula
Mean = \(\frac{Σfx}{Σf}\)
Given mean = 18
So, \(18 = \frac{752 + 20f}{44 + f}\)
Step 4: Cross multiply and solve
18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f
792 – 752 = 20f – 18f
40 = 2f
f = 20
Final Answer for Part B:
Missing frequency f = 20
📊 Important Statistical Formulas
Mean (Direct Method)
\(\bar{x} = \frac{Σf_ix_i}{Σf_i}\)
Where:
fᵢ = frequency of i-th class
xᵢ = mid-point of i-th class
Median (Grouped Data)
Median = \(l + \left(\frac{\frac{N}{2} – cf}{f}\right) × h\)
Where:
l = lower limit of median class
N = total frequency
cf = cumulative frequency before median class
f = frequency of median class
h = class width
Mode (Grouped Data)
Mode = \(l + \left(\frac{f₁ – f₀}{2f₁ – f₀ – f₂}\right) × h\)
Where:
l = lower limit of modal class
f₁ = frequency of modal class
f₀ = frequency of class before modal class
f₂ = frequency of class after modal class
h = class width
💡 Exam Strategy: For statistics questions, always follow these steps: 1) Create complete table with all required columns, 2) Double-check your summations, 3) Clearly identify median/modal class, 4) Apply formulas correctly, 5) Verify your answers make sense in context.