Class 10 Case Study Based Board Questions

Answer the following questions:

1. What is the probability that Rahul draws an even number? Show Solution ▶

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Step 1: Cards are numbered from 10 to 74.

Step 2: Total number of cards
= 74 − 10 + 1 = 65

Step 3: Even numbers between 10 and 74 are:
10, 12, 14, … , 74

Step 4: Number of even numbers
= (74 − 10) ÷ 2 + 1 = 33

Probability = Favorable outcomes / Total outcomes
= 33 / 65
2. What is the probability that Rahul draws an odd number less than 30? Show Solution ▶

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Step 1: The cards are numbered from 10 to 74.

Step 2: Total number of cards
= 74 − 10 + 1 = 65

Step 3: Odd numbers less than 30 between 10 and 74 are:
11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Step 4: Number of odd numbers less than 30
= 10

Probability = Favorable outcomes / Total outcomes
= 10 / 65
= 2 / 13
3. (a) Probability of prime number between 50 and 74 Show Solution ▶

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   Step 1: The cards are numbered from 10 to 74.

   Step 2: Total number of cards
   = 74 − 10 + 1 = 65

   Step 3: Prime numbers between 50 and 74 are:
   53, 59, 61, 67, 71, 73

   Step 4: Number of prime numbers between 50 and 74
   = 6

   Probability = Favorable outcomes / Total outcomes
   = 6 / 65
   
           OR
   (b) Probability of even number divisible by 5 Show Solution ▶

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Step 1: The cards are numbered from 10 to 74.

Step 2: Total number of cards
= 74 − 10 + 1 = 65

Step 3: Even numbers divisible by 5 are multiples of 10.

Step 4: Multiples of 10 between 10 and 74 are:
10, 20, 30, 40, 50, 60, 70

Step 5: Number of even numbers divisible by 5
= 7

Probability = Favorable outcomes / Total outcomes
= 7 / 65

Answer the following questions:

1. Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. Show Solution ▶

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Step 1: Distance of first round (a) = 300 m

Step 2: Increase in distance each round (d) = 50 m

Step 3: Formula for nth term of an A.P.
a_n = a + (n − 1)d

Step 4: 4th term = 300 + 3×50 = 450 m
5th term = 300 + 4×50 = 500 m
6th term = 300 + 5×50 = 550 m

Fourth, Fifth and Sixth terms are:
450 m, 500 m, 550 m
2. Determine the distance of the 8th round. Show Solution ▶

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Step 1: First term (a) = 300 m, Common difference (d) = 50 m

Step 2: Formula for nth term
a_n = a + (n − 1)d

Step 3: 8th term = 300 + 7×50

Step 4: = 650 m

Distance of the 8th round = 650 metres
3. (a) Find the total distance run after completing all 10 rounds. Show Solution ▶

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   Step 1: a = 300 m, d = 50 m, n = 10

   Step 2: Formula for sum of n terms
   S_n = n/2 [2a + (n − 1)d]

   Step 3: S₁₀ = 10/2 [2×300 + 9×50]

   Step 4: = 5 [600 + 450] = 5×1050

   Total distance run = 5250 metres
   
          OR
   (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? Show Solution ▶

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Step 1: a = 300 m, d = 50 m, n = 6

Step 2: Formula for sum of n terms
S_n = n/2 [2a + (n − 1)d]

Step 3: S₆ = 6/2 [2×300 + 5×50]

Step 4: = 3 [600 + 250] = 3×850

Total distance run = 2550 metres

Answer the following questions:

1. Find the central angle of each sector. Show Solution ▶

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Step 1: The circle is divided into 10 equal sectors.

Step 2: Total angle at the center of a circle = 360°.

Step 3: Central angle of each sector
= 360° ÷ 10

Central angle of each sector = 36°
2. Find the length of the arc ACB. Show Solution ▶

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Step 1: Diameter of the circle = 35 mm

Step 2: Radius (r) = 35 ÷ 2 = 17.5 mm

Step 3: Arc ACB covers 3 sectors
Central angle = 3 × 36° = 108°

Step 4: Formula for arc length
= (θ / 360) × 2πr

Step 5: = (108 / 360) × 2 × (22/7) × 17.5

Length of arc ACB = 33 mm
3. (a) Find the area of each sector of the brooch. Show Solution ▶

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   Step 1: Radius (r) = 17.5 mm

   Step 2: Central angle of one sector = 36°

   Step 3: Formula for area of a sector
   = (θ / 360) × πr²

   Step 4: = (36 / 360) × (22/7) × (17.5)²

   Area of each sector = 96.25 mm²
   
          OR
   (b) Find the total length of the silver wire used. Show Solution ▶

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Step 1: Radius of the circle = 17.5 mm

Step 2: Circumference of the circle
= 2πr = 2 × (22/7) × 17.5 = 110 mm

Step 3: Number of diameters used = 5

Step 4: Length of one diameter = 35 mm

Step 5: Total length of diameters
= 5 × 35 = 175 mm

Total length of silver wire used
= 110 + 175 = 285 mm

Based on the above given information, answer the following questions :

1. If $ CD$ is $ h$ metres, find the distance $ BD$ in terms of $'h'$. Show Solution ▶

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Step 1: At point B, angle of elevation to the top of lighthouse = 45°.

Step 2: In right-angled triangle BDC,
tan 45° = CD / BD

Step 3: Since tan 45° = 1,
1 = h / BD

BD = h metres
2. Find distance $ BC$ in terms of $'h'$. Show Solution ▶

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Step 1: From the figure,
$ BC^2 = BD^2 + DC^2$

Step 2: From Question 1,
BD = h and DC = h

$ BC = \sqrt{h^2 + h^2}$ = $\sqrt{2}h $metres
3. (a) Find the height $ CE$ of the lighthouse [use $\pi=1.73$] Show Solution ▶

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   Step 1: At point A, angle of elevation = 60°.

   Step 2: In right-angled triangle ACE,
   tan 60° = CE / AE

   Step 3: tan 60° = √3 = 1.73

   Step 4: From the figure,
   AE = BD = h

   Step 5:
   CE = 1.73 × h

   Height of the lighthouse, CE = 1.73h metres
   
          OR
   (b) Find distance $ AE$, if $ AC = 100~ m$. Show Solution ▶

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Step 1: In triangle ACE,
AC² = AE² + CE²

Step 2: Given:
AC = 100 m
CE = 1.73h

Step 3:
100² = AE² + (1.73h)²

Step 4:
10000 = AE² + 2.99h²

AE = √(10000 − 2.99h²) metres

If the school rents 'x' chairs and 'y' tables, answer the following questions :

1. Write down the pair of linear equations representing the given information. Show Solution ▶

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Step 1: Let the number of chairs rented be $$x$$ and the number of tables rented be $$y$$.

Step 2: Total number of items rented = 300
$$x + y = 300$$

Step 3: Rent of each chair = ₹50
Rent of each table = ₹200

Step 4: Total rent = ₹30,000
$$50x + 200y = 30000$$

Required pair of linear equations:
$$x + y = 300$$ $$50x + 200y = 30000$$
2. What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? Show Solution ▶

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Step 1: If no chairs are rented, then:
$$x = 0$$

Step 2: Rent per table = ₹200
Total amount = ₹30,000

Step 3:
$$200y = 30000$$

Step 4:
$$y = \frac{30000}{200} = 150$$

Maximum number of tables = 150
3. (a) Find the number of chairs and number of tables rented by the school. Show Solution ▶

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   Step 1: Given equations:
   $$x + y = 300 \quad (1)$$ $$50x + 200y = 30000 \quad (2)$$

   Step 2: Divide equation (2) by 50:
   $$x + 4y = 600 \quad (3)$$

   Step 3: Subtract equation (1) from equation (3):
   $$(x + 4y) - (x + y) = 600 - 300$$

   Step 4:
   $$3y = 300 \Rightarrow y = 100$$

   Step 5: Substitute $$y = 100$$ in equation (1):
   $$x + 100 = 300 \Rightarrow x = 200$$

   Number of chairs = 200
   Number of tables = 100
   
          OR
   (b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. Show Solution ▶

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Step 1: Let number of chairs = $$x$$ and number of tables = $$y$$.

Step 2: Total items:
$$x + y = 300$$

Step 3: Maximum total cost:
$$50x + 200y = 27000$$

Step 4: Divide second equation by 50:
$$x + 4y = 540$$

Step 5: Subtract first equation:
$$(x + 4y) - (x + y) = 540 - 300$$

Step 6:
$$3y = 240 \Rightarrow y = 80$$

Step 7:
$$x = 300 - 80 = 220$$

Chairs = 220
Tables = 80

Based on the above information, answer the following questions :

1. Find the volume of the bigger cylindrical part. Show Solution ▶

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Step 1: Diameter of bigger cylinder = $$7\text{ cm}$$
Radius $$r = \frac{7}{2} = 3.5\text{ cm}$$

Step 2: Height (length) of bigger cylinder = $$12\text{ cm}$$

Step 3: Formula for volume of a cylinder:
$$V = \pi r^2 h$$

Step 4:
$$V = \frac{22}{7} \times (3.5)^2 \times 12$$

Volume of the bigger cylindrical part = 462 cm³
2. Find the curved surface area of the bigger cylindrical part. Show Solution ▶

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Step 1: Radius $$r = 3.5\text{ cm}$$, Height $$h = 12\text{ cm}$$

Step 2: Formula for curved surface area:
$$\text{CSA} = 2\pi rh$$

Step 3:
$$\text{CSA} = 2 \times \frac{22}{7} \times 3.5 \times 12$$

Curved surface area of the bigger cylindrical part = 264 cm²
3. (a) Find the ratio of the volume of the bigger cylindrical part to the total volume of the two smaller (identical) cylindrical parts. Show Solution ▶

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   Step 1: Diameter of smaller cylinder = $$2.1\text{ cm}$$
   Radius $$r = \frac{2.1}{2} = 1.05\text{ cm}$$

   Step 2: Height of each smaller cylinder = $$5\text{ cm}$$

   Step 3: Volume of one smaller cylinder:
   $$V = \pi r^2 h$$
   $$= \frac{22}{7} \times (1.05)^2 \times 5 = 17.325\text{ cm}^3$$

   Step 4: Volume of two smaller cylinders:
   $$= 2 \times 17.325 = 34.65\text{ cm}^3$$

   Step 5: Volume of bigger cylinder = $$462\text{ cm}^3$$

   Required ratio = $$462 : 34.65 = \boxed{40 : 3}$$
   
          OR
   (b) Find the sum of the curved surface areas of the two identical smaller cylindrical parts. Show Solution ▶

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Step 1: Radius of smaller cylinder $$r = 1.05\text{ cm}$$
Height $$h = 5\text{ cm}$$

Step 2: Curved surface area of one smaller cylinder:
$$\text{CSA} = 2\pi rh$$

Step 3:
$$= 2 \times \frac{22}{7} \times 1.05 \times 5 = 33\text{ cm}^2$$

Step 4: Curved surface area of two smaller cylinders:
$$= 2 \times 33$$

Sum of curved surface areas = 66 cm²