CBSE Maths Standard Sample Paper Solution 2024-2025 (Solved)

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Solutions for all questions of CBSE Sample Paper 2024-2025

Question 1. The graph of a quadratic polynomial p(x) passes through the points (-6,0), (0, -30), (4,-20) and (6,0). The zeroes of the polynomial are 

Option: A) – 6,0 B) 4, 6 C) – 30,-20 D) – 6,6

Ans: We are given that the quadratic polynomial passes through the points (-6, 0), (0, -30), (4, -20), and (6, 0). The zeroes of the polynomial are the x-coordinates where the polynomial equals zero. From the points, the zeroes are at x = -6 and x = 6. So, the correct answer is D) -6, 6.

Question 2. The value of k for which the system of equations 3x-ky= 7 and 6x + 10y =3 is inconsistent, is

Option: A) -10 B) -5 C) 5 D) 7

Solutions: The given system of equations is:

• 3𝑥 − 𝑘𝑦 = 7
• 6𝑥 + 10𝑦 = 3

For the system to be inconsistent, the two lines must be parallel, meaning their slopes or (coefficient of x and y) must be equal.

• a1x + b1y + c1 = 0
• a2x + b2y + c2 = 0

a1/a2 = b1/b2 ≠ c1/c2

Hence: 

• 3/k = -3/5
• k = -5

Answer is B) -5

Question 3. Which of the following statements is not true?

A) A number of secants can be drawn at any point on the circle.

B) Only one tangent can be drawn at any point on a circle.

C) A chord is a line segment joining two points on the circle

D) From a point inside a circle only two tangents can be drawn.

The incorrect statement is A). 

Other statements are correct:

• B) Only one tangent can be drawn at any point on a circle. – This is the definition of a tangent.
• C) A chord is a line segment joining two points on the circle. – This is the definition of a chord.
• D) From a point inside a circle, only two tangents can be drawn. – This is incorrect. You cannot draw any tangents from a point inside a circle. Tangents can only be drawn from a point outside the circle.

4. If nth term of an A.P. is 7n – 4 then the common difference of the A.P. is 

Option: A) 7 B) 7n C) – 4 D) 4

Solution: The nth term of an A.P. is given by the formula: an = a1 + (n-1)d where:

• an is the nth term
• a1 is the first term
• n is the position of the term
• d is the common difference

We are given: an = 7n – 4

• Comparing this with the general formula, we can see:
  • a1 = 7 – 4 = 3 (by substituting n = 1)
  • d = 7 (the coefficient of n)

Therefore, the common difference of the A.P. is 7.

Answer: A) 7

Question 5. The radius of the base of a right circular cone and the radius of a sphere are each 5 cm in length. If the volume of the cone is equal to the volume of the sphere then the height of the cone is 

Option: A) 5 cm B) 20 cm C) 10 cm D) 4 cm

Solution: 

► Needed Formulas To solve the Question

• Volume of a Cone: V_cone = (1/3)πr²h
• Volume of a Sphere: V_sphere = (4/3)πr³

► Set Up the Equation

• Given:
  
  • Radius of cone (r_cone) = 5 cm
  • Radius of sphere (r_sphere) = 5 cm
  • Volume of cone = Volume of sphere
• Therefore: (1/3)π(5 cm)²h = (4/3)π(5 cm)³
  

► Solve for the Height (h)

• Simplify the equation: (25/3)πh = (500/3)π
• Divide both sides by (25/3)π: h = (500/3)π / (25/3)π
• Cancel out π and simplify: h = 20 cm

Therefore, the height of the cone is 20 cm.

Answer: B) 20 cm

Question 6) If tan𝜃 = 5/2, then: 

is equal to 

Option: A) 11/9  B)3/2 C) 9/11 D) 4

Step 1) Find sinθ and cosθ

• Given: tanθ = 5/2
• We know that tanθ = sinθ / cosθ
• Let sinθ = 5k and cosθ = 2k (where k is a constant)

Step 2) Substitute sinθ and cosθ in the given expression

• (4sinθ + cosθ) / (4sinθ – cosθ)
• = (4 × 5k + 2k) / (4 × 5k – 2k)
• = (20k + 2k) / (20k – 2k)
• = 22k / 18k
• = 11/9

Therefore, the value of (4sinθ + cosθ) / (4sinθ – cosθ) is 11/9.

Answer: A) 11/9

Question 7. In the given figure, a tangent has been drawn at a point P on the circle centred at O. 

A) 110^° B) 70^° C) 140^° D)55^°

Given:

• A circle with center O
• Tangent TP at point P
• ∠TPQ = 110°

To find: ∠POQ

Solution:

• Since TP is tangent to the circle at P, ∠OPT = 90° (tangent is perpendicular to radius at the point of contact).
  
• Now, in triangle TPQ, we have:
  
  • ∠TPQ = 110° (given)
  • ∠OPT = 90° (proven)
  • Therefore, ∠PQO = 180° – 110° – 90° = 20° (sum of angles in a triangle is 180°)
• In triangle POQ, we have:
  
  • ∠OPQ = 20°
  • ∠PQO = 20° 
  • Therefore, ∠POQ = 180° – 20° – 20° = 140°

So, the answer is C) 140°.

Question 8: A quadratic polynomial having zeroes: -√(5/2) and √(5/2)

A) 𝑥² − 5√2 x +1  B) 8𝑥² – 20 C) 15𝑥² – 6 D) 𝑥² – 2√5 x -1

Given:

• Zeroes of a quadratic polynomial are -√(5/2) and √(5/2)

To find: The quadratic polynomial

Solution:

• If the zeroes of a quadratic polynomial are α and β, then the polynomial can be written as:
  
  • (x – α)(x – β) = 0
• Substituting the given zeroes:
  
  • (x + √(5/2))(x – √(5/2)) = 0
  • x² – (5/2) = 0
  • 2x² – 5 = 0

So, the answer is B) 8x² – 20.

Question 9: Consider the frequency distribution of 45 observations. 

The upper limit of median class is Option: A) 20 B) 10 C) 30 D) 40

Answer: C) 30

Given:

To find the upper limit of the median class.

Total frequency (N) = 45.

The median class is identified as the class where the cumulative frequency is at least N/2 = 45/2 = 22.5.

The cumulative frequency first exceeds 22.5 in the 20-30 class.

The upper limit of the median class is the upper boundary of 20-30, which is 30.

Answer: C) 30

Question 10: O is the point of intersection of two chords AB and CD of a circle

Answer: C) isosceles but not similar

Given: 

• ∠BOC=80°
• OA = OD

We are to determine the relationship between △ODA and △OBC.

1. Key Observations:
   • OA=OD suggests that △ODA is isosceles.
   • Similarly, in △OBC, OB = OC, so △OBC is also isosceles.
   • ∠BOC=80° implies that the two triangles share symmetry but are not similar in all aspects (angles and sides vary).
2. Therefore, the triangles are isosceles but not similar.

Answer: C) isosceles but not similar

😊 Note: The Triangle Similarity Theorem states that two triangles are similar if their corresponding angles are equal, and their corresponding sides are in proportion.

Key Triangle Similarity Rules

1. AA (Angle-Angle) Similarity:
   If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
2. SAS (Side-Angle-Side) Similarity:
   If two sides of one triangle are in proportion to two sides of another triangle and the included angles are equal, the triangles are similar.
3. SSS (Side-Side-Side) Similarity:
   If the corresponding sides of two triangles are in the same ratio (proportion), then the triangles are similar.

Why Are Triangles OAC and OBD Similar?

In this case:

1. Equal Angles:
   • ∠AOC=∠BOD (vertically opposite angles).
   • ∠OAC=∠ODB (angles in the same segment of a circle).
2. Equal Sides:
   • OA = OC and OB = OD because all are radii of the circle.

Thus, by the AA Similarity Rule, triangles OAC and OBD are similar.

This is how the Triangle Similarity Theorem applies in this case! 😊

Question 11: The roots of the quadratic equation x² + x – 1 = 0 are

Option: A) Irrational and distinct B) not real C ) rational and distinct D) real and equal

Answer: A) Irrational and distinct

Solution:

1. Identify the coefficients:
   • a = 1 (coefficient of x²)
   • b = 1 (coefficient of x)
   • c = -1 (constant term)
2. Apply the quadratic formula:

The quadratic formula is:

Substitute the values of a, b, and c:
x = (-1 ± √(1² – 4 * 1 * -1)) / 2 * 1

x = (-1 ± √(1 + 4)) / 2

x = (-1 ± √5) / 2

3. Determine the nature of the roots:
   
   • The discriminant (b² – 4ac) is 5, which is positive and not a perfect square.
   • Therefore, the roots are irrational and distinct.

Hence Answer: A) Irrational and distinct

Question 12: If θ = 30°, then the value of 3tanθ is

A)1 B) 1/√3 C ) 3/√3 (D) not defined

Answer: C) 3/√3  or √3

Solution:

1. Recall the value of tan(30°):
   
   • tan(30°) = 1/√3  
2. Calculate 3tan(30°):
   
   • 3tan(30°) = 3 * (1/√3) = 3/√3  or √3 

Answer: C) 3/√3  or √3

Question 13: The volume of a solid hemisphere is 396/7 cm³. The total surface area of the solid hemisphere (in sq.cm) is

Option: A) 396/7 B) 594/7 C) 549/7 D) 604/7 

Answer: B) 594/7

Solution:

1. Recall the volume formula for a hemisphere:
   
   • Volume of a hemisphere = (2/3)πr³  
   • where r is the radius of the hemisphere
2. Set up the equation:
   
   • (2/3)πr³ = 396/7
3. Solve for the radius (r):
   
   • r³ = (396/7) * (3/2π)
   • r³ = (1188/14π)
   • r = ∛(1188/14π)
   • r = 3 cm (after simplifying)
4. Calculate the total surface area:
   
   • Total surface area of a hemisphere = 3πr²  
   • Substitute r = 3 cm:
     • Total surface area = 3π(3²) = 27π = 594/7 cm²

Answer: B) 594/7

Question 14: In a bag containing 24 balls, 4 are blue, 11 are green, and the rest are white. One ball is drawn at random. The probability that the drawn ball is white in color is

Option: 𝐴) 1/6 B) 3/8 C ) 11/24 D) ⅝

Answer: B) 3/8

Solution:

1. Calculate the number of white balls:
   
   • Number of white balls = Total balls – Blue balls – Green balls
   • Number of white balls = 24 – 4 – 11 = 9
2. Calculate the probability of drawing a white ball:
   
   • Probability = (Number of white balls) / (Total number of balls)
   • Probability = 9/24 = 3/8

Answer: B) ⅜

Question 15: The point on the x-axis nearest to the point (-4, -5) is

Option: A) (0, 0) B) (-4, 0) C ) (-5, 0) D) (√41, 0)

Answer: B) (-4, 0)

Solution:

1. Understand the x-axis:
   
   • The x-axis is the horizontal line where the y-coordinate is always 0.  
2. Find the point with the same x-coordinate as (-4, -5):
   
   • The point on the x-axis with the same x-coordinate as (-4, -5) is (-4, 0).

Answer: B) (-4, 0)

Question 16: Which of the following gives the middle-most observation of the data?

Option: A) Median B) Mean C) Range D) Mode

Solution:

• The median is the middle value in a dataset when the data is arranged in order.  

Answer: A) Median

Question 17: A point on the x-axis divides the line segment joining the points A(2, -3) and B(5, 6) in the ratio 1:2. The point is

Solution:

1. Use the section formula:
   
   • The section formula for dividing a line segment in the ratio m₁:m₂ is:

• where (x1, y1) and (x2, y2) are the coordinates of the endpoints, and  m₁:m₂ is the ratio.

2. Apply the section formula:
   
   • Let the point on the x-axis be (x, 0).
   •  m₁:m₂ = 1:2, so m = 1 and n = 2.
   • (x, 0) = ((2 * 2 + 5 * 1)/(1 + 2), (-3 * 2 + 6 * 1)/(1 + 2))
   • (x, 0) = (9/3, 0/3)
   • (x, 0) = (3, 0)

Answer: C) (3, 0)

Question 18: A card is drawn from a well-shuffled deck of playing cards. The probability of getting a red face card is

Option: 𝐴) 3/13 B) 1/2 C) 3/52 D) 3/26

Answer: D) 3/26

Solution:

1. Count the number of red face cards:
   
   • There are 6 red face cards in a deck: Jack, Queen, and King of Hearts and Diamonds.
2. Calculate the probability:
   
   • Probability = (Number of red face cards) / (Total number of cards)
   • Probability = 6/52 = 3/26

Answer: D) 3/26

Question 19: Assertion (A): HCF of any two consecutive even natural numbers is always 2. Reason (R): Even natural numbers are divisible by 2.

Answer:

• Reason (R) is true: Even natural numbers are indeed divisible by 2.
  
• Assertion (A) is true:
  
  • Let’s consider two consecutive even numbers: 2n and 2n + 2, where n is any natural number.
  • Their difference is 2.
  • Since the difference between two numbers is 2, their HCF can be at most 2.
  • And since both numbers are even, they are both divisible by 2.
  • Therefore, the HCF of any two consecutive even natural numbers is 2.
• Reason (R) explains Assertion (A): The fact that even numbers are divisible by 2 is the reason why their HCF is 2.

Therefore, the correct answer is (A): Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question 20: Assertion (A): If the radius of a sector of a circle is reduced to its half and the angle is doubled, then the perimeter of the sector remains the same. Reason (R): The perimeter of a sector is given by the formula: Perimeter = 2r + (θ/360°) × 2πr, where r is the radius and θ is the central angle.

Answer:

• Reason (R) is true: The formula for the perimeter of a sector is correct.
  
• Assertion (A) is false:
  
  • Let’s consider a sector with initial radius r and central angle θ.
  • Its initial perimeter is: P1 = 2r + (θ/360°) × 2πr
  • Now, let’s reduce the radius to half (r/2) and double the angle (2θ).
  • The new perimeter is: P2 = 2(r/2) + (2θ/360°) × 2π(r/2) = r + (θ/360°) × 2πr
  • Comparing P1 and P2, we can see that P1 is not equal to P2.

Therefore, the correct answer is (D): Assertion (A) is false but reason (R) is true.

Question 21: A) Find the H.C.F and L.C.M of 480 and 720 using the Prime factorization method.

Solution:

1. Prime Factorization:
   
   • 480 = 2 × 2 × 2 × 2 × 2 × 3 × 5
   • 720 = 2 × 2 × 2 × 3 × 3 × 5
2. H.C.F (Highest Common Factor):
   
   • Take the common prime factors with their lowest powers.
   • H.C.F = 2 × 2 × 2 × 3 × 5 = 120
3. L.C.M (Least Common Multiple):
   
   • Take all the prime factors with their highest powers.
   • L.C.M = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

B) The H.C.F of 85 and 238 is expressible in the form 85m – 238. Find the value of m.

Solution:

1. Prime Factorization:
   
   • 85 = 5 × 17
   • 238 = 2 × 7 × 17
2. H.C.F:
   
   • H.C.F = 17
3. Equate:
   
   • 85m – 238 = 17
   • 85m = 255
   • m = 3

Therefore, the value of m is 3.

Question 22 A) Two dice are rolled together bearing numbers 4, 6, 7, 9, 11, 12. Find the probability that the product of numbers obtained is an odd number.

Solution:

1. Possible outcomes:
   
   • The product of two numbers can only be odd if both numbers are odd.
   • Odd numbers on the dice: 7, 9, 11
2. Number of favorable outcomes:
   
   • There are 3 odd numbers.
   • Number of favorable outcomes = 3 × 3 = 9
3. Total number of possible outcomes:
   
   • Total outcomes = 6 × 6 = 36
4. Probability:
   
   • Probability = Number of favorable outcomes / Total number of possible outcomes
   • Probability = 9/36 = 1/4

B) How many positive three digit integers have the hundredths digit 8 and unit’s digit 5? Find the probability of selecting one such number out of all three digit numbers.

Solution:

1. Hundredths digit is 8 and unit’s digit is 5:
   
   • There are 10 possible values for the tens digit (0-9).
   • So, there are 10 such numbers.
2. Total three-digit numbers:
   
   • There are 900 three-digit numbers (from 100 to 999).
3. Probability:
   
   • Probability = Number of favorable outcomes / Total number of possible outcomes
   • Probability = 10/900 = 1/90

Question 23: Evaluate: 2(sin²60°)² × tan²30° / sec²45°

Solution:

1. Substitute values:
   
   • sin 60° = √3/2
   • tan 30° = 1/√3
   • sec 45° = √2
2. Evaluate:
   
   • 2 × (√3/2)^2 × (1/√3)^2 / (√2)^2
   • 2 × (3/4) × (1/3) / 2
   • 1/4

Therefore, the value of the expression is 1/4.

Question 24: Find the point(s) on the x-axis which is at a distance of √41 units from the point (8, -5).

Solution:

1. Let the point on the x-axis be (x, 0).
2. Distance formula:
   • √[(x – 8)² + (0 – (-5))² ] = √41
   • (x – 8)² + 25 = 41
   • (x – 8)² = 16
   • x – 8 = ±4
   • x = 8 ± 4

Therefore, the points on the x-axis are (4, 0) and (12, 0).

Question 25: Show that the points A(-5,6), B(3, 0) and C(9, 8) are the vertices of an isosceles triangle.

Solution:

1. Calculate distances:
   
   • AB = √[(-5 – 3)² + (6 – 0)² ] = √100 = 10
   • BC = √[(3 – 9)²  + (0 – 8)² ] = √100 = 10
   • AC = √[(-5 – 9)² + (6 – 8)² ] = √200 = 10√2
2. Compare distances:
   
   • AB = BC

Since two sides of the triangle have equal lengths, the points A, B, and C form an isosceles triangle.

Question 26) In 𝛥ABC, D, E and F are midpoints of BC,CA and AB respectively. Prove that △ 𝐹𝐵𝐷 ∼ △ DEF and △ DEF ∼ △ ABC

Solution: 

To Prove:

1. ΔFBD ~ ΔDEF
2. ΔDEF ~ ΔABC

Proof:

1. Proving ΔFBD ~ ΔDEF

• Step 1: Identify Corresponding Sides
  
  • In ΔFBD and ΔDEF, we can see that:
    • FB is corresponding to DE
    • BD is corresponding to EF
    • FD is common to both triangles
• Step 2: Apply Midsegment Theorem
  
  • According to the Midsegment Theorem, if a line segment connects the midpoints of two sides of a triangle, then it is parallel to the third side and half its length
  • Since F and D are midpoints, FD || EC and FD = 1/2 EC
  • Similarly, since D and E are midpoints, DE || FB and DE = 1/2 FB
• Step 3: Corresponding Sides Proportional
  
  • From the Midsegment Theorem, we know:
    • FD = 1/2 EC
    • DE = 1/2 FB
  • Therefore, FD/DE = 1/2 EC / 1/2 FB
  • FD/DE = FB/DE
• Step 4: Conclusion
  
  • We have shown that the corresponding sides of ΔFBD and ΔDEF are proportional.
  • Also, ∠FBD = ∠DEF (since FD || EC)
  • Therefore, by the Side-Angle-Side (SAS) similarity criterion, ΔFBD ~ ΔDEF.

2. Proving ΔDEF ~ ΔABC

• Step 1: Identify Corresponding Sides
  
  • In ΔDEF and ΔABC, we can see that:
    • DE is corresponding to AB
    • EF is corresponding to BC
    • FD is corresponding to AC
• Step 2: Apply Midsegment Theorem
  
  • As mentioned earlier, DE = 1/2 AB, EF = 1/2 BC, and FD = 1/2 AC.
• Step 3: Corresponding Sides Proportional
  
  • Therefore, DE/AB = EF/BC = FD/AC = 1/2
• Step 4: Conclusion
  
  • We have shown that the corresponding sides of ΔDEF and ΔABC are proportional.
  • Therefore, by the Side-Side-Side (SSS) similarity criterion, ΔDEF ~ ΔABC.

Hence, we have proven that ΔFBD ~ ΔDEF and ΔDEF ~ ΔABC.

Question 27. The sum of two numbers is 18 and the sum of their reciprocals is 9/40. Find the numbers. 

Find the numbers

• Let the two numbers be ‘x’ and ‘y’.
  
• Given:
  
  • x + y = 18
  • 1/x + 1/y = 9/40
• Simplify the second equation:
  
  • (y + x) / (xy) = 9/40
  • 18 / (xy) = 9/40
  • xy = 80
• Now we have a system of equations:
  
  • x + y = 18
  • xy = 80
• Solve for ‘x’ and ‘y’:
  
  • From the first equation, y = 18 – x
  • Substitute this into the second equation:
    • x(18 – x) = 80
    • 18x – x^2 = 80
    • x^2 – 18x + 80 = 0
• Solve the quadratic equation:
  
  • (x – 8)(x – 10) = 0
  • x = 8 or x = 10
• Find the corresponding values of ‘y’:
  
  • If x = 8, then y = 18 – 8 = 10
  • If x = 10, then y = 18 – 10 = 8

Therefore, the two numbers are 8 and 10.

28. If 𝛼 and 𝛽 are zeroes of a polynomial 6𝑥² -5𝑥+1 then form a quadratic polynomial whose zeroes are 𝛼² and 𝛽²

Form a quadratic polynomial

• Given:
  
  • α and β are zeroes of the polynomial 6x² – 5x + 1
• We know:
  
  • Sum of zeroes (α + β) = -b/a = 5/6
  • Product of zeroes (αβ) = c/a = 1/6
• Let α² and β² be the zeroes of the new polynomial.
  
• Find the sum of zeroes of the new polynomial:
  
  • α² + β² = (α + β)² – 2αβ
  • α² + β² = (5/6)² – 2(1/6)
  • α² + β² = 25/36 – 1/3
  • α² + β² = 13/36
• Find the product of zeroes of the new polynomial:
  
  • α²β² = (αβ)²
  • α²β² = (1/6)²
  • α²β² = 1/36
• The new quadratic polynomial is of the form:
  
  • x² – (sum of zeroes)x + (product of zeroes)
  • x² – (1/36)x + 1/36

Therefore, the quadratic polynomial whose zeroes are α² and β² is x² – (13/36)x + 1/36.

Question 29. If cosθ + sinθ = 1 , then prove that cosθ – sinθ = ±1

• Given:
  
  • cosθ + sinθ = 1
• Square both sides of the given equation:
  
  • (cosθ + sinθ)² = 1²
  • cos²θ + sin²θ + 2sinθcosθ = 1
• We know that cos²θ + sin²θ = 1:
  
  • 1 + 2sinθcosθ = 1
  • 2sinθcosθ = 0
• This implies:
  
  • sinθ = 0 or cosθ = 0
• Case 1: sinθ = 0
  
  • If sinθ = 0, then cosθ = 1 (from the given equation)
  • cosθ – sinθ = 1 – 0 = 1
• Case 2: cosθ = 0
  
  • If cosθ = 0, then sinθ = 1 (from the given equation)
  • cosθ – sinθ = 0 – 1 = -1

Therefore, if cosθ + sinθ = 1, then cosθ – sinθ = ±1.

Question 30. (A) The minute hand of a wall clock is 18 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

👉Solution Part A) 

 Find the angle swept by the minute hand in 35 minutes:

• In 60 minutes, the minute hand sweeps 360 degrees.
• In 35 minutes, it sweeps (360/60) * 35 = 210 degrees.
• Calculate the area of the sector:
  
  • The area of a sector is given by: (θ/360) * πr²
  • Here, θ = 210 degrees, r = 18 cm
  • Area = (210/360) * π * (18)²
  • Area ≈ 593.46 cm²

Therefore, the area of the face of the clock described by the minute hand in 35 minutes is approximately 593.46 square centimeters.

(B) AB is a chord of a circle centred at O such that ∠AOB=60˚. If OA = 14 cm then find the area of the minor segment. (take √3 =1.73)

👉 Solution for Option B)

• Find the area of the sector AOB:
  
  • Area of sector = (θ/360) * πr²
  • Here, θ = 60 degrees, r = 14 cm
  • Area = (60/360) * π * (14)²
  • Area = (1/6) * π * 196
  • Area = (32.67)π cm²
• Find the area of triangle AOB:
  
  • Since ∠AOB = 60 degrees, triangle AOB is an equilateral triangle.
  • Area of equilateral triangle = (√3/4) * side²
  • Area = (√3/4) * (14)²
  • Area = 49√3 cm²
  • Area ≈ 84.77 cm²
• Find the area of the minor segment:
  
  • Area of minor segment = Area of sector – Area of triangle
  • Area = (32.67)π – 84.77
  • Area ≈ 17.92 cm^2

Therefore, the area of the minor segment is approximately 17.92 square centimeters.

Question 31): Prove that √3 is an irrational number

Proof by Contradiction:

Let’s assume that √3 is a rational number.
► If √3 is rational, it can be expressed as a fraction in its simplest form (where the numerator and denominator have no common factors) as:

√3 = a/b where a and b are integers, and b ≠ 0.
Square both sides:

 ⟹ (√3)² = (a/b)² 

 ⟹ 3 = a²/b²

 ⟹ 3b² = a²
This equation shows that a² is divisible by 3.
If a² is divisible by 3, then ‘a’ itself must also be divisible by 3.
Since ‘a’ is divisible by 3, we can write it as: a = 3k where k is another integer.
Substitute a = 3k in the equation 

 ⟹ 3b² = a²

 ⟹ 3b² = (3k)² 

 ⟹ 3b² = 9k² 

 ⟹ b² = 3k²
This equation shows that b² is also divisible by 3.
⟹  if b² is divisible by 3, then ‘b’ itself must also be divisible by 3.
⟹ We have shown that both ‘a’ and ‘b’ are divisible by 3.

⟹ This contradicts our initial assumption that a/b is in its simplest form (no common factors).

⟹ Since our initial assumption that √3 is rational leads to a contradiction, it must be false.

Therefore, √3 is irrational.

Question 32. (A) Solve the following system of linear equations graphically: x+2y = 3, 2x-3y+8 = 0 

Solution: Option A: Solving the system of linear equations graphically 

Equation 1: x + 2y = 3

To graph this equation, we can find two points that satisfy it:

• When x = 0: 0 + 2y = 3 ➡ y = 1.5
• When y = 0: x + 2(0) = 3 ➡ x = 3

So, two points on the graph of this equation are (0, 1.5) and (3, 0).

Equation 2: 2x – 3y + 8 = 0

To graph this equation, we can find two points that satisfy it:

• When x = 0: 2(0) – 3y + 8 = 0 ➡ y = 8/3
• When y = 0: 2x – 3(0) + 8 = 0 ➡ x = -4

So, two points on the graph of this equation are (0, 8/3) and (-4, 0).

Plotting the graphs

1. Plot the points for each equation on the coordinate plane.
2. Draw a straight line through the points for each equation.

The point where the two lines intersect represents the solution to the system of equations.

To solve:

1. x+2y=3
2. 2x−3y+8=0
3. From x+2y=3, we get x = 3−2y.

• Substituting x=3−2y into 2x−3y+8=0, we find y = 2.
• Substituting y = 2 into x = 3 − 2y we get x = −1.

Final Answer: x = -1, y = 2

(B) Places A and B are 180 km apart on a highway. One car starts from A and another from B at the same time. If the car travels in the same direction at different speeds, they meet in 9 hours. If they travel towards each other with the same speeds as before, they meet in an hour. What are the speeds of the two cars?

Option B: Determining the speeds of the two cars

Let:

• Speed of car starting from A = x km/h
• Speed of car starting from B = y km/h

Case 1: Cars travel in the same direction

• Relative speed = (x – y) km/h (assuming x > y)
• Distance to cover = 180 km
• Time taken = 9 hours

Therefore, 

➡ (x – y) * 9 = 180 

➡ x – y = 20 

➡ x = y + 20

Case 2: Cars travel towards each other

• Relative speed = (x + y) km/h
• Distance to cover = 180 km
• Time taken = 1 hour

Therefore, (x + y) * 1 = 180 ➡ x + y = 180

Substitute x = y + 20 from Case 1 into Case 2:

➡ (y + 20) + y = 180 

➡ 2y + 20 = 180 

➡ 2y = 160 

➡ y = 80 km/h

Substitute y = 80 into x = y + 20:

x = 80 + 20 = 100 km/h

Therefore, the speeds of the two cars are 100 km/h and 80 km/h.

Question 33: Prove that the lengths of tangents drawn from an external point to a circle are equal. 

Solution: 

Proof

Let P be a point outside a circle with center O, and two tangents PAPAPA and PBPBPB are drawn to the circle. We need to prove that PA=PB.

• Join OA, OB, and OP, where O is the center of the circle.
• Since PAPAPA and PBPBPB are tangents, they meet the circle at AAA and BBB perpendicularly.
  Thus, OA⊥PA and OB⊥PB.
• Triangles △OPA and △OPB are right triangles.
  • OA=OB (radii of the circle).
  • OPOPOP is common.
  • ∠OPA=∠OPB=90°.

By the RHS (Right-angle Hypotenuse Side) congruence rule,

△OPA≅△OPB

Therefore, PA = PB

Hence, the lengths of tangents drawn from an external point to a circle are equal.

For Finding BC of ΔABC (circle inscribed in ΔABC)

Given:

• AB=10 cm, AQ=7 cm, CQ=5 cm.

To find BCBCBC:

• Tangents drawn from an external point to a circle are equal. Therefore,
  • AR = AQ=7 cm (tangents from A).
  • BP = BR = x (tangents from B).
  • CP = CQ = 5cm (tangents from CCC).

Now, for side BC:

BC = BP + PC = BR + CQ

Substitute values:

BC=x+5

Also, from the tangents AR + BR = AB,

7+x=10  ⟹  x = 3

Thus,

BC = 3 + 5 = 8 cm

Final Answer: BC = 8 cm.

Question 34: A boy whose eye level is 1.35 m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is 60𝑜 . After 12 seconds, the angle of elevation reduces to 30°. If the speed of the wind is 3m/s then find the height of the balloon from the ground. (Use √3= 1.73) 

1. Setup and Known Values:

• Height of boy’s eye level: 1.35 m
• Angle of elevation at t1: 60°
• Angle of elevation at t2 (after 12 seconds): 30°
• Horizontal speed of the balloon: 3 m/s
• Horizontal displacement in 12 seconds: 36 m
• Let the height of the balloon above the ground be H

2. Triangle Formulation:

• Let the initial horizontal distance from the boy to the balloon be ‘x’.
  
• Height of the balloon above the boy’s eyes: h = H – 1.35
  
• At θ1 = 60°:
  
  • tan(60°) = h/x
  • √3 = h/x
  • h = x√3 – – – – – – eq (i)
• At θ2 = 30°:
  
  • tan(30°) = h / (x + 36)
  • 1/√3 = h / (x + 36)
  • h = (x + 36) / √3  – – – – – – eq (ii)

3. Solve for x and h:

• Equate the two expressions eq (i) & eq (ii)for h:
  
  • x√3 = (x + 36) / √3
  • 3x = x + 36
  • 2x = 36
  • x = 18 m
• Substitute x = 18 into the equation h = x√3:
  
  • h = 18 * √3
  • h = 31.14 m
• Calculate the total height of the balloon:
  
  • H = h + 1.35 = 31.14 + 1.35 = 32.49 m

Final Answer: The height of the balloon from the ground is approximately 32.49 meters.

Question 35: Find the mean and median of the following data:

► To calculate the mean, we use the formula:

The given data is:

1. Finding the Mean:

To calculate the mean, we use the formula:

Here, x is the midpoint of each class interval. To calculate midpoints:

x = (Lower limit + Upper limit)/2

Let’s compute the midpoints (x) and corresponding f⋅x:

Now sum up f and f⋅x:

∑f=15+22+20+18+20+25=120

∑f⋅x=1312.5+2035+1950+1845+2150+2812.5=12105

Now calculate the mean:

Mean=∑f⋅x / ∑f=12105/120=100.875

Mean = 100.88 (approximately).

► To calculate the median, we use the formula:

Where:

• L: Lower boundary of the median class
• N: Total frequency (∑f\sum f∑f)
• F: Cumulative frequency before the median class
• f_m ​: Frequency of the median class
• h: Class width

Steps:

1. Find N/2 = 120/2 = 60.
2. Locate the median class where cumulative frequency just exceeds 60.

The cumulative frequency just exceeds 60 in the class 100−105. Hence:

• Median class = 100−105
• L=100 (lower boundary of median class)
• F=57 (cumulative frequency before median class)
• fm=18 (frequency of median class)
• h=5 (class width)

Substitute into the formula:

You will get Median = 100.83 (approximately).

We strive to provide accurate answers, but errors may occur. If you spot any mistakes, please let us know at support@6to12teach.co.in. Your feedback helps us improve! 

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