2024 – Mathematics 

SECTION – I

Each question carries 2 marks

Question 1:

Express 360 as a product of prime factors.

To express 360 as a product of prime factors, we can use the prime factorization method. Here’s the process:  

1. Divide 360 by the smallest prime number, which is 2. 360 ÷ 2 = 180  
2. Divide the result (180) by 2 again. 180 ÷ 2 = 90  
3. Continue dividing by 2 until you get an odd number. 90 ÷ 2 = 45
4. Now, divide the odd number (45) by the next smallest prime number, which is 3. 45 ÷ 3 = 15
5. Divide 15 by 3 again. 15 ÷ 3 = 5  
6. Finally, divide 5 by itself. 5 ÷ 5 = 1

Therefore, the prime factorization of 360 is:

360 = 2 * 2 * 2 * 3 * 3 * 5 = 2³ * 3² * 5

Question 2:

Is the pair of linear equations 3x – 5y = 7 and 6x – 10y = 13 inconsistent? Justify your answer.

To determine if the pair of linear equations is inconsistent, we can compare their coefficients. If the coefficients of x and y in one equation are proportional to the coefficients of x and y in the other equation, but the constant terms are not proportional, then the system is inconsistent.

Let’s compare the coefficients:

• Equation 1: 3x – 5y = 7
• Equation 2: 6x – 10y = 13

We can see that the coefficients of x and y in equation 2 are double the coefficients of x and y in equation 1 (6x = 2 * 3x and -10y = 2 * -5y). However, the constant term in equation 2 (13) is not double the constant term in equation 1 (7).

Therefore, the pair of linear equations is inconsistent.

Question 3:

A flag pole stands vertically on the ground. From a point which is 15 metres away from the foot of the tower, the angle of elevation of the top of the tower is 45°. Draw a suitable diagram for the given data.

Here’s a suitable diagram for the given data:

 

rightangled triangle with base 15 meters and angle of elevation 45 degrees

 

In the diagram:

• The vertical line represents the flag pole.
• The horizontal line represents the ground.
• The point 15 meters away from the foot of the flag pole is marked.
• The angle of elevation from this point to the top of the flag pole is 45 degrees.

Question 4:

AOB is the diameter of a circle with center ‘O’ and AC is a tangent to the circle at A. If ∠BOC = 130°, then find ∠ACO.

Answer  

1. Angle in a Semicircle: Since AOB is the diameter, ∠AOB = 90° (angle in a semicircle).
2. Angles in a Triangle: In triangle BOC, we have: ∠BOC + ∠OBC + ∠OCB = 180° (sum of angles in a triangle) 130° + ∠OBC + ∠OCB = 180° ∠OBC + ∠OCB = 50°
3. Tangents and Radii: The tangent AC is perpendicular to the radius OA at the point of contact A. Therefore, ∠OAC = 90°.
4. Angles in a Triangle: In triangle AOC, we have: ∠OAC + ∠ACO + ∠OCA = 180° (sum of angles in a triangle) 90° + ∠ACO + ∠OCA = 180° ∠ACO + ∠OCA = 90°
5. Relation between Angles: From steps 2 and 4, we can see that: ∠OBC + ∠OCB = 50° ∠ACO + ∠OCA = 90°

Therefore, ∠ACO = 90° – 50° = 40°

Hence, ∠ACO = 40°

Question 5:

Express sin θ in terms of tan θ.

Answer  

1. Trigonometric Identity: We know that: tan θ = sin θ / cos θ
2. Rearranging: sin θ = tan θ * cos θ
3. Using Pythagorean Identity: We also know that: sin² θ + cos² θ = 1 cos² θ = 1 – sin² θ cos θ = √(1 – sin² θ)
4. Substituting: sin θ = tan θ * √(1 – sin² θ)

Therefore, sin θ = tan θ * √(1 – sin² θ)

Question 6:

Construct a Quadratic equation having the roots log₂ 8 and log₁₀ 100.

Answer  

Simplify Roots:

log₂ 8 = 3 (since 2³ = 8)

log₁₀ 100 = 2 (since 10² = 100)

Forming the Equation: If α and β are the roots of a quadratic equation, then the equation can be written as: (x – α)(x – β) = 0

Substituting the values of α = 3 and β = 2: (x – 3)(x – 2) = 0

Expanding: x² – 3x – 2x + 6 = 0 x² – 5x + 6 = 0

Therefore, the quadratic equation is x² – 5x + 6 = 0

SECTION – II

Each question carries 4 marks

Question 7:

Write the formula for Mode of a grouped data and explain each term.

Formula:

Mode = L + (fm – f1) / (2 * fm – f1 – f2) * h

Explanation:

• L: Lower limit of the modal class (the class with the highest frequency).
• fm: Frequency of the modal class.

• f1: Frequency of the class preceding the modal class.

• f2: Frequency of the class succeeding the modal class
• h: Class width (the difference between the upper and lower limits of a class)

Question 8:

Prove that:

sin θ / (1 + cos θ) + (1 + cos θ) / sin θ = 2 csc θ

Solution:

1. Find a common denominator: (sin θ * sin θ + (1 + cos θ) * (1 + cos θ)) / ((1 + cos θ) * sin θ) = 2 csc θ (sin² θ + 1 + 2 cos θ + cos² θ) / ((1 + cos θ) * sin θ) = 2 csc θ
2. Use trigonometric identities: (1 + 2 cos θ) / ((1 + cos θ) * sin θ) = 2 csc θ (1 / sin θ) * (1 + cos θ) / (1 + cos θ) = 2 csc θ csc θ = 2 csc θ

Therefore, the equation is proven.

Question 9:

From the Venn diagram, find the following sets:

Given Venn Diagram:

Venn diagram

 

Solutions:

(i) X ∪ Y: The union of X and Y represents all elements that are in either set X or set Y or in both. In the diagram, it includes all the elements within the circles and the overlapping region.

(ii) X ∩ Y: The intersection of X and Y represents the elements that are common to both sets X and Y. In the diagram, it’s the overlapping region containing the letters “C”, “L”, “E”, “N”.

(iii) X – Y: This represents the elements that are in set X but not in set Y. In the diagram, it’s the region within circle X that doesn’t overlap with circle Y.

(iv) Y – X: This represents the elements that are in set Y but not in set X. In the diagram, it’s the region within circle Y that doesn’t overlap with circle X.

Question 10:

In an arithmetic progression, if 4 times of the fourth term is equal to 8 times of the eighth term, then prove that the twelfth term of the progression is zero.

Solution:

Let the first term of the arithmetic progression be ‘a’ and the common difference be ‘d’.

• Fourth term: a + 3d
• Eighth term: a + 7d

Given: 4 * (a + 3d) = 8 * (a + 7d) => 4a + 12d = 8a + 56d => 4a = -44d => a = -11d

Now, let’s find the twelfth term:

• Twelfth term: a + 11d = -11d + 11d = 0

Therefore, the twelfth term of the progression is zero.

Question 11:

In a bag, there are 5 Red balls, 2 Black balls, and 3 White balls. If one ball is selected randomly from the bag, then find the probability of:

(i) Getting a Red ball.

(ii) Getting not a Red ball.

Solution:

Total number of balls = 5 (Red) + 2 (Black) + 3 (White) = 10

(i) Probability of getting a Red ball:

Number of favorable outcomes (Red balls) = 5 Total number of possible outcomes = 10

Probability = Favorable outcomes / Total outcomes = 5/10 = 1/2

(ii) Probability of getting not a Red ball:

Number of favorable outcomes (not Red balls) = 2 (Black) + 3 (White) = 5 Total number of possible outcomes = 10

Probability = Favorable outcomes / Total outcomes = 5/10 = 1/2

Question 12:

In a rectangle ABCD, AB = 2x – y, BC = 15, CD = 2, and DA = x + 3y. Find the values of x and y.

Solution:

In a rectangle, opposite sides are equal. Therefore, AB = CD and BC = DA

Equation 1: AB = CD => 2x – y = 2

Equation 2: BC = DA => 15 = x + 3y

Solving these equations simultaneously:

1. Solve Equation 1 for y: y = 2x – 2
2. Substitute y in Equation 2: 15 = x + 3(2x – 2) 15 = x + 6x – 6 21 = 7x x = 3
3. Substitute x in Equation 1: 2(3) – y = 2 6 – y = 2 y = 4

Therefore, the values of x and y are 3 and 4, respectively.

SECTION – III

Each question carries 6 marks

Question 13:

Prove that 3√5 + √7 is an irrational number.

Solution:

Proof by Contradiction:

1. Assume that 3√5 + √7 is a rational number. This means it can be expressed as a fraction p/q, where p and q are integers with no common factors
2. Therefore, we can write: 3√5 + √7 = p/q
3. Isolate √7: √7 = (p/q) – 3√5
4. Square both sides: 7 = (p/q)² – 6(p/q)√5 + 45
5. Rearrange to isolate √5: √5 = [(p/q)² – 45 + 7q] / (6p/q)√5 = (p² – 38q² + 7q) / (6p)
6. Since p and q are integers, (p² – 38q² + 7q) / (6p) is also a rational number.
7. This contradicts the fact that √5 is an irrational number.
8. Our initial assumption that 3√5 + √7 is rational must be false.

Therefore, 3√5 + √7 is an irrational number.

Question 14:

Draw the graph of the Quadratic polynomial p(x) = x²+x-12 and find the zeroes of the polynomial from the graph.

Solution:

1.Find the vertex: The vertex of the parabola represented by p(x) = x² + x – 12 is given by (-b/2a, p(-b/2a)).

Here, a = 1, b = 1, c = -12. Vertex: (-1/2, p(-1/2)) = (-1/2, -12.25)

2.Find the x-intercepts (zeroes): Set p(x) = 0 and solve for x

: x² + x – 12 = 0

(x + 4)(x – 3) = 0

x = -4 and x = 3

3.Plot the points:

Vertex: (-1/2, -12.25)

x-intercepts: (-4, 0) and (3, 0)

y-intercept: (0, -12) (set x = 0)

4.Draw the parabola: Plot the points and sketch a smooth curve through them. The parabola opens upwards since the coefficient of x² is positive.

From the graph, the zeroes of the polynomial are -4 and 3.

Question 15:

Find the Arithmetic mean of the following data:

Solution:

1. Calculate the midpoints of each class interval:

2. Multiply each midpoint by its corresponding frequency:

3. Sum of (Midpoint x Frequency): 55 + 210 + 375 + 700 + 675 + 715 + 780 = 3510

4. Sum of Frequencies: 11 + 14 + 15 + 20 + 15 + 13 + 12 = 90

5. Calculate the Arithmetic Mean:

Arithmetic Mean = (Sum of (Midpoint x Frequency)) / (Sum of Frequencies) = 3510 / 90 = 39

Therefore, the Arithmetic Mean of the data is 39.

Question 16:

Construct a triangle ABC with AB = 5.6 cm, BC = 7.2 cm and CA = 4.8 cm. Construct another triangle similar to ΔABC, whose sides are 3/5 times of the corresponding sides of ΔABC.

Solution:

1.Construct ΔABC:

1. • Draw a line segment AB = 5.6 cm.
   • With B as center and radius 7.2 cm, draw an arc.
   • With A as center and radius 4.8 cm, draw another arc intersecting the previous arc at point C.
   • Join BC and AC to form ΔABC.

2.Construct a similar triangle:

• • Draw a ray AX making an acute angle with AB.
  • Mark 5 points A₁, A₂, A₃, A₄, A₅ on AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
  • Join A₅B and draw a line parallel to A₅B through A₃, intersecting AB at B’.
  • Draw a line parallel to BC through B’ intersecting AC at C’.

Triangle AB’C’ is the required similar triangle with sides 3/5 times the corresponding sides of ΔABC.

Question 17:

The three vertices of a parallelogram ABCD are A(-1, -2), B(4, -1) and C(6, 3). Find the coordinates of vertex D and find the area of parallelogram ABCD.

Solution:

1.Find the coordinates of D:

• Let D(x, y) be the coordinates of vertex D.
• In a parallelogram, diagonals bisect each other. Therefore, the midpoints of AC and BD are the same.
• Midpoint of AC = ((-1 + 6)/2, (-2 + 3)/2) = (5/2, 1/2)
• Midpoint of BD = ((4 + x)/2, (-1 + y)/2)

Equating the midpoints: (5/2, 1/2) = ((4 + x)/2, (-1 + y)/2)

=> 5/2 = (4 + x)/2 => x = 1 => 1/2 = (-1 + y)/2 => y = 2

Therefore, the coordinates of vertex D are (1, 2).

1.Find the area of parallelogram ABCD:

• Area of parallelogram = Base x Height
• Base = AB = √[(4 – (-1))² + (-1 – (-2))²] = √(5² + 1²) = √26
• Height = Distance between AB and CD

To find the height, we can use the formula for the distance between a point and a line.

• • Equation of line AB: (y + 2) = (1/5)(x + 1) => x – 5y – 9 = 0

  • Distance of C(6, 3) from AB = |(6 – 5(3) – 9)|/√(1² + (-5)²) = 12/√26

  • Area of parallelogram = √26 * (12/√26) = 12 square units

Therefore, the coordinates of vertex D are (1, 2) and the area of parallelogram ABCD is 12 square units.

Question 18:

Due to heavy floods in the state thousands were rendered homeless. The State Government decided to provide canvas for 1500 tents. The lower part of each tent is cylindrical of base radius 2.8 meters and height 3.5 meters with conical upper part of same base radius but of height 2.1 meters. If the canvas used to make the tent costs Rs 100 per square meter, find the total cost of canvas to construct the tents.

Solution:

1.Calculate the surface area of each tent:

1. • Surface area of cylindrical part = 2πrh = 2 * (22/7) * 2.8 * 3.5 = 61.6 m²
   • Slant height of conical part (l) = √(r² + h²) = √(2.8² + 2.1²) = 3.5 m
   • Surface area of conical part = πrl = (22/7) * 2.8 * 3.5 = 30.8 m²
   • Total surface area of one tent = 61.6 + 30.8 = 92.4 m²

2.Calculate the total surface area of 1500 tents:

1. • Total surface area = 1500 * 92.4 = 138600 m²

3.Calculate the total cost of canvas:

• • Cost of canvas per square meter = Rs 100
  • Total cost = 138600 * 100 = Rs 13860000

Therefore, the total cost of canvas to construct the tents is Rs 13860000.