CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 1- 30/2/1) with Answer Key

Solution: For a system of two linear equations to have infinitely many solutions, the ratios of the coefficients of x, y, and the constant terms must be equal. Mathematically:

^a₁⁄_a2 = ^b₁⁄_b2 = ^c₁⁄_c2

Here, the given equations are:
3x – y + 8 = 0 and 6x – ky + 16 = 0.

Compare the coefficients with the standard form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
a₁ = 3, b₁ = -1, c₁ = 8 and a₂ = 6, b₂ = -k, c₂ = 16.

From the condition for infinitely many solutions:
^a₁⁄_a2 = ^b₁⁄_b2 = ^c₁⁄_c2

Substitute the values:
³⁄₆ = ^-1⁄_-k = ⁸⁄₁₆

Simplify each ratio:
³⁄₆ = ¹⁄₂ and ⁸⁄₁₆ = ¹⁄₂

Thus:
^-1⁄_-k = ¹⁄₂

Solve for k:
¹⁄_k = ¹⁄₂ ⇒ k = 2

Quick Tip: For two equations to have infinitely many solutions, ensure that the ratios of corresponding coefficients ^a₁⁄_a2 = ^b₁⁄_b2 = ^c₁⁄_c2

Solution: The formula to find the coordinates of a point P(x, y) dividing the line segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m:n is:

P(^{mx₂ + nx₁}⁄_m+n, ^{my₂ + ny₁}⁄_m+n)

Here: x₁ = 4, y₁ = -5, x₂ = 1, y₂ = 2, m = 5, n = 2.

Step 1: Calculate the x-coordinate of P
Substitute the values in the formula:
x = ^{m ⋅ x₂ + n ⋅ x₁}⁄_m+n
x = ^{5 ⋅ 1 + 2 ⋅ 4}⁄₅₊₂
Simplify:
x = ^{5 + 8}⁄₇ = ¹³⁄₇

Step 2: Calculate the y-coordinate of P
Substitute the values in the formula:
y = ^{m ⋅ y₂ + n ⋅ y₁}⁄_m+n
y = ^{5 ⋅ 2 + 2 ⋅ (-5)}⁄₅₊₂
Simplify:
y = ^{10 - 10}⁄₇ = 0

Step 3: Conclusion
The coordinates of P are:
(¹³⁄₇, 0)

Quick Tip: To find the coordinates of a point dividing a line segment, apply the section formula P(x, y) = (^{mx₂ + nx₁}⁄_m+n, ^{my₂ + ny₁}⁄_m+n)

Solution: The formula for the n-th term of an arithmetic progression is:
a_n = a₁ + (n-1)d,
where d is the common difference.

Given: a₁₅ - a₁₁ = 48

Using the formula for a_n:
a₁₅ = a₁ + 14d and a₁₁ = a₁ + 10d

Subtract a₁₁ from a₁₅:
a₁₅ - a₁₁ = (a₁ + 14d) - (a₁ + 10d)

Simplify:
a₁₅ - a₁₁ = 4d

Given a₁₅ - a₁₁ = 48, we get:
4d = 48 ⇒ d = 12

Quick Tip: In an arithmetic progression, the difference between two terms depends on the number of terms between them multiplied by the common difference d.

Solution: To determine the nature of roots of a quadratic equation ax² + bx + c = 0, calculate the discriminant: D = b² - 4ac.

Here: a = 1, b = 1, c = 1

Substitute into the discriminant:
D = (1)² - 4(1)(1) = 1 - 4 = -3

Since D < 0, the roots are not real.

Quick Tip: If the discriminant D < 0, the roots of a quadratic equation are imaginary (not-real).

Solution: The relationship between HCF, LCM, and two numbers a and b is:
HCF × LCM = a × b.

Here: HCF = 40, LCM = 252 × k, a = 2520, b = 6600

Substitute into the equation:
40 × (252 × k) = 2520 × 6600

Simplify 2520 × 6600: 2520 × 6600 = 16632000

Simplify 40 × 252 × k: 40 × 252 = 10080

Thus: 10080 × k = 16632000

Solve for k: k = ^16632000⁄₁₀₀₈₀ = 1650

Quick Tip: Use the relationship HCF × LCM = Product of the two numbers to solve problems involving LCM and HCF.

Solution: Since DE || BC, by the Basic Proportionality Theorem (Thales' Theorem), the ratio of corresponding segments is equal. Therefore:
^AD⁄_AB = ^DE⁄_BC

Step 1: Find AB (total length of AD + DB). Given: AD = 5 cm, DB = 2.5 cm. Thus: AB = AD + DB = 5 + 2.5 = 7.5 cm

Step 2: Substitute the known values into the formula. We now substitute into the proportionality equation: ^AD⁄_AB = ^DE⁄_BC. Substitute AD = 5, AB = 7.5, and BC = 12: ⁵⁄_7.5 = ^DE⁄₁₂

Step 3: Simplify and solve for DE. Simplify ⁵⁄_7.5: ⁵⁄_7.5 = ²⁄₃. Thus: ²⁄₃ = ^DE⁄₁₂. Solve for DE by cross-multiplying: DE = ²⁄₃ × 12 = 8 cm

Step 4: Conclusion The length of DE is: 8 cm

Quick Tip: When a line is parallel to one side of a triangle, the Basic Proportionality Theorem divides the other two sides in the same ratio. Use proportions to find unknown lengths.

Solution: Given sin θ = cos θ, we know: sin² θ + cos² θ = 1

Since sin θ = cos θ, substitute: 2sin² θ = 1 ⇒ sin² θ = ¹⁄₂

Thus: sin θ = ¹⁄_√2 and cos θ = ¹⁄_√2

The secant of θ is: sec θ = ¹⁄_{cos θ} = √2

Now calculate sec θ ⋅ sin θ: sec θ ⋅ sin θ = √2 ⋅ ¹⁄_√2 = 1

Quick Tip: If sin θ = cos θ, the angle θ is 45° and trigonometric identities simplify calculations.

Solution: To calculate the probability, note that the sum of two dice can range from 2 to 12. We are tasked with finding the probability of the sum being more than 10.

The possible outcomes for a sum > 10 are: (5,6), (6,5), (6,6)

There are 3 favorable outcomes.

The total number of outcomes when two dice are rolled is: 6 × 6 = 36

Thus, the probability is:
P(Sum > 10) = ^{Favorable outcomes}⁄_{Total outcomes} = ³⁄₃₆ = ¹⁄₁₂

Quick Tip: To solve dice probability problems, count all favorable outcomes and divide by the total outcomes 6 × 6 = 36.

Solution: The polynomial is: 5x² + 3x - 7.

For any quadratic polynomial ax² + bx + c, the sum and product of the roots are:
α + β = -^b⁄_a and αβ = ^c⁄_a

Here: a = 5, b = 3, c = -7

Calculate the sum and product:
α + β = -³⁄₅, αβ = ^-7⁄₅

We are tasked to find ¹⁄_α + ¹⁄_β, which simplifies as:
¹⁄_α + ¹⁄_β = ^{α + β}⁄_αβ

Substitute the values:
¹⁄_α + ¹⁄_β = ^-3⁄₅⁄_^-7⁄5

Simplify:
¹⁄_α + ¹⁄_β = ^-3⁄_-7 = ³⁄₇

Quick Tip: For quadratic polynomials, use the relations α + β = -^b⁄_a and αβ = ^c⁄_a to simplify root-related expressions.

Solution: For two similar triangles, the ratio of their corresponding sides is equal to the ratio of their perimeters.

Given: ^{Perimeter of PQR}⁄_{Perimeter of ABC} = ^PQ⁄_AB

Substitute the values:
^PQ⁄_AB = ^{Perimeter of PQR}⁄_{Perimeter of ABC} = ⁴⁸⁄₅₆

Simplify the ratio: ^PQ⁄_AB = ^{48 ÷ 8}⁄_{56 ÷ 8} = ⁶⁄₇

Conclusion: The value of ^PQ⁄_AB is: ⁶⁄₇

Quick Tip: For similar triangles, the ratio of corresponding sides is equal to the ratio of their perimeters.

Solution: When two chords AB and CD intersect at a point P within a circle, the triangles formed, ∆ADP and ∆CBP, are similar.

Reason:

1. The angles opposite the intersecting chords are equal because they are subtended by the same arc:
   ∠ADP = ∠CBP
2. The vertical angles at P are equal:
   ∠DPA = ∠BPC

By the AA (Angle-Angle) Criterion, the two triangles are similar:
∆ADP ~ ∆CBP

Quick Tip: When two chords of a circle intersect, the triangles formed are similar by the AA criterion of similarity.

Solution: The median of a data set is the middle value when the data is arranged in ascending order.

Step 1: Effect of Adding a Constant If each observation in the data is increased by 2, all values shift upward by the same amount. This shift does not alter the position of the median but increases its value by 2.

Conclusion: The new median is: New Median = Old Median + 2.

Quick Tip: When a constant value is added to all observations in a data set, the mean, median, and mode all increase by the same constant value.

Solution: The total number of cards is: 55 - 6 + 1 = 50 cards.

Step 1: Identify the Perfect Squares between 6 and 55 The perfect squares in this range are: 9, 16, 25, 36, 49. Count the total number of perfect squares: 5 perfect squares.

Step 2: Probability of Drawing a Perfect Square The probability is given by: P(perfect square) = ^{Number of favorable outcomes}⁄_{Total outcomes}. Substitute the values: P(perfect square) = ⁵⁄₅₀ = ¹⁄₁₀

Conclusion: The probability that the drawn card has a number which is a perfect square is: ¹⁄₁₀

Quick Tip: To find probabilities, identify all favorable outcomes and divide by the total number of possible outcomes.

Solution: In the given figure, PA and PB are tangents to the circle and are perpendicular at point P. Thus, ∆OAP forms a right-angled triangle at A, where: OP (hypotenuse) = AB and PA = PB = 5 cm.

Step 1: Use Pythagoras Theorem For right-angled triangle OPA: OP² = PA² + PB². Substitute the values: OP² = 5² + 5². OP² = 25 + 25 = 50. Taking square root: OP = √50 = 5√2.

Conclusion: The length of AB is: 5√2 cm.

Quick Tip: In right-angled triangles involving tangents, always apply the Pythagoras theorem to calculate the hypotenuse or missing sides.

Solution: The length of the diagonal of a rectangle can be calculated using the distance formula: d = √(x₂ - x₁)² + (y₂ - y₁)².

Step 1: Identify Coordinates of XYZ Given vertices are: X(-3, 0), O(0, 0), Y(0, 4). The diagonal XY can be found as: XY = √(0 - (-3))² + (4 - 0)².

Step 2: Apply the Distance Formula Substitute the coordinates: XY = √(3)² + (4)². Simplify: XY = √9 + 16 = √25 = 5.

Conclusion: The length of the diagonal is: 5 units.

Quick Tip: The diagonal of a rectangle can be directly calculated using the distance formula between opposite vertices.

Solution: The general term of an A.P. is given by: a_n = a + (n-1)d, where a is the first term, d is the common difference, and n is the term number.

Step 1: Identify a, d, and a_n Given: a = -29, d = -26 - (-29) = 3, a_n = 61.

Step 2: Substitute in the Formula a_n = a + (n-1)d. Substitute the known values: 61 = -29 + (n-1) ⋅ 3. Simplify: 61 + 29 = (n-1) ⋅ 3. 90 = (n-1) ⋅ 3. Divide by 3: n-1 = 30 ⇒ n = 31.

Conclusion: The term number is: 16^th term.

Quick Tip: To find a specific term in an arithmetic progression, always use the general formula a_n = a + (n-1)d and substitute known values.

Solution: In the figure: AT is a tangent to the circle at A. The property of tangents states that the angle between the tangent and a chord through the point of contact is equal to the angle subtended by the chord at the opposite arc.

Step 1: Tangent-Chord Property Given: ∠CAT = 40°. By the tangent-chord property: ∠CBA = ∠CAT.

Conclusion: ∠CBA = 40°.

Quick Tip: The tangent-chord property states that the angle between a tangent and a chord is equal to the angle subtended by the chord at the opposite arc.

Solution: The mode of a data set refers to the value that occurs most frequently. Since the teacher wants to know the marks obtained by the maximum number of students, she needs to calculate the mode of marks.

Explanation:
- Median: Represents the middle value of a sorted data set.
- Mean: The average of all values.
- Range: The difference between the highest and lowest values.
- Mode: The most frequently occurring value in a data set.

Here, the mode is the most appropriate measure since it highlights the most common marks.

Quick Tip: The mode is useful in identifying the most frequently occurring value, which is ideal for determining trends or patterns in a data set.

Solution: Using the Pythagorean identity: sin² A + cos² A = 1. Given sin A = ¹⁄₃, substitute the value:
(¹⁄₃)² + cos² A = 1
¹⁄₉ + cos² A = 1
Simplify: cos² A = 1 - ¹⁄₉ = ⁸⁄₉
Taking the square root: cos A = √⁸⁄₉ = ^√8⁄₃ = ^2√2⁄₃

Thus, Assertion (A) is true, and Reason (R) correctly explains the result.

Quick Tip: The Pythagorean identity sin² θ + cos² θ = 1 is a fundamental trigonometric property.

Solution: To find the total surface area of a newly formed cuboid, note: 1. Each cube has an edge length of 10 cm. 2. When two cubes are joined along one face, the resulting cuboid has dimensions: Length = 20 cm (combined), Width = 10 cm, Height = 10 cm.

Step 1: Surface Area of the Cuboid The formula for the total surface area of a cuboid is: Surface Area = 2 (l ⋅ w + w ⋅ h + h ⋅ l). Substitute l = 20 cm, w = 10 cm, h = 10 cm: Surface Area = 2 (20 ⋅ 10 + 10 ⋅ 10 + 10 ⋅ 20). Simplify: Surface Area = 2 (200 + 100 + 200) = 2 ⋅ 500 = 1000 cm².

Step 2: Conclusion The actual total surface area is 1000 cm², not 1200 cm². The Reason (R) is correct as each surface area of the cube is 100 cm². Thus, Assertion (A) is false, but Reason (R) is true.

Quick Tip: When two cubes are joined, the surface area decreases because one face of each cube is no longer exposed.

Solution: The number 15ⁿ can be expressed as: 15ⁿ = 5ⁿ × 3ⁿ. To end with the digit 0, a number must have both 2 and 5 as its prime factors (since 2 × 5 = 10).

Step 1: Prime Factorization 15 = 5 × 3. Thus, 15ⁿ only includes prime factors 5 and 3.

Step 2: Absence of Factor 2 Since 15ⁿ does not include 2 as a prime factor, it cannot end with the digit 0.

Conclusion: A number can only end with 0 if it has 2 and 5 as factors. Therefore, 15ⁿ cannot end with 0.

Quick Tip: To check if a number ends with 0, ensure its prime factorization includes both 2 and 5.

Solution: To determine the type of triangle, calculate the distances between the vertices AB, BC, and CA using the distance formula: d = √(x₂ - x₁)² + (y₂ - y₁)².

Step 1: Calculate AB AB = √(-5 - 1)² + (0 - 0)² = √(-6)² = √36 = 6.

Step 2: Calculate BC BC = √(-5 + 2)² + (0 - 5)² = √(-3)² + (-5)² = √9 + 25 = √34.

Step 3: Calculate CA CA = √(1 + 2)² + (0 - 5)² = √(3)² + (-5)² = √9 + 25 = √34.

Step 4: Compare the Sides AB = 6, BC = √34, CA = √34. Since BC = CA, the triangle ABC has two equal sides.

Conclusion: The triangle ABC is isosceles.

Quick Tip: An isosceles triangle has two sides of equal length. Use the distance formula to compare side lengths.

Solution: We need to evaluate 2 sin² 30° sec 60° + tan² 60°.

Step 1: Substitute the Values sin 30° = ¹⁄₂, sec 60° = 2, tan 60° = √3. Substitute these values into the given expression: 2 sin² 30° sec 60° + tan² 60° = 2 (¹⁄₂)² × 2 + (√3)².

Step 2: Simplify the Expression 2 (¹⁄₂)² × 2 = 2 × ¹⁄₄ × 2 = 1. (√3)² = 3. Add the two results: 1 + 3 = 4.

Conclusion: The value of 2 sin² 30° sec 60° + tan² 60° is 4.

Quick Tip: Memorize the standard trigonometric values for 30°, 60°, and 45° to solve such problems quickly.

Solution: Given: 2 sin (A + B) = √3 and cos (A - B) = 1.

Step 1: Solve for A + B Divide both sides of 2 sin (A + B) = √3 by 2: sin (A + B) = ^√3⁄₂. From the standard trigonometric values: sin 60° = ^√3⁄₂. Thus: A + B = 60°.

Step 2: Solve for A - B Given cos (A - B) = 1, we know: cos 0° = 1. Thus: A - B = 0° ⇒ A = B.

Step 3: Find A and B From A + B = 60° and A = B, solve for A and B: A + A = 60° ⇒ 2A = 60° ⇒ A = 30°. Since A = B, we get: B = 30°.

Conclusion: The measures of angles A and B are: A = 30° and B = 30°.

Quick Tip: Use the properties of trigonometric functions and standard values to solve for unknown angles efficiently.

Solution: To prove: ∠BAC = ∠DCA.

Step 1: Join OA and OC Since AB and CD are tangents to the circle at points A and C, the radii OA and OC are perpendicular to the tangents: OA = OC (radii of the same circle).

Step 2: Analyze the angles The angles ∠OAC and ∠OCA are equal because they are opposite equal radii: ∠OAC = ∠OCA. Also, the angles ∠OAB and ∠OCD are equal because the tangents AB and CD subtend equal angles with the radii: ∠OAB = ∠OCD.

Step 3: Combine the results Now, add the angles: ∠OAC + ∠OAB = ∠OCA + ∠OCD. This gives: ∠BAC = ∠DCA.

Conclusion: Hence, ∠BAC is equal to ∠DCA.

Quick Tip: The tangents drawn to a circle from an external point are equal in length and subtend equal angles with the radii of the circle.

Solution: Let the required ratio be K : 1.

Step 1: Coordinates of the point of division P: The coordinates of P dividing the line segment joining (3, -5) and (-1, 6) in the ratio K : 1 are: P (^{-K + 3}⁄_{K + 1}, ^{6K - 5}⁄_{K + 1}).

Step 2: Point lies on the line y = x Since y = x, the x-coordinate and y-coordinate of P are equal: ^{-K + 3}⁄_{K + 1} = ^{6K - 5}⁄_{K + 1}. Simplify: -K + 3 = 6K - 5. 8K = 8 ⇒ K = 1.

Thus, the ratio is: Ratio = 8 : 7.

Final Answer: The required ratio is 8 : 7.

Solution: Step 1: Midpoint of AC The coordinates of the midpoint E of AC are: E = (^{x₁ + x₂}⁄₂, ^{y₁ + y₂}⁄₂) = (^{3 + (-1)}⁄₂, ^{0 + 3}⁄₂) = (1, ³⁄₂).

Step 2: Length of median BE The length of BE is given by: BE = √(x₂ - x₁)² + (y₂ - y₁)². Substitute B(6, 4) and E(1, ³⁄₂): BE = √(6 - 1)² + (4 - ³⁄₂)². Simplify: BE = √5² + (^{8 - 3}⁄₂)² = √25 + (⁵⁄₂)². BE = √25 + ²⁵⁄₄ = √¹⁰⁰⁄₄ + ²⁵⁄₄ = √¹²⁵⁄₄. BE = ^√125⁄₂ = ^5√5⁄₂.

Final Answer: The length of the median BE is ^5√5⁄₂.

Solution: The sum of the first m terms of an A.P. is: S_m = ^m⁄₂ [2a + (m-1)d]. Similarly, the sum of the first n terms is: S_n = ⁿ⁄₂ [2a + (n-1)d].

Step 1: Equating the two sums Since S_m = S_n, we have: ^m⁄₂ [2a + (m-1)d] = ⁿ⁄₂ [2a + (n-1)d]. Simplify by multiplying both sides by 2: m [2a + (m-1)d] = n [2a + (n-1)d].

Step 2: Solving for 2a Expand and simplify: 2am + m(m-1)d = 2an + n(n-1)d. Rearranging terms gives: 2a(m - n) = d[n² - m² - (n - m)]. Simplifying further: 2a = -d(m + n - 1).

Step 3: Sum of (m+n) terms The sum of the first (m + n) terms is: S_m+n = ^m+n⁄₂ [2a + (m+n-1)d]. Substitute 2a = -d(m+n-1) into the equation: S_m+n = ^m+n⁄₂ [-d(m+n-1) + (m+n-1)d]. Simplify: S_m+n = ^m+n⁄₂ [0] = 0.

Final Answer: Thus, the sum of the first (m+n) terms is zero.

Solution: Let the three consecutive terms in the A.P. be a-d, a, a+d.

Step 1: Sum of three terms The sum of the terms is given as: (a-d) + a + (a+d) = 24. Simplify: 3a = 24 ⇒ a = 8.

Step 2: Sum of squares of the terms The sum of the squares of the terms is: (a-d)² + a² + (a+d)² = 194. Substitute a = 8: (8-d)² + 8² + (8+d)² = 194. Simplify: (64 - 16d + d²) + 64 + (64 + 16d + d²) = 194. Combine like terms: 2d² + 192 = 194. 2d² = 2 ⇒ d² = 1 ⇒ d = ± 1.

Step 3: Find the terms If d = 1, the terms are: a-d, a, a+d = 7, 8, 9. If d = -1, the terms are: a-d, a, a+d = 9, 8, 7.

Final Answer: The numbers are 7, 8, 9 or 9, 8, 7.

Solution: Assume √5 is a rational number. Then it can be expressed as: √5 = ^p⁄_q, where p and q are co-prime integers and q ≠ 0.

Step 1: Square both sides Squaring both sides, we get: 5q² = p². This implies p² is divisible by 5. Hence, p must also be divisible by 5 (as 5 is prime). Let: p = 5a, where a is an integer.

Step 2: Substitute p = 5a Substitute p = 5a into 5q² = p²: 5q² = (5a)². Simplify: 5q² = 25a² ⇒ q² = 5a². Thus, q² is divisible by 5, which means q is also divisible by 5.

Step 3: Contradiction From steps 1 and 2, we conclude that both p and q are divisible by 5, which contradicts the assumption that p and q are co-prime.

Conclusion: Our assumption that √5 is rational is incorrect. Hence, √5 is an irrational number.

Solution: Step 1: Join OQ Since OQ and OA are radii of the circle: OQ = OA ⇒ ∠2 = 30°.

Step 2: Using tangent properties and angle sum At point Q, the tangent PQ is perpendicular to the radius OQ: ∠3 = 90° - ∠2 = 90° - 30° = 60°. At point B, the angle ∠4 is: ∠4 = 90° - ∠3 = 90° - 60° = 30°.

Step 3: Angle relationship and equality From the figure: ∠6 = ∠1 + ∠2 = 60°. Hence: ∠5 = 90° - ∠6 = 90° - 60° = 30° = ∠4.

Step 4: Conclusion Since ∠BPQ = ∠BQP, the two tangents BP and BQ are equal: BP = BQ.

Quick Tip: The tangents drawn to a circle from an external point are always equal in length.

Solution: Step 1: Join the radii and tangents Join OP, OQ, OR, and OS as shown in the figure. Since tangents from a point to a circle are equal, we observe the following: ∆POB ≅ ∆QOB (by symmetry) ⇒ ∠1 = ∠2.

Step 2: Angle relationships Similarly: ∠3 = ∠4, ∠5 = ∠6, and ∠7 = ∠8. The sum of all angles at O is 360°, so: ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°.

Step 3: Grouping angles Group the angles symmetrically: 2(∠1 + ∠3 + ∠5) = 360°. Simplify: ∠AOB + ∠COD = 180°.

Conclusion: Thus, it is proved that: ∠AOB + ∠COD = 180°.

Quick Tip: The tangents to a circle from external points form symmetrical triangles, which simplifies angle calculations.

Solution: Step 1: Simplify the Left-Hand Side (LHS) LHS = ^{(sec2 θ - tan2 θ) + (sec θ - tan θ)}⁄_{1 + sec θ + tan θ}. Using the identity sec² θ - tan² θ = 1, we get: LHS = ^{1 + sec θ - tan θ}⁄_{1 + sec θ + tan θ}.

Step 2: Factorize the denominator and numerator Using the form (a - b)(a + b) = a² - b², we write: 1 + sec θ - tan θ = ^{cos θ (1 - sin θ)}⁄_{cos θ}. Thus: LHS = ^{1 - sin θ}⁄_{cos θ} ⋅ cos θ = RHS.

Conclusion: Hence, it is proved that: ^{1 + sec θ - tan θ}⁄_{1 + sec θ + tan θ} = cos θ ⋅ ^{1 - sin θ}⁄_{cos θ}.

Quick Tip: Use trigonometric identities like sec² θ - tan² θ = 1 and factorize carefully to simplify complex expressions.

Solution: Step 1: Recall the formula for Mean: The formula for the mean is: Mean = ^{∑ f_ix_i}⁄_{∑ fi}.

Step 2: Substitute the given values: Here: ∑ f_ix_i = 2400 and ∑ f_i = 100. Thus: Mean = ²⁴⁰⁰⁄₁₀₀ = 24.

Conclusion: The mean marks of the students is: 24

Quick Tip: The mean is calculated by dividing the sum of all values (∑ f_ix_i) by the total number of observations (∑ f_i).

Solution: Step 1: Let the digit at the ten's place be x. Since the digit at the unit's place is 5 less than the ten's place, we have: Unit's place digit = x - 5.

Step 2: Form the equation. The product of the digits is given as 36: x(x - 5) = 36. Simplify: x² - 5x - 36 = 0.

Step 3: Solve the quadratic equation. Factorize x² - 5x - 36 = 0: (x - 9)(x + 4) = 0. Thus: x = 9 or x = -4. Since x represents a digit, it must be positive: x = 9.

Step 4: Find the number. If x = 9, the digit at the unit's place is: x - 5 = 9 - 5 = 4. Thus, the required number is: 94.

Conclusion: The required number is 94.

Quick Tip: When solving quadratic equations, ensure the solution aligns with the context of the problem, such as digits being non-negative and single-digit numbers.

Quick Tip: The intersection point of the two lines represents the solution of the system of equations.

Solution: Step 1: Define variables. Let the number of right answers be x and the number of wrong answers be y. The total number of questions is: x + y = N (where N is the total number of questions).

Step 2: Form equations based on given conditions. 1. From the first condition (3 marks for each correct and -1 for each wrong answer), Tara's score is: 3x - y = 40 (Equation 1). 2. From the second condition (4 marks for correct and -2 for wrong answers), Tara's score is: 4x - 2y = 50 (Equation 2).

Step 3: Solve the equations. From Equation 1: 3x - y = 40 ⇒ y = 3x - 40. Substitute y = 3x - 40 into Equation 2: 4x - 2(3x - 40) = 50. Simplify: 4x - 6x + 80 = 50. -2x + 80 = 50. -2x = -30 ⇒ x = 15.

Step 4: Find y. Substitute x = 15 into y = 3x - 40: y = 3(15) - 40 = 45 - 40 = 5.

Step 5: Total number of questions. The total number of questions is: N = x + y = 15 + 5 = 20.

Conclusion: The total number of questions in the test is 20.

Quick Tip: Always carefully frame equations based on the conditions given in the question and check the consistency of your solution.

Solution: Let ∆ABC be a triangle, and let a line DE be drawn parallel to BC, intersecting AB at D and AC at E.

Step 1: Prove that ^AD⁄_DB = ^AE⁄_EC 1. Since DE ∥ BC, by the Basic Proportionality Theorem (Thales' Theorem), we have: ^AD⁄_DB = ^AE⁄_EC. 2. The theorem states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally.

Step 2: Conclusion Thus, the line DE divides the sides AB and AC in the same ratio: ^AD⁄_DB = ^AE⁄_EC.

Quick Tip: The Basic Proportionality Theorem (Thales' Theorem) is a fundamental result in geometry used to prove proportional divisions of triangle sides.

Solution: Given two triangles ∆ABC and ∆PQR: 1. It is given that: ^AB⁄_PQ = ^AC⁄_PR = ^AD⁄_PM. 2. In ∆ABC, AD is the median, which divides BC into two equal parts. Similarly, in ∆PQR, PM is the median, which divides QR into two equal parts.

Step 1: Prove similarity of triangles using the proportionality condition Since: ^AB⁄_PQ = ^AC⁄_PR and ^AD⁄_PM, the corresponding sides of triangles ∆ABC and ∆PQR are proportional. By the Side-Side-Side (SSS) Similarity Criterion, two triangles are similar if their corresponding sides are proportional.

Step 2: Conclusion From the given condition and the SSS similarity criterion, we conclude that: ∆ABC ~ ∆PQR.

Quick Tip: To prove two triangles are similar, use the SSS Similarity Criterion when the ratios of corresponding sides are equal.

Solution: Let AB = 45 m be the height of the lighthouse. Let the ships be positioned at C and D, on opposite sides of the lighthouse. ∠ACB = 30° (depression angle of ship C). ∠ADB = 60° (depression angle of ship D). Let BC = x and BD = y be the horizontal distances of ships C and D from the base of the lighthouse.

Step 1: Use trigonometric relations From ∆ABC, using tan θ = ^Opposite⁄_Adjacent: tan 30° = ^AB⁄_BC ⇒ ¹⁄_√3 = ⁴⁵⁄_x. Solve for x: x = 45 √3 = 45 × 1.73 = 77.85 m. From ∆ABD, using tan 60° = ^Opposite⁄_Adjacent: tan 60° = ^AB⁄_BD ⇒ √3 = ⁴⁵⁄_y. Solve for y: y = ⁴⁵⁄_√3 = ⁴⁵⁄_1.73 = 25.98 m.

Step 2: Find the distance between the ships CD Since the ships C and D are on opposite sides of the lighthouse, the total distance between them is: CD = BC + BD = x + y. Substitute the values of x and y: CD = 77.85 + 25.98 = 103.83 m.

Final Answer: The distance between the ships is approximately 103.83 m.

Quick Tip: To solve angle of depression problems, always use trigonometric ratios like tan θ and carefully analyze the triangle formed with horizontal distances.

Solution: The perimeter of a sector is given by: Perimeter = 2r + ^{2π r θ}⁄₃₆₀. Substitute the known values r = 5.6 m and Perimeter = 20 m: 2(5.6) + ^{2 × 22⁄₇ × 5.6 × θ}⁄₃₆₀ = 20. Simplify: 11.2 + ^{2 × 22 × 5.6 × θ}⁄_{7 × 360} = 20. 11.2 + ^{246.4 θ}⁄₂₅₂₀ = 20. Simplify further: ^{246.4 θ}⁄₂₅₂₀ = 8.8. Solve for θ: θ = ^{8.8 × 2520}⁄_246.4 = 90°.

Step 2: Find the area of the sector The area of a sector is given by: Area = ^{π r2 θ}⁄₃₆₀. Substitute r = 5.6 m, θ = 90°, and π = ²²⁄₇: Area = ^{22⁄₇ × 5.6 × 5.6 × 90}⁄₃₆₀. Simplify: Area = ²²⁄₇ × 5.6 × 5.6 × ¹⁄₄. Area = ^{22 × 31.36}⁄₂₈. Area = 24.64 m².

Final Answer: The area of the sector is 24.64 m².

Quick Tip: The perimeter of a sector combines the arc length and the two radii. Always simplify step by step to find the unknown angle or area.

(i) Write zeroes of the given polynomial: The given polynomial is: h = 25t - 5t² ⇒ h = -5t(t - 5). Setting h = 0: -5t(t - 5) = 0 ⇒ t = 0 and t = 5. Thus, the zeroes of the polynomial are t = 0 and t = 5.

(ii) Maximum height achieved by the ball: The height is given as: h = 25t - 5t². This is a quadratic equation in the form h = at² + bt + c where a = -5, b = 25, and c = 0. The time at which the maximum height occurs is: t = -^b⁄_2a. Substitute a = -5 and b = 25: t = -²⁵⁄_2(-5) = ²⁵⁄₁₀ = ⁵⁄₂. At t = ⁵⁄₂, the height is: h = 25(⁵⁄₂) - 5(⁵⁄₂)². Simplify: h = 25 × ⁵⁄₂ - 5 × ²⁵⁄₄. h = ¹²⁵⁄₂ - ¹²⁵⁄₄. Take the LCM of 4: h = ²⁵⁰⁄₄ - ¹²⁵⁄₄ = ¹²⁵⁄₄. h = 31.25 m. Thus, the maximum height achieved by the ball is 31.25 m.

(iii) (A) Time taken to reach the height of 30 m: Given: h = 25t - 5t² and h = 30. Substitute: 30 = 25t - 5t². Rearrange: -5t² + 25t - 30 = 0. Divide through by -5: t² - 5t + 6 = 0. Factorize: (t - 2)(t - 3) = 0. Thus: t = 2 and t = 3. Therefore, the ball takes t = 2 seconds and t = 3 seconds to reach the height of 30 m.

(iii) (b) Time when height was 20 m: Given h = 20: 20 = 25t - 5t². Rearrange: -5t² + 25t - 20 = 0. Divide through by -5: t² - 5t + 4 = 0. Factorize: (t - 4)(t - 1) = 0. Thus: t = 4 and t = 1.

Final Answers:

• (i) Zeroes are t = 0 and t = 5.
• (ii) Maximum height is 31.25 m.
• (iii) (A) Time to reach 30 m is t = 2 seconds and t = 3 seconds.
• (iii) (b) Time when height was 20 m is t = 1 second and t = 4 seconds.

Quick Tip: For quadratic motion problems, identify the maximum height using t = -^b⁄_2a and solve for specific heights using factorization or the quadratic formula.

(i) Find slant height of the conical part: The height of the conical part is: Height of cone = 18.5 - 8 = 10.5 m. The radius of the base is: Radius = ^Diameter⁄₂ = ²⁸⁄₂ = 14 m. Using the Pythagoras theorem, the slant height l is: l = √(Height)² + (Radius)². l = √(10.5)² + (14)². Simplify: l = √110.25 + 196 = √306.25 = 17.5 m.

(ii) Determine the floor area of the tent: The floor area is the area of the circular base: Floor area = π r². Substitute r = 14 m and π = ²²⁄₇: Floor area = ²²⁄₇ × 14 × 14. Simplify: Floor area = 616 m².

(iii) (A) Find area of the cloth used for making the tent: The area of cloth includes the curved surface area of the cylinder and cone. 1. Curved surface area of cylinder: CSA of cylinder = 2π r h. Substitute r = 14 m and h = 8 m: CSA of cylinder = 2 × ²²⁄₇ × 14 × 8. CSA of cylinder = 704 m². 2. Curved surface area of cone: CSA of cone = π r l. Substitute r = 14 m and l = 17.5 m: CSA of cone = ²²⁄₇ × 14 × 17.5. CSA of cone = 770 m². Total area of cloth: Total area = CSA of cylinder + CSA of cone. Total area = 704 + 770 = 1474 m².

(iii) (b) Find total volume of air inside the empty tent: The volume of the tent includes the volume of the cylinder and cone. 1. Volume of cylinder: Volume of cylinder = π r² h. Substitute r = 14 m and h = 8 m: Volume of cylinder = ²²⁄₇ × 14 × 14 × 8. Volume of cylinder = 4928 m³. 2. Volume of cone: Volume of cone = ¹⁄₃ π r² h. Substitute r = 14 m and h = 10.5 m: Volume of cone = ¹⁄₃ × ²²⁄₇ × 14 × 14 × 10.5. Volume of cone = 2156 m³. Total volume: Total volume = Volume of cylinder + Volume of cone. Total volume = 4928 + 2156 = 7084 m³.

Final Answers:

• (i) Slant height of conical part = 17.5 m.
• (ii) Floor area of the tent = 616 m².
• (iii) (A) Area of cloth used = 1474 m².
• (iii) (b) Total volume of air = 7084 m³.

Quick Tip: To calculate the total area and volume of structures combining a cylinder and cone, solve for each part separately and add the results.

(i) Find the probability of travelling by bus or ship: The angle representing: Bus = 36° and Ship = 33°. The total angle of the pie chart is 360°. Therefore, the probability is: P(travelling by bus or ship) = ^{Bus angle + Ship angle}⁄_{Total angle}. P = ^{36° + 33°}⁄_360° = ⁶⁹⁄₃₆₀ = ²³⁄₁₂₀.

(ii) Find the most favourite mode of transport and the number of people: The angle for Car is 177°, which is the largest angle. The number of people who used car is: Number of people = ^{Angle for Car}⁄_{Total angle} × Total surveyed people. Number of people = ¹⁷⁷⁄₃₆₀ × 120. Number of people = 59.

(iii) (A) Find the number of people who used train: The probability that a person did not use train is: P(Not using train) = ⁴⁄₅. The probability that a person used train is: P(Using train) = 1 - ⁴⁄₅ = ¹⁄₅. The number of people who used train is: Number of people = P(Using train) × Total surveyed people. Number of people = ¹⁄₅ × 120 = 24.

(iii) (b) Find the revenue collected for aeroplane users: The probability of using an aeroplane is ⁷⁄₆₀. The number of people who used an aeroplane is: Number of people = P(Using aeroplane) × Total surveyed people. Number of people = ⁷⁄₆₀ × 120. Number of people = 14. The revenue collected at the rate of 5000 ₹ per person is: Revenue = Number of people × Rate per person. Revenue = 14 × 5000 = 70,000 ₹.

Final Answers:

• (i) Probability of travelling by bus or ship = ²³⁄₁₂₀.
• (ii) Most favourite mode of transport: Car, with 59 people.
• (iii) (A) Number of people who used train = 24.
• (b) Revenue collected = ₹ 70,000.

Quick Tip: To calculate probabilities using a pie chart, divide the specific angle by the total 360°. For revenue-related problems, multiply the number of users by the rate per person.