CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 3- 30/5/3) with Answer Key

Solution: The smallest prime number is 2, and the smallest odd composite number is 9. The LCM of two numbers is calculated by finding the smallest number divisible by both.

Prime factorization of 2: 2.

Prime factorization of 9: 3 x 3.

The LCM is the product of the highest powers of all prime factors:

LCM = 2 x 3² = 18.

Conclusion: The LCM of 2 and 9 is 18.

Quick Tip: For finding the LCM of two numbers, use the highest power of each prime factor involved.

Solution: The formula for the mean of the first n natural numbers is:

Mean = ^{Sum of the first n natural numbers}/_n = ^n(n+1)/2/_n = ⁿ⁺¹/₂.

Equating this to ⁵ⁿ/₉:

ⁿ⁺¹/₂ = ⁵ⁿ/₉.

Cross-multiply:

9(n+1) = 10n.

Simplify:

9n + 9 = 10n.

n = 9.

Conclusion: The value of n is 9.

Quick Tip: For mean calculations involving natural numbers, always simplify the equation step-by-step for accuracy.

Solution: From the given equation:

5 tan θ - 12 = 0 ⇒ tan θ = ¹²/₅.

Using the Pythagorean identity:

sec² θ = 1 + tan² θ.

Substitute tan θ = ¹²/₅:

sec² θ = 1 + (¹²/₅)² = 1 + ¹⁴⁴/₂₅ = ¹⁶⁹/₂₅.

sec θ = ¹³/₅.

The reciprocal gives:

cos θ = ⁵/₁₃.

Using sin² θ + cos² θ = 1:

sin² θ = 1 - cos² θ = 1 - (⁵/₁₃)² = 1 - ²⁵/₁₆₉ = ¹⁴⁴/₁₆₉.

sin θ = √¹⁴⁴/₁₆₉ = ¹²/₁₃.

Conclusion: The value of sin θ is ¹²/₁₃.

Quick Tip: Remember, tan θ = ^Opposite/_Adjacent, and use the Pythagorean theorem for finding sin θ and cos θ.

Solution: The given sequence √18, √50, √98, ... is in arithmetic progression (A.P.) since the difference between consecutive terms is constant.

Step 1: Calculate the common difference (d):

d = √50 - √18.

Simplify each term:

√50 = √(25 ⋅ 2) = 5√2, √18 = √(9 ⋅ 2) = 3√2.

Thus:

d = 5√2 - 3√2 = 2√2.

Step 2: Find the 4^th term:

The general term of an A.P. is given by:

a_n = a + (n - 1)d,

where a = √18 = 3√2, n = 4, and d = 2√2. Substituting:

a₄ = 3√2 + (4 - 1)(2√2) = 3√2 + 6√2 = 9√2.

Simplify:

a₄ = √(81 ⋅ 2) = √162.

Conclusion: The 4^th term is √162.

Quick Tip: To find the next term in an A.P., calculate the common difference and apply the formula for the n-th term.

Solution: Given DE || BC, by the Basic Proportionality Theorem (Thales' theorem), we have:

^AD/_DB = ^AE/_EC.

Step 1: Express EC in terms of AE, AD, and DB:

^AD/_DB = ^AE/_EC.

Substitute AD = 2.4 cm, DB = 4 cm, AE = 2 cm:

^2.4/₄ = ²/_EC.

Step 2: Solve for EC:

EC = ^{2 × 4}/_2.4 = ⁸/_2.4 = ¹⁰/₃ cm.

Step 3: Find AC:

AC = AE + EC = 2 + ¹⁰/₃ = ⁶/₃ + ¹⁰/₃ = ¹⁶/₃ cm.

Conclusion: The length of AC is:

AC = ¹⁶/₃ cm.

Quick Tip: Use the Basic Proportionality Theorem (^AD/_DB = ^AE/_EC) to solve problems involving parallel lines in triangles.

Solution: Given:

∠RJL = 42°, RJ and RL are tangents.

In a circle, the angle formed by the tangents at the external point (∠RJL) is half the angle subtended by the chord at the center (∠JOL):

∠JOL = 2 ⋅ ∠RJL.

Substitute the given value:

∠JOL = 2 ⋅ 42° = 84°.

Conclusion: The measure of ∠JOL is 84°.

Quick Tip: For tangents and angles in circles, the angle subtended at the center is twice the angle between the tangents.

Solution: The perimeter of the sector is given by:

Perimeter = 2r + Arc length.

The arc length is:

Arc length = ^θ/_360° ⋅ 2πr.

Substitute θ = 60°, r = 21, and π = ²²/₇:

Arc length = ⁶⁰/₃₆₀ ⋅ 2 ⋅ ²²/₇ ⋅ 21 = ¹/₆ ⋅ 2 ⋅ ²²/₇ ⋅ 21 = 22 cm.

The perimeter is:

Perimeter = 2(21) + 22 = 42 + 22 = 64 cm.

Conclusion: The perimeter is 64 cm.

Quick Tip: For sector problems, always add the arc length and twice the radius to find the total perimeter.

Solution: For a quadratic equation of the form:

ax² + bx + c = 0,

the sum of the roots is:

^-b/_a.

The product of the roots is:

^c/_a.

Substitute the values a = 5, b = -6, and c = 21:

Sum of the roots: ^-(-6)/₅ = ⁶/₅.

Product of the roots: ²¹/₅.

Find the ratio of the sum to the product:

Ratio = ^6/5/_21/5 = ⁶/₂₁ = ²/₇.

Conclusion: The ratio of the sum and product of the roots is 2 : 7.

Quick Tip: For quadratic equations, use the relationships between coefficients and roots: Sum of roots = ^-b/_a, Product of roots = ^c/_a.

Solution: The nth term of an arithmetic progression (A.P.) is given by:

a_n = a + (n-1)d.

For terms from the end, the sequence reverses, and the new first term (a') becomes 49, with a common difference of -3 (reverse of the original d = 3).

The 14th term from the end is:

a₁₄ = 49 + (14-1)(-3).

Simplify:

a₁₄ = 49 - 39 = 10.

Conclusion: The 14th term from the end is 10.

Quick Tip: To find terms from the end of an A.P., reverse the sequence and recalculate the term using the modified common difference.

Solution: The relationship between the height of the tower (h), the length of the shadow (l), and the angle of elevation (θ) is given by:

tan θ = ^{Height of the tower}/_{Length of the shadow} = ^h/_l.

Given:

l = √3h ⇒ tan θ = ^h/_√3h = ¹/_√3.

The angle θ such that tan θ = ¹/_√3 is:

θ = 30°.

Conclusion: The angle of elevation of the Sun is 30°.

Quick Tip: For shadow problems, remember standard trigonometric values: tan 30° = ¹/_√3, tan 45° = 1, and tan 60° = √3.

Solution: The multiples of 4 between 1 and 15 are:

4, 8, 12.

Thus, there are 3 favorable outcomes.

The total number of outcomes is:

15 (numbers from 1 to 15).

The probability is given by:

Probability = ^{Number of favorable outcomes}/_{Total number of outcomes} = ³/₁₅.

Conclusion: The probability of selecting a multiple of 4 is ³/₁₅.

Quick Tip: To calculate probability, always divide the count of favorable outcomes by the total number of possible outcomes.

Solution: From the given equations:

^x/₃ = 2 sin A ⇒ x = 6 sin A,

^y/₃ = 2 cos A ⇒ y = 6 cos A.

We need to find:

x² + y².

Substitute the values of x and y:

x² + y² = (6 sin A)² + (6 cos A)².

Simplify:

x² + y² = 36 (sin² A + cos² A).

Using the trigonometric identity sin² A + cos² A = 1:

x² + y² = 36 ⋅ 1 = 36.

Conclusion: The value of x² + y² is 36.

Quick Tip: Always remember the trigonometric identity: sin² A + cos² A = 1.

Solution: For a quadratic polynomial p(x) = ax² + bx + c, the sum and product of the roots are given by:

α + β = ^-b/_a, αβ = ^c/_a.

Given:

p(x) = kx² - 30x + 45k ⇒ a = k, b = -30, c = 45k.

Substitute into the formulas:

α + β = ^-(-30)/_k = ³⁰/_k,

αβ = ^45k/_k = 45.

It is also given that:

α + β = αβ ⇒ ³⁰/_k = 45.

Solve for k:

30 = 45k ⇒ k = ³⁰/₄₅ = ²/₃.

Conclusion: The value of k is ²/₃.

Quick Tip: For polynomials, always use the sum and product of roots formulas: α + β = ^-b/_a and αβ = ^c/_a.

Solution: The length of an arc is given by:

Arc length = ^θ/_360° ⋅ 2πr.

Substitute Arc length = 10π, r = 12, and solve for θ:

10π = ^θ/₃₆₀ ⋅ 2π ⋅ 12.

Simplify:

10 = ^θ/₃₆₀ ⋅ 24 ⇒ ^θ/₃₆₀ = ¹⁰/₂₄.

θ = ¹⁰/₂₄ ⋅ 360 = 150°.

Conclusion: The angle subtended by the arc is 150°.

Quick Tip: For arc length problems, substitute known values into the formula and solve for the unknown step-by-step.

Solution: To find the LCM of 28, 44, 132:

1. Perform prime factorization:

28 = 2² ⋅ 7, 44 = 2² ⋅ 11, 132 = 2² ⋅ 3 ⋅ 11.

2. Take the highest powers of all prime factors:

LCM = 2² ⋅ 3 ⋅ 7 ⋅ 11 = 924.

Conclusion: The LCM of 28, 44, 132 is 924.

Quick Tip: For LCM, take the highest powers of all prime factors common or unique to the given numbers.

Solution: Given:

• Radius r = 10 cm.
• The chord subtends a 90° angle at the center.

In △OAB, O is the center, and ∠AOB = 90°. Using the Pythagoras theorem:

Chord length (AB) = √(OA² + OB²).

Substitute OA = OB = r = 10:

Chord length (AB) = √(10² + 10²) = √200 = 10√2.

Conclusion: The length of the chord is 10√2.

Quick Tip: For chords subtending 90° at the center, use the Pythagoras theorem to calculate the chord length.

Solution: The given equations are:

1. 3x + 4y = 5 and 2. 6x + 8y = 7

Rewrite Equation 2 to check if it is a multiple of Equation 1:

6x + 8y = 7 ⇒ 2(3x + 4y) = 7

Since the constant terms (5 and 7) are not in the same ratio as the coefficients, the lines are not coincident. The coefficients of x and y are in the same ratio:

³/₆ = ⁴/₈.

This implies the lines are parallel.

Conclusion: The lines represented by the equations are parallel.

Quick Tip: For a system of linear equations, lines are parallel if the ratios of the coefficients of x and y are equal but differ in constant terms.

Solution: If a number N divides 281 leaving a remainder of 5 and 1249 leaving a remainder of 7, then:

281 - 5 = 276 and 1249 - 7 = 1242

must be divisible by N.

Thus, N is the greatest common divisor (GCD) of 276 and 1242.

Step 1: Find the GCD of 276 and 1242.

Using the Euclidean algorithm:

1242 = 276 ⋅ 4 + 138

276 = 138 ⋅ 2 + 0

The GCD is 138.

Conclusion: The greatest number which divides 281 and 1249, leaving the specified remainders, is 138.

Quick Tip: For such problems, subtract the remainders from the given numbers and find the GCD of the resulting values.

Solution:

Both Assertion (A) and Reason (R) are true. However, Reason (R) is not the correct explanation for Assertion (A). The degree of a zero polynomial is undefined because it does not have any terms, while the degree of a non-zero constant polynomial is defined as 0.

Quick Tip: Remember: The degree of a zero polynomial is undefined, while the degree of any non-zero constant polynomial is 0.

Solution:

Both Assertion (A) and Reason (R) are true. The line EF, being parallel to the parallel sides of the trapezium (AB and DC), divides the non-parallel sides proportionally. The Reason (R) provides the correct explanation for Assertion (A).

Quick Tip: Use the property of proportionality in trapeziums: A line parallel to the parallel sides divides the non-parallel sides proportionally.

Solution:

Total cards left after removing the king, queen, and ace of clubs and diamonds:

52 - 3 - 3 = 46.

(i) Probability of getting a card of clubs:

The total number of club cards is originally 13. After removing 3 (king, queen, and ace), the remaining number of club cards is:

13 - 3 = 10.

The probability is:

P(card of clubs) = ^{Number of club cards left}/_{Total number of cards left} = ¹⁰/₄₆ = ⁵/₂₃.

(ii) Probability of getting a red-colored card:

The total number of red cards (hearts and diamonds) is originally 26. After removing 3 red cards (king, queen, and ace of diamonds), the remaining number of red cards is:

26 - 3 = 23.

The probability is:

P(red-colored card) = ^{Number of red cards left}/_{Total number of cards left} = ²³/₄₆ = ¹/₂.

Conclusion:

(i) The probability of getting a card of clubs is ⁵/₂₃.

(ii) The probability of getting a red-colored card is ¹/₂.

Quick Tip: Always adjust the total number of cards based on what has been removed, and use the ratio of favorable outcomes to total outcomes for probability calculations.

Solution:

Given:

• Radius of the circle: r = 3 cm
• Angle between the tangents: ∠APB = 60°

Step 1: Analyze the geometry.

The tangents form two right triangles with the center of the circle. In △APO, where O is the center of the circle:

∠APO = ^∠APB/₂ = ^60°/₂ = 30°.

Step 2: Use trigonometric ratios.

From △APO, using tan 30°:

tan 30° = ^{Opposite side (radius)}/_{Adjacent side (tangent length)}.

Substitute the values:

tan 30° = ¹/_√3, ¹/_√3 = ³/_AP.

Solve for AP:

AP = 3√3 cm.

Step 3: Finalize the result.

Since the tangents are symmetrical, the length of each tangent is:

AP = 3√3 cm.

Conclusion: The length of each tangent is 3√3 cm.

Quick Tip: When two tangents are drawn to a circle, use trigonometric ratios in the formed right triangles to find tangent lengths.

Solution:

Let AB be the diameter of a circle with center O, and let P and Q be the tangents at A and B, respectively.

Step 1: Analyze the geometry.

The tangent to a circle is perpendicular to the radius at the point of tangency. Therefore:

∠OAY = 90° and ∠OBP = 90°.

Step 2: Prove parallelism.

The angles ∠OAY and ∠OBP are equal and form alternate interior angles between the lines PQ and XY. Hence, by the property of alternate interior angles:

PQ || XY.

Conclusion: The tangents drawn at the ends of a diameter of a circle are parallel.

Diagram:

Quick Tip: The tangents at the ends of the diameter are always parallel because they form equal alternate interior angles with the line joining the ends of the diameter.

Solution:

Let the ratio in which P divides the line AB be k:1. Using the section formula, the coordinates of the dividing point P(x, y) are:

x = ^{kx₂ + x₁}/_k+1, y = ^{ky₂ + y₁}/_k+1.

Substitute the coordinates of P(-4, 6), A(-6, 10), and B(3, -8):

-4 = ^{3k - 6}/_k+1, 6 = ^{-8k + 10}/_k+1.

Solve the first equation:

-4(k+1) = 3k - 6 ⇒ -4k - 4 = 3k - 6 ⇒ -7k = -2 ⇒ k = ²/₇.

Thus, the ratio is k:1 = 2:7.

Conclusion: The required ratio in which P(-4, 6) divides AB is:

2:7.

Quick Tip: For points dividing a line segment, use the section formula:

x = ^{kx₂ + x₁}/_k+1, y = ^{ky₂ + y₁}/_k+1.

Solution:

Let the points be A(3, 0), B(6, 4), and C(-1, 3). Calculate the lengths of the sides using the distance formula:

d = √( (x₂ - x₁)² + (y₂ - y₁)² ).

1. Length of AB:

AB = √( (6-3)² + (4-0)² ) = √( 3² + 4² ) = √(9 + 16) = √25 = 5.

2. Length of BC:

BC = √( (6 - (-1))² + (4 - 3)² ) = √( (6+1)² + 1² ) = √( 7² + 1² ) = √(49 + 1) = √50.

3. Length of CA:

CA = √( (3 - (-1))² + (0 - 3)² ) = √( (3+1)² + (-3)² ) = √( 4² + 3² ) = √(16 + 9) = √25 = 5.

Since AB = CA, the triangle is isosceles.

Conclusion: The points A(3, 0), B(6, 4), and C(-1, 3) form an isosceles triangle as:

AB = CA = 5.

Quick Tip: To prove a triangle is isosceles, calculate the lengths of all sides using the distance formula:

d = √( (x₂ - x₁)² + (y₂ - y₁)² ).

Check if any two sides are equal.

Solution:

For a polynomial ax² + bx + c, the sum and product of the roots are:

α + β = ^{-Coefficient of x}/_{Coefficient of x²} = ^-(-6)/₅ = ⁶/₅.

αβ = ^{Constant term}/_{Coefficient of x²} = ¹/₅.

Add α + β and αβ:

α + β + αβ = ⁶/₅ + ¹/₅ = ⁷/₅.

Conclusion: The value of α + β + αβ is:

⁷/₅.

Quick Tip: For zeroes of a polynomial, use the relationships:

α + β = ^-b/_a, αβ = ^c/_a.

Solution:

Substitute the trigonometric values:

tan 30° = ¹/_√3, sec 60° = 2, tan 45° = 1, sin² 60° = (^√3/₂)² = ³/₄.

Simplify the numerator:

2 ⋅ ¹/_√3 ⋅ 2 ⋅ 1 = ⁴/_√3.

Simplify the denominator:

1 - sin² 60° = 1 - ³/₄ = ¹/₄.

The entire expression becomes:

^4/√3/_1/4 = ⁴/_√3 ⋅ 4 = ¹⁶/_√3.

Rationalize the denominator:

¹⁶/_√3 ⋅ ^√3/_√3 = ^16√3/₃.

Conclusion: The value of the expression is:

^16√3/₃.

Quick Tip: Always simplify trigonometric functions step-by-step and rationalize the denominator when necessary.

Solution:

Start with the left-hand side (LHS):

LHS = ^{tan θ - cot θ}/_{sin θ cos θ}.

Substitute tan θ = ^{sin θ}/_{cos θ} and cot θ = ^{cos θ}/_{sin θ}:

LHS = ^{(sin θ/cos θ) - (cos θ/sin θ)}/_{sin θ cos θ}.

Simplify the numerator:

^{sin θ}/_{cos θ} - ^{cos θ}/_{sin θ} = ^{sin2 θ - cos2 θ}/_{sin θ cos θ}.

Now substitute back into the LHS:

LHS = ^{(sin2 θ - cos2 θ)/(sin θ cos θ)}/_{sin θ cos θ}.

Simplify further:

LHS = ^{sin2 θ - cos2 θ}/_{sin² θ cos² θ}.

Split the terms:

LHS = ¹/_{cos² θ} - ¹/_{sin² θ}.

Using the identities sec² θ = ¹/_{cos² θ} and csc² θ = ¹/_{sin² θ}:

LHS = sec² θ - csc² θ.

Thus:

LHS = RHS.

Conclusion: The given identity is proved:

^{tan θ - cot θ}/_{sin θ cos θ} = sec² θ - csc² θ.

Quick Tip: For proving trigonometric identities, always convert terms to their sine and cosine equivalents to simplify.

Solution:

The length of the arc of a sector is given by:

Arc length = 2 π r ⋅ ^θ/_360°.

Substitute r = 21, θ = 150°, and π = ²²/₇:

Arc length = 2 ⋅ ²²/₇ ⋅ 21 ⋅ ¹⁵⁰/₃₆₀.

Simplify:

Arc length = ^{2 ⋅ 22 ⋅ 21 ⋅ 150}/_{7 ⋅ 360} = 55 cm.

The area of the sector is given by:

Area of sector = π r² ⋅ ^θ/_360°.

Substitute r = 21, θ = 150°, and π = ²²/₇:

Area of sector = ²²/₇ ⋅ 21 ⋅ 21 ⋅ ¹⁵⁰/₃₆₀.

Simplify:

Area of sector = ^{22 ⋅ 21 ⋅ 21 ⋅ 150}/_{7 ⋅ 360} = 577.5 cm².

Conclusion: The length of the arc is 55 cm, and the area of the sector is 577.5 cm².

Quick Tip: For sector calculations, use:

1. Arc length = 2 π r ⋅ ^θ/_360°.
2. Area of sector = π r² ⋅ ^θ/_360°.

Solution:

Let us assume, for the sake of contradiction, that √3 is a rational number. Then it can be expressed as:

√3 = ^p/_q,

where p and q are integers, q ≠ 0, and p and q are coprime (have no common factors other than 1).

Step 1: Square both sides.

3 = ^p2/_q² ⇒ p² = 3q².

Step 2: Analyze divisibility of p.

Since p² is divisible by 3, it follows that p must also be divisible by 3 (property of prime numbers). Let:

p = 3a, where a is an integer.

Step 3: Substitute p = 3a into the equation.

p² = 3q² ⇒ (3a)² = 3q² ⇒ 9a² = 3q² ⇒ q² = 3a².

Step 4: Analyze divisibility of q.

Since q² is divisible by 3, it follows that q must also be divisible by 3.

Step 5: Contradiction.

If both p and q are divisible by 3, then p and q are not coprime, which contradicts our assumption that ^p/_q is in its simplest form.

Conclusion: The assumption that √3 is a rational number leads to a contradiction.

Therefore, √3 is an irrational number.

Quick Tip: To prove a number is irrational, assume it is rational and derive a contradiction using properties of divisibility.

Solution:

Expand (√2 + √3)²:

(√2 + √3)² = 2 + 3 + 2√6 = 5 + 2√6.

Step 1: Assume, for contradiction, that 5 + 2√6 is a rational number.

Let:

5 + 2√6 = ^a/_b,

where a, b are integers, and b ≠ 0.

Step 2: Rearrange to isolate √6.

2√6 = ^a/_b - 5 ⇒ √6 = ^{a - 5b}/_2b.

Step 3: Analyze rationality.

Since a and b are integers, ^{a - 5b}/_2b is a rational number. However, it is given that √6 is an irrational number. This leads to a contradiction.

Step 4: Conclude irrationality.

The assumption that 5 + 2√6 is rational is incorrect. Therefore:

5 + 2√6 = (√2 + √3)²

is an irrational number.

Conclusion: (√2 + √3)² is an irrational number.

Quick Tip: When proving irrationality, assume rationality, isolate the square root term, and demonstrate that it contradicts the given property of irrationality.

Solution:

Step 1: Total number of outcomes.

When three coins are tossed, the possible outcomes are:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Thus, the total number of outcomes is:

n(S) = 8.

Step 2: Calculate probabilities.

Outcomes with at least one head:

{HHH, HHT, HTH, HTT, THH, THT, TTH}.

Number of favorable outcomes:

n(E) = 7.

Probability:

P(At least one head) = ^n(E)/_n(S) = ⁷/₈.

Outcomes with exactly one tail:

{HHT, HTH, THH}.

Number of favorable outcomes:

n(E) = 3.

Probability:

P(Exactly one tail) = ^n(E)/_n(S) = ³/₈.

Outcomes with two heads and one tail:

{HHT, HTH, THH}.

Number of favorable outcomes:

n(E) = 3.

Probability:

P(Two heads and one tail) = ^n(E)/_n(S) = ³/₈.

1. At least one head:
2. Exactly one tail:
3. Two heads and one tail:

Conclusion:

1. Probability of at least one head: ⁷/₈.
2. Probability of exactly one tail: ³/₈.
3. Probability of two heads and one tail: ³/₈.

Quick Tip: List all possible outcomes for clarity when calculating probabilities involving multiple events.

Solution:

Let the parallelogram ABCD circumscribe a circle, touching the sides AB, BC, CD, and DA at points P, Q, R, and S, respectively.

Step 1: Use the property of tangents.

The lengths of tangents drawn from an external point to a circle are equal. Therefore:

AP = AS, BP = BQ, CR = CQ, DR = DS.

Step 2: Add the tangent pairs.

Adding all the equal tangents:

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ).

Simplify:

AB + CD = AD + BC.

Step 3: Use the properties of a parallelogram.

In a parallelogram, opposite sides are equal:

AB = CD and AD = BC.

Substitute these values:

AB + AB = AB + AB ⇒ 2AB = 2BC.

Divide by 2:

AB = BC.

Step 4: Conclude that the parallelogram is a rhombus.

Since all sides of the parallelogram are equal (AB = BC = CD = DA), ABCD is a rhombus.

Conclusion: The parallelogram circumscribing a circle is a rhombus.

Diagram:

Quick Tip: The key to solving this problem is using the property that the sum of tangents from opposite sides of a circumscribed circle is equal.

Solution:

The sum of the first n terms of an A.P. is given by:

S_n = ⁿ/₂ [2a + (n-1)d].

Here:

S₁₄ = 1050, a = 10, n = 14.

Substitute the values:

¹⁴/₂ [2(10) + 13d] = 1050.

Simplify:

7 [20 + 13d] = 1050 ⇒ 20 + 13d = 150 ⇒ d = 10.

Find the 20th term (a₂₀):

The n-th term of an A.P. is given by:

a_n = a + (n-1)d.

For n = 20:

a₂₀ = 10 + (20-1)10 = 10 + 190 = 200.

Find the general n-th term (a_n):

Substitute a = 10 and d = 10:

a_n = 10 + (n-1)10 = 10n.

Conclusion: The 20th term is 200 and the n-th term is 10n.

Quick Tip: Use the formula S_n = ⁿ/₂[2a + (n-1)d] to calculate the sum of n terms and a_n = a + (n-1)d for specific terms.

Solution:

The sum of the first n terms is given by:

S_n = ⁿ/₂(a + l),

where a = 5, l = 45, and S_n = 400.

Substitute the values:

ⁿ/₂(5 + 45) = 400.

Simplify:

ⁿ/₂(50) = 400 ⇒ 25n = 400 ⇒ n = 16.

Find the common difference (d):

The last term of an A.P. is given by:

a_n = a + (n-1)d.

Substitute a_n = 45, a = 5, and n = 16:

45 = 5 + (16-1)d ⇒ 45 = 5 + 15d ⇒ 15d = 40 ⇒ d = ⁴⁰/₁₅ = ⁸/₃.

Conclusion: The number of terms is 16 and the common difference is ⁸/₃.

Quick Tip: For problems involving the sum of an A.P., use S_n = ⁿ/₂(a + l) when the first and last terms are given.

Solution:

From the given information:

△FEC ≅ △GDB.

This implies:

∠3 = ∠4.

In △ABC:

∠3 = ∠4 (as given).

Thus:

AB = AC (i).

In △ADE:

∠1 = ∠2 (as given).

Thus:

AD = AE (ii).

Now, divide equation (ii) by equation (i):

^AD/_AB = ^AE/_AC.

This implies:

DE || BC (By Basic Proportionality Theorem).

Since:

∠1 = ∠2 and ∠3 = ∠4,

it follows that:

△ADE ∼ △ABC (by the AA similarity criterion).

Conclusion:

△ADE ∼ △ABC.

Quick Tip: To prove similarity between triangles, look for proportional sides and equal angles. The AA similarity criterion is one of the simplest methods.

Solution:

To prove △ABC ∼ △PQR, extend AD to E such that AD = DE, and join EC. Similarly, extend PM to L such that PM = ML, and join LR.

1. Since △ABD ≅ △ECD (by construction), we have:

AB = EC.

2. Similarly, in △PQR, extend PM, and by symmetry, we have:

PQ = LR.

Now, by the given proportionality condition:

^AB/_PQ = ^AC/_PR = ^AD/_PM.

Since AD = DE and PM = ML, the proportionality extends:

^EC/_LR = ^AC/_PR = ^AD/_PM.

Thus:

△AEC ∼ △PLR (by the proportionality criterion).

3. Since △AEC ∼ △PLR, we have:

∠BAC = ∠QPR.

4. Similarly, ∠1 = ∠3 and ∠2 = ∠4, proving △ABC ∼ △PQR by the AA similarity criterion.

Conclusion:

△ABC ∼ △PQR.

Quick Tip: To prove triangle similarity using medians:

• Extend the median symmetrically and use congruence or similarity of smaller triangles.
• Check proportionality and equal angles to apply the AA similarity criterion.

Solution:

For real and equal roots, the discriminant (D) of the quadratic equation must be zero:

D = b² - 4ac = 0.

Here, a = (k+1), b = -6(k+1), and c = 3(k+9).

Substitute these values into the discriminant:

D = [-6(k+1)]² - 4(k+1)[3(k+9)].

Simplify:

D = 36(k+1)² - 12(k+1)(k+9).

Factorize:

D = 12(k+1)[3(k+1) - (k+9)].

Simplify further:

D = 12(k+1)[3k + 3 - k - 9].

D = 12(k+1)(2k - 6).

Set D = 0:

12(k+1)(2k-6) = 0.

Solve for k:

k+1 = 0 ⇒ k = -1 (not allowed, as k ≠ -1).

2k-6 = 0 ⇒ k = 3.

Conclusion: The value of k is k = 3.

Quick Tip: For real and equal roots of a quadratic equation, always use the condition D = b² - 4ac = 0, and solve systematically for unknown parameters.

Solution:

Let the present age of the son be x years. Then the present age of the man is 2x² years.

Step 1: Form an equation based on the given condition.

After 8 years, the age of the man will be 2x² + 8, and the age of the son will be x + 8. According to the problem:

2x² + 8 = 4 + 3(x + 8).

Simplify:

2x² + 8 = 4 + 3x + 24.

2x² + 8 = 3x + 28.

2x² - 3x - 20 = 0.

Step 2: Solve the quadratic equation.

Factorize:

2x² - 8x + 5x - 20 = 0.

(2x+5)(x-4) = 0.

Solve for x:

x = ^-5/₂ (not possible, as age cannot be negative),

x = 4.

Step 3: Calculate the man's age.

Substitute x = 4 into 2x²:

Man's age = 2(4²) = 2 × 16 = 32 years.

Conclusion: The son's age is 4 years, and the man's age is 32 years.

Quick Tip: When solving age-related problems, carefully form equations based on given conditions and ensure the solutions are realistic (e.g., positive ages).

Solution:

Let the height of the opposite house be CD = h + 15, where 15 m is the height of the window above the ground. Let the horizontal distance between the house and the window be x.

Diagram:

Step 1: Calculate x using tan 45°:

In △BPC:

tan 45° = ^Height/_Base = ¹⁵/_x.

Since tan 45° = 1, we have:

x = 15.

Step 2: Calculate h using tan 30°:

In △BPD:

tan 30° = ^{Height above window (h)}/_{Base (x)}.

Substitute tan 30° = ¹/_√3 and x = 15:

¹/_√3 = ^h/₁₅.

Solve for h:

h = ¹⁵/_√3 = 5√3.

Using √3 = 1.732:

h = 5 ⋅ 1.732 = 8.66 m.

Step 3: Total height of the opposite house:

CD = h + 15 = 8.66 + 15 = 23.66 m.

Conclusion: The height of the opposite house is 23.66 m.

Quick Tip: For problems involving angles of elevation and depression, always break the problem into right-angled triangles and use trigonometric ratios step-by-step.

Solution:

The radius of the cylindrical glass is:

r = ^Diameter/₂ = ^5.6/₂ = 2.8 cm.

Step 1: Calculate the apparent capacity of the cylindrical glass:

The volume of the cylinder is given by:

Volume of cylinder = π r² h.

Substitute π = ²²/₇, r = 2.8, and h = 10:

Apparent capacity = ²²/₇ ⋅ 2.8 ⋅ 2.8 ⋅ 10.

Simplify:

Apparent capacity = 246.4 cm³.

Step 2: Calculate the volume of the hemispherical part:

The volume of a hemisphere is given by:

Volume of hemisphere = ²/₃ π r³.

Substitute π = ²²/₇ and r = 2.8:

Volume of hemisphere = ²/₃ ⋅ ²²/₇ ⋅ 2.8 ⋅ 2.8 ⋅ 2.8.

Simplify:

Volume of hemisphere = 45.9 cm³.

Step 3: Calculate the actual capacity of the glass:

The actual capacity is the apparent capacity minus the volume of the hemispherical part:

Actual capacity = 246.4 - 45.9 = 200.5 cm³.

Convert to milliliters (1 cm³ = 1 ml):

Actual capacity = 200.5 ml.

Conclusion: The apparent capacity of the glass is 246.4 cm³, and the actual capacity of the glass is 200.5 cm³ or 200.5 ml.

Diagram:

Quick Tip: To calculate volumes of combined shapes, always handle each shape separately and adjust the final result accordingly.

Solution:

(i) Coordinates of the vertices of △PQR:

From the figure, the coordinates are:

P(4, 6), Q(3, 2), R(6, 5).

(ii) Find distances and coordinates:

(A) Distance between P and Q:

PQ = √( (4 - 3)² + (6 - 2)² ) = √( 1² + 4² ) = √(1 + 16) = √17.

Distance between Q and R:

QR = √( (3 - 6)² + (2 - 5)² ) = √( (-3)² + (-3)² ) = √(9 + 9) = √18.

(b) The coordinates of the point dividing PR in the ratio 2:1:

Using section formula: (^{2x₂ + 1x₁}/_{2 + 1}, ^{2y₂ + 1y₁}/_{2 + 1}).

Substitute P(4, 6) and R(6, 5):

(^{2 × 6 + 1 × 4}/₃, ^{2 × 5 + 1 × 6}/₃) = (^{12 + 4}/₃, ^{10 + 6}/₃) = (¹⁶/₃, ¹⁶/₃).

(iii) Check if △PQR is an isosceles triangle:

Distance PR:

PR = √( (4 - 6)² + (6 - 5)² ) = √( (-2)² + 1² ) = √(4 + 1) = √5.

Since PQ ≠ QR ≠ PR, △PQR is not an isosceles triangle.

Conclusion:

(i) Coordinates of the vertices: P(4, 6), Q(3, 2), R(6, 5).

(ii) (A) PQ = √17, QR = √18.
(b) The coordinates of the point dividing PR are (¹⁶/₃, ¹⁶/₃).

(iii) △PQR is not isosceles.

Quick Tip: To determine the nature of a triangle, compute the lengths of all sides using the distance formula and compare them.

Solution:

(i) Median Class:

The cumulative frequency is calculated as follows:

The total number of students is 40. Since N/2 = 20, the median class is 40 - 60.

(ii) Mean Time:

The table for x_i (midpoints) and f_ix_i is as follows:

Mean = ^{∑ f_i x_i}/_{∑ fi} = ¹⁷²⁰/₄₀ = 43.

(ii) Mode:

The modal class is 40 - 60. Using the formula:

Mode = L + (^{f₁ - f₀}/_{2f1 - f0 - f2}) × h

where L = 40, f₁ = 13, f₀ = 10, f₂ = 6, h = 20:

Mode = 40 + (^{13 - 10}/_{2(13) - 10 - 6}) × 20 = 40 + (³/_{26 - 16}) × 20 = 40 + 6 = 46.

(iii) Students Taking Less than 60 Seconds:

From the cumulative frequency table, the number of students taking less than 60 seconds is 31.

Conclusion:

(i) Median class: 40 - 60.

(ii) Mean time: 43, Mode: 46.

(iii) Number of students: 31.

Quick Tip: To calculate the mean, use midpoints and summations. The modal class is identified as the class with the highest frequency.

Solution:

(i) Formulate the equations:

Let the number of children be x and the number of adults be y. The given conditions can be written as:

x + y = 300 … (i)

150x + 250y = 55000 … (ii)

(ii) Solve for the number of children and adults:

(A) From equations (i) and (ii), solve for x:

Substitute y = 300 - x into equation (ii):

150x + 250(300 - x) = 55000.

Simplify:

150x + 75000 - 250x = 55000.

-100x + 75000 = 55000 ⇒ -100x = -20000 ⇒ x = 200.

Therefore, the number of children is x = 200.

(b) Substituting x = 200 into equation (i):

y = 300 - 200 = 100.

Therefore, the number of adults is y = 100.

(iii) Calculate the amount collected if 250 children and 100 adults visit the park:

Amount collected = 150 × 250 + 250 × 100.

Amount collected = 37500 + 25000 = 62500.

Conclusion:

(i) The algebraic equations are x + y = 300 and 150x + 250y = 55000.

(ii) Number of children: 200, Number of adults: 100.

(iii) Total amount collected: 62500.

Quick Tip: Use substitution or elimination methods for solving linear equations systematically. Ensure accurate substitution and simplification in word problems.