CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 1- 30/1/1) Available: Download Solution PDF with Answer Key

Solution: The sum of zeroes of a quadratic polynomial ax² + bx + c is given by ^-b⁄_a. For p(x) = 2x² - k√2x + 1, the sum of zeroes is:

^-k√2⁄₂ = ^k√2⁄₂

Equating this to √2:

^k√2⁄₂ = √2
k = 2

Quick Tip: For quadratic equations, use ^{coefficient of x}⁄_{coefficient of x²} to find the sum of roots.

Solution: The probability of losing is given by:
1 - Probability of winning = 1 – 0.79 = 0.21

Quick Tip: The sum of probabilities of all possible outcomes is always 1. Use this property to find the complementary probability.

Solution: For real and equal roots, the discriminant of the quadratic equation must be zero:
b²-4ac = 0
b² = 4ac
ac = ^b2⁄₄

Quick Tip: For real and equal roots, always set Δ = 0 and simplify to derive the correct relationship.

Solution: The n^th term of an A.P. is:
a_n = a + (n − 1)d
84 = 7 + (n - 1)d => 77 = (n − 1)d => d = ⁷⁷⁄_(n-1)

The sum of the first n terms is:
S_n = ⁿ⁄₂[2a + (n-1)d]
2093 = ⁿ⁄₂[2*7 + 77]
2093 = ⁿ⁄₂[91]
n=46/2=23

Quick Tip: Use the formulas a_n = a + (n − 1)d and S_n = ⁿ⁄₂[2a + (n-1)d] to solve A.P. problems systematically.

Solution: The LCM is determined by taking the highest powers of all prime factors:
p = 18a²b⁴ = 2 * 3² * a² * b⁴
q = 20a³b^k = 2² * 5 * a³ * b^k

The LCM is: LCM(p, q) = 2² * 3² * 5 * a³ * b⁴ = 180a³b⁴

Quick Tip: To find the LCM, always take the highest powers of all prime factors present in the given numbers.

Solution: To find the length of AD:
1. First, find the midpoint of BC:
M = (⁶⁺⁰⁄₂, ⁴⁺⁰⁄₂) = (3, 2)
2. Calculate the distance between A(5, −6) and M(3, 2) using the distance formula:

AD = √(5-3)² + (-6 - 2)² = √2² + (-8)² = √4 + 64 = √68

Quick Tip: The median connects a vertex of a triangle to the midpoint of the opposite side. Use the midpoint and distance formulas systematically for accurate results.

Solution:
Given: secθ - tanθ = m
We know: (secθ - tanθ)(secθ + tanθ) = sec²θ - tan²θ = 1
Thus:
secθ + tanθ = ¹⁄_m

Quick Tip: Use the identity sec²θ – tan²θ = 1 to quickly solve problems involving secθ and tanθ.

Solution:
1. Remove all even numbers: 4, 16.
2. Remaining numbers: 1, 7, 9, 21, 25.
3. Prime numbers in the remaining set: 7.
4. Probability: P(prime) = ^{Number of primes}⁄_{Total numbers} = ¹⁄₅

Quick Tip: Remove all non-prime numbers systematically. Remember, 1 is neither prime nor composite.

Solution: We know that the sum of deviations from the mean, weighted by frequencies, is always zero:

Σⁿ_i=1 f_i(x_i - x̄) = 0

Quick Tip: The sum of deviations from the mean is zero because the mean balances all values in a dataset.

Solution:
The given polynomial 4x² – 5x – 6 can be solved to find its zeroes using the quadratic formula:

x = ^{-b ± √(b2 - 4ac)}⁄_2a
a = 4, b = -5, c = -6

Substitute the values:
x = ^{-(-5) ± √((-5)2 – 4(4)(-6))}⁄₂₍₄₎ = ^{5 ± √(25 + 96)}⁄₈ = ^{5 ± √121}⁄₈

Simplify:
x = ^{5 + 11}⁄₈ or x = ^{5 - 11}⁄₈
x = 2 or x = ^-3⁄₄

Thus, the zeroes of 4x² – 5x – 6 are 2 and ^-3⁄₄.
The zeroes of the polynomial x² - px + q are twice these zeroes. Therefore, the zeroes are:
2 × 2 = 4 and 2 × (^-3⁄₄) = ^-3⁄₂

The sum of the zeroes is:
Sum of zeroes = 4 + (^-3⁄₂) = ⁵⁄₂

For a polynomial x² - px + q, the sum of the zeroes is given by:
Sum of zeroes = -(-p) = p

Thus: p=⁵⁄₂

Quick Tip: For finding the roots of quadratics, use the quadratic formula and remember the relationships between roots and coefficients.

Solution: The distance formula is:
Distance = √(x₂ - x₁)² + (y₂ - y₁)²

Given points are (3, -5) and (x, −5). Using the formula:
15 = √(x - 3)² + (−5 − (-5))² = √(x − 3)²

Squaring both sides:
15² = (x - 3)²
225 = (x - 3)²

Taking square roots:
x - 3 = 15 or x - 3 = -15
x = 18 or x = -12

Quick Tip: For horizontal or vertical distances, the distance formula simplifies to the absolute difference of coordinates in one direction.

Solution: Given: cos(α + β) = 0
This implies: α + β = (2n + 1)^π⁄₂ for n ∈ Z

Thus: ^α+β⁄₂ = (2n + 1)^π⁄₄

The value of cos(^α+β⁄₂) for (2n+1)^π⁄₄ alternates between:
cos(^π⁄₄) = ¹⁄_√2

Quick Tip: For trigonometric equations, simplify step by step and remember periodicity to identify values.

Solution: The surface area of a sphere is: 4πr²
Each hemisphere has:
2πr² (curved surface area) + πr² (base area) = 3πr²
Two hemispheres together: 2 × 3πr² = 6πr²

Ratio of the sphere to the two hemispheres: 4πr² : 6πr² = 2:3

Quick Tip: Always add the base area to the curved surface area when dealing with hemispheres.

Solution: The median is defined as the middle value of a dataset when arranged in ascending or descending order. It divides the data into two equal halves.

Quick Tip: To find the median, arrange the data and pick the middle value or average of two middle values for even-sized datasets.

Solution: The largest cone will have its height equal to the edge of the cube (h = 2) and the diameter of the base also equal to the edge (r = 1):
Volume of cone = ¹⁄₃πr²h = ¹⁄₃π(1)²(2) = ^2π⁄₃ cu cm

Quick Tip: For the largest cone in a cube, use the edge of the cube as the cone's height and base diameter.

Solution: List all combinations for sums 2, 3, and 5:
1. Sum 2: (1, 1) (1 way)
2. Sum 3: (1, 2), (2, 1) (2 ways)
3. Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) (4 ways)
Total favorable outcomes: 1 + 2 + 4 = 7
Total outcomes when rolling two dice: 6 × 6 = 36

Probability: P = ^{Favorable outcomes}⁄_{Total outcomes} = ⁷⁄₃₆

Quick Tip: For dice problems, list all combinations carefully for the desired sum and count the favorable outcomes.

Solution: The center of the circle is the midpoint of the diameter.

Midpoint formula: (^{x₁ + x₂}⁄₂, ^{y₁ + y₂}⁄₂)

Given one end (6, 0), let the other end be (x, 0). Using the formula:
(^6+x⁄₂, 0) = (2, 0)
^6+x⁄₂ = 2 => x = -2
The other end is: (-2, 0)

Quick Tip: For diameter problems, use the midpoint formula and substitute the center coordinates to find the unknown endpoint.

Solution: From the figure, the two lines intersect at a single point. This indicates a consistent system with a unique solution.

Quick Tip: Intersecting lines represent a consistent system with a unique solution. Parallel lines indicate an inconsistent system, and coincident lines represent a consistent system with infinite solutions.

Solution: Tangents drawn at the endpoints of a diameter are parallel because they are perpendicular to the radius, and the radii are collinear. While the diameter is indeed the longest chord, it does not explain the parallelism of the tangents.

Quick Tip: Tangents at the ends of a diameter are parallel because they are perpendicular to the same straight line (the diameter).

Solution: The graph of a quadratic polynomial *can* touch the x-axis at one point (e.g., x² has one repeated root at x = 0). Thus, Assertion (A) is false. Reason (R) is correct, as it describes the fundamental property of polynomials.

Quick Tip: If a polynomial graph touches the x-axis, it implies a repeated root, but it can still be a quadratic polynomial.

Solution: Using substitution or elimination method:
1. Solve 7x – 2y = 5 => y = ^{(7x - 5)}⁄₂
2. Substitute y in 8x + 7y = 15:
8x + 7(^{(7x - 5)}⁄₂) = 15 => 16x + 49x – 35 = 30 => 65x = 65 => x = 1
3. Substitute x = 1 in y = ^{(7x - 5)}⁄₂:
y = ^{(7(1) - 5)}⁄₂ = 1
4. Verification: 7(1) - 2(1) = 5 and 8(1) + 7(1) = 15

Solution: x = 1, y = 1

Quick Tip: To verify, substitute the values back into the original equations. If satisfied, your solution is correct.

Solution: If the lost card is black, the total remaining cards are: 52 - 1 = 51
The queen of hearts is still in the pack, so the probability is:
P(Queen of Hearts) = ^{Favorable outcomes}⁄_{Total outcomes} = ¹⁄₅₁

Answer: ¹⁄₅₁

Quick Tip: For probability questions, count the favorable outcomes and divide by the total possible outcomes after the given condition.

Solution: Substitute values of trigonometric functions:
cos 45° = ¹⁄_√2, sin 30° = ¹⁄₂, cos 30° = ^√3⁄₂

2√2 cos 45° sin 30° + 2√3 cos 30° = 2√2 * (¹⁄_√2) * (¹⁄₂) + 2√3 * (^√3⁄₂)
= 1 + 3 = 4

Answer: 4

Quick Tip: Simplify trigonometric expressions by substituting known values step by step.

Solution: The left-hand side (LHS) of the equation is:
sin(A + B) = sin(60° + 30°) = sin 90° = 1.

The right-hand side (RHS) of the equation is: sin A cos B + cos A sin B.

Substitute A = 60° and B = 30°:
sin A = sin 60° = ^√3⁄₂, cos B = cos 30° = ^√3⁄₂, cos A = cos 60° = ¹⁄₂, sin B = sin 30° = ¹⁄₂

Substitute the values: RHS = (^√3⁄₂ * ^√3⁄₂) + (¹⁄₂ * ¹⁄₂) = ³⁄₄ + ¹⁄₄ = 1.
Since LHS = RHS, the equation is verified.

Answer: Verified.

Quick Tip: The sum of angles identity for sine is sin(A + B) = sin A cos B + cos A sin B. Use known trigonometric values for standard angles to simplify.

Solution:
(i) Since BD bisects ∠B and ∠D, we have:
∠ABD = ∠CBD and ∠ADB = ∠CDB
Thus, by AA similarity criterion: ΔABD ~ ΔCBD

(ii) From ΔABD ~ ΔCBD, the sides opposite the equal angles are proportional:
^AB⁄_BC = ^AD⁄_DC
Proof Complete.

Quick Tip: For similarity, use AA, SSS, or SAS criteria and compare corresponding sides to prove equalities.

Solution: Assume 5 - 2√3 is rational. Then:
2√3 = 5 - (rational number)

Since the sum or difference of a rational and an irrational number is irrational, 5 – 2√3 must be irrational, contradicting our assumption.
Proof Complete.

Quick Tip: The sum or difference of a rational number and an irrational number is always irrational.

Solution: Factorize the given expression:
5 × 11 × 17 + 3 × 11 = 11(5 × 17 + 3) = 11(85 + 3) = 11 × 88

Since 11 and 88 are both greater than 1, the given number is a composite number.
Proof Complete.

Quick Tip: A composite number is one that has more than two distinct factors. Factorization helps identify them quickly.

Solution: Let the ratio be m : n. Using the section formula for the x-coordinate:
x = ^{(mx₂ + nx₁)}⁄_{(m + n)}

Substitute x = ⁸⁄₅, x₁ = 1, x₂ = 2:
⁸⁄₅ = ^{(m(2) + n(1))}⁄_{(m + n)} => ⁸⁄₅(m + n) = 2m + n => 8m + 8n = 10m + 5n => 3n = 2m => ^m⁄_n = ³⁄₂

Thus, the ratio is: m : n = 3 : 2

For the y-coordinate:
y = ^{(my₂ + ny₁)}⁄_{(m + n)}
Substitute m : n = 3 : 2, y₁ = 2, y₂ = 3:
y = ^{(3(3) + 2(2))}⁄₍₃₊₂₎ = ¹³⁄₅

Answer: The point divides the line segment in the ratio 3 : 2 and y = ¹³⁄₅

Quick Tip: Use the section formula (^{(mx₂ + nx₁)}⁄_{(m + n)}, ^{(my₂ + ny₁)}⁄_{(m + n)}) for any division point.

Solution: Midpoints of the sides are:
P = (^{(-1 + (-1))}⁄₂, ^{(-1 + 6)}⁄₂) = (-1, ⁵⁄₂)
Q = (^{(-1 + 3)}⁄₂, ^{(6 + 6)}⁄₂) = (1, 6)
R = (^{(3 + 3)}⁄₂, ^{(6 + (-1))}⁄₂) = (3, ⁵⁄₂)
S = (^{(3 + (-1))}⁄₂, ^{(-1 + (-1))}⁄₂) = (1, -1)

Diagonals of PQRS:
• Diagonal PR: Midpoint: (^{(-1 + 3)}⁄₂, ^{(5/2 + 5/2)}⁄₂) = (1, ⁵⁄₂)
• Diagonal QS: Midpoint: (^{(1 + 1)}⁄₂, ^{(6 + (-1))}⁄₂) = (1, ⁵⁄₂)

Since the midpoints of both diagonals are the same, the diagonals bisect each other.
Proof Complete.

Quick Tip: For midpoint calculations, use (^{(x₁ + x₂)}⁄₂, ^{(y₁ + y₂)}⁄₂) and compare midpoints of diagonals.

Solution: To find the minimum number of rooms, we need to calculate the greatest common divisor (GCD) of 48, 80, and 144. Perform prime factorization:
48 = 2⁴ * 3, 80 = 2⁴ * 5, 144 = 2⁴ * 3²
The GCD is: 2⁴ = 16

Thus, each room can accommodate 16 teachers. The number of rooms required for each subject is:
⁴⁸⁄₁₆ = 3, ⁸⁰⁄₁₆ = 5, ¹⁴⁴⁄₁₆ = 9

Total number of rooms: 3 + 5 + 9 = 17

Answer: 17 rooms.

Quick Tip: For similar problems, use the GCD to determine the maximum number of teachers per room and divide the total accordingly.

Solution: Start with the left-hand side:
LHS = ^tanθ⁄_{(1 - cotθ)} + ^cotθ⁄_{(1 - tanθ)}

Substitute tanθ = ^sinθ⁄_cosθ and cotθ = ^cosθ⁄_sinθ
LHS = ^{(sinθ/cosθ)}⁄_{(1 - (cosθ/sinθ))} + ^{(cosθ/sinθ)}⁄_{(1 - (sinθ/cosθ))}

Simplify each term:
LHS = ^sin2θ⁄_{(cosθ(sinθ - cosθ))} + ^cos2θ⁄_{(sinθ(cosθ - sinθ))}

Combine terms:
LHS = ^{(sin2θ - cos2θ)}⁄_{(cosθsinθ(1-(cosθ/sinθ)))} = ^{(sin2θ + cos2θ)}⁄_sinθcosθ = ¹⁄_sinθcosθ = 1 + secθcscθ
Proof Complete.

Quick Tip: Simplify using trigonometric identities like tanθ = ^sinθ⁄_cosθ and sin²θ + cos²θ = 1.

Solution: Let the present ages of Rashmi and Nazma be R and N respectively. From the problem:
R - 3 = 3(N − 3) (1)
R + 10 = 2(N + 10) (2)

Simplify equation (1): R - 3 = 3N - 9 => R = 3N - 6
Substitute R = 3N – 6 into equation (2):
3N - 6 + 10 = 2N + 20 => N = 22
Substitute N = 22 into R = 3N – 6: R = 3(22) - 6 = 60

Answer: Rashmi is 60 years old, and Nazma is 22 years old.

Quick Tip: Translate word problems into equations carefully and solve step by step.

Solution: Step 1: AB is the diameter, so by the property of a circle, ∠APB = 90° (angle subtended by the diameter in a semicircle).
Step 2: Since AQ and BP are tangents to the circle, the radius drawn to the tangents at A and B is perpendicular to the tangents. Hence:
∠OAQ = 90° and ∠OBP = 90°
Step 3: In ΔPOQ, ∠OAQ and ∠OBP are perpendicular, and both lines meet at P and Q, forming a right angle at O. Therefore, ∠POQ = 90°

Correct Answer: ∠POQ = 90° is proved.

Quick Tip: To prove such geometry problems, use circle theorems like "Angle subtended by the diameter is a right angle" and "Tangents are perpendicular to the radius."

Solution:
Step 1: Understanding the properties of tangents. For tangents drawn to a circle from an external point: AS = AP, BS = BR, CR = CQ, DP = DQ.
The radius of the circle is given as: OP = OQ = OR = OS = 8 cm.
Step 2: Calculate the lengths. From the given data: BC = BR + RC, and BS = BR.
Substitute BS = 24 cm: RC = BC – BS = 30 – 24 = 6 cm. Thus: QC = 6 cm. (i)
Step 3: Analyze quadrilateral OPDQ. Given AD ⊥ DC, quadrilateral OPDQ forms a square because: ∠POQ = ∠QOD = ∠ODP = 90° (angle between radius and tangent).
Hence: DP = DQ = OP = OQ = 8 cm. … (ii)
Step 4: Find the length of DC. The length of DC is: DC = DQ + QC.
Substitute values from equations (i) and (ii): DC = 8 + 6 = 14 cm.

Final Answer: The length of DC is 14 cm.

Quick Tip: For quadrilaterals with an inscribed circle, use tangent properties and radius relations to calculate unknown lengths.

Solution: Let the outer and inner radii be R and r, respectively. Given R - r = 1 and the volume of the metal is:
Volume = πh(R² – r²)

Substitute h = 14 and Volume = 176:
176 = π(14)(R² – r²) => R² - r² = ¹⁷⁶⁄_14π = 4

From R² – r² = (R - r)(R + r):
4 = (1)(R + r) => R + r = 4

Solve R - r = 1 and R + r = 4:
2R = 5 => R = ⁵⁄₂, 2r = 3 => r = ³⁄₂

Answer: Outer radius R = ⁵⁄₂ cm, Inner radius r = ³⁄₂ cm.

Quick Tip: Use the relationship R² – r² = (R – r)(R + r) to simplify calculations.

Solution:
(i) Length of the arc: The length of an arc is given by:
Length of arc = (^θ⁄_360°) * 2πr

Substitute θ = 60° and r = 21:
Length of arc = (⁶⁰⁄₃₆₀) * 2π(21) = 22π cm

(ii) Area of the minor segment:
The area of the sector is given by: Area of sector = (^θ⁄_360°) * πr²
Area of sector = (⁶⁰⁄₃₆₀) * π(21)² = 73.5π cm²
The area of the triangle formed by the chord and the two radii can be calculated using the formula:
Area of triangle = ¹⁄₂r²sinθ

Substitute r = 21 and θ = 60°:
Area of triangle = ¹⁄₂(21)²sin60° = 220.5√3 cm²
The area of the minor segment is:
Area of minor segment = Area of sector – Area of triangle = 73.5π – 220.5√3 cm²

Answer:
Length of arc = 22π cm
Area of minor segment = 73.5π – 220.5√3 cm²

Quick Tip: Use Length of arc = (^θ⁄₃₆₀) * 2πr and Area of sector = (^θ⁄_360°) * πr² for arc-related calculations.

Solution: Let the first term of the A.P. be 'a' and the common difference be 'd'.
The general formula for the n-th term of an A.P. is: a_n = a + (n − 1)d.

Step 1: Write the equations for the given conditions. The first term is a₁ = a, and the eighth term is: a₈ = a + 7d.
From the problem: a₁ + a₈ = 32 and a₁a₈ = 60.
Substitute a₁ = a and a₈ = a + 7d:
a + (a + 7d) = 32 => 2a + 7d = 32 … (1),
a(a + 7d) = 60 => a² + 7ad = 60 … (2).

Step 2: Solve the equations. From equation (1):
2a + 7d = 32 => a = ^{(32 - 7d)}⁄₂……(3).
Substitute 'a' from equation (3) into equation (2):
(^{(32 - 7d)}⁄₂)² + 7(^{(32 - 7d)}⁄₂)d = 60

Simplify:
(^{(32 - 7d)2}⁄₄) + (^{7(32 - 7d)d}⁄₂) = 60
Multiply through by 4 to eliminate fractions: (32 - 7d)² + 14(32 – 7d)d = 240.
Expand: 1024 - 448d + 49d² + 448d – 98d² = 240.
Combine terms: 49d² – 98d² + 1024 = 240 => -49d² + 1024 = 240.
Simplify: -49d² = -784 => d² = 16 => d = 4 or d = -4.

Step 3: Find 'a'. Substitute d = 4 into equation (3): a = ^{(32 - 7(4))}⁄₂ = 2
If d = -4, then: a = ^{(32 - 7(-4))}⁄₂ = 30

Thus, the possible values are: a = 2, d = 4 or a = 30, d = -4.

Step 4: Sum of the first 20 terms. The sum of the first n terms of an A.P. is: S_n = ⁿ⁄₂[2a + (n-1)d].
Substitute n = 20, a = 2, and d = 4: S₂₀ = ²⁰⁄₂[2(2) + 19(4)] = 10[4 + 76] = 800.
If a = 30 and d = −4: S₂₀ = ²⁰⁄₂[2(30) + 19(-4)] = 10[60 – 76] = -160.

Answer: First term = a = 2, Common difference = d = 4, Sum of first 20 terms = 800. (Alternatively: First term a=30, common difference d=-4, Sum of first 20 terms = -160)

Quick Tip: For solving A.P. problems, use the general term formula a_n = a + (n - 1)d and the sum formula S_n = ⁿ⁄₂[2a + (n − 1)d].

Solution: Let the first term of the A.P. be 'a' and the common difference be 'd'.
The formula for the sum of n terms of an A.P. is: S_n = ⁿ⁄₂[2a + (n - 1)d].

Step 1: Use the given condition for S₉. For the first 9 terms: S₉ = ⁹⁄₂[2a + (9 − 1)d] = ⁹⁄₂[2a + 8d].
Substitute S₉ = 153:
153 = ⁹⁄₂[2a + 8d] => 2a + 8d = 34 ……(i).

Step 2: Use the given condition for the sum of the last 6 terms. The sum of the last 6 terms of the A.P. is given by:
S_{last 6} = ⁶⁄₂[2a_40-5+5d] = 3[2a₃₅ + 5d], where the last term is a₄₀ = a + 39d and a₃₅ = a₄₀ - 5d = a+34d
Substitute a₃₅ into equation S_{last 6}=687 => 687 = 3[2(a+34d)+5d] = 6a + 213d => 2a + 71d = 229 ...(ii)

Step 3: Solve equations (i) and (ii). Solving these two equations we get a=5 and d=3
Step 4: Find the sum of all terms. The sum of 40 terms is: S₄₀ = ⁴⁰⁄₂[2a + (40 − 1)d].
Substitute a = 5 and d = 3: S₄₀ = 20[2(5) + 39(3)] = 2540.

Final Answer: First term: a = 5, Common difference: d = 3, Sum of all terms: S₄₀ = 2540.

Quick Tip: To solve A.P. problems, use the sum formula S_n = ⁿ⁄₂[2a + (n − 1)d] and systematically eliminate variables by solving equations.

Solution: Let ΔABC have a line DE || BC intersecting AB at D and AC at E.
From the property of similar triangles:
If a line is parallel to one side of a triangle, it divides the other two sides proportionally.
Thus: ^AD⁄_DB = ^AE⁄_EC

Proof:
Since DE || BC, by the Basic Proportionality Theorem (or Thales' theorem): ΔADE ~ ΔABC
Hence: ^AD⁄_DB = ^AE⁄_EC

This proves that DE divides AB and AC in the same ratio.
Proof Complete.

Quick Tip: When solving similar triangle problems, identify corresponding sides and apply proportionality rules.

Solution: From the given figure, triangles ΔPAQ, ΔQBR, ΔRCA are similar because they all share the same right angles and a common angle.
For ΔPAQ and ΔQBR:
^x⁄_y = ^AQ⁄_BR
For ΔQBR and ΔRCA:
^y⁄_z = ^BR⁄_RC

Substitute BR into RC:
RC = ^{z * AQ}⁄_x
Since AB+BC = AC, we have AP + QB*AB/AP + RC = AC, leading to 1/y = 1/x + 1/z

Using the proportionality of segments:
¹⁄_x + ¹⁄_z = ¹⁄_y
Proof Complete.

Quick Tip: Use similarity properties of triangles to express ratios between the given sides and simplify accordingly.

Solution:
Let the height of the tower be 'h' m and the distance of point P from the foot of the tower be PR. The total height of the pole and tower is 6 + h m.

Step 1: Using the angle of depression (45°) From ΔQAR:
tan 45° = ^QR⁄_AR
Since tan 45° = 1:
1 = ^h⁄_AR => AR = h. … (i)

Step 2: Using the angle of elevation (60°) From ΔPAR:
tan 60° = ^PR⁄_AR
Substitute PR = 6 + h and AR = h (from equation (i)):
√3 = ^{(6 + h)}⁄_h
Rearrange: √3h = 6 + h => h(√3 − 1) = 6
Simplify: h = ⁶⁄_{(√3 - 1)}

Step 3: Rationalize the denominator. Rationalize: h = ^{6(√3 + 1)}⁄_{((√3 - 1)(√3 + 1))} = ^{6(√3 + 1)}⁄₂ = 3(√3+1)
Simplify: h = 3(√3 + 1).
Substitute √3 = 1.73: h = 3(1.73 + 1) = 8.19 m ≈ 8.20m

Thus, the height of the tower is: h ≈ 8.20 m.

Step 4: Find the distance of point P from the foot of the tower. The distance PR is: PR = 6 + h.
Substitute h = 8.20: PR = 6 + 8.20 = 14.20 m.

Final Answer: Height of the tower: 8.20 m. Distance of point P from the foot of the tower: 14.20 m.

Quick Tip: For problems involving elevation and depression, use trigonometric ratios (tan θ = ^opposite⁄_adjacent) and simplify using given relationships step by step.

Solution:
(i) Forming the quadratic equation:
Let the side of the original square tile be x units. The total area of the floor is: Area of the floor = 200 * x²
If the side length of the tile is increased by 1 unit, the side becomes (x + 1), and the number of tiles required is 128. Hence:
Area of the floor = 128 * (x + 1)²
Equating both expressions for the area: 200x² = 128(x + 1)²

(ii) Simplifying to standard form:
Expand and simplify:
200x² = 128(x² + 2x + 1) => 200x² = 128x² + 256x + 128 => 200x² - 128x² – 256x – 128 = 0 => 72x² – 256x – 128 = 0
Divide through by 8 to simplify:
9x² - 32x - 16 = 0

Standard form: 9x² - 32x - 16 = 0

(iii) (a) Solving by factorization:
Multiply 9(-16) = −144. Find two numbers whose product is -144 and sum is -32: -36 and 4. Rewrite the middle term:
9x² - 36x + 4x - 16 = 0
Group terms:
(9x² - 36x) + (4x - 16) = 0 => 9x(x - 4) + 4(x - 4) = 0
Factorize: (9x + 4)(x - 4) = 0
Solutions: x = ^-4⁄₉ (not valid, side length cannot be negative), x = 4

Answer: The original side length of the tile is x = 4 units.

(iii) (b) Solving using the quadratic formula:
The quadratic equation is: 9x² - 32x - 16 = 0
Using the quadratic formula: x = ^{-b ± √(b² - 4ac)}⁄_2a
Here, a = 9, b = −32, c = -16. Substitute:
x = ^{-(-32) ± √((-32)² – 4(9)(-16))}⁄₂₍₉₎ = ^{(32 ± √1600)}⁄₁₈ = ^{(32 ± 40)}⁄₁₈

Calculate: x = ⁷²⁄₁₈ = 4, x = ^-8⁄₁₈ = ^-4⁄₉
Reject x = ^-4⁄₉ as it is not valid. Thus: x = 4.

Final Answer: The original side length of the tile is 4 units.

Quick Tip: For problems involving quadratic equations, use both factorization and the quadratic formula to cross-verify results.

Solution:
(i) Median Class:
The total frequency is: N = 8 + 9 + 10 + 12 + 9 = 48
The cumulative frequency (CF) is:

Solution:
(i) AR = x m (Tangents from an external point are equal)
(ii) Quadrilateral BQOR is a square because all angles are 90° (tangent and radius are perpendicular) and adjacent sides are equal (OQ = OR = radius, BQ = BR = tangents from same point).
(iii)(a) Find the length PC in terms of x and hence find the value of x:
PC = AC - AP = AC-x
Using Pythagoras theorem in ΔABC:
AC² = AB² + BC²
Substituting the values:
AC² = 7² + 15² = 274 => AC = √274
Also: AP + PC = AC
Substituting AP = x and PC = AC-x => x + AC - x = AC. Since PC= CQ and AR=x, BR = 7-x. Also BR=BQ=radius of incircle which is CQ = 15-CP= 15-x, Thus 7-x=15-AC+x
AC = 8 +2x. So 8 + 2x = √274. Solving for x gives x≈4.28m

(iii) (b) Find x and hence find the radius 'r' of the circle: Since BQOR is a square, hence the radius is equal to the side BQ or OR. Thus r=BQ
BQ = 7-x => BQ= 7-4.28m = 2.72m. So radius is approximately 2.72m.

Final Answer:
• PC = 8 + x ≈ 12.28 m.
• Radius of the circle r ≈ 2.72 m.

Quick Tip: For problems involving inradius and tangents:
• Use the inradius formula r = ^{(AB + BC - AC)}⁄₂
• Apply tangent properties to find segment lengths.
• Use the Pythagoras theorem to compute side lengths.