CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 1- 30/4/1) with Answer Key

Solution:

We are given two equations:
1. ax + by = a² - b²
2. bx + ay = 0

From equation (2), solve for x or y in terms of the other variable. Assume ay = -bx, so: y = -^b⁄_ax (substitute this into equation 1).

Substituting y = -^b⁄_ax into equation (1):
ax + b(-^b⁄_ax) = a² - b²

Simplify:
ax - ^b2⁄_ax = a² - b²

Combine terms:
^{a2x - b2x}⁄_a = a² - b²

Factorize x:
x(a² - b²) = a(a² - b²)

Divide both sides by a² - b² (assuming a ≠ b):
x = a

Substitute x = a into y = -^b⁄_ax:
y = -^b⁄_aa = -b

Finally, calculate x + y:
x + y = a - b

Conclusion:
The value of x + y is a - b.

Quick Tip: When solving simultaneous equations, substitute one variable in terms of the other to simplify and solve step-by-step.

Solution:

The relationship between the HCF and LCM of two numbers is given by:
HCF × LCM = Product of the two numbers.

Here, the two numbers are 65 and 104, with HCF = 13 and LCM = 40x. Substituting into the equation:
13 × (40x) = 65 × 104.

Calculate 65 × 104:
65 × 104 = 6760.

Now solve for x:
520x = 6760 ⇒ x = ⁶⁷⁶⁰⁄₅₂₀ = 13.

Conclusion:
The value of x is 13.

Quick Tip: The product of HCF and LCM of two numbers is always equal to the product of the numbers.

Solution:

We are given p(x) = x² - 5x + 6. First, calculate p(1):
p(1) = (1)² - 5(1) + 6 = 1 - 5 + 6 = 2.

Now calculate p(4):
p(4) = (4)² - 5(4) + 6 = 16 - 20 + 6 = 2.

Add the two values:
p(1) + p(4) = 2 + 2 = 4.

Conclusion:
The value of p(1) + p(4) is 4.

Quick Tip: For polynomial problems, substitute the given values into the expression and simplify step-by-step.

Solution:

The discriminant of a quadratic equation ax² + bx + c = 0 is given by:
Δ = b² - 4ac.

For the equation 3x² - 2x + c = 0, we have a = 3, b = -2, and c = c. Substituting into the discriminant formula:
Δ = (-2)² - 4(3)(c).

Simplify:
Δ = 4 - 12c.

We are given Δ = 16, so:
4 - 12c = 16.

Solve for c:
-12c = 16 - 4 ⇒ -12c = 12 ⇒ c = -1.

Conclusion:
The value of c is -1.

Quick Tip: For quadratic equations, use the discriminant formula Δ = b² - 4ac to determine the nature of the roots or unknown coefficients.

Solution:

The length of an arc is proportional to the angle it subtends at the centre of the circle. The total circumference of a circle corresponds to an angle of 360°. Hence, the ratio of the arc length to the circumference is:
Ratio = ^{Angle subtended by arc}⁄_{Total angle of circle} = ^90°⁄_360°.

Simplify:
Ratio = ⁹⁰⁄₃₆₀ = ¹⁄₄.

Conclusion:
The ratio of the arc length to the circumference is 1:4.

Quick Tip: The arc length ratio can be determined by dividing the angle subtended by the arc by 360°.

Solution:

The formula for the area of a sector of a circle is:
Area of sector = ^θ⁄_360° * πr²,
where θ is the central angle in degrees and r is the radius of the circle.

Substitute the given values (Area = 60π cm², r = 12 cm):
60π = ^θ⁄_360° * π(12)².

Simplify:
60π = ^θ⁄_360° * 144π.

Cancel π and solve for θ:
60 = ^θ⁄_360° * 144 ⇒ ^θ⁄_360° = ⁶⁰⁄₁₄₄.

Simplify:
^θ⁄_360° = ⁵⁄₁₂ ⇒ θ = ⁵⁄₁₂ * 360° = 150°.

Conclusion:
The central angle of the sector is 150°.

Quick Tip: The formula for the area of a sector relates the angle to the total area of the circle. Always substitute known values and simplify systematically.

Solution:

For a moderately skewed distribution, the relationship between the mode, median, and mean is approximated by:
Mode - Mean = 3(Median - Mean).

We are given:
Mode - Median = 24.

Substitute into the equation:
24 = 3(Median - Mean).

Solve for Median - Mean:
Median - Mean = ²⁴⁄₃ = 12.

Conclusion:
The difference between the median and the mean is 12.

Quick Tip: For skewed distributions, use the empirical relationship between mode, median, and mean to solve for missing values.

Solution:

Each die has 6 faces: 1, 2, 3, 4, 5, 6. Odd numbers are 1, 3, 5, so the probability of getting an odd number on one die is:
P(odd on one die) = ³⁄₆ = ¹⁄₂.

The two dice are independent, so the probability of getting odd numbers on both dice is:
P(odd on both dice) = P(odd on first die) * P(odd on second die) = ¹⁄₂ * ¹⁄₂ = ¹⁄₄.

Convert to common fractions:
P(odd on both dice) = ⁹⁄₃₆.

Conclusion:
The probability of getting odd numbers on both dice is 9/36.

Quick Tip: For independent events, multiply individual probabilities to find the combined probability.

Solution:

The total surface area of a solid hemisphere is the sum of its curved surface area and the area of its circular base:
Total Surface Area = 2πr² + πr² = 3πr².

The square of the radius is:
Square of radius = r².

The ratio of total surface area to the square of the radius is:
Ratio = ^3πr2⁄_r² = 3π : 1.

Conclusion:
The ratio of total surface area to the square of the radius is 3π : 1.

Quick Tip: For solid hemispheres, remember that the total surface area includes both the curved surface and the base.

Solution:

We are given sin θ = 1. This implies:
θ = 90° (since sin 90° = 1).

Substitute θ = 90° into the expression ¹⁄₂ sin(^θ⁄₂):
¹⁄₂ sin(^θ⁄₂) = ¹⁄₂ sin(^90°⁄₂) = ¹⁄₂ sin(45°).

We know sin 45° = ¹⁄_√2, so:
¹⁄₂ sin(45°) = ¹⁄₂ * ¹⁄_√2 = ¹⁄_2√2.

Conclusion:
The value of ¹⁄₂ sin(^θ⁄₂) is ¹⁄_2√2.

Quick Tip: Remember that sin 45° = ¹⁄_√2. Use trigonometric identities and angle relationships to simplify such expressions.

Solution:

Two lines are parallel if their slopes are equal. The general equation of a line is given by:
Ax + By + C = 0.

The slope of the line is:
Slope = -^A⁄_B.

For the given line 5x - 3y = 2, the slope is:
Slope = -⁵⁄_-3 = ⁵⁄₃.

Now check the options for the second line to find which has the same slope:
- For -15x + 9y = 5:
Slope = -^-15⁄₉ = ⁵⁄₃.

This matches the slope of the given line, so the two lines are parallel.

Conclusion:
The equation of the second line is -15x + 9y = 5.

Quick Tip: Parallel lines have the same slope. To verify, compare -^A⁄_B for both equations.

Solution:

Let the three numbers in A.P. be (a - d), a, and (a + d). The sum of these numbers is:
(a - d) + a + (a + d) = 30.

Simplify:
3a = 30 ⇒ a = 10.

Thus, the middle term of the sequence is a = 10.

Conclusion:
The middle term of the sequence is 10.

Quick Tip: In an arithmetic progression, the middle term is equal to the average of the sum of all terms.

Solution:

Since DE || BC, by the basic proportionality theorem (Thales theorem), we have:
^AD⁄_DB = ^AE⁄_EC.

Given:
AD = 4 cm, AB = 9 cm, AC = 13.5 cm.

Calculate DB and AE:
DB = AB - AD = 9 - 4 = 5 cm.
AE = AC - EC.

From the proportionality relation:
⁴⁄₅ = ^AE⁄_EC.

Rewriting AE = AC - EC:
⁴⁄₅ = ^{13.5 - EC}⁄_EC.

Cross multiply:
4EC = 5(13.5 - EC).

Simplify:
4EC = 67.5 - 5EC ⇒ 9EC = 67.5 ⇒ EC = ^67.5⁄₉ = 7.5 cm.

Conclusion:
The length of EC is 7.5 cm.

Quick Tip: Use the basic proportionality theorem to solve problems involving parallel lines in triangles.

Solution:

The shadow of the tower is equal to its height. Let the height of the tower be h and the length of its shadow also be h. The altitude of the Sun forms a right triangle, where:
tan(Sun's altitude) = ^{Height of tower}⁄_{Length of shadow}.

Substitute:
tan(Sun's altitude) = ^h⁄_h = 1.

We know:
tan(45°) = 1.

Thus, the Sun's altitude is 45°.

Conclusion:
The Sun's altitude is 45°.

Quick Tip: When the height and shadow of an object are equal, the angle of elevation is always 45°.

Solution:

In the given figure, AB and AC are tangents to the circle from point A. The tangents from a common external point to a circle are equal in length, and the angle between the tangents is bisected by the line segment joining the center of the circle to the external point.

Here, ∠ABC = 42° is the angle between the tangent AB and the chord BC. By the properties of tangents and circles:
∠BAC = 180° - 2 * ∠ABC.

Substitute the given value ∠ABC = 42°:
∠BAC = 180° - 2 * 42° = 180° - 84° = 96°.

Conclusion:
The measure of ∠BAC is 96°.

Quick Tip: For tangents drawn from a common external point, use the property that the angles formed are related to the supplementary angles of the triangle.

Solution:

In a parallelogram, the diagonals bisect each other. Let the coordinates of the fourth vertex D be (x, y). Using the midpoint formula, the midpoint of diagonal AC must be the same as the midpoint of diagonal BD.

Midpoint of AC:
((^{-2 + 8}⁄₂), (^{3 + 3}⁄₂)) = (3, 3).

Midpoint of BD:
((^{6 + x}⁄₂), (^{7 + y}⁄₂)).

Equating the midpoints:
^{6 + x}⁄₂ = 3 ⇒ 6 + x = 6 ⇒ x = 0,
^{7 + y}⁄₂ = 3 ⇒ 7 + y = 6 ⇒ y = -1.

Thus, the coordinates of D are (0, -1).

Conclusion:
The fourth vertex D is at (0, -1).

Quick Tip: In a parallelogram, use the midpoint formula to find the unknown vertex by equating the midpoints of the diagonals.

Solution:

The probability of an event P(E) and its complement P(Ē) always satisfy:
P(E) + P(Ē) = 1.

Given:
P(E) + P(Ē) = q.

This implies:
q = 1.

Now, calculate q² - 4:
q² - 4 = 1² - 4 = 1 - 4 = -3.

Conclusion:
The value of q² - 4 is -3.

Quick Tip: The sum of the probabilities of an event and its complement is always 1. Use this relationship to solve related problems.

Solution:

In the given figure, the tangent QR is divided into two equal segments at P because the circles are symmetric, and the tangent passes through the point of contact A. Thus:
QR = 2 * AP.

Substitute the given value AP = 4.2 cm:
QR = 2 * 4.2 = 8.4 cm.

Conclusion:
The length of QR is 8.4 cm.

Quick Tip: For tangents between two externally touching circles, the tangent length is twice the distance from the point of contact to the tangent's midpoint.

Solution:

1. The mid-point of a line segment always divides the segment into two equal parts, which means the ratio is 1 : 1. Therefore, Assertion (A) is true.

2. To verify Reason (R), let the point (-3, k) divide the segment joining (-5, 4) and (-2, 3). Using the section formula:
x = (^{m₁x₂ + m₂x₁}⁄_{m1 + m2}), y = (^{m₁y₂ + m₂y₁}⁄_{m1 + m2}).

For the x-coordinate:
-3 = (^{1 * (-2) + 2 * (-5)}⁄_{1 + 2}) = (^{-2 - 10}⁄₃) = (^-12⁄₃) = -4 (which does not match -3).

Thus, the point does not divide the line segment in the ratio 1 : 2. Reason (R) is false.

Conclusion:
Assertion (A) is true, but Reason (R) is false.

Quick Tip: Use the section formula to determine the ratio in which a point divides a line segment.

Solution:

1. The formula for the circumference of a circle is:
Circumference = 2πr.

2. Given the circumference is 176 cm, solve for the radius:
176 = 2πr ⇒ r = ¹⁷⁶⁄_2π = ⁸⁸⁄_π.

Approximating π ≈ 3.14:
r = ⁸⁸⁄_3.14 = 28 cm.

3. Both the Assertion (A) and the Reason (R) are true, and Reason (R) correctly explains the Assertion (A).

Conclusion:
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Quick Tip: For problems involving circles, always use the formula Circumference = 2πr and substitute known values to find unknown parameters.

Find the LCM of the intervals 9, 12, and 15:
9 = 3², 12 = 2² * 3, 15 = 3 * 5.

The LCM is the product of the highest powers of all prime factors:
LCM = 2² * 3² * 5 = 180.

Thus, the three bells will toll together after 180 minutes.

Conclusion:
The three bells will next toll together after 180 minutes.

Quick Tip: For periodic events, calculate the least common multiple (LCM) of the time intervals to determine when they coincide.

The angle subtended by the minute hand in 5 minutes is:
Angle = (^360°⁄₆₀) * 5 = 30°.

The area described by the minute hand is a sector of a circle, given by:
Area = (^θ⁄_360°) * πr²,
where θ = 30° and r = 14 cm.

Substitute the values:
Area = (³⁰⁄₃₆₀) * (²²⁄₇) * 14 * 14.

Simplify:
Area = (¹⁄₁₂) * (²²⁄₇) * 196 = ¹⁵⁴⁄₃ cm² ≈ 51.33 cm².

Conclusion:
The area described by the minute hand is approximately 51.33 cm².

Quick Tip: The area of a sector is proportional to the angle subtended by it at the center of the circle. Always express the angle in degrees when using the formula.

The formula for the length of an arc is:
Length of arc = 2πr * (^θ⁄₃₆₀),
where r is the radius and θ is the central angle in degrees.

Substitute the given values r = 42 cm and θ = 60°:
Length of arc = 2 * (²²⁄₇) * 42 * (⁶⁰⁄₃₆₀).

Simplify step-by-step:
Length of arc = 2 * (²²⁄₇) * 42 * (¹⁄₆)
Length of arc = (^{2 * 22 * 42}⁄_{7 * 6})
Length of arc = (¹⁸⁴⁸⁄₄₂) = 44 cm.

Conclusion:
The length of the arc is 44 cm.

Quick Tip: The length of an arc is proportional to the central angle it subtends. Use Length of arc = 2πr * (^θ⁄₃₆₀) for quick calculations.

Substitute the trigonometric values:
cos 60° = ¹⁄₂, sec 30° = ²⁄_√3, tan 45° = 1, sin 30° = ¹⁄₂, sin 60° = ^√3⁄₂.

The numerator is:
5 cos² 60° + 4 sec² 30° - tan² 45° = 5(¹⁄₂)² + 4(²⁄_√3)² - 1.

Simplify:
5(¹⁄₄) + 4(⁴⁄₃) - 1 = ⁵⁄₄ + ¹⁶⁄₃ - 1 = ¹⁵⁄₁₂ + ⁶⁴⁄₁₂ - ¹²⁄₁₂ = ⁶⁷⁄₁₂.

The denominator is:
sin² 30° + sin² 60° = (¹⁄₂)² + (^√3⁄₂)² = ¹⁄₄ + ³⁄₄ = 1.

Thus:
^Numerator⁄_Denominator = (⁶⁷⁄₁₂) / 1 = ⁶⁷⁄₁₂.

Conclusion:
The value of the expression is ⁶⁷⁄₁₂.

Quick Tip: Substitute known trigonometric values into the expression and simplify step-by-step.

1. From sin (A - B) = ¹⁄₂, we know:
sin (A - B) = sin 30° ⇒ A - B = 30° (i).

2. From cos (A + B) = ¹⁄₂, we know:
cos (A + B) = cos 60° ⇒ A + B = 60° (ii).

3. Solve equations (i) and (ii):
A - B = 30°, A + B = 60°.

Add the equations:
2A = 90° ⇒ A = 45°.

Subtract the equations:
2B = 30° ⇒ B = 15°.

Conclusion:
The values of ∠A and ∠B are:
A = 45°, B = 15°.

Quick Tip: For problems involving sin and cos equations, use reference angles and solve the resulting linear equations systematically.

Since △AHK ~ △ABC, the corresponding sides are proportional:
^AK⁄_AC = ^HK⁄_BC.

Substitute the known values:
⁸⁄_AC = ^6.4⁄_3.2.

Simplify:
⁸⁄_AC = 2 ⇒ AC = ⁸⁄₂ = 4 cm.

Conclusion:
The length of AC is 4 cm.

Quick Tip: For similar triangles, use the property that corresponding sides are proportional to find unknown lengths.

Start with the L.H.S:
L.H.S = (^{sin θ - cos θ + 1}⁄_{sin θ + cos θ - 1}).

Divide both the numerator and denominator by cos θ:
L.H.S = (^{(sin θ / cos θ) - (cos θ / cos θ) + (1 / cos θ)}⁄_{(sin θ / cos θ) + (cos θ / cos θ) - (1 / cos θ)}) = (^{tan θ - 1 + sec θ}⁄_{tan θ + 1 - sec θ}).

Simplify further:
L.H.S = (^{tan θ - 1 + sec θ}⁄_{(tan θ - sec θ) + (sec² θ - tan² θ)}).

Using the identity sec² θ - tan² θ = 1, this becomes:
L.H.S = (^{tan θ - 1 + sec θ}⁄_{(sec θ - tan θ)(tan θ + sec θ - 1)}).

Cancel (tan θ + sec θ - 1) in the numerator and denominator:
L.H.S = (¹⁄_{sec θ - tan θ}).

Thus:
L.H.S = R.H.S.

Conclusion:
The given identity is proven:
(^{sin θ - cos θ + 1}⁄_{sin θ + cos θ - 1}) = (¹⁄_{sec θ - tan θ}).

Quick Tip: Divide terms by a common trigonometric function (e.g., cos θ) to simplify complex expressions and apply trigonometric identities.

The total number of outcomes when three coins are tossed is:
Total outcomes = 2³ = 8 (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT).

(i) Probability of at least one head:
P(at least one head) = 1 - P(no heads) = 1 - (¹⁄₈) = ⁷⁄₈.

(ii) Probability of exactly two tails:
P(exactly two tails) = (^{Number of favorable outcomes}⁄_{Total outcomes}) = ³⁄₈ (favorable outcomes: HTT, THT, TTH).

(iii) Probability of at most one tail:
P(at most one tail) = (^{Number of favorable outcomes}⁄_{Total outcomes}) = ⁴⁄₈ = ¹⁄₂ (favorable outcomes: HHH, HHT, HTH, THH).

Conclusion:

• (i) P(at least one head) = ⁷⁄₈.
• (ii) P(exactly two tails) = ³⁄₈.
• (iii) P(at most one tail) = ¹⁄₂.

Quick Tip: For coin toss problems, count all possible outcomes carefully and identify favorable cases for each condition.

The total number of outcomes is:
Total outcomes = 90.

(i) Probability of a two-digit number less than 40:
Two-digit numbers less than 40 are 10 to 39 (inclusive). The total count is:
Count = 30 (10, 11, 12, ..., 39).
P(two-digit number less than 40) = ³⁰⁄₉₀ = ¹⁄₃.

(ii) Probability of a number divisible by 5 and greater than 50:
Numbers divisible by 5 and greater than 50 are 55, 60, 65, 70, 75, 80, 85, 90. The total count is:
Count = 8.
P(divisible by 5 and greater than 50) = ⁸⁄₉₀ = ⁴⁄₄₅.

(iii) Probability of a perfect square number:
Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81. The total count is:
Count = 9.
P(perfect square number) = ⁹⁄₉₀ = ¹⁄₁₀.

Conclusion:

• (i) P(two-digit number less than 40) = ¹⁄₃.
• (ii) P(divisible by 5 and greater than 50) = ⁴⁄₄₅.
• (iii) P(perfect square number) = ¹⁄₁₀.

Quick Tip: For probability problems, list all favorable outcomes systematically and divide by the total possible outcomes.

Let the number of ₹50 notes be x and the number of ₹100 notes be y.

From the problem, we are given:
x + y = 25 (i).

The total amount is ₹2000:
50x + 100y = 2000.

Simplify the second equation:
x + 2y = 40 (ii).

Solve equations (i) and (ii):

1. From equation (i): x = 25 - y.
2. Substitute x = 25 - y into equation (ii):
   25 - y + 2y = 40 ⇒ 25 + y = 40 ⇒ y = 15.
3. Substitute y = 15 into equation (i):
   x + 15 = 25 ⇒ x = 10.

Thus, Rehana received:
x = 10 (₹50 notes) and y = 15 (₹100 notes).

Conclusion:
Rehana received 10 notes of ₹50 and 15 notes of ₹100.

Quick Tip: For such problems, use a system of linear equations to represent the constraints and solve using substitution or elimination methods.

The given polynomial is:
P(x) = 4x² + 4x - 3 = (2x + 3)(2x - 1).

Zeroes of the polynomial:
x = -³⁄₂, x = ¹⁄₂.

1. Verify the sum of the zeroes:
Sum of zeroes = -³⁄₂ + ¹⁄₂ = -⁴⁄₂ = -2.
This matches:
Sum of zeroes = - (^{(coefficient of x)}⁄_{coefficient of x²}).

2. Verify the product of the zeroes:
Product of zeroes = -³⁄₂ * ¹⁄₂ = -³⁄₄.
This matches:
Product of zeroes = (^{constant term}⁄_{coefficient of x²}).

Conclusion:
The zeroes are -³⁄₂ and ¹⁄₂, and they satisfy the relationship between zeroes and coefficients of the polynomial.

Quick Tip: To verify zeroes of a polynomial, use the relationships: Sum of zeroes = - (^{(coefficient of x)}⁄_{coefficient of x²}) and Product of zeroes = (^{constant term}⁄_{coefficient of x²}).

From the given polynomial:
α + β = -1, αβ = -2.

We need to find:
(^α⁄_β) + (^β⁄_α) = (^{α2 + β2}⁄_αβ).

1. Using the identity α² + β² = (α + β)² - 2αβ:
α² + β² = (-1)² - 2(-2) = 1 + 4 = 5.

2. Substitute into the expression for (^α⁄_β) + (^β⁄_α):
(^α⁄_β) + (^β⁄_α) = (^{α2 + β2}⁄_αβ) = ⁵⁄_-2 = -⁵⁄₂.

Conclusion:
The value of (^α⁄_β) + (^β⁄_α) is -⁵⁄₂.

Quick Tip: For problems involving zeroes of polynomials, use the relationships: α + β = - (^{(coefficient of x)}⁄_{coefficient of x²}) and αβ = (^{constant term}⁄_{coefficient of x²}).

Assume (^{2 - √3}⁄₅) to be a rational number.

Let:
(^{2 - √3}⁄₅) = ^p⁄_q,
where p and q are integers, and q ≠ 0.

Rearrange the equation to isolate √3:
√3 = ^{2q - 5p}⁄_q.

Here:

• p and q are integers, so 2q - 5p and q are also integers.
• Therefore, (^{2q - 5p}⁄_q) is a rational number.

However, √3 is known to be an irrational number. This creates a contradiction.

Thus, our assumption that (^{2 - √3}⁄₅) is rational is false.

Conclusion:
(^{2 - √3}⁄₅) is an irrational number.

Quick Tip: To prove a number is irrational, assume it to be rational and show that this assumption leads to a contradiction.

From the given figure:

1. AP = AS (i),
2. BP = BQ (ii),
3. CR = CQ (iii),
4. DR = DS (iv).

Adding equations (i), (ii), (iii), and (iv):
AP + BP + CR + DR = AS + BQ + CQ + DS.

This simplifies to:
AB + CD = AD + BC.

Since ABCD is a parallelogram, the opposite sides are equal:
AB = CD and AD = BC.

Substitute these equalities:
2AB = 2AD ⇒ AB = AD.

Thus, all sides of ABCD are equal, which proves it is a rhombus.

Conclusion:
The parallelogram ABCD circumscribing a circle is a rhombus.

Quick Tip: A parallelogram circumscribing a circle always has equal sides due to the tangential property of the circle, making it a rhombus.

Let AB and CD be the two pillars of equal height h m, and let P be the point on the road x m away from pillar CD.

In △CDP:
tan 60° = ^h⁄_x, so h = √3x (i).

In △ABP:
tan 30° = ^h⁄_{100 - x}, so h = (^{100 - x}⁄_√3) (ii).

Equate equations (i) and (ii):
√3x = (^{100 - x}⁄_√3).

Multiply through by √3:
3x = 100 - x.

Solve for x:
4x = 100, x = 25 m.

Substitute x = 25 into equation (i):
h = √3x = 25 * 1.732 = 43.3 m.

Conclusion:
The height of each pillar is 43.3 m, and the distances of the point P from the pillars are:
Distance from CD = 25 m, Distance from AB = 100 - 25 = 75 m.

Quick Tip: For such problems, use trigonometric ratios to establish equations for the unknowns and solve them systematically.

In △ABE and △CFB:

1. ∠EAB = ∠BCF (corresponding angles, since AB || CD).
2. ∠AEB = ∠CBF (vertically opposite angles).

Thus, by the AA similarity criterion:
△ABE ~ △CFB.

Conclusion:
△ABE ~ △CFB by the AA similarity criterion.

Quick Tip: To prove triangles similar, identify pairs of equal angles using properties like corresponding angles, vertically opposite angles, or parallel line properties.

From the problem, it is given that:
^AB⁄_PQ = ^BC⁄_QR = ^AD⁄_PM.

Thus, by the SSS similarity criterion (corresponding sides proportional):
△ABD ~ △PQM (i).

From the property of medians and similarity:
∠B = ∠Q (ii).

Now, consider △ABC and △PQR. The sides are proportional, and the corresponding angles are equal:
^AB⁄_PQ = ^BC⁄_QR, ∠B = ∠Q.

By the SSS similarity criterion:
△ABC ~ △PQR.

Conclusion:
△ABC ~ △PQR by the SSS similarity criterion.

Quick Tip: To prove two triangles are similar using medians, ensure the sides and medians are proportional and verify one pair of corresponding angles is equal.

Let the original speed of the train be x km/h.
The new speed = (x + 15) km/h.

According to the problem:
(⁹⁰⁄_x) - (⁹⁰⁄_{x + 15}) = ¹⁄₂.

Simplify the equation:
90((¹⁄_x) - (¹⁄_{x + 15})) = ¹⁄₂,
(^{90(x + 15) - 90x}⁄_{x(x + 15)}) = ¹⁄₂,
(^{90 * 15}⁄_{x(x + 15)}) = ¹⁄₂.

Cross-multiply:
180x(x + 15) = 2700.

Simplify:
x² + 15x - 2700 = 0.

Factorize:
(x + 60)(x - 45) = 0.

Since speed cannot be negative:
x = 45.

Conclusion:
The original speed of the train is 45 km/h.

Quick Tip: For time-speed-distance problems, set up equations using the relationship: Time = ^Distance⁄_Speed, and solve step-by-step.

For real and equal roots, the discriminant (Δ) of the quadratic equation must be zero:
Δ = b² - 4ac = 0.

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

Substitute these into the discriminant formula:
[-6(c + 1)]² - 4(c + 1)[3(c + 9)] = 0.

Simplify:
36(c + 1)² - 12(c + 1)(c + 9) = 0,
12(c + 1)[3(c + 1) - (c + 9)] = 0,
12(c + 1)(2c - 6) = 0.

Since c ≠ -1, solve:
2c - 6 = 0 ⇒ c = 3.

Conclusion:
The value of c is 3.

Quick Tip: For quadratic equations, check the discriminant (b² - 4ac) to determine the nature of the roots. If Δ = 0, the roots are real and equal.

The table with calculated midpoints (x_i) and x_if_i is:

1. Mean Calculation:
Mean = (^{Σ f_ix_i}⁄_{Σ fi}) = ²⁸³⁰⁄₈₀ = 35.375 years.

2. Mode Calculation:
The modal class is 35–45, as it has the highest frequency (f = 23).

Using the formula for mode:
Mode = l + ((^{f_m - f₁}⁄_{2fm - f1 - f2}))h,
where:

• l = 35,
• f_m = 23,
• f₁ = 21,
• f₂ = 14,
• h = 10.

Substitute the values:
Mode = 35 + ((^{23 - 21}⁄_{2(23) - 21 - 14})) * 10 = 35 + (²⁄₁₁) * 10 = 36.81 years.

Conclusion:
The mode and mean of the data are 36.81 years and 35.375 years, respectively.

Quick Tip: For grouped data, use the formulas for mean and mode:
Mean = (^{Σ f_ix_i}⁄_{Σ fi}), Mode = l + ((^{f_m - f₁}⁄_{2fm - f1 - f2}))h.

1. Mid-point of F and G:
   The coordinates of F and G are (-3, 0) and (1, 4), respectively.
   Using the mid-point formula:
   Mid-point = ((^{x₁ + x₂}⁄₂), (^{y₁ + y₂}⁄₂)),
   Mid-point = ((^{-3 + 1}⁄₂), (^{0 + 4}⁄₂)) = (-1, 2).
   Hence, the mid-point is (-1, 2).
2. (A) Distance between A and C:
   The coordinates of A and C are (3, -1) and (-3, -4), respectively.
   Using the distance formula:
   Distance = √((x₂ - x₁)² + (y₂ - y₁)²),
   Distance = √((-3 - 3)² + (-4 - (-1))²) = √((-6)² + (-3)²) = √(36 + 9) = √45 = 3√5
   The previous answer was calculating distance to B which was incorrect. The correct Distance is √45 = 3√5
   Hence, the distance is √45 = 3√5.
   • (B) Coordinates of the point dividing A and B in the ratio 1:3:
     The coordinates of A and B are (3, -1) and (6,7), respectively.
     Using the section formula:
     Point = ((^{m₁x₂ + m₂x₁}⁄_{m1 + m2}), (^{m₁y₂ + m₂y₁}⁄_{m1 + m2})),
     where m₁ = 1, m₂ = 3. Substitute the values:
     Point = ((^{1 * 6 + 3 * 3}⁄_{1 + 3}), (^{1 * 7 + 3 * -1}⁄_{1 + 3})) = ((^{6 + 9}⁄₄), (^{7 - 3}⁄₄)) = (¹⁵⁄₄, 1). Corrected answer
3. Coordinates of D:
   From the graph, the coordinates of D are (-2, -5).

Conclusion:

• The mid-point of F and G is (-1, 2).
• The distance between A and C is √45 = 3√5.
• The point dividing A and B in the ratio 1:3 is (¹⁵⁄₄, 1).
• The coordinates of D are (-2, -5).

Quick Tip: For solving coordinate geometry problems, use the mid-point formula, distance formula, and section formula systematically to compute the required values.

1. Number on the first spot:
   Substitute n = 1 into the formula 20 + 4n:
   Number on the first spot = 20 + 4 * 1 = 24.
2. (A) Spot numbered 112:
   Solve 20 + 4n = 112:
   4n = 112 - 20 = 92 ⇒ n = ⁹²⁄₄ = 23.
   Hence, the 23rd spot is numbered 112.
   • (B) Sum of the first 10 spots:
     The A.P. has first term a = 24 and common difference d = 4.
     The sum of the first n terms of an A.P. is:
     S_n = (ⁿ⁄₂) [2a + (n - 1)d].
     Substitute n = 10, a = 24, d = 4:
     S₁₀ = (¹⁰⁄₂) [2 * 24 + 9 * 4] = 5 [48 + 36] = 5 * 84 = 420.
     Hence, the sum of the first 10 spots is 420.
3. Number on the (n - 2)th spot:
   Substitute n - 2 into the formula 20 + 4n:
   Number on the (n - 2)th spot = 20 + 4(n - 2) = 20 + 4n - 8 = 12 + 4n.
   Hence, the number on the (n - 2)th spot is 12 + 4n.

Conclusion:

• The number on the first spot is 24.
• The spot numbered 112 is the 23rd spot.
• The sum of the first 10 spots is 420.
• The number on the (n - 2)th spot is 12 + 4n.

Quick Tip: For arithmetic progression problems, use the formulas for the nth term and sum of n terms:
a_n = a + (n - 1)d, S_n = (ⁿ⁄₂) [2a + (n - 1)d].

1. Volume of the cuboidal carton:
   The dimensions of the carton are 30 cm * 32 cm * 15 cm.
   Using the volume formula for a cuboid:
   Volume = length * breadth * height = 30 * 32 * 15 = 14400 cm³.
2. (A) Total surface area of the milk packet:
   The dimensions of the milk packet are 15 cm * 8 cm * 5 cm.
   Using the total surface area formula for a cuboid:
   Total Surface Area = 2(lb + bh + hl),
   Total Surface Area = 2(15 * 8 + 8 * 5 + 5 * 15) = 2(120 + 40 + 75) = 2 * 235 = 470 cm².
   • (B) Number of milk packets in the carton:
     Volume of one milk packet:
     Volume = length * breadth * height = 15 * 8 * 5 = 600 cm³.
     Number of packets:
     Number of packets = ^{Volume of carton}⁄_{Volume of one packet} = ¹⁴⁴⁰⁰⁄₆₀₀ = 24.
     Hence, 24 packets can be filled in the carton.
3. Capacity of the cup:
   The cup is cylindrical with radius r = 5 cm and height h = 7 cm.
   Using the volume formula for a cylinder:
   Volume = πr²h,
   Volume = (²²⁄₇) * 5² * 7 = (²²⁄₇) * 25 * 7 = 550 cm³.
   Hence, the cup can hold 550 mL of milk.

Conclusion:

• The volume of the cuboidal carton is 14400 cm³.
• The total surface area of the milk packet is 470 cm².
• 24 milk packets can be filled in the carton.
• The cup can hold 550 mL of milk.

Quick Tip: For 3D geometry problems, use the volume and surface area formulas specific to the given shape (cuboid or cylinder). Pay attention to unit conversions if required.