CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 1- 30/3/1) with Answer Key

Solution: The given pair of equations are:
1. x + 2y + 5 = 0 or x + 2y = -5,
2. -3x = 6y - 1 or 3x + 6y = 1.

Convert both equations into the standard form ax + by + c = 0:
1. x + 2y + 5 = 0,
2. 3x + 6y - 1 = 0.

Find the ratios of coefficients:
^a₁⁄_a2 = ¹⁄₃, ^b₁⁄_b2 = ²⁄₆ = ¹⁄₃, ^c₁⁄_c2 = ⁵⁄_-1 = -5.

Since ^a₁⁄_a2 = ^b₁⁄_b2 ≠ ^c₁⁄_c2, the lines are parallel and do not intersect. Hence, the system of equations has no solution.

Conclusion: The given pair of equations is inconsistent, and there is no solution.

Quick Tip: For a pair of linear equations, if ^a₁⁄_a2 = ^b₁⁄_b2 ≠ ^c₁⁄_c2, the lines are parallel, and there is no solution.

Solution: The terms of the A.P. are:
T₁ = ¹⁄_2x, T₂ = ^{1 - 4x}⁄_2x, T₃ = ^{1 - 8x}⁄_2x.

The common difference d is:
d = T₂ - T₁ = ^{1 - 4x}⁄_2x - ¹⁄_2x.

Combine terms:
d = ^{(1 - 4x) - 1}⁄_2x = ^-4x⁄_2x = -2.

Conclusion: The common difference of the A.P. is -2.

Quick Tip: To find the common difference of an A.P., subtract any term from the preceding term.

Solution: When two dice are thrown, the total number of outcomes is:
6 x 6 = 36.

For the two dice to show the same number (e.g., (1, 1), (2, 2), ..., (6, 6)), there are 6 favorable outcomes. The probability of the same number is:
P(same) = ⁶⁄₃₆ = ¹⁄₆.

The probability of showing different numbers is the complement:
P(different) = 1 - P(same) = 1 - ¹⁄₆ = ⁵⁄₆.

Conclusion: The probability that the two dice show different numbers is ⁵⁄₆.

Quick Tip: The complement rule states that P(A') = 1 - P(A). Use it for quick probability calculations.

Solution: The total probability for an event and its complement is:
P(correct answer) + P(not correct answer) = 1.

Substitute the given probabilities:
^x⁄₆ + ²⁄₃ = 1.

Simplify the equation:
^x⁄₆ = 1 - ²⁄₃.

Convert 1 - ²⁄₃ to a common denominator:
^x⁄₆ = ³⁄₃ - ²⁄₃ = ¹⁄₃.

Multiply through by 6:
x = 6 * ¹⁄₃ = 2.

Conclusion: The value of x is 2.

Quick Tip: The sum of probabilities for an event and its complement is always 1. Use this relationship to solve such questions.

Solution: The LCM of multiple numbers is determined by taking the highest powers of all prime factors.

Prime factorization of 3780:
3780 = 2² x 3³ x 5 x 7.

From the given expressions for a, b, c:
The highest power of 2 is 2² (common in all terms).
The highest power of 3 is 3^x. This must match the 3³ in the LCM.
Factors 5 and 7 appear in b and c, respectively.

Thus, x = 3, as the power of 3 in a must match the LCM.

Conclusion: The value of x is 3.

Quick Tip: To find the LCM of numbers, take the highest power of each prime factor across all terms.

Solution: The given quadratic polynomial is:
2x² - 3x - 9.

The formula for finding the roots of a quadratic equation ax² + bx + c = 0 is:
x = ^{-b ± √(b2 - 4ac)}⁄_2a.

Here, a = 2, b = -3, and c = -9. Substitute into the formula:
x = ^{-(-3) ± √((-3)2 - 4(2)(-9))}⁄₂₍₂₎.

Simplify:
x = ^{3 ± √(9 + 72)}⁄₄ = ^{3 ± √81}⁄₄.

x = ^{3 ± 9}⁄₄.

The roots are:
x = ^{3 + 9}⁄₄ = 3 and x = ^{3 - 9}⁄₄ = ^-3⁄₂.

Conclusion: The zeroes of the quadratic polynomial are 3 and ^-3⁄₂.

Quick Tip: To find the zeroes of a quadratic polynomial, use the quadratic formula: x = ^{-b ± √(b2 - 4ac)}⁄_2a.

Solution: Let the height of the tower be h. Using trigonometry, we have:
tan θ = ^{Opposite side}⁄_{Adjacent side}.

Here, θ = 60°, and the adjacent side is 30 m. Substitute:
tan 60° = ^h⁄₃₀.

The value of tan 60° is √3:
√3 = ^h⁄₃₀.

Solve for h:
h = 30√3.

Conclusion: The height of the tower is 30√3 m.

Quick Tip: In problems involving angles of elevation, use trigonometric ratios like tan θ to find unknown heights or distances.

Solution: We are given:
cos θ = ^√3⁄₂, sin φ = ¹⁄₂.

From trigonometric identities:
sin θ = √(1 - cos² θ) = √(1 - (^√3⁄₂)²) = √(¹⁄₄) = ¹⁄₂,
cos φ = √(1 - sin² φ) = √(1 - (¹⁄₂)²) = √(³⁄₄) = ^√3⁄₂.

The formula for tan(θ + φ) is:
tan(θ + φ) = ^{tan θ + tan φ}⁄_{1 - tan θ tan φ}.

Compute tan θ and tan φ:
tan θ = ^{sin θ}⁄_{cos θ} = ^1⁄₂⁄_^√3⁄2 = ¹⁄_√3, tan φ = ^{sin φ}⁄_{cos φ} = ^1⁄₂⁄_^√3⁄2 = ¹⁄_√3.

Substitute into the formula:
tan(θ + φ) = ^{1⁄_√3 + 1⁄_√3}⁄_{1 - ¹⁄√3 * ¹⁄√3} = ^2⁄_√3⁄_{1 - ¹⁄3} = ^2⁄_√3⁄_²⁄3 = √3.

Conclusion: The value of tan(θ + φ) is √3.

Quick Tip: To compute tan(θ + φ), use the formula tan(θ + φ) = ^{tan θ + tan φ}⁄_{1 - tan θ tan φ} and substitute the values.

Solution: When two circles intersect at two distinct points, the following cases arise:
- Direct tangents: These are tangents that do not pass through the region of intersection. There are 2 such tangents.
- Common internal tangents: These are tangents that pass through the region of intersection. In this case, there are no internal tangents, as the circles intersect.

Thus, the maximum number of common tangents is:
2 (direct tangents only).

Conclusion: The maximum number of common tangents for two circles intersecting at two distinct points is 2.

Quick Tip: The number of common tangents between two circles depends on their relative positions (disjoint, intersecting, or touching).

Solution: In the given figure:
- PT is a tangent to the circle at T.
- ∠TPO = 35°.

The angle ∠x is an exterior angle of the triangle ΔOPT. The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Thus:
∠x = ∠O + ∠TPO.

Since ∠O = 90° (radius OT is perpendicular to the tangent PT):
∠x = 90° + 35° = 125°.

Conclusion: The measure of ∠x is 125°.

Quick Tip: For a tangent to a circle, the angle between the radius and the tangent is always 90°.

Solution: For a quadrilateral:
- If the diagonals divide each other proportionally, it is a trapezium.

- This property arises because the triangles formed by the diagonals are similar, which is a characteristic feature of trapeziums.

In other types of quadrilaterals like parallelograms, rectangles, or squares, the diagonals either bisect each other or have specific geometric constraints, but they do not divide each other proportionally.

Conclusion: If the diagonals of a quadrilateral divide each other proportionally, it is a trapezium.

Quick Tip: A trapezium's diagonals divide each other proportionally because it forms pairs of similar triangles.

Solution: By the Basic Proportionality Theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Hence:
^AD⁄_AB = ^DE⁄_BC.

First, find AB (the total length of side AB):
AB = AD + BD = 2 + 3 = 5 cm.

Now substitute the known values:
^AD⁄_AB = ^DE⁄_BC.

²⁄₅ = ^DE⁄_7.5.

Solve for DE:
DE = ²⁄₅ x 7.5 = 3 cm.

Conclusion: The length of DE is 3 cm.

Quick Tip: The Basic Proportionality Theorem states that if a line is parallel to one side of a triangle, it divides the other sides proportionally.

Solution: The product of HCF and LCM of two numbers is equal to the product of the numbers. This can be expressed as:
HCF x LCM = Number 1 x Number 2.

Substitute the given values:
40 x (252 x k) = 2520 x 6600.

Simplify:
252 x k = ^{2520 x 6600}⁄₄₀.

Calculate:
252 x k = 415800.

Solve for k:
k = ⁴¹⁵⁸⁰⁰⁄₂₅₂ = 1650.

Conclusion: The value of k is 1650.

Quick Tip: For any two numbers, the product of their HCF and LCM equals the product of the numbers.

Solution: The product of two irrational numbers can be rational if their product simplifies to a perfect square or rational number.

Here:
√3 * √27 = √(3 * 27) = √81 = 9,

which is a rational number.

For the other pairs:
(√16, √4) simplifies to √64, but √16 and √4 are not irrational.

(√5, √2) results in √10, which is irrational.

(√36, √2) results in √72, which is also irrational.

Conclusion: The correct pair is (√3, √27).

Quick Tip: The product of two irrational numbers is rational if the resulting value simplifies to a perfect square or a rational number.

Solution: The given digits are:
1, 2, 3, 4, 5, 6, 7, 8, 9.

The odd prime numbers from this set are:
3, 5, 7.

The total number of digits is 9, and the number of odd prime numbers is 3. Hence, the probability is:
P = ^{Number of favorable outcomes}⁄_{Total number of outcomes} = ³⁄₉ = ¹⁄₃.

Conclusion: The probability that the chosen digit is an odd prime number is ¹⁄₃.

Quick Tip: Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves.

Solution: The mean of five observations is 15. Therefore, the sum of all observations is:
15 x 5 = 75.

The mean of the first three observations is 14. Hence, the sum of the first three observations is:
14 x 3 = 42.

The mean of the last three observations is 17. Hence, the sum of the last three observations is:
17 x 3 = 51.

The third observation is common to both the first and last three observations. Let the third observation be x. Then:
Sum of all observations = Sum of first three observations + Sum of last three observations - x.

Substitute the values:
75 = 42 + 51 - x.

Solve for x:
75 = 93 - x => x = 93 - 75 = 18.

Conclusion: The third observation is 18.

Quick Tip: To solve such problems, use the relationship Mean = ^{Sum of observations}⁄_{Number of observations}.

Solution: The perimeter of a sector of a circle is the sum of:
1. The length of the arc, and
2. Twice the radius of the circle.

The formula for the arc length is:
Arc length = 2πr * ^θ⁄_360°.

Here:
r = 7 cm, θ = 90°.

Substitute the values:
Arc length = 2 * ²²⁄₇ * 7 * ⁹⁰⁄₃₆₀.

Simplify step-by-step:
Arc length = ^{2 * 22 * 7 * 90}⁄_{7 * 360} = ^{22 * 90}⁄₁₈₀ = 11 cm.

Now, calculate the perimeter:
Perimeter = Arc length + 2r = 11 + 2 * 7 = 11 + 14 = 25 cm.

Conclusion: The perimeter of the sector is 25 cm.

Quick Tip: To calculate the perimeter of a sector, add the arc length to twice the radius.

Solution: The angle between a tangent and a chord drawn at the point of tangency is equal to the angle subtended by the chord at the centre of the circle.

From the figure:
∠TML = ∠MON.

Here:
∠TML = 70°.

Since ∠MON is subtended at the centre of the circle, it is twice the angle at the tangent:
∠MON = 2 * ∠TML = 2 * 70° = 140°.

Conclusion: The measure of ∠MON is 140°.

Quick Tip: The angle subtended by a chord at the centre of a circle is twice the angle subtended at the tangent.

Solution: To find the point dividing the line segment joining A(1, 2) and B(-1, 1) in the ratio 1 : 2, we use the formula:
(^{m₁x₂ + m₂x₁}⁄_{m1 + m2}, ^{m₁y₂ + m₂y₁}⁄_{m1 + m2}).

Substitute m₁ = 1, m₂ = 2, A(1, 2), B(-1, 1):
x = ^{(1)(-1) + (2)(1)}⁄_{1 + 2} = ^{-1 + 2}⁄₃ = ¹⁄₃, y = ^{(1)(1) + (2)(2)}⁄_{1 + 2} = ^{1 + 4}⁄₃ = ⁵⁄₃.

The point is (¹⁄₃, ⁵⁄₃), not (^-1⁄₃, ⁵⁄₃).

Therefore, Assertion (A) is false, but Reason (R) is correct.

Solution: The probability of hitting a boundary is:
P(E) = ^{Number of times boundary is hit}⁄_{Total number of balls} = ⁹⁄₄₅ = ¹⁄₅.

The probability of not hitting a boundary is:
P(not E) = 1 - P(E) = 1 - ¹⁄₅ = ⁴⁄₅.

The Reason (R) correctly states that P(E) + P(not E) = 1, which is used to calculate P(not E). Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) explains Assertion (A).

Solution:
Total number of outcomes = 52.

(i) The probability of drawing the queen of hearts is:
P(queen of hearts) = ¹⁄₅₂.

(ii) There are 4 jacks in the deck, so the number of cards that are not jacks is:
52 - 4 = 48.

The probability of not drawing a jack is:
P(not a jack) = ⁴⁸⁄₅₂ = ¹²⁄₁₃.

Conclusion:
(i) P(queen of hearts) = ¹⁄₅₂.
(ii) P(not a jack) = ¹²⁄₁₃.

Quick Tip: In probability, P(not A) = 1 - P(A). For cards, ensure you account for the total number of specific cards (e.g., 4 jacks, 4 queens).

Solution: The given equations are:
2x + y = 13 (i),
4x - y = 17 (ii).

Add equations (i) and (ii) to eliminate y:
(2x + y) + (4x - y) = 13 + 17.

Simplify:
6x = 30 => x = 5.

Substitute x = 5 into equation (i):
2(5) + y = 13 => 10 + y = 13 => y = 3.

The value of x - y is:
x - y = 5 - 3 = 2.

Conclusion: The value of x - y is 2.

Quick Tip: For solving linear equations, try eliminating one variable by adding or subtracting equations, then solve for the other variable.

Solution: Let the two numbers be x and y, where x > y.

The given equations are:
x + y = 105 (i),
x - y = 45 (ii).

Add equations (i) and (ii) to eliminate y:
(x + y) + (x - y) = 105 + 45.

Simplify:
2x = 150 => x = 75.

Substitute x = 75 into equation (i):
75 + y = 105 => y = 105 - 75 = 30.

Thus, the two numbers are:
x = 75, y = 30.

Conclusion: The numbers are 75 and 30.

Quick Tip: To solve problems involving the sum and difference of two numbers, add and subtract the equations to eliminate one variable and solve for the other.

Solution: The condition for equidistance is:
PA = PB => PA² = PB².

Substitute the coordinates:
(x - 7)² + (y - 1)² = (x - 3)² + (y - 5)².

Expand both sides:
(x² - 14x + 49) + (y² - 2y + 1) = (x² - 6x + 9) + (y² - 10y + 25).

Simplify:
-14x - 2y + 50 = -6x - 10y + 34.

Rearrange:
-8x + 8y + 16 = 0 => -8(x - y - 2) = 0.

Therefore, the relation is:
x - y - 2 = 0.

Conclusion: The required relation between x and y is:
x - y - 2 = 0.

Quick Tip: To find the relation for equidistant points, equate the squared distance from each point and simplify the equation.

Solution: The centre O(2, -3y) is the midpoint of AB. Using the midpoint formula:
(^{-1 + 5}⁄₂, ^{y + 7}⁄₂) = (2, -3y).

Equating the x-coordinates:
^{-1 + 5}⁄₂ = 2 => ⁴⁄₂ = 2 (satisfied).

Equating the y-coordinates:
^{y + 7}⁄₂ = -3y.

Simplify:
y + 7 = -6y => 7y = -7 => y = -1.

Now calculate the radius of the circle. Since AB is the diameter, the radius is half the length of AB.

The distance AB is:
AB = √((5 - (-1))² + (7 - (-1))²).

Simplify:
AB = √((5 + 1)² + (7 + 1)²) = √(6² + 8²) = √(36 + 64) = √100 = 10.

The radius is:
Radius = ^AB⁄₂ = ¹⁰⁄₂ = 5.

Conclusion: The value of y is -1, and the radius of the circle is 5.

Quick Tip: When the diameter is given, the radius is half its length. Use the distance formula to calculate the diameter and divide by 2 for the radius.

Solution: In ΔEAB and ΔECD, we are given:
^EA⁄_EC = ^EB⁄_ED.

Also, ∠AEB = ∠CED (vertically opposite angles).

By the Side-Angle-Side (SAS) similarity criterion:
ΔEAB ~ ΔECD.

Conclusion: It is proven that ΔEAB ~ ΔECD.

Quick Tip: For two triangles to be similar by SAS similarity, the ratios of two corresponding sides must be equal, and the included angles must be equal.

Solution:
cos 45° = ¹⁄_√2, sin 60° = ^√3⁄₂, sec 30° = ²⁄_√3, csc 30° = 2.

The given expression becomes:
^{cos 45° + sin 60°}⁄_{sec 30° + csc 30°} = ^{1⁄_√2 + √3⁄₂}⁄_{²⁄√3 + 2}.

Simplify the numerator:
¹⁄_√2 + ^√3⁄₂ = ^{2 + √6}⁄_2√2.

Simplify the denominator:
²⁄_√3 + 2 = ^{2 + 2√3}⁄_√3.

The expression now becomes:
^{2 + √6⁄_2√2}⁄_{^{2 + 2√3}⁄√3}.

Invert and multiply:
^{2 + √6}⁄_2√2 * ^√3⁄_{2(1 + √3)}.

Simplify further:
^{(2 + √6)√3}⁄_{4√2(1 + √3)}.

Solution: Let a be the first term and d be the common difference.

The sum of n terms of an A.P. is given by:
S_n = ⁿ⁄₂ [2a + (n-1)d].

For the first 7 terms (S₇ = 49):
⁷⁄₂ [2a + 6d] = 49.

Simplify:
7a + 21d = 49 => a + 3d = 7 (i).

For the first 17 terms (S₁₇ = 289):
¹⁷⁄₂ [2a + 16d] = 289.

Simplify:
17a + 136d = 289 => a + 8d = 17 (ii).

Solve equations (i) and (ii):
a + 3d = 7,
a + 8d = 17.

Subtract (i) from (ii):
(8d - 3d) = (17 - 7) => 5d = 10 => d = 2.

Substitute d = 2 into (i):
a + 3(2) = 7 => a = 1.

Now, find the sum of the first 20 terms (S₂₀):
S₂₀ = ²⁰⁄₂ [2a + 19d].

Substitute a = 1 and d = 2:
S₂₀ = 10 [2(1) + 19(2)] = 10 [2 + 38] = 10 * 40 = 400.

Conclusion: The sum of the first 20 terms of the A.P. is 400.

Quick Tip: For sums of A.P., use the formula S_n = ⁿ⁄₂ [2a + (n-1)d] and solve equations systematically to find a and d.

Solution: Let a be the first term and d be the common difference.

The general term of an A.P. is given by:
T_n = a + (n-1)d.

For the 10th term:
T₁₀ = a + 9d.

For the 30th term:
T₃₀ = a + 29d.

Given:
^T₁₀⁄_T30 = ¹⁄₃.

Substitute T₁₀ and T₃₀:
^{a + 9d}⁄_{a + 29d} = ¹⁄₃.

Cross-multiply:
3(a + 9d) = (a + 29d).

Simplify:
3a + 27d = a + 29d => 2a = 2d => a = d.

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The sum of the first 6 terms is given as S₆ = 42. The formula for the sum of the first n terms is:
S_n = ⁿ⁄₂ [2a + (n-1)d].

For S₆:
⁶⁄₂ [2a + 5d] = 42.

Simplify:
3(2a + 5d) = 42 => 2a + 5d = 14 (ii).

Substitute a = d into (ii):
2d + 5d = 14 => 7d = 14 => d = 2.

Since a = d, we have:
a = 2.

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Conclusion: The first term (a) is 2, and the common difference (d) is 2.

Quick Tip: When ratios of terms are given, equate their formula and simplify. Use the sum formula to find unknowns like a and d.

Solution: Let P(x) = x² - 15.

Factorize P(x):
P(x) = (x - √15)(x + √15).

Thus, the zeroes of P(x) are:
-√15 and √15.

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Verification:

1. Sum of zeroes:
-√15 + √15 = 0.

Compare with:
Sum of zeroes = ^{-coefficient of x}⁄_{coefficient of x²} = ⁰⁄₁ = 0.

2. Product of zeroes:
(-√15) * (√15) = -15.

Compare with:
Product of zeroes = ^{constant term}⁄_{coefficient of x²} = ^-15⁄₁ = -15.

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Conclusion: The sum and product of the zeroes are verified to match the relationships:
Sum of zeroes = ^{-coefficient of x}⁄_{coefficient of x²}, Product of zeroes = ^{constant term}⁄_{coefficient of x²}.

Quick Tip: For a quadratic polynomial ax² + bx + c, the sum of zeroes is -^b⁄_a and the product of zeroes is ^c⁄_a.

Solution: The given equations are:
1. x - y + 1 = 0, or equivalently x - y = -1 (i).
2. x + y = 5 (ii).

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Step 1: Find points for each line.

For equation (i) (x - y = -1):

If x = 0, -y = -1 => y = 1.

If x = -1, -y = -2 => y = 2.

If x = 1, -y = 0 => y = 0.

Thus, the points for x - y = -1 are:
(0, 1), (-1, 2), (1, 0).

For equation (ii) (x + y = 5):

If x = 0, y = 5.

If x = 5, y = 0.

If x = 2, y = 3.

Thus, the points for x + y = 5 are:
(0, 5), (5, 0), (2, 3).

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Step 2: Plot the lines on the graph.

- Plot the line x - y = -1 using the points (0, 1), (-1, 2), (1, 0).
- Plot the line x + y = 5 using the points (0, 5), (5, 0), (2, 3).

Note: Embedding an image of the graph is not supported directly here. The code below helps to plot the graph directly in an HTML context. In a HTML file, you need to embed the following code with a script tag.

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Step 3: Find the point of intersection.

The two lines intersect at the point (2, 3). This is the solution to the system of equations.

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Conclusion: The solution to the given system of equations is:
x = 2, y = 3.

Quick Tip: To solve linear equations graphically, plot each equation as a straight line and find their point of intersection. This point gives the solution.

Solution: Let the line segment divide the Y-axis at the point (0, y).

Let the required ratio be k : 1. Using the section formula, the x-coordinate of the point dividing the line segment is:
x = ^{k(-1) + 1(5)}⁄_{k + 1}.

Since the point lies on the Y-axis, x = 0. Substitute x = 0:
^{k(-1) + 5}⁄_{k + 1} = 0.

Simplify:
-k + 5 = 0 => k = 5.

Hence, the required ratio is:
5 : 1.

Conclusion: The line segment is divided by the Y-axis in the ratio 5 : 1.

Quick Tip: For a line segment divided by an axis, use the section formula and set the relevant coordinate (x for the Y-axis, y for the X-axis) to 0.

Solution: Let the coordinates of R be (x, y).

Given:
PR : RQ = 2 : 1.

Using the section formula:
x = ^{m₂x₁ + m₁x₂}⁄_{m1 + m2}, y = ^{m₂y₁ + m₁y₂}⁄_{m1 + m2}.

Here:
m₁ = 2, m₂ = 1, P(-2, 5), Q(3, 2).

Substitute into the formulas for x and y:
x = ^{1(-2) + 2(3)}⁄_{2 + 1} = ^{-2 + 6}⁄₃ = ⁴⁄₃.
y = ^{1(5) + 2(2)}⁄_{2 + 1} = ^{5 + 4}⁄₃ = 3.

Thus, the coordinates of R are:
(⁴⁄₃, 3).

Conclusion: The coordinates of the point R are:
(⁴⁄₃, 3).

Quick Tip: Use the section formula for dividing a line segment in a given ratio:
( ^{m₂x₁ + m₁x₂}⁄_{m1 + m2}, ^{m₂y₁ + m₁y₂}⁄_{m1 + m2} ).
Ensure the correct substitution of ratios and coordinates.

Solution: We are given:
^{sin θ - 2 sin3 θ}⁄_{2 cos³ θ - cos θ}.

Factorize the numerator and denominator.

1. Numerator:
sin θ - 2 sin³ θ = sin θ (1 - 2 sin² θ).

Using the identity sin² θ = 1 - cos² θ, rewrite:
1 - 2 sin² θ = 1 - 2(1 - cos² θ) = 2 cos² θ - 1.

Thus, the numerator becomes:
sin θ (2 cos² θ - 1).

2. Denominator:
2 cos³ θ - cos θ = cos θ (2 cos² θ - 1).

Simplify the fraction:
^{sin θ (2 cos2 θ - 1)}⁄_{cos θ (2 cos² θ - 1)}.

Cancel (2 cos² θ - 1) (valid if 2 cos² θ - 1 ≠ 0):
^{sin θ}⁄_{cos θ}.

This simplifies to:
tan θ.

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Conclusion: It is proven that:
^{sin θ - 2 sin3 θ}⁄_{2 cos³ θ - cos θ} = tan θ.

Quick Tip: When simplifying trigonometric fractions, use factorization and Pythagorean identities like sin² θ + cos² θ = 1.

Solution: Let the chord AB of a circle have endpoints A and B, and let O be the center of the circle. Draw tangents at A and B, and let the chord AB intersect the tangents at angles ∠OAT and ∠OBT.

Note: Embedding an image of the graph is not supported directly here. The code below helps to plot the graph directly in an HTML context. In a HTML file, you need to embed the following code with a script tag.

1. Properties of the circle:
- The radius OA is perpendicular to the tangent at A.
- The radius OB is perpendicular to the tangent at B.

2. Triangles involved:
- In ΔOAT, ∠OAT is the angle between the tangent at A and the chord AB.
- In ΔOBT, ∠OBT is the angle between the tangent at B and the chord AB.

3. Prove equality:
Since the chord AB subtends equal angles at the center (∠OAB = ∠OBA) and the radii OA and OB are equal, the triangles ΔOAT and ΔOBT are congruent (by RHS criterion).

Thus:
∠OAT = ∠OBT.

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Conclusion: The tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Quick Tip: For chord-tangent problems, remember that the radius is perpendicular to the tangent, and congruence of triangles can be used to prove angle relationships.

Solution: Let the original speed of the aircraft be x km/h.

1. Original Time of Flight:
Time = ^Distance⁄_Speed = ²⁸⁰⁰⁄_x.

2. New Speed and Time:
When the speed is reduced by 100 km/h, the new speed is (x - 100), and the new time is:
²⁸⁰⁰⁄_{x - 100}.

The time difference between the original and new times is 30 minutes, or ¹⁄₂ hour. Therefore:
²⁸⁰⁰⁄_{x - 100} - ²⁸⁰⁰⁄_x = ¹⁄₂.

3. Simplify the Equation:
Take the LCM of x(x - 100):
^{2800x - 2800(x - 100)}⁄_{x(x - 100)} = ¹⁄₂.

Simplify:
^{2800 * 100}⁄_{x(x - 100)} = ¹⁄₂.

Multiply through by 2x(x - 100):
560000 = x(x - 100).

Expand:
x² - 100x - 560000 = 0.

4. Solve the Quadratic Equation:
Using the quadratic formula:
x = ^{-b ± √(b2 - 4ac)}⁄_2a, a = 1, b = -100, c = -560000.
x = ^{-(-100) ± √((-100)2 - 4(1)(-560000))}⁄₂₍₁₎.
x = ^{100 ± √(10000 + 2240000)}⁄₂.
x = ^{100 ± √2250000}⁄₂.
x = ^{100 ± 1500}⁄₂.

Select the positive root:
x = ^{100 + 1500}⁄₂ = 800.

5. Find the Original Time:
The original time of the flight is:
²⁸⁰⁰⁄₈₀₀ = 3.5 hours.

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Conclusion: The original duration of the flight is 3.5 hours.

Quick Tip: For time-speed-distance problems, relate the difference in times to the change in speeds and solve using algebraic equations.

Solution: Let the numerator of the fraction be x. Then the denominator is:
2x + 1.

The fraction is:
^x⁄_2x+1.

The reciprocal is:
^2x+1⁄_x.

Given:
^x⁄_2x+1 + ^2x+1⁄_x = 2 ¹⁶⁄₂₁.

Convert 2 ¹⁶⁄₂₁ to an improper fraction:
2 ¹⁶⁄₂₁ = ^{42 + 16}⁄₂₁ = ⁵⁸⁄₂₁.

Equate:
^x⁄_2x+1 + ^2x+1⁄_x = ⁵⁸⁄₂₁.

1. Simplify the Left Side:
Take the LCM of x(2x+1):
^{x2 + (2x+1)2}⁄_x(2x+1) = ⁵⁸⁄₂₁.

Expand (2x+1)²:
^{x2 + 4x2 + 4x + 1}⁄_x(2x+1) = ⁵⁸⁄₂₁.

Simplify:
^{5x2 + 4x + 1}⁄_x(2x+1) = ⁵⁸⁄₂₁.

2. Cross Multiply:
21(5x² + 4x + 1) = 58x(2x+1).

Expand both sides:
105x² + 84x + 21 = 116x² + 58x.

Simplify:
116x² - 105x² + 58x - 84x - 21 = 0.
11x² - 26x - 21 = 0.

3. Solve the Quadratic Equation:
Using the quadratic formula:
x = ^{-b ± √(b2 - 4ac)}⁄_2a, a = 11, b = -26, c = -21.
x = ^{-(-26) ± √((-26)2 - 4(11)(-21))}⁄₂₍₁₁₎.
x = ^{26 ± √(676 + 924)}⁄₂₂.
x = ^{26 ± √1600}⁄₂₂.
x = ^{26 ± 40}⁄₂₂.

Select the positive root:
x = ^{26 + 40}⁄₂₂ = ⁶⁶⁄₂₂ = 3.

4. Find the Fraction:
The fraction is:
^x⁄_2x+1 = ³⁄₂₍₃₎₊₁ = ³⁄₇.

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Conclusion: The fraction is ³⁄₇.

Quick Tip: For fraction-reciprocal problems, set up the equation, clear the denominators using LCM, and solve the resulting quadratic equation.

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

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Given: In ΔABC, a line DE ∥ BC intersects AB at D and AC at E.

To Prove: ^AD⁄_DB = ^AE⁄_EC.

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Note: Embedding an image of the graph is not supported directly here. The code below helps to plot the graph directly in an HTML context. In a HTML file, you need to embed the following code with a script tag.

Note: You must define a function called "BPTCanvas()" if you are using multiple of this type of graph generator

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Proof:
1. In ΔADE and ΔDBE:
- ∠ADE = ∠DBE (corresponding angles, as DE ∥ BC).
- ∠AED = ∠EBD (corresponding angles, as DE ∥ BC).

By the AA similarity criterion:
ΔADE ~ ΔDBE.

Therefore:
^AD⁄_DB = ^AE⁄_EB.

2. Similarly, in ΔAEC and ΔBEC:
- ∠AEC = ∠BEC (corresponding angles, as DE ∥ BC).
- ∠EAC = ∠EBC (corresponding angles, as DE ∥ BC).

By the AA similarity criterion:
ΔAEC ~ ΔBEC.

Therefore:
^AE⁄_EC = ^AD⁄_DB.

Combining the two results:
^AD⁄_DB = ^AE⁄_EC.

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Conclusion: If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides the two sides in the same ratio.

Quick Tip: In problems involving parallel lines in triangles, use similarity criteria (e.g., AA) to establish proportional relationships between segments.

Solution: Let h be the height of the transmission tower and x be the horizontal distance from the point of observation P to the base of the building A.

1. In ΔABP:
tan 45° = ²⁰⁄_x.

Simplify:
1 = ²⁰⁄_x => x = 20 m. (i)

2. In ΔCAP:
tan 60° = ^{h + 20}⁄_x.

Substitute x = 20:
√3 = ^{h + 20}⁄₂₀.

Simplify:
h + 20 = 20√3 => h = 20√3 - 20. (ii)

3. Final Answer:
h = 20(√3 - 1) m.

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Conclusion: The height of the transmission tower is:
20(√3 - 1) m.

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Note: Embedding an image of the graph is not supported directly here. The code below helps to plot the graph directly in an HTML context. In a HTML file, you need to embed the following code with a script tag.

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Quick Tip: Use trigonometric ratios (tan) and the given angles of elevation to set up equations for height and horizontal distance, then solve systematically.

Solution: The pole consists of two cylindrical parts. Let us calculate the volume of each cylinder separately and then find the total mass.

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1. Volume of the lower cylinder:
Height (h₁) = 200 cm,
Radius (r₁) = ^diameter⁄₂ = ²⁸⁄₂ = 14 cm.

The volume of a cylinder is given by:
V = πr²h.

For the lower cylinder:
V₁ = πr₁²h₁ = π (14)² (200).

Simplify:
V₁ = π (196)(200) = 39200π cm³.

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2. Volume of the upper cylinder:
Height (h₂) = 50 cm,
Radius (r₂) = 7 cm.

For the upper cylinder:
V₂ = πr₂²h₂ = π (7)² (50).

Simplify:
V₂ = π (49)(50) = 2450π cm³.

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3. Total volume of the pole:
V_total = V₁ + V₂ = 39200π + 2450π = 41650π cm³.

Substitute π ≈ 3.1416:
V_total = 41650 * 3.1416 = 130881.55 cm³.

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4. Mass of the pole:
Given that 1 cm³ of iron has a mass of 8 g:
Mass = V_total * 8 = 130881.55 * 8 = 1047052.4 g.

Convert to kilograms:
Mass = ^1047052.4⁄₁₀₀₀ = 1047.05 kg.

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Conclusion: The mass of the iron pole is approximately 1047.05 kg.

Quick Tip: To find the mass of a solid object, calculate its volume using geometric formulas, then multiply by the given density (mass per unit volume).

The capsule consists of:

1. A cylindrical part with two hemispheres attached at the ends.
2. Radius (r) of the hemispheres and the cylinder:
r = ^diameter⁄₂ = ⁴⁄₂ = 2 mm.
3. Length of the cylindrical part:
Length of cylinder = Total length of capsule - 2(Radius of hemispheres) = 14 - 4 = 10 mm.

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1. Surface Area of the Capsule:

The surface area consists of:
1. Curved surface area (CSA) of the cylinder:
CSA of cylinder = 2πrh = 2 * ²²⁄₇ * 2 * 10 = 40 * ²²⁄₇ = 125.71 mm².
2. CSA of the two hemispheres:
CSA of one hemisphere = 2πr² = 2 * ²²⁄₇ * 2² = 2 * ²²⁄₇ * 4 = ¹⁷⁶⁄₇ = 25.14 mm².

For two hemispheres:
CSA of both hemispheres = 2 * 25.14 = 50.28 mm².

Total surface area:
Surface Area = CSA of cylinder + CSA of hemispheres = 125.71 + 50.28 = 176 mm².

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2. Volume of the Capsule:

The volume consists of:
1. Volume of the cylinder:
Volume of cylinder = πr²h = ²²⁄₇ * 2² * 10 = ²²⁄₇ * 4 * 10 = ⁸⁸⁰⁄₇ = 125.71 mm³.
2. Volume of the two hemispheres:
Volume of one hemisphere = ²⁄₃ πr³ = ²⁄₃ * ²²⁄₇ * 2³ = ²⁄₃ * ²²⁄₇ * 8 = ³⁵²⁄₂₁ = 16.76 mm³.

For two hemispheres:
Volume of both hemispheres = 2 * 16.76 = 33.52 mm³.

Total volume:
Volume = Volume of cylinder + Volume of hemispheres = 125.71 + 33.52 = 159.24 mm³.

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Conclusion:
1. The surface area of the capsule is:
176 mm².
2. The volume of the capsule is:
159.24 mm³.

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Quick Tip: For composite shapes, break them into basic geometric shapes (cylinder, hemisphere, etc.) and calculate their properties individually. Combine the results for the total.

Solution:

1. Area of the square-shaped grass field:
Side of the square = 20 m.
Area of the square field = side² = 20 * 20 = 400 m².

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2. (A) Area of the total field grazed by the horses:
Each horse grazes a quarter-circle area (due to the rope length forming a circular section). The area grazed by one horse is:
Area of one horse's grazing region = ¹⁄₄ πr².

Here, r = 7 m, and π = ²²⁄₇:
Area of one grazing region = ¹⁄₄ * ²²⁄₇ * 7 * 7 = ¹⁄₄ * 154 = 38.5 m².

For four horses:
Total grazing area = 4 * 38.5 = 154 m².

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2. (b) If the rope length is increased to 10 m:
For one horse:
Area grazed by one horse = ¹⁄₄ πr², r = 10 m, π = 3.14.
Area grazed by one horse = ¹⁄₄ * 3.14 * 10 * 10 = ¹⁄₄ * 314 = 78.5 m².

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3. Area of the field left ungrazed:
Area left ungrazed = Area of square field - Area grazed by all horses.

Substitute:
Area left ungrazed = 400 - 154 = 246 m².

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Conclusion:
1. The area of the square-shaped grass field is 400 m².
2. (A) The total grazing area for all horses is 154 m².
(b) If the rope length is increased to 10 m, the grazing area for one horse is 78.5 m².
3. The area left ungrazed is 246 m².

Quick Tip: For problems involving grazing regions or circular segments, use the formula for the area of a sector or fractional part of a circle: ^θ⁄_360° πr².

Solution:

1. Lower Limit of the Modal Class:

Modal class is 20 - 24.

The lower limit is 20.

• Lower Limit of the Modal Class is 20

2. (A) Median Class:

The cumulative frequency (CF) is calculated as follows:

Note: Embed this inside table tags. This tag is only required to display the table

The total number of participants is N = 365. The median class corresponds to ^N⁄₂ = ³⁶⁵⁄₂ = 182.5, which lies in the cumulative frequency 194.

Therefore, the median class is 20 - 24.

2. (b) Number of participants below 50 years:

Participants below 50 years correspond to the classes 15-19 to 45-49.

Total frequency below 50 years = 62 + 132 + 96 + 37 + 13 + 11 + 10 = 361.

3. Empirical Relationship:

The empirical relationship between mean, median, and mode is given by:
Mode = 3(Median) - 2(Mean).

Conclusion:
1. The lower limit of the modal class is 20.
2. (A) The median class is 20 - 24.
(b) The number of participants below 50 years is 361.
3. The empirical relationship is Mode = 3(Median) - 2(Mean).

Quick Tip: When solving grouped frequency problems, compute the cumulative frequency to locate the median class and sum the required frequencies for other queries.

Solution:

The given number is 173250. Perform the prime factorization of 173250:
173250 = 2 * 5³ * 3² * 7 * 11.

1. Least Prime Number:

2. (A) Number of Students in the Class:

Each student multiplies the number by one prime number. The total prime factors used are:
1 + 3 + 2 + 1 + 1 = 8.

Thus, the total number of students is 8.

2. (b) Highest Prime Number:

3. Prime Number Used Maximum Times:

Conclusion:
1. The least prime number used is 3.
2. (A) The total number of students in the class is 8.
(b) The highest prime number used is 11.
3. The prime number used maximum times is 5.

Quick Tip: For prime factorization problems, break the number into its smallest prime factors to analyze patterns or answer related queries.