CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 2- 30/3/2) with Answer Key

In the given figure:
- \( PT \) is a tangent to the circle at \( T \).
- \( \angle TPO = 35^\circ \).

The angle \( \angle x \) is an exterior angle of the triangle \( \triangle OPT \). The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Thus: \[ \angle x = \angle O + \angle TPO. \]

Since \( \angle O = 90^\circ \) (radius \( OT \) is perpendicular to the tangent \( PT \)): \[ \angle x = 90^\circ + 35^\circ = 125^\circ. \]

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Conclusion:

The measure of \( \angle x \) is \( 125^\circ \). Quick Tip: For a tangent to a circle, the angle between the radius and the tangent is always \( 90^\circ \).

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

Substitute the given probabilities: \[ \frac{x}{6} + \frac{2}{3} = 1. \]

Simplify the equation: \[ \frac{x}{6} = 1 - \frac{2}{3}. \]

Convert \(1 - \frac{2}{3}\) to a common denominator: \[ \frac{x}{6} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}. \]

Multiply through by 6: \[ x = 6 \times \frac{1}{3} = 2. \]

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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.

Let the height of the tower be \( h \). Using trigonometry, we have: \[ \tan \theta = \frac{Opposite side}{Adjacent side}. \]

Here, \( \theta = 60^\circ \), and the adjacent side is \( 30 \, m \). Substitute: \[ \tan 60^\circ = \frac{h}{30}. \]

The value of \( \tan 60^\circ \) is \( \sqrt{3} \): \[ \sqrt{3} = \frac{h}{30}. \]

Solve for \( h \): \[ h = 30\sqrt{3}. \]

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Conclusion:

The height of the tower is \( 30\sqrt{3} \, m \). Quick Tip: In problems involving angles of elevation, use trigonometric ratios like \( \tan \theta \) to find unknown heights or distances.

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: \[ \angle TML = \angle MON. \]

Here: \[ \angle TML = 70^\circ. \]

Since \( \angle MON \) is subtended at the centre of the circle, it is twice the angle at the tangent: \[ \angle MON = 2 \cdot \angle TML = 2 \cdot 70^\circ = 140^\circ. \]

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Conclusion:

The measure of \( \angle MON \) is \( 140^\circ \). Quick Tip: The angle subtended by a chord at the centre of a circle is twice the angle subtended at the tangent.

For a consistent pair of linear equations, there is always at least one solution. This can occur in the following cases:


1. Intersecting Lines: The lines intersect at a single point, giving a unique solution.


2. Coincident Lines: The lines overlap entirely, resulting in infinitely many solutions.


Thus, the lines represented by the equations are either intersecting or coincident.


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Conclusion:

The lines represented by the equations are either intersecting or coincident, making the correct answer \( (D) \). Quick Tip: For consistent systems of linear equations, either the lines intersect at one point (unique solution) or they coincide (infinitely many solutions).

The area of a sector of a circle is given by: \[ Area of sector = \frac{\theta}{360^\circ} \times Area of circle, \]
where \( \theta \) is the angle at the center.

It is given that the area of the sector is \( \frac{7}{20} \) of the area of the circle: \[ \frac{\theta}{360^\circ} = \frac{7}{20}. \]

Simplify to find \( \theta \): \[ \theta = \frac{7}{20} \times 360^\circ. \]

Calculate: \[ \theta = \frac{7 \times 360}{20} = \frac{2520}{20} = 126^\circ. \]

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Conclusion:

The angle at the center is \( 126^\circ \). Quick Tip: For sector problems, use the formula \( Area of sector = \frac{\theta}{360^\circ} \times Area of circle \) to relate the fraction of the area to the central angle.

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: \[ P = \frac{Number of favorable outcomes}{Total number of outcomes} = \frac{3}{9} = \frac{1}{3}. \]

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Conclusion:

The probability that the chosen digit is an odd prime number is \( \frac{1}{3} \). Quick Tip: Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves.

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.

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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.

The LCM is obtained by taking the highest powers of all prime factors across \( a \), \( b \), and \( c \). From the given LCM: \[ 3780 = 2^2 \times 3^3 \times 5 \times 7. \]

In \( a = 2^2 \times 3^x \), the power of 3 must match the highest power in the LCM, which is \( 3 \). Therefore: \[ x = 3. \]

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Conclusion:

The value of \( x \) is \( 3 \). Quick Tip: A trapezium's diagonals divide each other proportionally because it forms pairs of similar triangles.

The sample space for tossing two coins is: \[ \{ HH, HT, TH, TT \}. \]

The outcomes with at most one tail are: \[ \{ HH, HT, TH \}. \]

The total number of favorable outcomes is \( 3 \), and the total number of outcomes is \( 4 \). Therefore, the probability is: \[ P(at most one tail) = \frac{favorable outcomes}{total outcomes} = \frac{3}{4}. \]

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Conclusion:

The probability of getting at most one tail is \( \frac{3}{4} \). Quick Tip: A trapezium's diagonals divide each other proportionally because it forms pairs of similar triangles.

The mean of five observations is given by: \[ Mean = \frac{Sum of observations}{Number of observations}. \]

Substitute the given values: \[ 11 = \frac{x + (x+2) + (x+4) + (x+6) + (x+8)}{5}. \]

Simplify the numerator: \[ 11 = \frac{5x + 20}{5}. \]

Multiply through by 5: \[ 55 = 5x + 20. \]

Solve for \( x \): \[ 5x = 55 - 20 = 35 \quad \implies \quad x = 7. \]

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Conclusion:

The value of \( x \) is \( 7 \). Quick Tip: To find the zeroes of a quadratic polynomial, use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

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The given quadratic polynomial is: \[ 2x^2 - 3x - 9. \]

The formula for finding the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

Here, \( a = 2 \), \( b = -3 \), and \( c = -9 \). Substitute into the formula: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-9)}}{2(2)}. \]

Simplify: \[ x = \frac{3 \pm \sqrt{9 + 72}}{4} = \frac{3 \pm \sqrt{81}}{4}. \]
\[ x = \frac{3 \pm 9}{4}. \]

The roots are: \[ x = \frac{3 + 9}{4} = 3 \quad and \quad x = \frac{3 - 9}{4} = -\frac{3}{2}. \]

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Conclusion:

The zeroes of the quadratic polynomial are \( 3 \) and \( -\frac{3}{2} \). Quick Tip: To find the zeroes of a quadratic polynomial, use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

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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: \[ 2 \quad (direct tangents only). \]

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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).

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: \[ \frac{AD}{AB} = \frac{DE}{BC}. \]

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

Now substitute the known values: \[ \frac{AD}{AB} = \frac{DE}{BC}. \]
\[ \frac{2}{5} = \frac{DE}{7.5}. \]

Solve for \( DE \): \[ DE = \frac{2}{5} \times 7.5 = 3 \, cm. \]

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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.

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We are given: \[ \cos \theta = \frac{\sqrt{3}}{2}, \quad \sin \phi = \frac{1}{2}. \]

From trigonometric identities: \[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4}} = \frac{1}{2}, \] \[ \cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}. \]

The formula for \( \tan(\theta + \phi) \) is: \[ \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}. \]

Compute \( \tan \theta \) and \( \tan \phi \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}, \quad \tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}. \]

Substitute into the formula: \[ \tan(\theta + \phi) = \frac{\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \sqrt{3}. \]

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Conclusion:

The value of \( \tan(\theta + \phi) \) is \( \sqrt{3} \). Quick Tip: To compute \( \tan(\theta + \phi) \), use the formula \( \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} \) and substitute the values.

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The product of HCF and LCM of two numbers is equal to the product of the numbers. This can be expressed as: \[ HCF \times LCM = Number 1 \times Number 2. \]

Substitute the given values: \[ 40 \times (252 \times k) = 2520 \times 6600. \]

Simplify: \[ 252 \times k = \frac{2520 \times 6600}{40}. \]

Calculate: \[ 252 \times k = 415800. \]

Solve for \( k \): \[ k = \frac{415800}{252} = 1650. \]

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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.

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The sum of the first \( n \) terms of an A.P. is given as: \[ S_n = 3n^2 + 4n. \]

The first term \( a_1 \) is: \[ a_1 = S_1 = 3(1)^2 + 4(1) = 3 + 4 = 7. \]

The common difference \( d \) is: \[ d = a_2 - a_1. \]

Find \( a_2 \) by calculating \( S_2 - S_1 \): \[ S_2 = 3(2)^2 + 4(2) = 3(4) + 8 = 12 + 8 = 20. \]

Thus: \[ a_2 = S_2 - S_1 = 20 - 7 = 13, \quad d = a_2 - a_1 = 13 - 7 = 6. \]

The first term \( a_1 = 7 \).

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Conclusion:

The first term of the A.P. is \( 7 \). Quick Tip: To find the zeroes of a quadratic polynomial, use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

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The polynomial is: \[ P(x) = x^2 - 16x + 30. \]

For \( x = 15 \) to be a zero of the resulting polynomial: \[ P(15) = 0. \]

Substitute \( x = 15 \) in \( P(x) \): \[ P(15) = (15)^2 - 16(15) + 30. \]

Simplify: \[ P(15) = 225 - 240 + 30 = 225 - 210 = 15. \]

To make \( P(15) = 0 \), we need to subtract \( 15 \) from the polynomial.

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Conclusion:

The value to be subtracted is \( 15 \). Quick Tip: To find the zeroes of a quadratic polynomial, use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

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The probability of hitting a boundary is: \[ P(E) = \frac{Number of times boundary is hit}{Total number of balls} = \frac{9}{45} = \frac{1}{5}. \]

The probability of not hitting a boundary is: \[ P(not E) = 1 - P(E) = 1 - \frac{1}{5} = \frac{4}{5}. \]

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). Quick Tip: For probabilities, the sum of the probability of an event and its complement is always 1: \[ P(E) + P(not E) = 1. \] Use this rule to find the complement probability quickly.

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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: \[ \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right). \]

Substitute \( m_1 = 1, m_2 = 2, A(1, 2), B(-1, 1) \): \[ x = \frac{(1)(-1) + (2)(1)}{1 + 2} = \frac{-1 + 2}{3} = \frac{1}{3}, \quad y = \frac{(1)(1) + (2)(2)}{1 + 2} = \frac{1 + 4}{3} = \frac{5}{3}. \]

The point is \( \left(\frac{1}{3}, \frac{5}{3}\right) \), not \( \left(-\frac{1}{3}, \frac{5}{3}\right) \).


Therefore, Assertion (A) is false, but Reason (R) is correct. Quick Tip: When dividing a line segment in a ratio, use the section formula: \[ \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right). \] Make sure the ratio and coordinates are substituted correctly.

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We start with the given expression: \[ \frac{\sec^2 45^\circ - \tan^2 45^\circ}{\sin^2 45^\circ}. \]

1. Use the trigonometric identities:
- \( \sec^2 \theta - \tan^2 \theta = 1 \),
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).

Substitute these values: \[ \frac{\sec^2 45^\circ - \tan^2 45^\circ}{\sin^2 45^\circ} = \frac{1}{\sin^2 45^\circ}. \]

2. Calculate \( \sin^2 45^\circ \): \[ \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}. \]

Substitute \( \sin^2 45^\circ = \frac{1}{2} \) into the expression: \[ \frac{1}{\sin^2 45^\circ} = \frac{1}{\frac{1}{2}} = 2. \]

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Conclusion:

The value of the given expression is \( 2 \). Quick Tip: Remember that \( \sec^2 \theta - \tan^2 \theta = 1 \) and use the square of trigonometric values for accurate simplifications.

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The condition for equidistance is: \[ PA = PB \quad \implies \quad PA^2 = PB^2. \]

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

Expand both sides: \[ (x^2 - 14x + 49) + (y^2 - 2y + 1) = (x^2 - 6x + 9) + (y^2 - 10y + 25). \]

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

Rearrange: \[ -8x + 8y + 16 = 0 \quad \implies \quad -8(x - y - 2) = 0. \]

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

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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.

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The centre \( O(2, -3y) \) is the midpoint of \( AB \). Using the midpoint formula: \[ \left(\frac{-1 + 5}{2}, \frac{y + 7}{2}\right) = (2, -3y). \]

Equating the \( x \)-coordinates: \[ \frac{-1 + 5}{2} = 2 \quad \implies \quad \frac{4}{2} = 2 \quad (satisfied). \]

Equating the \( y \)-coordinates: \[ \frac{y + 7}{2} = -3y. \]

Simplify: \[ y + 7 = -6y \quad \implies \quad 7y = -7 \quad \implies \quad y = -1. \]

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

The distance \( AB \) is: \[ AB = \sqrt{(5 - (-1))^2 + (7 - (-1))^2}. \]

Simplify: \[ AB = \sqrt{(5 + 1)^2 + (7 + 1)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10. \]

The radius is: \[ Radius = \frac{AB}{2} = \frac{10}{2} = 5. \]

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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.

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Total number of outcomes = \( 52 \).

(i) The probability of drawing the queen of hearts is: \[ P(queen of hearts) = \frac{1}{52}. \]

(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: \[ P(not a jack) = \frac{48}{52} = \frac{12}{13}. \]

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Conclusion:

(i) \( P(queen of hearts) = \frac{1}{52} \).

(ii) \( P(not a jack) = \frac{12}{13} \). 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).

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

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

Simplify: \[ 6x = 30 \quad \implies \quad x = 5. \]

Substitute \( x = 5 \) into equation (i): \[ 2(5) + y = 13 \quad \implies \quad 10 + y = 13 \quad \implies \quad y = 3. \]

The value of \( x - y \) is: \[ x - y = 5 - 3 = 2. \]

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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.

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Let the two numbers be \( x \) and \( y \), where \( x > y \).

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

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

Simplify: \[ 2x = 150 \quad \implies \quad x = 75. \]

Substitute \( x = 75 \) into equation (i): \[ 75 + y = 105 \quad \implies \quad y = 105 - 75 = 30. \]

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

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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.

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In \( \triangle EAB \) and \( \triangle ECD \), we are given: \[ \frac{EA}{EC} = \frac{EB}{ED}. \]

Also, \( \angle AEB = \angle CED \) (vertically opposite angles).

By the Side-Angle-Side (SAS) similarity criterion: \[ \triangle EAB \sim \triangle ECD. \]

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Conclusion:

It is proven that \( \triangle EAB \sim \triangle ECD \). 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.

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

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Step 1: Find points for each line.

For equation (i) (\( x - y = -1 \)):
\[ \text{If x = 0, \, -y = -1 \quad \implies \quad y = 1. \] \[ If x = -1, \, -y = -2 \quad \implies \quad y = 2. \] \[ If x = 1, \, -y = 0 \quad \implies \quad y = 0. \]

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 \( (0, 5), (5, 0), (2, 3) \).


\begin{tikzpicture[scale=1.2]
% Axes
\draw[thick,->] (-2,0) -- (6,0) node[right] {\(x\);
\draw[thick,->] (0,-1) -- (0,6) node[above] {\(y\);

% Grid
\draw[very thin, gray] (-2,-1) grid (6,6);

% Line 1: x - y = -1
\draw[thick,blue] (-1,2) -- (3,-1) node[right] {\(x - y = -1\);

% Line 2: x + y = 5
\draw[thick,red] (-1,6) -- (6,-1) node[right] {\(x + y = 5\);

% Points of intersection and specific points
\filldraw[black] (2,3) circle (2pt) node[above right] {\((2, 3)\);
\filldraw[black] (0,1) circle (2pt) node[left] {\((0, 1)\);
\filldraw[black] (-1,2) circle (2pt) node[left] {\((-1, 2)\);
\filldraw[black] (0,5) circle (2pt) node[above left] {\((0, 5)\);
\filldraw[black] (5,0) circle (2pt) node[below right] {\((5, 0)\);

% Labels for axes
\node at (6.2, 0) {\(x\);
\node at (0, 6.2) {\(y\);
\end{tikzpicture


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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.

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Let the given expression be: \[ \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}. \]

Take a common denominator: \[ Expression = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)}. \]

Simplify the numerator using the identity \( (x + y)^2 + (x - y)^2 = 2x^2 + 2y^2 \): \[ (\sin A + \cos A)^2 + (\sin A - \cos A)^2 = 2\sin^2 A + 2\cos^2 A. \]

Using \( \sin^2 A + \cos^2 A = 1 \), the numerator becomes: \[ 2\sin^2 A + 2\cos^2 A = 2(1) = 2. \]

Simplify the denominator: \[ (\sin A - \cos A)(\sin A + \cos A) = \sin^2 A - \cos^2 A. \]

The expression now becomes: \[ \frac{2}{\sin^2 A - \cos^2 A}. \]

Using the identity \( \sin^2 A - \cos^2 A = 2\sin^2 A - 1 \), the denominator is: \[ \sin^2 A - \cos^2 A = 2\sin^2 A - 1. \]

Thus, the expression becomes: \[ \frac{2}{2\sin^2 A - 1}. \]

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Conclusion:
\[ \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2\sin^2 A - 1}. \] Quick Tip: For trigonometric proofs, simplify each term step by step and use standard identities like \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 A - \cos^2 A = 2\sin^2 A - 1 \).

N/A Quick Tip: Use the section formula \( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \) and \( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \) to find points of division. For medians, the midpoint formula helps simplify the calculations.

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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 \( \angle OAT \) and \( \angle OBT \).


\begin{tikzpicture[scale=1.5]
% Draw the circle
\draw[thick] (0,0) circle(2);

% Points on the circle
\coordinate [label=above left:\(O\)] (O) at (0,0);
\coordinate [label=below left:\(A\)] (A) at (-1.5,-1.34);
\coordinate [label=below right:\(B\)] (B) at (1.5,-1.34);
\coordinate [label=below:\(T_A\)] (TA) at (-2,-1.34);
\coordinate [label=below:\(T_B\)] (TB) at (2,-1.34);

% Draw the chord and radii
\draw[thick] (A) -- (B) node[midway, above] {\(AB\);
\draw[thick] (O) -- (A);
\draw[thick] (O) -- (B);

% Tangents
\draw[dashed] (TA) -- (A);
\draw[dashed] (TB) -- (B);

% Mark angles
\node at (-0.8, -0.5) {\small \(\angle OAT\);
\node at (0.8, -0.5) {\small \(\angle OBT\);
\end{tikzpicture


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 \( \triangle OAT \), \( \angle OAT \) is the angle between the tangent at \( A \) and the chord \( AB \).

- In \( \triangle OBT \), \( \angle OBT \) 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 (\( \angle OAB = \angle OBA \)) and the radii \( OA \) and \( OB \) are equal, the triangles \( \triangle OAT \) and \( \triangle OBT \) are congruent (by RHS criterion).

Thus: \[ \angle OAT = \angle OBT. \]

---

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.

\normalfont
Let \( P(x) = x^2 - 15 \).

Factorize \( P(x) \): \[ P(x) = (x - \sqrt{15})(x + \sqrt{15}). \]

Thus, the zeroes of \( P(x) \) are: \[ -\sqrt{15} \quad and \quad \sqrt{15}. \]

---

Verification:

1. Sum of zeroes: \[ -\sqrt{15} + \sqrt{15} = 0. \]

Compare with: \[ Sum of zeroes = \frac{-coefficient of x}{coefficient of x^2} = \frac{0}{1} = 0. \]

2. Product of zeroes: \[ (-\sqrt{15}) \times (\sqrt{15}) = -15. \]

Compare with: \[ Product of zeroes = \frac{constant term}{coefficient of x^2} = \frac{-15}{1} = -15. \]

---

Conclusion:

The sum and product of the zeroes are verified to match the relationships: \[ Sum of zeroes = \frac{-coefficient of x}{coefficient of x^2}, \quad Product of zeroes = \frac{constant term}{coefficient of x^2}. \] Quick Tip: For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes is \( -\frac{b}{a} \) and the product of zeroes is \( \frac{c}{a} \).

\normalfont
Let \( a \) be the first term and \( d \) be the common difference.

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

For the first 7 terms (\( S_7 = 49 \)): \[ \frac{7}{2} [2a + 6d] = 49. \]

Simplify: \[ 7a + 21d = 49 \quad \implies \quad a + 3d = 7 \quad (i). \]

For the first 17 terms (\( S_{17} = 289 \)): \[ \frac{17}{2} [2a + 16d] = 289. \]

Simplify: \[ 17a + 136d = 289 \quad \implies \quad a + 8d = 17 \quad (ii). \]

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

Subtract (i) from (ii): \[ (8d - 3d) = (17 - 7) \quad \implies \quad 5d = 10 \quad \implies \quad d = 2. \]

Substitute \( d = 2 \) into (i): \[ a + 3(2) = 7 \quad \implies \quad a = 1. \]

Now, find the sum of the first 20 terms (\( S_{20} \)): \[ S_{20} = \frac{20}{2} [2a + 19d]. \]

Substitute \( a = 1 \) and \( d = 2 \): \[ S_{20} = 10 [2(1) + 19(2)] = 10 [2 + 38] = 10 \cdot 40 = 400. \]

---

Conclusion:

The sum of the first 20 terms of the A.P. is \( 400 \). Quick Tip: For sums of A.P., use the formula \( S_n = \frac{n}{2} [2a + (n-1)d] \) and solve equations systematically to find \( a \) and \( d \).

\normalfont
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: \[ T_{10} = a + 9d. \]

For the 30th term: \[ T_{30} = a + 29d. \]

Given: \[ \frac{T_{10}}{T_{30}} = \frac{1}{3}. \]

Substitute \( T_{10} \) and \( T_{30} \): \[ \frac{a + 9d}{a + 29d} = \frac{1}{3}. \]

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

Simplify: \[ 3a + 27d = a + 29d \quad \implies \quad 2a = 2d \quad \implies \quad a = d. \]

---

The sum of the first 6 terms is given as \( S_6 = 42 \). The formula for the sum of the first \( n \) terms is: \[ S_n = \frac{n}{2} [2a + (n-1)d]. \]

For \( S_6 \): \[ \frac{6}{2} [2a + 5d] = 42. \]

Simplify: \[ 3(2a + 5d) = 42 \quad \implies \quad 2a + 5d = 14 \quad (ii). \]

Substitute \( a = d \) into (ii): \[ 2d + 5d = 14 \quad \implies \quad 7d = 14 \quad \implies \quad d = 2. \]

Since \( a = d \), we have: \[ a = 2. \]

---

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 \( d \).

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

---

1. Volume of the lower cylinder:

Height (\( h_1 \)) = 200 cm,

Radius (\( r_1 \)) = \( \frac{diameter}{2} = \frac{28}{2} = 14 \, cm \).

The volume of a cylinder is given by: \[ V = \pi r^2 h. \]

For the lower cylinder: \[ V_1 = \pi r_1^2 h_1 = \pi (14)^2 (200). \]

Simplify: \[ V_1 = \pi (196)(200) = 39200\pi \, cm^3. \]

---

2. Volume of the upper cylinder:

Height (\( h_2 \)) = 50 cm,

Radius (\( r_2 \)) = 7 cm.

For the upper cylinder: \[ V_2 = \pi r_2^2 h_2 = \pi (7)^2 (50). \]

Simplify: \[ V_2 = \pi (49)(50) = 2450\pi \, cm^3. \]

---

3. Total volume of the pole:
\[ V_{total} = V_1 + V_2 = 39200\pi + 2450\pi = 41650\pi \, cm^3. \]

Substitute \( \pi \approx 3.1416 \): \[ V_{total} = 41650 \times 3.1416 = 130881.55 \, cm^3. \]

---

4. Mass of the pole:

Given that 1 cm\(^3\) of iron has a mass of 8 g: \[ Mass = V_{total} \times 8 = 130881.55 \times 8 = 1047052.4 \, g. \]

Convert to kilograms: \[ Mass = \frac{1047052.4}{1000} = 1047.05 \, kg. \]

---

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).

\normalfont
The capsule consists of:
1. A cylindrical part with two hemispheres attached at the ends.
2. Radius (\( r \)) of the hemispheres and the cylinder:
\[ r = \frac{diameter}{2} = \frac{4}{2} = 2 \, mm. \]
3. Length of the cylindrical part:
\[ Length of cylinder = Total length of capsule - 2(Radius of hemispheres) = 14 - 4 = 10 \, mm. \]

---

1. Surface Area of the Capsule:

The surface area consists of:
1. Curved surface area (CSA) of the cylinder:
\[ CSA of cylinder = 2\pi r h = 2 \times \frac{22}{7} \times 2 \times 10 = 40 \times \frac{22}{7} = 125.71 \, mm^2. \]
2. CSA of the two hemispheres:
\[ CSA of one hemisphere = 2\pi r^2 = 2 \times \frac{22}{7} \times 2^2 = 2 \times \frac{22}{7} \times 4 = \frac{176}{7} = 25.14 \, mm^2. \]

For two hemispheres: \[ CSA of both hemispheres = 2 \times 25.14 = 50.28 \, mm^2. \]

Total surface area: \[ Surface Area = CSA of cylinder + CSA of hemispheres = 125.71 + 50.28 = 176 \, mm^2. \]

---

2. Volume of the Capsule:

The volume consists of:
1. Volume of the cylinder:
\[ Volume of cylinder = \pi r^2 h = \frac{22}{7} \times 2^2 \times 10 = \frac{22}{7} \times 4 \times 10 = \frac{880}{7} = 125.71 \, mm^3. \]
2. Volume of the two hemispheres:
\[ Volume of one hemisphere = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 2^3 = \frac{2}{3} \times \frac{22}{7} \times 8 = \frac{352}{21} = 16.76 \, mm^3. \]

For two hemispheres: \[ Volume of both hemispheres = 2 \times 16.76 = 33.52 \, mm^3. \]

Total volume: \[ Volume = Volume of cylinder + Volume of hemispheres = 125.71 + 33.52 = 159.24 \, mm^3. \]

---

Conclusion:

1. The surface area of the capsule is: \[ 176 \, mm^2. \]
2. The volume of the capsule is: \[ 159.24 \, mm^3. \]

--- 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.

\normalfont
Let the original speed of the aircraft be \( x \, km/h \).

1. Original Time of Flight: \[ Time = \frac{Distance}{Speed} = \frac{2800}{x}. \]

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: \[ \frac{2800}{x - 100}. \]

The time difference between the original and new times is 30 minutes, or \( \frac{1}{2} \, hour \). Therefore: \[ \frac{2800}{x - 100} - \frac{2800}{x} = \frac{1}{2}. \]

3. Simplify the Equation:
Take the LCM of \( x(x - 100) \): \[ \frac{2800x - 2800(x - 100)}{x(x - 100)} = \frac{1}{2}. \]

Simplify: \[ \frac{2800 \cdot 100}{x(x - 100)} = \frac{1}{2}. \]

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

Expand: \[ x^2 - 100x - 560000 = 0. \]

4. Solve the Quadratic Equation:
Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, \, b = -100, \, c = -560000. \] \[ x = \frac{-(-100) \pm \sqrt{(-100)^2 - 4(1)(-560000)}}{2(1)}. \] \[ x = \frac{100 \pm \sqrt{10000 + 2240000}}{2}. \] \[ x = \frac{100 \pm \sqrt{2250000}}{2}. \] \[ x = \frac{100 \pm 1500}{2}. \]

Select the positive root: \[ x = \frac{100 + 1500}{2} = 800. \]

5. Find the Original Time:
The original time of the flight is: \[ \frac{2800}{800} = 3.5 \, hours. \]

---

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.

\normalfont
Let the numerator of the fraction be \( x \). Then the denominator is: \[ 2x + 1. \]

The fraction is: \[ \frac{x}{2x+1}. \]

The reciprocal is: \[ \frac{2x+1}{x}. \]

Given: \[ \frac{x}{2x+1} + \frac{2x+1}{x} = 2 \frac{16}{21}. \]

Convert \( 2 \frac{16}{21} \) to an improper fraction: \[ 2 \frac{16}{21} = \frac{42 + 16}{21} = \frac{58}{21}. \]

Equate: \[ \frac{x}{2x+1} + \frac{2x+1}{x} = \frac{58}{21}. \]

1. Simplify the Left Side:
Take the LCM of \( x(2x+1) \): \[ \frac{x^2 + (2x+1)^2}{x(2x+1)} = \frac{58}{21}. \]

Expand \( (2x+1)^2 \): \[ \frac{x^2 + 4x^2 + 4x + 1}{x(2x+1)} = \frac{58}{21}. \]

Simplify: \[ \frac{5x^2 + 4x + 1}{x(2x+1)} = \frac{58}{21}. \]

2. Cross Multiply: \[ 21(5x^2 + 4x + 1) = 58x(2x+1). \]

Expand both sides: \[ 105x^2 + 84x + 21 = 116x^2 + 58x. \]

Simplify: \[ 116x^2 - 105x^2 + 58x - 84x - 21 = 0. \] \[ 11x^2 - 26x - 21 = 0. \]

3. Solve the Quadratic Equation:
Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 11, \, b = -26, \, c = -21. \] \[ x = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(11)(-21)}}{2(11)}. \] \[ x = \frac{26 \pm \sqrt{676 + 924}}{22}. \] \[ x = \frac{26 \pm \sqrt{1600}}{22}. \] \[ x = \frac{26 \pm 40}{22}. \]

Select the positive root: \[ x = \frac{26 + 40}{22} = \frac{66}{22} = 3. \]

4. Find the Fraction:
The fraction is: \[ \frac{x}{2x+1} = \frac{3}{2(3)+1} = \frac{3}{7}. \]

---

Conclusion:

The fraction is \( \frac{3}{7} \). Quick Tip: For fraction-reciprocal problems, set up the equation, clear the denominators using LCM, and solve the resulting quadratic equation.

\normalfont
Let \( ABCD \) be a parallelogram with \( M \) as the mid-point of \( CD \). The diagonals of a parallelogram bisect each other, so \( AC \) is divided equally by the point of intersection \( O \) of the diagonals.


1. Step 1: Using the properties of a parallelogram:
- Since \( M \) is the mid-point of \( CD \), \( BM \) divides \( AC \) in the ratio \( 2:1 \) at \( L \) (from the mid-point theorem).


2. Step 2: Extend \( BM \) to meet \( AD \) produced at \( E \). Using the concept of proportionality:
- The triangle \( \Delta BLM \) and \( \Delta ELM \) are similar because they share the same base \( LM \) and are cut by parallel lines.


3. Step 3: From similarity, \( EL = 2BL \):
- Since \( L \) divides \( AC \) in the ratio \( 2:1 \), the extended line \( EL \) satisfies \( EL = 2BL \).


---

Conclusion:

Thus, it is proved that \( EL = 2BL \) using the mid-point theorem and similarity of triangles. Quick Tip: Use the mid-point theorem to determine proportional relationships in parallelograms, and apply similarity of triangles to solve extended line segment problems.

\normalfont

Refer to the diagram for clarity. Let the point \( P \) represent the jet plane's position initially, and \( Q \) represent its position after 30 seconds. The height of the plane is constant at \( 3600\sqrt{3} \, m \).

Step 1: Calculate \( x \) in \( \triangle APB \)
Using \( \tan 60^\circ = \sqrt{3} \), we have: \[ \tan 60^\circ = \frac{Height of plane}{Base (AB)} = \frac{3600\sqrt{3}}{x}. \]

Simplify: \[ x = 3600 \, m. \]

Step 2: Calculate \( x + y \) in \( \triangle AQC \)
Using \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we have: \[ \tan 30^\circ = \frac{Height of plane}{Base (AC)} = \frac{3600\sqrt{3}}{x + y}. \]

Substitute \( x = 3600 \, m \): \[ \frac{1}{\sqrt{3}} = \frac{3600\sqrt{3}}{3600 + y}. \]

Simplify: \[ 3600 + y = 3 \times 3600 = 10800 \quad \implies \quad y = 7200 \, m. \]

Step 3: Calculate the speed of the jet plane
The total horizontal distance covered by the plane in 30 seconds is: \[ y = 7200 \, m. \]

The speed of the plane is: \[ Speed = \frac{Distance}{Time} = \frac{7200}{30} = 240 \, m/s. \]

---

Conclusion:

The speed of the jet plane is \( 240 \, m/s \). Quick Tip: Remember that for height and angle of elevation problems, use trigonometric identities \( \tan \theta = \frac{Height}{Base} \). Calculate distances step-by-step and substitute known values carefully.

\normalfont

The given number is \( 173250 \). Perform the prime factorization of \( 173250 \): \[ 173250 = 2 \times 5^3 \times 3^2 \times 7 \times 11. \]

1. (i) Least Prime Number:

The least prime number used by students is \( 3 \).

---

2. (ii)(a) Number of Students in the Class:

Each student multiplies the number by one prime number. The total prime factors used are: \[ 2 + 3 + 2 + 1 + 1 = 7. \]

Thus, the total number of students is \( 7 \).

---

2. (ii)(b) Highest Prime Number:

The highest prime number in the factorization is \( 11 \).

---

3. (iii) Prime Number Used Maximum Times:

The prime number \( 5 \) appears \( 3 \) times, which is the maximum frequency.

---

Conclusion:


[(i)] The least prime number used is \( 3 \).
[(ii)(a)] The total number of students in the class is \( 7 \).

(b) The highest prime number used is \( 11 \).
[(iii)] 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.

\normalfont

1. Area of the square-shaped grass field:
\[ Side of the square = 20 \, m. \] \[ Area of the square field = side^2 = 20 \times 20 = 400 \, m^2. \]

---

2. (ii)(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: \[ Area of one horse's grazing region = \frac{1}{4} \pi r^2. \]

Here, \( r = 7 \, m \), and \( \pi = \frac{22}{7} \): \[ Area of one grazing region = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{4} \times 154 = 38.5 \, m^2. \]

For four horses: \[ Total grazing area = 4 \times 38.5 = 154 \, m^2. \]

---

2. (ii)(b) If the rope length is increased to 10 m:

For one horse: \[ Area grazed by one horse = \frac{1}{4} \pi r^2, \quad r = 10 \, m, \, \pi = 3.14. \] \[ Area grazed by one horse = \frac{1}{4} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5 \, m^2. \]

---

3. (iii) Area of the field left ungrazed:
\[ Area left ungrazed = Area of square field - Area grazed by all horses. \]

Substitute: \[ Area left ungrazed = 400 - 0.0154 = 399.9846 \, m^2. \]

---

Conclusion:


[(i)] The area of the square-shaped grass field is \( 400 \, m^2 \).
[(ii)(a)] The total grazing area for all horses is \( 154 \, m^2 \).
[(ii)(b)] If the rope length is increased to 10 m, the grazing area for one horse is \( 78.5 \, m^2 \).
[(iii)] The area left ungrazed is \( 399.9846 \, m^2 \). 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: \( \frac{\theta}{360^\circ} \pi r^2 \).

1. (i) Lower Limit of the Modal Class:

The modal class is the class with the highest frequency. From the table, the highest frequency is \( 132 \), corresponding to the class \( 20-24 \).
\[ Lower limit of the modal class = 20. \]

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2. (ii)(a) Median Class:

The cumulative frequency (CF) is calculated as follows: \[ \begin{array}{|c|c|c|} \hline Age (in years) & Frequency (f) & Cumulative Frequency (CF)
\hline 15-19 & 62 & 62
\hline 20-24 & 132 & 194
\hline 25-29 & 96 & 290
\hline 30-34 & 37 & 327
\hline 35-39 & 13 & 340
\hline 40-44 & 11 & 351
\hline 45-49 & 10 & 361
\hline 50-54 & 4 & 365
\hline \end{array} \]

The total number of participants is \( N = 365 \). The median class corresponds to \( \frac{N}{2} = \frac{365}{2} = 182.5 \), which lies in the cumulative frequency \( 194 \). Therefore, the median class is \( 20-24 \).

---

2. (ii)(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. (iii) Empirical Relationship:

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

---

Conclusion:


[(i)] The lower limit of the modal class is \( 20 \).
[(ii)(a)] The median class is \( 20-24 \).

(b) The number of participants below 50 years is \( 361 \).
[(iii)] 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.