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

The total number of balls is: \[ 3 + 5 + 7 = 15. \]

The number of balls that are neither red nor black (white balls) is: \[ 5. \]

The probability is: \[ P(neither red nor black) = \frac{Number of white balls}{Total number of balls} = \frac{5}{15} = \frac{1}{3}. \]

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Conclusion:

The probability that the ball will be neither red nor black is \( \frac{1}{3} \). Quick Tip: To calculate probabilities, divide the favorable outcomes by the total outcomes and simplify.

The probability of getting a bad egg is: \[ P(bad egg) = 0.045. \]

The total number of eggs is \( 400 \), so the number of bad eggs is: \[ Number of bad eggs = 0.045 \times 400 = 18. \]

The number of good eggs is: \[ Number of good eggs = 400 - 18 = 382. \]

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Conclusion:

The number of good eggs in the lot is \( 382 \). Quick Tip: For probability problems, multiply the total outcomes by the given probability to find specific counts.

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: \[ Arc length = 2\pi r \cdot \frac{\theta}{360^\circ}. \]

Here: \[ r = 7 \, cm, \quad \theta = 90^\circ. \]

Substitute the values: \[ Arc length = 2 \cdot \frac{22}{7} \cdot 7 \cdot \frac{90}{360}. \]

Simplify step-by-step: \[ Arc length = \frac{2 \cdot 22 \cdot 7 \cdot 90}{7 \cdot 360} = \frac{22 \cdot 90}{180} = 11 \, cm. \]

Now, calculate the perimeter: \[ Perimeter = Arc length + 2r = 11 + 2 \cdot 7 = 11 + 14 = 25 \, cm. \]

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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.

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

Here: \[ \sqrt{3} \cdot \sqrt{27} = \sqrt{3 \times 27} = \sqrt{81} = 9, \]
which is a rational number.

For the other pairs:
- \( (\sqrt{16}, \sqrt{4}) \) simplifies to \( \sqrt{64} \), but \( \sqrt{16} \) and \( \sqrt{4} \) are not irrational.
- \( (\sqrt{5}, \sqrt{2}) \) results in \( \sqrt{10} \), which is irrational.
- \( (\sqrt{36}, \sqrt{2}) \) results in \( \sqrt{72} \), which is also irrational.

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Conclusion:

The correct pair is \( (\sqrt{3}, \sqrt{27}) \). Quick Tip: The product of two irrational numbers is rational if the resulting value simplifies to a perfect square or a rational number.

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).

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 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.

The total of five numbers is given by: \[ Total of 5 numbers = Mean \times Count = 15 \times 5 = 75. \]

When one more number is included, the mean becomes 17. Therefore, the total of six numbers is: \[ Total of 6 numbers = Mean \times Count = 17 \times 6 = 102. \]

The included number is: \[ Included number = Total of 6 numbers - Total of 5 numbers = 102 - 75 = 27. \]

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Conclusion:

The included number is \( 27 \). Quick Tip: For mean-related problems, calculate the total sum using \( Total = Mean \times Count \). Use this relationship to find missing values.

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

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: \[ \frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{5}{-1} = -5. \]

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel and do not intersect. Hence, the system of equations has no solution.

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Conclusion:

The given pair of equations is inconsistent, and there is no solution. Quick Tip: For a pair of linear equations, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the lines are parallel, and there is no 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: \[ \angle TML = \angle MON. \]

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.

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.

When two dice are thrown, the total number of outcomes is: \[ 6 \times 6 = 36. \]

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: \[ P(same) = \frac{6}{36} = \frac{1}{6}. \]

The probability of showing different numbers is the complement: \[ P(different) = 1 - P(same) = 1 - \frac{1}{6} = \frac{5}{6}. \]

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Conclusion:

The probability that the two dice show different numbers is \( \frac{5}{6} \). Quick Tip: The complement rule states that \(P(A') = 1 - P(A)\). Use it for quick probability calculations.

1. For \( \sin \alpha = \frac{\sqrt{3}}{2} \):
\[ \alpha = 60^\circ \quad \implies \quad \tan \alpha = \sqrt{3}. \]

2. For \( \cos \beta = \frac{\sqrt{3}}{2} \):
\[ \beta = 30^\circ \quad \implies \quad \tan \beta = \frac{1}{\sqrt{3}}. \]

3. Multiply \( \tan \alpha \) and \( \tan \beta \):
\[ \tan \alpha \cdot \tan \beta = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1. \]

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Conclusion:

The value of \( \tan \alpha \cdot \tan \beta \) is \( 1 \). Quick Tip: Remember the trigonometric values for standard angles \( 30^\circ, 45^\circ, and 60^\circ \). They simplify calculations involving sine, cosine, and tangent.

For \( 3 \) to be a zero of the resulting polynomial, substituting \( x = 3 \) in the polynomial \( x^2 - 5x + 4 + k \) should give: \[ 3^2 - 5(3) + 4 + k = 0. \]

Simplify: \[ 9 - 15 + 4 + k = 0 \quad \implies \quad -2 + k = 0 \quad \implies \quad k = 2. \]

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Conclusion:

The value that should be added to the polynomial is \( 2 \). Quick Tip: To make a specific number a zero of a polynomial, substitute it into the polynomial and solve for the unknown constant.

To make \( \sqrt{20} \) a rational number, it should be multiplied by an irrational number that removes the square root. The simplest way is to simplify: \[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}. \]

Multiplying \( \sqrt{20} \) by \( \sqrt{5} \): \[ \sqrt{20} \times \sqrt{5} = 2\sqrt{5} \times \sqrt{5} = 2 \times 5 = 10. \]

Thus, \( \sqrt{5} \) is the smallest irrational number that makes \( \sqrt{20} \) rational.

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Conclusion:

The smallest irrational number is \( \sqrt{5} \). Quick Tip: For square root simplifications, multiply by a term that completes the square under the radical to eliminate it.

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 terms of the A.P. are: \[ T_1 = \frac{1}{2x}, \quad T_2 = \frac{1 - 4x}{2x}, \quad T_3 = \frac{1 - 8x}{2x}. \]

The common difference \(d\) is: \[ d = T_2 - T_1 = \frac{1 - 4x}{2x} - \frac{1}{2x}. \]

Combine terms: \[ d = \frac{(1 - 4x) - 1}{2x} = \frac{-4x}{2x} = -2. \]

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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.

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 \( 90^\circ \).

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.

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.

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.

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.

Step 1: Simplify the denominator. Using the Pythagorean identity: \[ \sin^2 60^\circ + \cos^2 60^\circ = 1. \]

Step 2: Substitute values of \( \tan 60^\circ \) and \( \tan 30^\circ \): \[ \tan 60^\circ = \sqrt{3}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}. \]

Step 3: Substitute into the expression: \[ \frac{5 \tan 60^\circ}{(\sin^2 60^\circ + \cos^2 60^\circ) \tan 30^\circ} = \frac{5 \cdot \sqrt{3}}{1 \cdot \frac{1}{\sqrt{3}}}. \]

Step 4: Simplify the fraction: \[ \frac{5 \cdot \sqrt{3}}{\frac{1}{\sqrt{3}}} = 5 \cdot \sqrt{3} \cdot \sqrt{3} = 5 \cdot 3 = 15. \]

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Conclusion:

The value of the expression is \( 15 \). Quick Tip: For trigonometric simplifications, use the Pythagorean identities \( \sin^2 x + \cos^2 x = 1 \) and standard angle values for quick calculations.

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.

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).

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.

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.

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: \[ x = \frac{k(-1) + 1(5)}{k + 1}. \]

Since the point lies on the Y-axis, \( x = 0 \). Substitute \( x = 0 \): \[ \frac{k(-1) + 5}{k + 1} = 0. \]

Simplify: \[ -k + 5 = 0 \quad \implies \quad k = 5. \]

Hence, the required ratio is: \[ 5 : 1. \]

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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.

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

Given: \[ PR : RQ = 2 : 1. \]

Using the section formula: \[ x = \frac{m_2x_1 + m_1x_2}{m_1 + m_2}, \quad y = \frac{m_2y_1 + m_1y_2}{m_1 + m_2}. \]

Here: \[ m_1 = 2, \, m_2 = 1, \, P(-2, 5), \, Q(3, 2). \]

Substitute into the formulas for \( x \) and \( y \): \[ x = \frac{1(-2) + 2(3)}{2 + 1} = \frac{-2 + 6}{3} = \frac{4}{3}. \] \[ y = \frac{1(5) + 2(2)}{2 + 1} = \frac{5 + 4}{3} = 3. \]

Thus, the coordinates of \( R \) are: \[ \left(\frac{4}{3}, 3\right). \]

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Conclusion:

The coordinates of the point \( R \) are: \[ \left(\frac{4}{3}, 3\right). \] Quick Tip: Use the section formula for dividing a line segment in a given ratio: \[ \left( \frac{m_2x_1 + m_1x_2}{m_1 + m_2}, \frac{m_2y_1 + m_1y_2}{m_1 + m_2} \right). \] Ensure the correct substitution of ratios and coordinates.

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

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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 \( d \).

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

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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: \[ a = 2. \]

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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 \( d \).

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


---

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.

---

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.

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 \( \frac{c}{a} \).

1. Given:A circle with centre \( O \), and two tangents \( PA \) and \( PB \) drawn from an external point \( P \). \( A \) and \( B \) are points of tangency, and \( AB \) is a chord.


2. To Prove: \( OP \) is the perpendicular bisector of \( AB \).

3. Proof:

- Join \( OA \), \( OB \), and \( OP \).
- In \( \triangle OPA \) and \( \triangle OPB \):
- \( OP = OP \) (common side),
- \( PA = PB \) (tangents from an external point are equal),
- \( OA = OB \) (radii of the same circle).


Thus, \( \triangle OPA \cong \triangle OPB \) (by SSS congruence).


- Since \( \triangle OPA \cong \triangle OPB \), \( \angle OPA = \angle OPB \).


- Line \( OP \) is perpendicular to \( AB \) because the tangents \( PA \) and \( PB \) are symmetrical about \( OP \).


4. Hence, \( OP \) is the perpendicular bisector of \( AB \).


---

Conclusion:

The line \( OP \) bisects \( AB \) at right angles. Quick Tip: When two tangents are drawn from an external point, they are equal, and the line joining the external point to the circle's center is always perpendicular to the chord joining the tangency points.

To prove the given identity, simplify step by step:

1. Simplify \( \csc \theta - \sin \theta \): \[ \csc \theta - \sin \theta = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta}. \]

Using \( 1 - \sin^2 \theta = \cos^2 \theta \), we get: \[ \csc \theta - \sin \theta = \frac{\cos^2 \theta}{\sin \theta}. \]

2. Simplify \( \sec \theta - \cos \theta \): \[ \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta}. \]

Using \( 1 - \cos^2 \theta = \sin^2 \theta \), we get: \[ \sec \theta - \cos \theta = \frac{\sin^2 \theta}{\cos \theta}. \]

3. Simplify \( \tan \theta + \cot \theta \): \[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}. \]

Taking the LCM: \[ \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}. \]

Using \( \sin^2 \theta + \cos^2 \theta = 1 \), we get: \[ \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}. \]

4. Multiply the three expressions: \[ (\csc \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta) = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} \cdot \frac{1}{\sin \theta \cos \theta}. \]

Simplify: \[ = \frac{\cos^2 \theta \sin^2 \theta}{\sin \theta \cos \theta} \cdot \frac{1}{\sin \theta \cos \theta} = \frac{\cos^2 \theta \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} = 1. \]

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Conclusion:

The given expression is equal to \( 1 \). Quick Tip: Use trigonometric identities like \( 1 - \sin^2 \theta = \cos^2 \theta \) and \( \sin^2 \theta + \cos^2 \theta = 1 \) for simplifications. Simplify each term before multiplying.

1. Let the height of the cliff be \( h \), the distance of the boat from the shore at the first observation be \( x \), and the distance moved by the boat in 6 minutes be \( y \).

2. From \( \triangle QAB \): \[ \tan 60^\circ = \frac{h}{x} \implies h = \sqrt{3}x \tag{1}. \]

3. From \( \triangle PAB \): \[ \tan 30^\circ = \frac{h}{y + x} \implies \frac{1}{\sqrt{3}} = \frac{h}{y + x} \implies y + x = \sqrt{3}h \tag{2}. \]

4. Substitute \( h = \sqrt{3}x \) from (1) into (2): \[ y + x = \sqrt{3}(\sqrt{3}x) \implies y + x = 3x \implies y = 2x. \]

5. The time taken by the boat to cover distance \( y \) is 6 minutes. Since \( y = 2x \), the time taken by the boat to travel distance \( x \) is: \[ \frac{1}{2} \times 6 = 3 \, minutes. \]

---

Conclusion:

The time taken by the boat to reach the shore is \( 3 \, minutes \). Quick Tip: Use trigonometric ratios like \( \tan \) in right triangles to relate height and distances. Substitution simplifies calculations when multiple triangles share common parameters.

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 (\( h_1 \)) = 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: \[ V_1 = \pi (196)(200) = 39200\pi \, cm^3. \]

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

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

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

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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).

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

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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.

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.

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: \[ \frac{x}{2x+1} = \frac{3}{2(3)+1} = \frac{3}{7}. \]

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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.

1. In \( \triangle ABC \), let \( AD \perp BC \). Since \( AD^2 = BD \times DC \), this implies a specific geometric relationship between the segments of \( BC \).


2. From the given condition: \[ AD^2 = BD \cdot DC \]


3. Using the property of a right triangle, if the altitude from the vertex opposite the hypotenuse satisfies \( AD^2 = BD \cdot DC \), then \( \triangle ABC \) is a right triangle with \( \angle BAC = 90^\circ \).


4. To verify, recall the geometric mean theorem:


\text{In a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments

\text{it divides the hypotenuse into.



5. Here, \( AD^2 = BD \cdot DC \) confirms the above theorem. Thus, \( \triangle ABC \) must be a right triangle, and \( \angle BAC = 90^\circ \).


---

Conclusion:

The given condition proves that \( \triangle ABC \) is a right triangle with \( \angle BAC = 90^\circ \). Quick Tip: The geometric mean theorem is a powerful tool to prove right triangles. If the altitude to the hypotenuse satisfies \( h^2 = p \cdot q \), then the triangle is right-angled.

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 \( 20-24 \).

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

Participants below 50 years correspond to the classes \( 15-19 \) to \( 45-49 \).
\[ Total frequency below 50 years = 62 + 132 + 96 + 37 + 13 + 11 + 10 = 361. \]

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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.

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 \( 3 \).

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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 \( 7 \).

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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.

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

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

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

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

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

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

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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: \( \frac{\theta}{360^\circ} \pi r^2 \).