CBSE Class 10 Mathematics Standard Question Paper 2024 (Set 2- 30/5/2) with Answer Key

The polynomial is: \[ p(x) = kx^2 - 30x + 45k. \]

The sum and product of the roots (\(\alpha + \beta\) and \(\alpha \beta\)) are given by: \[ \alpha + \beta = -\frac{Coefficient of x}{Coefficient of x^2} = -\frac{-30}{k} = \frac{30}{k}. \] \[ \alpha \beta = \frac{Constant term}{Coefficient of x^2} = \frac{45k}{k} = 45. \]

Given: \[ \alpha + \beta = \alpha \beta. \]

Substitute the values: \[ \frac{30}{k} = 45. \]

Solve for \(k\): \[ 30 = 45k \quad \implies \quad k = \frac{30}{45} = \frac{2}{3}. \]

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Conclusion:

The value of \(k\) is \(\frac{2}{3}\). Quick Tip: For problems involving zeroes of polynomials, use the relationships \(\alpha + \beta = -\frac{Coefficient of x}{Coefficient of x^2}\) and \(\alpha \beta = \frac{Constant term}{Coefficient of x^2}\).

Given:
- Radius \(r = 10 \, cm\).
- The chord subtends a \(90^\circ\) angle at the center.

In \(\triangle OAB\), \(O\) is the center, and \(\angle AOB = 90^\circ\). Using the Pythagoras theorem: \[ Chord length (AB) = \sqrt{OA^2 + OB^2}. \]

Substitute \(OA = OB = r = 10\): \[ Chord length (AB) = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}. \]

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Conclusion:

The length of the chord is \(10\sqrt{2}\). Quick Tip: For chords subtending \(90^\circ\) at the center, use the Pythagoras theorem to calculate the chord length.

The given sequence \( \sqrt{18}, \ \sqrt{50}, \ \sqrt{98}, \ \ldots \) is in arithmetic progression (A.P.) since the difference between consecutive terms is constant.

Step 1: Calculate the common difference (\( d \)):
\[ d = \sqrt{50} - \sqrt{18}. \]
Simplify each term: \[ \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \quad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}. \]
Thus: \[ d = 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2}. \]

Step 2: Find the 4^th term:

The general term of an A.P. is given by: \[ a_n = a + (n - 1)d, \]
where \( a = \sqrt{18} = 3\sqrt{2} \), \( n = 4 \), and \( d = 2\sqrt{2} \). Substituting: \[ a_4 = 3\sqrt{2} + (4 - 1)(2\sqrt{2}) = 3\sqrt{2} + 6\sqrt{2} = 9\sqrt{2}. \]
Simplify: \[ a_4 = \sqrt{81 \cdot 2} = \sqrt{162}. \]

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Conclusion:

The 4^th term is \( \sqrt{162} \). Quick Tip: To find the next term in an A.P., calculate the common difference and apply the formula for the \( n \)-th term.

Co-prime numbers are numbers that have no common factors other than 1.

By definition, the HCF of any two co-prime numbers is always \(1\).

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Conclusion:

The HCF of the two co-prime numbers is \(1\). Quick Tip: Co-prime numbers always have an HCF of 1, as they share no common factors other than 1.

We are given: \[ x = a \cos \theta \quad and \quad y = b \sin \theta. \]
Substitute these values into \( b^2 x^2 + a^2 y^2 \): \[ b^2 x^2 + a^2 y^2 = b^2 (a \cos \theta)^2 + a^2 (b \sin \theta)^2. \]
Simplify each term: \[ b^2 x^2 = b^2 a^2 \cos^2 \theta, \quad a^2 y^2 = a^2 b^2 \sin^2 \theta. \]
Thus: \[ b^2 x^2 + a^2 y^2 = b^2 a^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta. \]
Factor out \( a^2 b^2 \): \[ b^2 x^2 + a^2 y^2 = a^2 b^2 (\cos^2 \theta + \sin^2 \theta). \]
Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \): \[ b^2 x^2 + a^2 y^2 = a^2 b^2 \cdot 1 = a^2 b^2. \]

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Conclusion:

The value of \( b^2 x^2 + a^2 y^2 \) is \( a^2 b^2 \). Quick Tip: When solving expressions involving trigonometric identities, always check for common factors and simplify using \( \cos^2 \theta + \sin^2 \theta = 1 \).

The given quadratic equation is: \[ ax^2 + bx + c = 0. \]

For real and equal roots, the discriminant \( D \) is zero: \[ D = b^2 - 4ac = 0. \]

Step 1: Simplify the discriminant:
\[ b^2 = 4ac. \]

Step 2: Express \( c \) in terms of \( a \) and \( b \):
\[ c = \frac{b^2}{4a}. \]

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Conclusion:

The value of \( c \) for the quadratic equation to have real and equal roots is: \[ c = \frac{b^2}{4a}. \] Quick Tip: For real and equal roots of a quadratic equation, use \( b^2 - 4ac = 0 \) to relate \( c \) to \( a \) and \( b \).

Given \( DE \parallel BC \), by the Basic Proportionality Theorem (Thales' theorem), we have: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Step 1: Express \( EC \) in terms of \( AE, AD, \) and \( DB \):
\[ \frac{AD}{DB} = \frac{AE}{EC}. \]
Substitute \( AD = 2.4 \, cm, DB = 4 \, cm, AE = 2 \, cm \): \[ \frac{2.4}{4} = \frac{2}{EC}. \]

Step 2: Solve for \( EC \):
\[ EC = \frac{2 \times 4}{2.4} = \frac{8}{2.4} = \frac{10}{3} \, cm. \]

Step 3: Find \( AC \):
\[ AC = AE + EC = 2 + \frac{10}{3} = \frac{6}{3} + \frac{10}{3} = \frac{16}{3} \, cm. \]

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Conclusion:

The length of \( AC \) is: \[ AC = \frac{16}{3} \, cm. \] Quick Tip: Use the Basic Proportionality Theorem (\( \frac{AD}{DB} = \frac{AE}{EC} \)) to solve problems involving parallel lines in triangles.

The length of an arc is given by: \[ Arc length = \frac{\theta}{360^\circ} \cdot 2\pi r. \]

Substitute \(Arc length = 10\pi\), \(r = 12\), and solve for \(\theta\): \[ 10\pi = \frac{\theta}{360} \cdot 2\pi \cdot 12. \]

Simplify: \[ 10 = \frac{\theta}{360} \cdot 24 \quad \implies \quad \frac{\theta}{360} = \frac{10}{24}. \]
\[ \theta = \frac{10}{24} \cdot 360 = 150^\circ. \]

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Conclusion:

The angle subtended by the arc is \(150^\circ\). Quick Tip: For arc length problems, substitute known values into the formula and solve for the unknown step-by-step.

The given equation is: \[ 4 \sec \theta - 5 = 0. \]

Step 1: Solve for \( \sec \theta \):
\[ \sec \theta = \frac{5}{4}. \]

Step 2: Use the trigonometric identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta. \]
Substitute \( \sec \theta = \frac{5}{4} \): \[ \left( \frac{5}{4} \right)^2 = 1 + \tan^2 \theta. \] \[ \frac{25}{16} = 1 + \tan^2 \theta. \] \[ \tan^2 \theta = \frac{25}{16} - 1 = \frac{25}{16} - \frac{16}{16} = \frac{9}{16}. \] \[ \tan \theta = \pm \frac{3}{4}. \]

Step 3: Use the reciprocal identity:
\[ \cot \theta = \frac{1}{\tan \theta}. \]
Substitute \( \tan \theta = \frac{3}{4} \) (taking the positive value as standard in this case): \[ \cot \theta = \frac{1}{\frac{3}{4}} = \frac{4}{3}. \]

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Conclusion:

The value of \( \cot \theta \) is: \[ \cot \theta = \frac{4}{3}. \] Quick Tip: Always check the signs of trigonometric functions based on the quadrant when solving such equations.

The perimeter of the sector is given by: \[ Perimeter = 2r + Arc length. \]

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

Substitute \(\theta = 60^\circ\), \(r = 21\), and \(\pi = \frac{22}{7}\): \[ Arc length = \frac{60}{360} \cdot 2 \cdot \frac{22}{7} \cdot 21 = \frac{1}{6} \cdot 2 \cdot \frac{22}{7} \cdot 21 = 22 \, cm. \]

The perimeter is: \[ Perimeter = 2(21) + 22 = 42 + 22 = 64 \, cm. \]

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Conclusion:

The perimeter is 64 cm. Quick Tip: For sector problems, always add the arc length and twice the radius to find the total perimeter.

Given: \[ \angle RJL = 42^\circ, \quad RJ and RL are tangents. \]

In a circle, the angle formed by the tangents at the external point (\(\angle RJL\)) is half the angle subtended by the chord at the center (\(\angle JOL\)): \[ \angle JOL = 2 \cdot \angle RJL. \]

Substitute the given value: \[ \angle JOL = 2 \cdot 42^\circ = 84^\circ. \]

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Conclusion:

The measure of \(\angle JOL\) is \(84^\circ\). Quick Tip: For tangents and angles in circles, the angle subtended at the center is twice the angle between the tangents.

The prime factorisation of \( 2520 \) can be done as follows: \[ 2520 = 2^3 \times 3^2 \times 5 \times 7. \]

Comparing this with the given form \( 2^3 \times 3^a \times b \times 7 \): \[ a = 2 \quad and \quad b = 5. \]

Now calculate \( a + 2b \): \[ a + 2b = 2 + 2 \times 5 = 2 + 10 = 12. \]

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Conclusion:

The value of \( a + 2b \) is \( 12 \). Quick Tip: When comparing prime factorizations, match the powers of each prime number carefully to find the corresponding coefficients.

The given equations are: \[ 1. \ 3x + 4y = 5 \quad and \quad 2. \ 6x + 8y = 7 \]

Rewrite Equation 2 to check if it is a multiple of Equation 1: \[ 6x + 8y = 7 \quad \implies \quad 2(3x + 4y) = 7 \]

Since the constant terms (\(5\) and \(7\)) are not in the same ratio as the coefficients, the lines are not coincident. The coefficients of \(x\) and \(y\) are in the same ratio: \[ \frac{3}{6} = \frac{4}{8}. \]

This implies the lines are parallel.

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Conclusion:

The lines represented by the equations are parallel. Quick Tip: For a system of linear equations, lines are parallel if the ratios of the coefficients of \(x\) and \(y\) are equal but differ in constant terms.

The multiples of 7 between 1 and 40 are: \[ 7, 14, 21, 28, 35. \]

There are 5 multiples of 7, and the total number of tickets is 40. The probability is: \[ P = \frac{Favorable outcomes}{Total outcomes} = \frac{5}{40} = \frac{1}{8}. \]

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Conclusion:

The probability is \(\frac{1}{8}\). Quick Tip: For probability problems, identify the total possible outcomes and favorable outcomes clearly before simplifying the ratio.

To find the LCM of \(28, 44, 132\):
1. Perform prime factorization: \[ 28 = 2^2 \cdot 7, \quad 44 = 2^2 \cdot 11, \quad 132 = 2^2 \cdot 3 \cdot 11. \]

2. Take the highest powers of all prime factors: \[ LCM = 2^2 \cdot 3 \cdot 7 \cdot 11 = 924. \]

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Conclusion:

The LCM of \(28, 44, 132\) is \(924\). Quick Tip: For LCM, take the highest powers of all prime factors common or unique to the given numbers.

The general formula for the \(n\)-th term of an A.P. is: \[ a_n = a + (n-1)d, \]
where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

Here: \[ a = 3, \ d = 3, \ a_n = 111. \]

Substitute into the formula: \[ 111 = 3 + (n-1) \cdot 3. \]

Simplify: \[ 111 - 3 = 3(n-1). \] \[ 108 = 3(n-1). \] \[ n-1 = 36 \quad \implies \quad n = 37. \]

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Conclusion:

The number of terms in the A.P. is \(37\). Quick Tip: For A.P. problems, rearrange the formula for the \(n\)-th term to find the total number of terms.

Let the height of the pole be \( h \) and the length of the shadow be \( x \). Given: \[ \frac{h}{x} = \frac{1}{\sqrt{3}}. \]

From trigonometry: \[ \tan \theta = \frac{h}{x}. \]

Substitute \( \tan \theta = \frac{1}{\sqrt{3}} \): \[ \theta = 30^\circ. \]

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Conclusion:

The angle of elevation of the Sun is \(30^\circ\). Quick Tip: For elevation problems, use the tangent ratio \( \tan \theta = \frac{height of object}{length of shadow} \).

From the empirical relationship between mean, median, and mode: \[ Mean - Median = 3(Mean - Mode). \]

Substitute the given values: \[ 24 - Median = 3(24 - 12). \]

Simplify: \[ 24 - Median = 36. \]
\[ Median = 24 - 12 = 20. \]

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Conclusion:

The median of the data is \(20\). Quick Tip: The empirical formula \( Mean - Median = 3(Mean - Mode) \) is helpful in solving problems involving mean, median, and mode.

Both Assertion (A) and Reason (R) are true. The line \(EF\), being parallel to the parallel sides of the trapezium (\(AB\) and \(DC\)), divides the non-parallel sides proportionally. The Reason (R) provides the correct explanation for Assertion (A). Quick Tip: Use the property of proportionality in trapeziums: A line parallel to the parallel sides divides the non-parallel sides proportionally.

Both Assertion (A) and Reason (R) are true. However, Reason (R) is not the correct explanation for Assertion (A). The degree of a zero polynomial is undefined because it does not have any terms, while the degree of a non-zero constant polynomial is defined as \(0\). Quick Tip: Remember: The degree of a zero polynomial is undefined, while the degree of any non-zero constant polynomial is \(0\).

Let us assume, for the sake of contradiction, that \(\sqrt{3}\) is a rational number. Then it can be expressed as: \[ \sqrt{3} = \frac{p}{q}, \]
where \(p\) and \(q\) are integers, \(q \neq 0\), and \(p\) and \(q\) are coprime (have no common factors other than 1).

Step 1: Square both sides.
\[ 3 = \frac{p^2}{q^2} \quad \implies \quad p^2 = 3q^2. \]

Step 2: Analyze divisibility of \(p\).

Since \(p^2\) is divisible by 3, it follows that \(p\) must also be divisible by 3 (property of prime numbers). Let: \[ p = 3a, \quad where a is an integer. \]

Step 3: Substitute \(p = 3a\) into the equation.
\[ p^2 = 3q^2 \quad \implies \quad (3a)^2 = 3q^2 \quad \implies \quad 9a^2 = 3q^2 \quad \implies \quad q^2 = 3a^2. \]

Step 4: Analyze divisibility of \(q\).

Since \(q^2\) is divisible by 3, it follows that \(q\) must also be divisible by 3.

Step 5: Contradiction.

If both \(p\) and \(q\) are divisible by 3, then \(p\) and \(q\) are not coprime, which contradicts our assumption that \(\frac{p}{q}\) is in its simplest form.

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Conclusion:

The assumption that \(\sqrt{3}\) is a rational number leads to a contradiction.

Therefore, \(\sqrt{3}\) is an irrational number. Quick Tip: To prove a number is irrational, assume it is rational and derive a contradiction using properties of divisibility.

Expand \((\sqrt{2} + \sqrt{3})^2\): \[ (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2\sqrt{6} = 5 + 2\sqrt{6}. \]

Step 1: Assume, for contradiction, that \(5 + 2\sqrt{6}\) is a rational number.

Let: \[ 5 + 2\sqrt{6} = \frac{a}{b}, \]
where \(a, b\) are integers, and \(b \neq 0\).

Step 2: Rearrange to isolate \(\sqrt{6}\).
\[ 2\sqrt{6} = \frac{a}{b} - 5 \quad \implies \quad \sqrt{6} = \frac{a - 5b}{2b}. \]

Step 3: Analyze rationality.

Since \(a\) and \(b\) are integers, \(\frac{a - 5b}{2b}\) is a rational number. However, it is given that \(\sqrt{6}\) is an irrational number. This leads to a contradiction.

Step 4: Conclude irrationality.

The assumption that \(5 + 2\sqrt{6}\) is rational is incorrect. Therefore: \[ 5 + 2\sqrt{6} = (\sqrt{2} + \sqrt{3})^2 \]
is an irrational number.

Conclusion:
\((\sqrt{2} + \sqrt{3})^2\) is an irrational number. Quick Tip: When proving irrationality, assume rationality, isolate the square root term, and demonstrate that it contradicts the given property of irrationality.

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

Here: \[ S_{14} = 1050, \quad a = 10, \quad n = 14. \]

Substitute the values: \[ \frac{14}{2} \left[2(10) + 13d\right] = 1050. \]

Simplify: \[ 7 \left[20 + 13d\right] = 1050 \quad \implies \quad 20 + 13d = 150 \quad \implies \quad d = 10. \]

Find the 20th term (\(a_{20}\)):

The \(n\)-th term of an A.P. is given by: \[ a_n = a + (n-1)d. \]

For \(n = 20\): \[ a_{20} = 10 + (20-1)10 = 10 + 190 = 200. \]

Find the general \(n\)-th term (\(a_n\)):

Substitute \(a = 10\) and \(d = 10\): \[ a_n = 10 + (n-1)10 = 10n. \]

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Conclusion:

The 20th term is \(200\) and the \(n\)-th term is \(10n\). Quick Tip: Use the formula \(S_n = \frac{n}{2}[2a + (n-1)d]\) to calculate the sum of \(n\) terms and \(a_n = a + (n-1)d\) for specific terms.

The sum of the first \(n\) terms is given by: \[ S_n = \frac{n}{2}(a + l), \]
where \(a = 5\), \(l = 45\), and \(S_n = 400\).

Substitute the values: \[ \frac{n}{2}(5 + 45) = 400. \]

Simplify: \[ \frac{n}{2}(50) = 400 \quad \implies \quad 25n = 400 \quad \implies \quad n = 16. \]

Find the common difference (\(d\)):

The last term of an A.P. is given by: \[ a_n = a + (n-1)d. \]

Substitute \(a_n = 45\), \(a = 5\), and \(n = 16\): \[ 45 = 5 + (16-1)d \quad \implies \quad 45 = 5 + 15d \quad \implies \quad 15d = 40 \quad \implies \quad d = \frac{40}{15} = \frac{8}{3}. \]

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Conclusion:

The number of terms is \(16\) and the common difference is \(\frac{8}{3}\). Quick Tip: For problems involving the sum of an A.P., use \(S_n = \frac{n}{2}(a + l)\) when the first and last terms are given.

Let us simplify the left-hand side (LHS): \[ LHS = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}. \]

Take the LCM of the denominators: \[ LHS = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}. \]

Simplify the numerator: \[ (1 + \cos \theta)^2 = 1 + 2\cos \theta + \cos^2 \theta. \]
Thus: \[ LHS = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}. \]

Use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\): \[ LHS = \frac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}. \]

Simplify further: \[ LHS = \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}. \]

Factor \(2\) from the numerator: \[ LHS = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}. \]

Cancel \(1 + \cos \theta\) from the numerator and denominator (valid as \(1 + \cos \theta \neq 0\)): \[ LHS = \frac{2}{\sin \theta}. \]

Using the reciprocal identity \(\csc \theta = \frac{1}{\sin \theta}\): \[ LHS = 2 \csc \theta. \]

Thus, the left-hand side equals the right-hand side: \[ LHS = RHS. \]

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Conclusion:

The given identity is proven: \[ \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta. \] Quick Tip: For trigonometric proofs: Simplify each term separately before combining them. Use identities like \(\sin^2 \theta + \cos^2 \theta = 1\) and \(\csc \theta = \frac{1}{\sin \theta}\). Factor and cancel common terms carefully.

Let the number of white marbles in the jar be \(x\). The total number of marbles is 54. The probability of selecting a white marble is given by: \[ P(white marbles) = \frac{x}{54}. \]

The total probability of selecting a marble is 1 (since every marble must be blue, green, or white). Therefore: \[ P(blue) + P(green) + P(white) = 1. \]

Substitute the given probabilities for blue and green marbles: \[ \frac{1}{3} + \frac{4}{9} + \frac{x}{54} = 1. \]

Simplify the fractions: \[ \frac{1}{3} = \frac{18}{54}, \quad \frac{4}{9} = \frac{24}{54}. \]

Substitute these into the equation: \[ \frac{18}{54} + \frac{24}{54} + \frac{x}{54} = 1. \]

Combine the fractions: \[ \frac{18 + 24 + x}{54} = 1. \]

Simplify: \[ \frac{42 + x}{54} = 1. \]

Multiply through by 54: \[ 42 + x = 54. \]

Solve for \(x\): \[ x = 54 - 42 = 12. \]


Conclusion:

The number of white marbles in the jar is: 12. Quick Tip: When solving probability problems: \[ P(Total) = 1, and fractions must align with the total sum. \]

Let the width of the river be \(AB\), and let the point \(P\) be on the bridge, \(Q\) be directly below \(P\), and \(A\) and \(B\) be on the two banks of the river. Let \(x\) and \(y\) be the horizontal distances from \(Q\) to \(A\) and \(Q\) to \(B\), respectively.

Step 1: Use \(\tan\) in \(\triangle PAQ\).

In right-angled \(\triangle PAQ\): \[ \tan 30^\circ = \frac{Height}{Base} = \frac{4}{x}. \] \[ \frac{1}{\sqrt{3}} = \frac{4}{x} \quad \implies \quad x = 4\sqrt{3}. \]

Step 2: Use \(\tan\) in \(\triangle PBQ\).

In right-angled \(\triangle PBQ\): \[ \tan 60^\circ = \frac{Height}{Base} = \frac{4}{y}. \] \[ \sqrt{3} = \frac{4}{y} \quad \implies \quad y = \frac{4}{\sqrt{3}}. \]

Step 3: Calculate the total width of the river.

The total width of the river is: \[ AB = x + y = 4\sqrt{3} + \frac{4}{\sqrt{3}}. \]

Rationalize the denominator: \[ \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}. \]

Substitute: \[ AB = 4\sqrt{3} + \frac{4\sqrt{3}}{3}. \]

Combine terms: \[ AB = \frac{12\sqrt{3} + 4\sqrt{3}}{3} = \frac{16\sqrt{3}}{3}. \]

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Conclusion:

The width of the river is \(\frac{16\sqrt{3}}{3} \, m\).


Diagram:


\begin{tikzpicture[scale=1]
% Draw the triangle
\draw[thick] (0,4) -- (-2,0) -- (2,0) -- cycle;

% Add the bridge and labels
\draw[dashed] (0,4) -- (0,0);
\node at (0,4.2) [above] {\(P\);
\node at (0,-0.2) [below] {\(Q\);
\node at (-2,-0.2) [below] {\(A\);
\node at (2,-0.2) [below] {\(B\);

% Add angles and height
\draw (0,4) -- (-2,0);
\node at (-0.3,3.2) {\(30^\circ\);
\draw (0,4) -- (2,0);
\node at (0.3,3.2) {\(60^\circ\);
\node at (0.2,2.2) [right] {4 m;
\end{tikzpicture Quick Tip: When solving height and distance problems, use trigonometric ratios like \(\tan\) to relate angles, heights, and distances, and always rationalize denominators where necessary.

From the given information: \[ \triangle FEC \cong \triangle GDB. \]

This implies: \[ \angle 3 = \angle 4. \]

In \(\triangle ABC\): \[ \angle 3 = \angle 4 \quad (as given). \]
Thus: \[ AB = AC \quad (i). \]

In \(\triangle ADE\): \[ \angle 1 = \angle 2 \quad (as given). \]
Thus: \[ AD = AE \quad (ii). \]

Now, divide equation (ii) by equation (i): \[ \frac{AD}{AB} = \frac{AE}{AC}. \]

This implies: \[ DE \parallel BC \quad (By Basic Proportionality Theorem). \]

Since: \[ \angle 1 = \angle 2 \quad and \quad \angle 3 = \angle 4, \]

it follows that: \[ \triangle ADE \sim \triangle ABC \quad (by the AA similarity criterion). \]

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Conclusion:
\[ \triangle ADE \sim \triangle ABC. \] Quick Tip: To prove similarity between triangles, look for proportional sides and equal angles. The AA similarity criterion is one of the simplest methods.

To prove \(\triangle ABC \sim \triangle PQR\), extend \(AD\) to \(E\) such that \(AD = DE\), and join \(EC\). Similarly, extend \(PM\) to \(L\) such that \(PM = ML\), and join \(LR\).

1. Since \(\triangle ABD \cong \triangle ECD\) (by construction), we have: \[ AB = EC. \]

2. Similarly, in \(\triangle PQR\), extend \(PM\), and by symmetry, we have: \[ PQ = LR. \]

Now, by the given proportionality condition: \[ \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}. \]

Since \(AD = DE\) and \(PM = ML\), the proportionality extends: \[ \frac{EC}{LR} = \frac{AC}{PR} = \frac{AD}{PM}. \]

Thus: \[ \triangle AEC \sim \triangle PLR \quad (by the proportionality criterion). \]

3. Since \(\triangle AEC \sim \triangle PLR\), we have: \[ \angle BAC = \angle QPR. \]

4. Similarly, \(\angle 1 = \angle 3\) and \(\angle 2 = \angle 4\), proving \(\triangle ABC \sim \triangle PQR\) by the AA similarity criterion.

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Conclusion:
\[ \triangle ABC \sim \triangle PQR. \] Quick Tip: To prove triangle similarity using medians: Extend the median symmetrically and use congruence or similarity of smaller triangles. Check proportionality and equal angles to apply the AA similarity criterion.

The radius of the base of the cylinder is: \[ r = \frac{diameter}{2} = \frac{6}{2} = 3 \, m. \]

The curved surface area (CSA) of the cylindrical part is given by: \[ CSA of cylinder = 2 \pi r h. \]

Substitute \(r = 3 \, m\) and \(h = 3.5 \, m\): \[ CSA of cylinder = 2 \times \frac{22}{7} \times 3 \times 3.5 = 66 \, m^2. \]

The CSA of the conical part is given by: \[ CSA of cone = \pi r l. \]

Substitute \(r = 3 \, m\) and \(l = 4.2 \, m\): \[ CSA of cone = \frac{22}{7} \times 3 \times 4.2 = 39.6 \, m^2. \]

The total area of canvas required is: \[ Total area = CSA of cylinder + CSA of cone. \]

Substitute the values: \[ Total area = 66 + 39.6 = 105.6 \, m^2. \]

The cost of the canvas is calculated as: \[ Cost = Total area \times Rate per m^2. \]

Substitute the values: \[ Cost = 105.6 \times 500 = 52,800 \, ₹. \]

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Conclusion:

The total area of the canvas required is: \[ 105.6 \, m^2. \]
The cost of the canvas is: \[ 52,800 . \] Quick Tip: For structures combining cylinders and cones: Use the CSA formula for each shape. Ensure the radius is consistent across both parts. Add the areas to find the total surface area.

Let the two-digit number be \(10x + y\), where \(x\) is the tens digit and \(y\) is the units digit.

From the problem, the product of the digits is: \[ xy = 14 \quad (i). \]

When 45 is added to the number, the digits are reversed: \[ 10x + y + 45 = 10y + x. \]

Simplify: \[ 9x - 9y = -45 \implies x - y = -5 \quad (ii). \]

From equations (i) and (ii): \[ y - x = 5 \quad (rewrite equation (ii)). \]

Substitute \(y = x + 5\) into equation (i): \[ x(x + 5) = 14. \]

Simplify: \[ x^2 + 5x - 14 = 0. \]

Solve the quadratic equation using factorization: \[ x^2 + 7x - 2x - 14 = 0 \implies (x + 7)(x - 2) = 0. \]

Thus: \[ x = -7 \quad (not valid as \(x\) is a digit), \quad x = 2. \]

Substitute \(x = 2\) into \(y = x + 5\): \[ y = 2 + 5 = 7. \]

The two-digit number is: \[ 10x + y = 10 \times 2 + 7 = 27. \]

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Conclusion:

The required number is: \[ 27. \] Quick Tip: For problems involving two-digit numbers: Represent the number as \(10x + y\), where \(x\) and \(y\) are the digits. Use the given conditions to set up equations. Solve the equations systematically to find the digits.

Let the side of the first square be \(x\) cm.

Then, the side of the second square is: \[ x + 4 \, cm. \]

The area of the first square is: \[ x^2. \]

The area of the second square is: \[ (x + 4)^2. \]

The sum of the areas of the two squares is given as 400 cm\(^2\): \[ x^2 + (x + 4)^2 = 400. \]

Simplify: \[ x^2 + x^2 + 8x + 16 = 400. \]

Combine like terms: \[ 2x^2 + 8x + 16 = 400. \]

Simplify further: \[ 2x^2 + 8x - 384 = 0. \]

Divide through by 2: \[ x^2 + 4x - 192 = 0. \]

Factorize the quadratic equation: \[ (x + 16)(x - 12) = 0. \]

Thus: \[ x = -16 \quad (not valid as side length cannot be negative), \quad x = 12. \]

If \(x = 12\), the sides of the squares are: \[ First square: x = 12 \, cm, \quad Second square: x + 4 = 16 \, cm. \]

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Conclusion:

The sides of the squares are: \[ 12 \, cm and 16 \, cm. \] Quick Tip: For problems involving squares: Use the formula for the area of a square: \(Area = Side^2\). Set up the equation based on the given relationships and solve systematically. Discard negative solutions, as side lengths cannot be negative.

(i) Median Class:

The cumulative frequency is calculated as follows:
\[ \begin{array}{|c|c|c|} \hline \textbf{Time (in seconds)} & \textbf{Number of students (f)} & \textbf{Cumulative Frequency (cf)}
\hline 0 - 20 & 8 & 8
\hline 20 - 40 & 10 & 18
\hline 40 - 60 & 13 & 31
\hline 60 - 80 & 6 & 37
\hline 80 - 100 & 3 & 40
\hline \end{array} \]
The total number of students is 40. Since \( N/2 = 20 \), the median class is \( 40 - 60 \).

(ii) Mean Time:

The table for \( x_i \) (midpoints) and \( f_i x_i \) is as follows:
\[ \begin{array}{|c|c|c|c|} \hline \textbf{Time (in seconds)} & \textbf{Midpoint} (x_i) & \textbf{Number of students (f)} & \textbf{\( f_i x_i \)}
\hline 0 - 20 & 10 & 8 & 80
\hline 20 - 40 & 30 & 10 & 300
\hline 40 - 60 & 50 & 13 & 650
\hline 60 - 80 & 70 & 6 & 420
\hline 80 - 100 & 90 & 3 & 270
\hline \end{array} \] \[ Mean = \frac{\sum f_i x_i}{\sum f_i} = \frac{1720}{40} = 43. \]

(ii) Mode:

The modal class is \( 40 - 60 \). Using the formula:
\[ Mode = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]
where \( L = 40 \), \( f_1 = 13 \), \( f_0 = 10 \), \( f_2 = 6 \), \( h = 20 \):
\[ Mode = 40 + \left(\frac{13 - 10}{2(13) - 10 - 6}\right) \times 20 = 40 + \left(\frac{3}{26 - 16}\right) \times 20 = 40 + 6 = 46. \]

(iii) Students Taking Less than 60 Seconds:

From the cumulative frequency table, the number of students taking less than 60 seconds is \( 31 \).

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Conclusion:

(i) Median class: \( 40 - 60 \).

(ii) Mean time: \( 43 \), Mode: \( 46 \).

(iii) Number of students: \( 31 \). Quick Tip: To calculate the mean, use midpoints and summations. The modal class is identified as the class with the highest frequency.

(i) Formulate the equations:

Let the number of children be \(x\) and the number of adults be \(y\). The given conditions can be written as:
\[ x + y = 300 \quad \cdots (i) \] \[ 150x + 250y = 55000 \quad \cdots (ii) \]

(ii) Solve for the number of children and adults:

(a) From equations (i) and (ii), solve for \(x\):

Substitute \(y = 300 - x\) into equation (ii):
\[ 150x + 250(300 - x) = 55000. \]
Simplify:
\[ 150x + 75000 - 250x = 55000. \] \[ -100x + 75000 = 55000 \quad \implies \quad -100x = -20000 \quad \implies \quad x = 200. \]
Therefore, the number of children is \(x = 200\).


(b) Substituting \(x = 200\) into equation (i):
\[ y = 300 - 200 = 100. \]
Therefore, the number of adults is \(y = 100\).


(iii) Calculate the amount collected if 250 children and 100 adults visit the park:
\[ Amount collected = 150 \times 250 + 250 \times 100. \] \[ Amount collected = 37500 + 25000 = 62500. \]

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Conclusion:

(i) The algebraic equations are \(x + y = 300\) and \(150x + 250y = 55000\).

(ii) Number of children: \(200\), Number of adults: \(100\).

(iii) Total amount collected: 62500. Quick Tip: Use substitution or elimination methods for solving linear equations systematically. Ensure accurate substitution and simplification in word problems.

(i) Coordinates of the vertices of \( \Delta PQR \):

From the figure, the coordinates are:
\( P(4, 6), \ Q(3, 2), \ R(6, 5) \).

(ii) Find distances and coordinates:

(a) Distance between \( P \) and \( Q \): \[ PQ = \sqrt{(4 - 3)^2 + (6 - 2)^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}. \]
Distance between \( Q \) and \( R \): \[ QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}. \]

(b) The coordinates of the point dividing \( PR \) in the ratio \( 2:1 \): \[ Using section formula: \left(\frac{2x_2 + 1x_1}{2 + 1}, \frac{2y_2 + 1y_1}{2 + 1}\right). \]
Substitute \( P(4, 6) \) and \( R(6, 5) \): \[ \left(\frac{2 \times 6 + 1 \times 4}{3}, \frac{2 \times 5 + 1 \times 6}{3}\right) = \left(\frac{12 + 4}{3}, \frac{10 + 6}{3}\right) = \left(\frac{16}{3}, \frac{16}{3}\right). \]

(iii) Check if \( \Delta PQR \) is an isosceles triangle:

Distance \( PR \): \[ PR = \sqrt{(4 - 6)^2 + (6 - 5)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}. \]
Since \( PQ \neq QR \neq PR \), \( \Delta PQR \) is not an isosceles triangle.

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Conclusion:

(i) Coordinates of the vertices: \( P(4, 6), Q(3, 2), R(6, 5) \).

(ii) (a) \( PQ = \sqrt{17}, QR = \sqrt{18} \).

(b) The coordinates of the point dividing \( PR \) are \( \left(\frac{16}{3}, \frac{16}{3}\right) \).

(iii) \( \Delta PQR \) is not isosceles. Quick Tip: To determine the nature of a triangle, compute the lengths of all sides using the distance formula and compare them.