UP Board Class 10 Mathematics Question Paper 2024 (Code 822 HW) with Solutions

Step 1: Use the relation between LCM and HCF.

The formula is: \[ LCM \times HCF = Product of the numbers \]

Step 2: Substitute the given values.
\[ 84 \times HCF = 12 \times 21 \] \[ 84 \times HCF = 252 \] \[ HCF = \frac{252}{84} = 3 \]

Step 3: Verify.

Wait — this seems incorrect; checking again:
The correct relation gives HCF = 3? But actual HCF of (12, 21) is 3, not 6.
LCM(12, 21) = 84 → yes.
Hence: \[ HCF = 3 \]

Step 4: Conclusion.

The HCF is 3.
Quick Tip: Use the relation \( LCM \times HCF = Product of the numbers \) to easily find one when the other is known.

Step 1: Total number of marbles.
\[ 6 + 4 + 8 = 18 \]

Step 2: Number of blue marbles = 6.


Step 3: Probability formula.
\[ P(blue) = \frac{favorable outcomes}{total outcomes} = \frac{6}{18} = \frac{1}{3} \]

Step 4: Conclusion.

The probability of drawing a blue marble is \( \frac{1}{3} \).
Quick Tip: Always divide favorable outcomes by total outcomes when calculating probability.

Step 1: Use the empirical relation.
\[ Mode = 3 \times Median - 2 \times Mean \]

Step 2: Substitute values.
\[ 42 = 3(38.1) - 2 \times Mean \] \[ 42 = 114.3 - 2 \times Mean \] \[ 2 \times Mean = 114.3 - 42 = 72.3 \] \[ Mean = 36.15 \]

Wait, this gives 36.15 — but that’s *less* than median, which seems inconsistent with the given pattern (mode > median).
Check relation again:
Correct formula → \( Mode = 3 \times Median - 2 \times Mean \).
Yes, solving for mean: \[ Mean = \frac{3 \times Median - Mode}{2} = \frac{3(38.1) - 42}{2} = \frac{114.3 - 42}{2} = \frac{72.3}{2} = 36.15 \]

Step 3: Conclusion.

The mean is \( 36.15 \). (Hence correct option (B).) Quick Tip: Remember the empirical formula: \( Mode = 3 \times Median - 2 \times Mean \).

Step 1: Identify the highest frequency.

The maximum frequency is 18.

Step 2: Corresponding class interval.

The class interval corresponding to the highest frequency (18) is 15–20.

Step 3: Conclusion.

Hence, the modal class is 15–20.
Quick Tip: The modal class is the class interval with the highest frequency.

Step 1: Total cards = 52.


Step 2: Face cards in a deck.

Each suit (hearts, spades, clubs, diamonds) has 3 face cards (Jack, Queen, King).
Total = \( 4 \times 3 = 12 \).

Step 3: Probability formula.
\[ P(face card) = \frac{12}{52} = \frac{3}{13} \]

Step 4: Conclusion.

The probability of drawing a face card is \( \frac{3}{13} \).
Quick Tip: In a standard deck, there are 12 face cards (J, Q, K of each suit).

Step 1: Recall the definitions.

A rational number is one that can be expressed as \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).

An irrational number cannot be expressed as a simple fraction (for example, \( \sqrt{2}, \pi, e \)).


Step 2: Consider the difference.

Let the rational number be \( r \) and the irrational number be \( i \).

Then, the difference is \( r - i \).


Step 3: Analyze the result.

Suppose \( r - i \) were rational. Then, rearranging: \[ i = r - (rational number) \]
Since the subtraction of two rational numbers is always rational, this would make \( i \) rational — which is a contradiction.


Step 4: Conclusion.

Therefore, the difference of a rational and an irrational number is always irrational.
Quick Tip: Rational ± irrational = irrational (always). The rational part never cancels the non-repeating, non-terminating nature of the irrational part.

Step 1: Recall the identity used.
\[ (a + b)(a - b) = a^2 - b^2 \]

Step 2: Apply it to option (C).
\[ (\sqrt{5} + \sqrt{7})(\sqrt{5} - \sqrt{7}) = (\sqrt{5})^2 - (\sqrt{7})^2 = 5 - 7 = -2 \]

Step 3: Identify the nature of the result.

The result is \(-2\), which is a rational number.


Step 4: Verify other options.

(A) \( \frac{\sqrt{3}}{\sqrt{5}} = \sqrt{\frac{3}{5}} \) — irrational because the square root of a non-perfect fraction is irrational.

(B) \( \sqrt{2} \times \sqrt{7} = \sqrt{14} \) — irrational.

(D) \( \sqrt{12} = 2\sqrt{3} \) — also irrational.


Step 5: Conclusion.

Hence, the only rational result is from option (C).
Quick Tip: Use the identity \( (a+b)(a-b)=a^2-b^2 \) to simplify radical expressions quickly. If the radicals cancel completely, the result is rational.

Step 1: Recall the distance formula.

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the given points.

Let \( (x_1, y_1) = (a, b) \) and \( (x_2, y_2) = (b, -a) \).
\[ Distance = \sqrt{(b - a)^2 + (-a - b)^2} \]

Step 3: Expand both squares.
\[ (b - a)^2 = b^2 - 2ab + a^2 \] \[ (-a - b)^2 = a^2 + 2ab + b^2 \]

Step 4: Add them.
\[ (b - a)^2 + (-a - b)^2 = (b^2 - 2ab + a^2) + (a^2 + 2ab + b^2) = 2a^2 + 2b^2 \]

Step 5: Take square root.
\[ Distance = \sqrt{2a^2 + 2b^2} \]

Step 6: Conclusion.

The required distance is \( \boxed{\sqrt{2a^2 + 2b^2}} \).
Quick Tip: In coordinate geometry, always expand step-by-step using the distance formula to avoid sign mistakes, especially with negative coordinates.

Step 1: Recall the formula for the sum of zeroes.

For a quadratic polynomial \( ax^2 + bx + c \), the sum of its zeroes is given by \[ \alpha + \beta = -\frac{b}{a} \]

Step 2: Identify coefficients.

In \( 4x^2 - 4x + 1 \), we have: \[ a = 4, \quad b = -4, \quad c = 1 \]

Step 3: Apply the formula.
\[ \alpha + \beta = -\frac{-4}{4} = \frac{4}{4} = 1 \]

Step 4: Conclusion.

Hence, the sum of the zeroes of the given quadratic polynomial is \( 1 \).
Quick Tip: For any quadratic equation \( ax^2 + bx + c \), the sum of roots \( = -\frac{b}{a} \) and the product \( = \frac{c}{a} \).

Step 1: Express the equations in standard form.

Equation (1): \( x - y - 8 = 0 \) → \( a_1 = 1, b_1 = -1, c_1 = -8 \)

Equation (2): \( 3x - 3y - 16 = 0 \) → \( a_2 = 3, b_2 = -3, c_2 = -16 \)


Step 2: Compare ratios.
\[ \frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2} \]

Step 3: Analyze the ratios.

We have \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
This condition means the pair of lines are parallel and distinct.

Step 4: Conclusion.

Hence, there is no solution to this pair of linear equations.
Quick Tip: For two linear equations, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), then the lines are parallel and have no solution.

Step 1: Substitute the known root in the equation.

Given that \( x = 2 \) is a root of \( x^2 - kx - 8 = 0 \), substitute \( x = 2 \): \[ (2)^2 - k(2) - 8 = 0 \]

Step 2: Simplify.
\[ 4 - 2k - 8 = 0 \] \[ -2k - 4 = 0 \] \[ -2k = 4 \] \[ k = -2 \]

Wait—check the signs: The given equation is \( x^2 - kx - 8 = 0 \). If \( x = 2 \): \[ 4 - 2k - 8 = 0 \Rightarrow -2k = 4 \Rightarrow k = -2 \]
Hence, the correct answer is \( -2 \).

Step 3: Conclusion.

The value of \( k \) is \( \boxed{-2} \).
Quick Tip: Always substitute the root directly into the equation to find unknown coefficients. Be careful with negative signs.

Step 1: Identify the first term and common difference.

First term, \( a = 10 \)

Common difference, \( d = 7 - 10 = -3 \)


Step 2: Recall the nth term formula.
\[ a_n = a + (n - 1)d \]

Step 3: Substitute the given values.
\[ a_{20} = 10 + (20 - 1)(-3) \] \[ a_{20} = 10 + 19(-3) = 10 - 57 = -47 \]

Step 4: Conclusion.

Hence, the 20th term of the A.P. is \( \boxed{-47} \).
Quick Tip: In an A.P., if the common difference is negative, terms will decrease successively. Use \( a_n = a + (n-1)d \) to find any term.

Step 1: Concept used.

A tangent to a circle is always perpendicular to the radius at the point of contact. Thus, \(\triangle OPQ\) is a right-angled triangle at \( P \).


Step 2: Apply Pythagoras theorem.
\[ OQ^2 = OP^2 + PQ^2 \] \[ PQ^2 = OQ^2 - OP^2 = 12^2 - 10^2 = 144 - 100 = 44 \] \[ PQ = \sqrt{44} = 2\sqrt{11} cm \]

Step 3: Verify the options.

Hence, \( PQ = 2\sqrt{11} \) cm, which is not in simplified radical form equal to \( 3\sqrt{5} \). Wait — since \( \sqrt{44} = 2\sqrt{11} \), option (C) is correct, not (D).

Step 4: Conclusion.

The correct length of \( PQ \) is \( \boxed{2\sqrt{11}\ cm} \).
Quick Tip: For tangents, always use \( PQ^2 = OQ^2 - OP^2 \). The tangent, radius, and line through the centre form a right-angled triangle.

Step 1: Find the side of one cube.

Volume of cube \( = a^3 = 8 \Rightarrow a = 2 cm \).


Step 2: When two cubes are joined end-to-end.

The cuboid formed will have dimensions:
Length = \( 2a = 4 cm \), Breadth = \( a = 2 cm \), Height = \( a = 2 cm \).


Step 3: Use surface area formula.
\[ Surface Area = 2(lb + bh + hl) \] \[ = 2(4\times2 + 2\times2 + 2\times4) = 2(8 + 4 + 8) = 2(20) = 40 cm^2 \]

Wait — but when two cubes are joined, one face (area = \(2\times2=4\)) of each cube becomes internal and is not exposed.
Hence, total area = \( 48 - 4 = 44 cm^2 \).


Step 4: Conclusion.

Therefore, the surface area of the cuboid is \( \boxed{44 cm^2} \).
Quick Tip: When solids are joined, always subtract the area of the common face(s) to avoid double counting.

Step 1: Recall the formula.
\[ Area of sector = \frac{\theta}{360} \times \pi r^2 \]

Step 2: Substitute known values.
\[ 154 = \frac{\theta}{360} \times \frac{22}{7} \times 14^2 \] \[ 154 = \frac{\theta}{360} \times \frac{22}{7} \times 196 \] \[ 154 = \frac{\theta}{360} \times 616 \]

Step 3: Simplify.
\[ \theta = \frac{154 \times 360}{616} = \frac{55440}{616} = 90 \]

Wait — recalculate carefully: \[ \frac{616}{4} = 154 \Rightarrow \theta = 360/4 = 90° \]
Hence, the angle is \( \boxed{90°} \).


Step 4: Conclusion.

The required angle of the sector is \( 90° \).
Quick Tip: Use \( Area = \frac{\theta}{360} \pi r^2 \). Cross-check using simple ratios — a 90° sector is one-fourth of a circle.

Step 1: Recall the identity.
\[ 2\sin A \cos A = \sin(2A) \]

Step 2: Substitute \( A = 30° \).
\[ 2\sin30° \cos30° = \sin(60°) \]

Step 3: Value of \( \sin60° \).
\[ \sin60° = \frac{\sqrt{3}}{2} \]

Step 4: Conclusion.

Hence, \( 2\sin30°\cos30° = \boxed{\frac{\sqrt{3}}{2}} \).
Quick Tip: Always remember: \( 2\sin A \cos A = \sin(2A) \). It helps simplify many trigonometric expressions quickly.

Step 1: Recall the identity.
\[ \sin^2\theta + \cos^2\theta = 1 \]

Step 2: Find \( \cos\theta \).
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

Step 3: Find \( \tan\theta \).
\[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} \]

Step 4: Conclusion.

Hence, the value of \( \tan\theta \) is \( \boxed{\dfrac{3}{\sqrt{7}}} \).
Quick Tip: When \( \sin\theta \) is given, use \( \cos\theta = \sqrt{1 - \sin^2\theta} \) and \( \tan\theta = \frac{\sin\theta}{\cos\theta} \).

Step 1: Write in terms of sine and cosine.
\[ (\csc A + \cot A)(1 - \cos A) = \left(\frac{1}{\sin A} + \frac{\cos A}{\sin A}\right)(1 - \cos A) \] \[ = \frac{(1 + \cos A)(1 - \cos A)}{\sin A} \]

Step 2: Simplify the numerator using an identity.
\[ (1 + \cos A)(1 - \cos A) = 1 - \cos^2 A = \sin^2 A \]

Step 3: Substitute back.
\[ (\csc A + \cot A)(1 - \cos A) = \frac{\sin^2 A}{\sin A} = \sin A \]

Step 4: Conclusion.

Hence, the value of \( (\csc A + \cot A)(1 - \cos A) \) is \( \boxed{\sin A} \).
Quick Tip: Always convert trigonometric terms into sine and cosine to simplify easily. Use \( 1 - \cos^2A = \sin^2A \) wherever applicable.

Step 1: Write cot in terms of tan.
\[ \cot A = \frac{1}{\tan A} \]

Step 2: Substitute in the expression.
\[ \dfrac{1 - \tan^2 A}{1 - \cot^2 A} = \dfrac{1 - \tan^2 A}{1 - \dfrac{1}{\tan^2 A}} \] \[ = \dfrac{1 - \tan^2 A}{\dfrac{\tan^2 A - 1}{\tan^2 A}} = \dfrac{(1 - \tan^2 A) \times \tan^2 A}{\tan^2 A - 1} \]

Step 3: Simplify.
\[ 1 - \tan^2 A = -(\tan^2 A - 1) \]
Substitute this: \[ = \dfrac{- (\tan^2 A - 1) \tan^2 A}{\tan^2 A - 1} = -\tan^2 A \]

Wait — simplifying numerators cancels \((\tan^2 A - 1)\), giving: \[ \dfrac{1 - \tan^2 A}{1 - \cot^2 A} = -1 \]

Step 4: Conclusion.

The value of the expression is \( \boxed{-1} \).
Quick Tip: Always convert all trigonometric functions to a single ratio (like tan or sin) to simplify complex fractions.

Step 1: Concept used (Basic Proportionality Theorem).

According to the BPT (Thales’ Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those sides in the same ratio.


Step 2: Apply BPT.
\[ \frac{QS}{SR} = \frac{PT}{TR} \]

Step 3: Substitute given values.
\[ \frac{3}{1.5} = \frac{2.8}{TR} \] \[ 2 = \frac{2.8}{TR} \]

Step 4: Solve for \( TR \).
\[ TR = \frac{2.8}{2} = 1.4 cm \]

Step 5: Conclusion.

Therefore, the value of \( TR \) is \( \boxed{1.4\ cm} \).
Quick Tip: In problems involving parallel lines in triangles, use the Basic Proportionality Theorem: ratios of corresponding sides are equal.

Step 1: Assume the contrary.

Let us assume, to the contrary, that \(\sqrt{2}\) is a rational number. Then it can be expressed as \[ \sqrt{2} = \frac{p}{q} \]
where \( p \) and \( q \) are co-prime integers (having no common factors other than 1), and \( q \neq 0 \).


Step 2: Square both sides.
\[ 2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2 \]

Step 3: Analyze divisibility.

Since \( p^2 \) is even, \( p \) must also be even. Let \( p = 2r \).

Step 4: Substitute and simplify.
\[ (2r)^2 = 2q^2 \Rightarrow 4r^2 = 2q^2 \Rightarrow q^2 = 2r^2 \]
Thus, \( q^2 \) is also even, and therefore \( q \) is even.


Step 5: Contradiction.

If both \( p \) and \( q \) are even, they have a common factor 2, contradicting the assumption that \( p \) and \( q \) are co-prime.


Step 6: Conclusion.

Hence, our assumption is false, and therefore \( \sqrt{2} \) is an irrational number.
Quick Tip: To prove irrationality, always start by assuming the opposite and reach a contradiction using the properties of even and odd numbers.

Step 1: Start with LHS.
\[ LHS = \frac{1 + \sec A}{\sec A} = \frac{1}{\sec A} + 1 = \cos A + 1 \]

Step 2: Simplify the RHS.
\[ RHS = \frac{\sin^2 A}{1 - \cos A} \]
Use the identity \( \sin^2 A = 1 - \cos^2 A \): \[ RHS = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A \]

Step 3: Compare both sides.
\[ LHS = RHS = 1 + \cos A \]

Step 4: Conclusion.

Hence, \[ \boxed{\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}} \] Quick Tip: Use trigonometric identities like \( \sin^2A = 1 - \cos^2A \) and cancel common terms carefully to prove equalities.

Step 1: Given data.

Radius \( r = 21 \) cm, central angle \( \theta = 90° \).


Step 2: Area of the sector.
\[ Area of sector = \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 21^2 \] \[ = \frac{1}{4} \times \frac{22}{7} \times 441 = \frac{22 \times 63}{4} = 346.5 cm^2 \]

Step 3: Area of the triangle formed by the two radii.
\[ Area of triangle = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 21^2 \times \sin90° = \frac{1}{2} \times 441 = 220.5 cm^2 \]

Step 4: Area of minor segment.
\[ Area of segment = Area of sector - Area of triangle \] \[ = 346.5 - 220.5 = 126 cm^2 \]

Step 5: Conclusion.

Hence, the area of the minor segment is \( \boxed{126\ cm^2} \).
Quick Tip: Area of a segment = (Area of sector) – (Area of triangle). For a 90° angle, the sector is one-fourth of the circle.

Step 1: Identify the modal class.

The highest frequency is 10 for the class interval 30–35.
So, the modal class is 30–35.


Step 2: Use the formula for mode.
\[ Mode = l + \frac{(f_1 - f_0)}{2f_1 - f_0 - f_2} \times h \]
where \( l = 30 \), \( f_1 = 10 \), \( f_0 = 9 \), \( f_2 = 3 \), and \( h = 5 \).


Step 3: Substitute the values.
\[ Mode = 30 + \frac{(10 - 9)}{2(10) - 9 - 3} \times 5 \] \[ = 30 + \frac{1}{20 - 12} \times 5 = 30 + \frac{1}{8} \times 5 = 30 + 0.625 = 30.625 \]

Step 4: Conclusion.

Hence, the mode of the data is \( \boxed{30.625} \).
Quick Tip: The modal class is always the one with the highest frequency. Use the mode formula carefully with correct substitution.

Step 1: Recall the section formula.

If a point \( P(x, y) \) divides the line segment joining \( A(x_1, y_1) \) and \( B(x_2, y_2) \) internally in the ratio \( m : n \), then \[ P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \]

Step 2: Substitute the given values.
\[ x_1 = 4, \; y_1 = -3, \; x_2 = 8, \; y_2 = 5, \; m = 3, \; n = 1 \]

Step 3: Find the coordinates.
\[ x = \frac{(3)(8) + (1)(4)}{3 + 1} = \frac{24 + 4}{4} = 7 \] \[ y = \frac{(3)(5) + (1)(-3)}{3 + 1} = \frac{15 - 3}{4} = 3 \]

Step 4: Conclusion.

Hence, the coordinates of the required point are \( \boxed{(7, 3)} \).
Quick Tip: Always substitute carefully in the section formula: \( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \).

Step 1: Recall the distance formula.
\[ Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the given values.
\[ 10 = \sqrt{(13 - 5)^2 + (y - (-3))^2} \] \[ 10 = \sqrt{8^2 + (y + 3)^2} \]

Step 3: Simplify.
\[ 10 = \sqrt{64 + (y + 3)^2} \] \[ 100 = 64 + (y + 3)^2 \] \[ (y + 3)^2 = 36 \]

Step 4: Solve for \( y \).
\[ y + 3 = \pm 6 \]
Case 1: \( y + 3 = 6 \Rightarrow y = 3 \)

Case 2: \( y + 3 = -6 \Rightarrow y = -9 \)


Step 5: Conclusion.

Hence, the values of \( y \) are \( \boxed{3 and -9} \).
Quick Tip: When using the distance formula, always square both sides to eliminate the square root and solve for the variable.

Step 1: Given polynomial.
\[ p(x) = 3x^2 - x - 4 \]
Here, \( a = 3, \; b = -1, \; c = -4 \).


Step 2: Use the quadratic formula to find the zeroes.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substitute the values: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)} = \frac{1 \pm \sqrt{1 + 48}}{6} = \frac{1 \pm \sqrt{49}}{6} \] \[ x = \frac{1 \pm 7}{6} \]

Step 3: Calculate the two roots.
\[ x_1 = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3} \] \[ x_2 = \frac{1 - 7}{6} = \frac{-6}{6} = -1 \]

Step 4: Verify the relationships between the zeroes and coefficients.

Sum of zeroes \( = x_1 + x_2 = \frac{4}{3} - 1 = \frac{1}{3} \).

Product of zeroes \( = x_1 \times x_2 = \frac{4}{3} \times (-1) = -\frac{4}{3} \).


Now, \[ -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}, \quad \frac{c}{a} = \frac{-4}{3} \]
Hence verified: \[ Sum of zeroes = -\frac{b}{a}, \quad Product of zeroes = \frac{c}{a} \]

Step 5: Conclusion.

The zeroes are \( \boxed{\frac{4}{3} and -1} \), and the relationship between zeroes and coefficients is verified.
Quick Tip: Always verify the sum and product of zeroes using the standard relations \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \).

Step 1: Eliminate decimals for simplicity.

Multiply both equations by 10: \[ 2x + 3y = 13 \quad (i) \] \[ 4x - 5y = -7 \quad (ii) \]

Step 2: Use the elimination method.

Multiply (i) by 2 to make the coefficients of \( x \) equal: \[ 4x + 6y = 26 \quad (iii) \]
Now subtract (ii) from (iii): \[ (4x + 6y) - (4x - 5y) = 26 - (-7) \] \[ 4x + 6y - 4x + 5y = 33 \] \[ 11y = 33 \Rightarrow y = 3 \]

Step 3: Substitute \( y = 3 \) in equation (i).
\[ 2x + 3(3) = 13 \] \[ 2x + 9 = 13 \Rightarrow 2x = 4 \Rightarrow x = 2 \]

Step 4: Conclusion.

Hence, the solution of the given pair of linear equations is \[ \boxed{x = 2, \; y = 3} \] Quick Tip: Always remove decimals first when solving equations — it simplifies calculations and reduces chances of error.

Step 1: Construction and given information.

Let \( P \) be an external point from which two tangents \( PA \) and \( PB \) are drawn to a circle with centre \( O \).
We need to prove that \( PA = PB \).


Step 2: Join \( OA \) and \( OB \).

Since \( OA \) and \( OB \) are radii, and \( PA \) and \( PB \) are tangents, \[ OA \perp PA \quad and \quad OB \perp PB \]

Step 3: Consider triangles \( \triangle OAP \) and \( \triangle OBP \).

In both triangles: \[ OA = OB \quad (radii of the same circle) \] \[ OP = OP \quad (common side) \] \[ \angle OAP = \angle OBP = 90^\circ \]

Step 4: Apply RHS congruence criterion.

By the RHS congruence rule, \[ \triangle OAP \cong \triangle OBP \]

Step 5: Conclusion.

Hence, \( PA = PB \) (corresponding sides of congruent triangles). \[ \boxed{PA = PB} \] Quick Tip: Tangents drawn from an external point to a circle are equal in length — always prove this using RHS congruence.

Step 1: Construction and concept.

In \( \triangle ABC \), draw \( AD \) such that \( \angle ADC = \angle BAC \).
We need to prove: \( CA^2 = CB \times CD \).


Step 2: Use the property of similar triangles.

Since \( \angle ADC = \angle BAC \) and \( \angle ACD = \angle ACB \) (common), \[ \triangle CAD \sim \triangle CBA \]

Step 3: Write the ratio of corresponding sides.
\[ \frac{CA}{CB} = \frac{CD}{CA} \]

Step 4: Cross multiply.
\[ CA^2 = CB \times CD \]

Step 5: Conclusion.

Hence proved that \( \boxed{CA^2 = CB \times CD} \).
Quick Tip: Whenever two triangles have two equal angles, they are similar — use side ratios to derive relationships between sides.

Step 1: Calculate cumulative frequency.


\begin{tabular{|c|c|c|
\hline
Class Interval & Frequency (f) & Cumulative Frequency (cf)

\hline
15–20 & 14 & 14

20–25 & 56 & 70

25–30 & 60 & 130

30–35 & 86 & 216

35–40 & 74 & 290

40–45 & 62 & 352

45–50 & 48 & 400

\hline
\end{tabular

Step 2: Identify median class.

Total frequency \( N = 400 \). \[ \frac{N}{2} = 200 \]
The cumulative frequency just greater than 200 is 216, so the median class is \( 30–35 \).


Step 3: Use the median formula.
\[ Median = l + \left(\frac{\frac{N}{2} - cf_{before}}{f}\right) \times h \]
Here, \( l = 30, \; cf_{before} = 130, \; f = 86, \; h = 5 \).


Step 4: Substitute the values.
\[ Median = 30 + \left(\frac{200 - 130}{86}\right) \times 5 = 30 + \frac{70}{86} \times 5 \] \[ = 30 + 4.07 = 34.07 \]

Step 5: Conclusion.

Hence, the median of the data is \( \boxed{34.07} \).
Quick Tip: To find the median class, locate where \( \frac{N}{2} \) lies in the cumulative frequency table.

Step 1: Write the sample space.

When a die is thrown, the sample space is \[ S = \{1, 2, 3, 4, 5, 6\} \]
Total outcomes \( = 6 \).


Step 2: (i) Probability of getting a prime number.

Prime numbers between 1 and 6 are \( \{2, 3, 5\} \). \[ P(prime) = \frac{3}{6} = \frac{1}{2} \]

Step 3: (ii) Probability of getting a number lying between 2 and 6.

Numbers between 2 and 6 are \( \{3, 4, 5\} \). \[ P(number between 2 and 6) = \frac{3}{6} = \frac{1}{2} \]

Step 4: Conclusion.

(i) Probability of a prime number = \( \boxed{\frac{1}{2}} \)

(ii) Probability of a number between 2 and 6 = \( \boxed{\frac{1}{2}} \)
Quick Tip: Always write the sample space first. Each outcome of a fair die is equally likely, so probability = \(\frac{favourable outcomes}{total outcomes}\).

Step 1: Recall the formula for the sum of n terms of an A.P.
\[ S_n = \frac{n}{2} [2a + (n - 1)d] \]

Step 2: Use the given information.

For \( n = 8 \): \[ S_8 = 64 \Rightarrow 64 = \frac{8}{2} [2a + 7d] \Rightarrow 4(2a + 7d) = 64 \] \[ 2a + 7d = 16 \quad (i) \]

For \( n = 17 \): \[ S_{17} = 289 \Rightarrow 289 = \frac{17}{2} [2a + 16d] \] \[ 34a + 272d = 578 \Rightarrow 2a + 16d = 34 \quad (ii) \]

Step 3: Solve equations (i) and (ii).

From (i): \( 2a + 7d = 16 \)
From (ii): \( 2a + 16d = 34 \)
Subtract (i) from (ii): \[ (2a + 16d) - (2a + 7d) = 34 - 16 \] \[ 9d = 18 \Rightarrow d = 2 \]

Step 4: Substitute \( d = 2 \) in (i).
\[ 2a + 7(2) = 16 \Rightarrow 2a + 14 = 16 \Rightarrow 2a = 2 \Rightarrow a = 1 \]

Step 5: Conclusion.

The first term \( a = 1 \) and the common difference \( d = 2 \). \[ \boxed{a = 1, \; d = 2} \] Quick Tip: To find unknown terms in an A.P., use the sum formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \) for different values of \( n \) to form simultaneous equations.

Step 1: Let the speed of the train be \( x \) km/h.

Distance \( = 180 \) km.
Time taken \( = \frac{180}{x} \) hours.

Step 2: Write the equation for the second condition.

If the speed is 5 km/h more, then the time taken is \( \frac{180}{x + 5} \) hours.
According to the question: \[ \frac{180}{x} - \frac{180}{x + 5} = \frac{1}{2} \]

Step 3: Simplify.
\[ 180 \left(\frac{(x + 5) - x}{x(x + 5)}\right) = \frac{1}{2} \] \[ 180 \times \frac{5}{x(x + 5)} = \frac{1}{2} \] \[ \frac{900}{x(x + 5)} = \frac{1}{2} \] \[ x(x + 5) = 1800 \] \[ x^2 + 5x - 1800 = 0 \]

Step 4: Solve the quadratic equation.
\[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1800)}}{2} \] \[ x = \frac{-5 \pm \sqrt{25 + 7200}}{2} = \frac{-5 \pm \sqrt{7225}}{2} \] \[ x = \frac{-5 \pm 85}{2} \]
Taking the positive value, \[ x = \frac{80}{2} = 40 \]

Step 5: Conclusion.

The speed of the train is \( \boxed{40\ km/h} \).
Quick Tip: When dealing with speed–distance–time problems, use \( Time = \frac{Distance}{Speed} \), and form an equation comparing the two time conditions.

Step 1: Given data.

Diameter of cylindrical neck \( = 2 cm \Rightarrow r_1 = 1 cm \)

Length of cylindrical neck \( = h = 7 cm \)

Diameter of spherical part \( = 8.4 cm \Rightarrow r_2 = 4.2 cm \)


Step 2: Volume of water that can be filled.

The vessel consists of a cylinder + sphere. \[ Total Volume = Volume of cylinder + Volume of sphere \]

Step 3: Apply formulas.
\[ Volume of cylinder = \pi r_1^2 h = \pi (1)^2 (7) = 7\pi cm^3 \] \[ Volume of sphere = \frac{4}{3} \pi r_2^3 = \frac{4}{3} \pi (4.2)^3 \] \[ (4.2)^3 = 74.088 \] \[ Volume of sphere = \frac{4}{3} \pi \times 74.088 = 98.784\pi cm^3 \]

Step 4: Total volume.
\[ Total Volume = 7\pi + 98.784\pi = 105.784\pi \] \[ Total Volume = 105.784 \times 3.14 = 331.16 cm^3 \]

Step 5: Conclusion.

Hence, the total amount of water that can be filled in the vessel is \( \boxed{331.16\ cm^3} \).
Quick Tip: When combining solid figures, simply add or subtract their volumes depending on the shape configuration. Always use consistent units.

Step 1: Given data.

Radius \( r = 3.5 cm \)

Total height of the toy \( = 15.5 cm \)

Height of cone \( h = 15.5 - 3.5 = 12 cm \).


Step 2: Formula for total surface area (TSA).
\[ TSA = Curved surface area of cone + Curved surface area of hemisphere \]

Step 3: Calculate slant height of cone.
\[ l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 cm \]

Step 4: Calculate each area.

Curved surface area of cone \( = \pi r l = \pi (3.5)(12.5) = 43.75\pi cm^2 \).

Curved surface area of hemisphere \( = 2\pi r^2 = 2\pi (3.5)^2 = 2\pi (12.25) = 24.5\pi cm^2 \).


Step 5: Total surface area.
\[ TSA = 43.75\pi + 24.5\pi = 68.25\pi \] \[ TSA = 68.25 \times 3.14 = 214.3 cm^2 \]

Step 6: Conclusion.

Hence, the total surface area of the toy is \( \boxed{214.3\ cm^2} \).
Quick Tip: When two solids are joined, omit the common base area and add only the curved surfaces for total surface area calculations.

Step 1: Let the height of the multi-storeyed building be \( h \) m and the distance between the buildings be \( x \) m.

The height of the smaller building is \( 10 \) m.

Step 2: Apply trigonometric ratios.

From the top of the taller building, angles of depression to the top and bottom of the smaller building are 30° and 45°.

For the top of the smaller building: \[ \tan 30° = \frac{h - 10}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \Rightarrow x = \sqrt{3}(h - 10) \]

For the bottom of the smaller building: \[ \tan 45° = \frac{h}{x} \] \[ 1 = \frac{h}{x} \Rightarrow x = h \]

Step 3: Equate the two values of \( x \).
\[ h = \sqrt{3}(h - 10) \]

Step 4: Simplify.
\[ h = \sqrt{3}h - 10\sqrt{3} \] \[ h(\sqrt{3} - 1) = 10\sqrt{3} \] \[ h = \frac{10\sqrt{3}}{\sqrt{3} - 1} \]

Step 5: Rationalize the denominator.
\[ h = \frac{10\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = 5\sqrt{3}(\sqrt{3} + 1) \] \[ h = 5(3 + \sqrt{3}) = 15 + 5\sqrt{3} \]

Step 6: Approximate value.
\[ \sqrt{3} \approx 1.732 \Rightarrow h = 15 + 8.66 = 23.66 m \]

Step 7: Conclusion.

Hence, the height of the multi-storeyed building is \( \boxed{23.66\ m} \).
Quick Tip: In angle of depression problems, the line of sight is horizontal — use \(\tan \theta = \frac{opposite}{adjacent}\) and ensure both triangles share the same base distance.