CBSE Class 10 Mathematics Basic Question Paper 2024 (Set 1- 430/5/1) with Answer Key

1. The product of HCF and LCM of two numbers is equal to the product of the numbers:
\[ HCF \times LCM = 40 \times 30 = 1200 \] Quick Tip: Use the formula \(HCF \times LCM = Product of the numbers\) to quickly calculate the answer.

1. Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \(x^2 + 3x - 10 = 0\), \(a = 1\), \(b = 3\), \(c = -10\). Substitute these values:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} \]
2. Solve for the two roots:
\[ x = \frac{-3 + 7}{2} = 2, \quad x = \frac{-3 - 7}{2} = -5 \]
3. Therefore, the roots are \(-5\) and \(2\). Quick Tip: For quadratic equations, always use the quadratic formula and verify by substitution.

1. For two linear equations to have a unique solution, their slopes must not be equal:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
2. Compare coefficients:
\[ \frac{2k}{6} \neq \frac{5}{5} \quad \Rightarrow \quad \frac{k}{3} \neq 1 \quad \Rightarrow \quad k \neq 3 \] Quick Tip: For a unique solution, check that the ratios of coefficients of \(x\) and \(y\) are not equal.

1. Use the empirical relationship between mean, median, and mode:
\[ Mean - Mode = 3(Mean - Median) \]
2. Substitute the given values:
\[ 28 - 16 = 3(28 - Median) \quad \Rightarrow \quad 12 = 84 - 3 \times Median \]
\[ 3 \times Median = 84 - 12 = 72 \quad \Rightarrow \quad Median = \frac{72}{3} = 24 \] Quick Tip: Remember the formula \(Mean - Mode = 3(Mean - Median)\) for frequency distributions.

1. Odd prime numbers on a die are \(3\) and \(5\).

2. Total possible outcomes when rolling a die are \(6\).

3. Probability of getting an odd prime number:
\[ Probability = \frac{Favorable outcomes}{Total outcomes} = \frac{2}{6} = \frac{1}{3} \] Quick Tip: Prime numbers on a die are \(2\), \(3\), and \(5\). For "odd primes," only \(3\) and \(5\) qualify.

1. The area of a sector is proportional to its central angle. Since the area of the sector is \(\frac{1}{8}\) of the total circle:
\[ \frac{\theta}{360^\circ} = \frac{1}{8} \]
2. Solving for \(\theta\):
\[ \theta = \frac{360^\circ}{8} = 45^\circ \] Quick Tip: For sectors, use the formula \(\frac{Sector Area}{Circle Area} = \frac{\theta}{360^\circ}\) to relate area and central angle.

1. Perform prime factorization of 424:
\[ 424 \div 2 = 212, \quad 212 \div 2 = 106, \quad 106 \div 2 = 53 \]
2. Therefore:
\[ 424 = 2^3 \times 53 \] Quick Tip: Divide the number successively by the smallest primes (2, 3, 5, etc.) until only a prime factor remains.

1. For \(x = 0\), \(8^x = 8^0 = 1\), which ends in an odd digit.

2. For all other values of \(x\), \(8^x\) results in powers of 8, which always end in even digits.

3. Hence, the exception is when \(x = 0\). Quick Tip: Check special cases like \(x = 0\) in problems involving patterns of powers.

1. For an angle of elevation of \(45^\circ\), the height of the tree is equal to the length of the shadow:
\[ Height = Shadow Length = 7 \, m \] Quick Tip: For \(45^\circ\), height and shadow length are always equal in right triangles.

1. Rewrite \(\cot^2 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):
\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]
2. Substitute into the expression:
\[ \frac{5}{\cot^2 \theta} - \frac{5}{\cos^2 \theta} = 5 \times \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{5}{\cos^2 \theta} \]
3. Factorize:
\[ \frac{5(\sin^2 \theta - 1)}{\cos^2 \theta} \]
4. Using the identity \(\sin^2 \theta - 1 = -\cos^2 \theta\):
\[ \frac{5(-\cos^2 \theta)}{\cos^2 \theta} = -5 \] Quick Tip: Always use trigonometric identities like \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify expressions.

1. Given \(\tan^2 \theta = 3\), take the square root to find \(\tan \theta\):
\[ \tan \theta = \sqrt{3} \]
2. The angle whose tangent is \(\sqrt{3}\) is \(\theta = 60^\circ\) (since \(\theta\) is acute).

3. Therefore, \(\theta = 60^\circ\). Quick Tip: Remember the standard trigonometric values: \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), \(\tan 45^\circ = 1\), and \(\tan 60^\circ = \sqrt{3}\).

1. Use the formula for the length of an arc:
\[ Length of arc = \frac{\theta}{360^\circ} \times 2\pi r \]
Substituting the given values:
\[ 10\pi = \frac{\theta}{360^\circ} \times 2\pi \times 12 \]
2. Simplify:
\[ 10\pi = \frac{\theta}{360^\circ} \times 24\pi \quad \Rightarrow \quad 10 = \frac{\theta \times 24}{360} \]
\[ \theta = \frac{10 \times 360}{24} = 150^\circ \] Quick Tip: To calculate the central angle, rearrange the arc length formula and solve for \(\theta\).

1. The angle between the tangents is supplementary to the angle between the radii:
\[ Angle between tangents = 180^\circ - Angle between radii \]
2. Substituting the given angle:
\[ Angle between tangents = 180^\circ - 130^\circ = 50^\circ \] Quick Tip: The angle between tangents at the endpoints of two radii is always the supplement of the angle between the radii.

1. The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
2. Substituting \(a = 1\), \(b = -4\), \(c = 3\):
\[ D = (-4)^2 - 4(1)(3) = 16 - 12 = 4 \] Quick Tip: The discriminant determines the nature of roots: \(D > 0\) (real and distinct), \(D = 0\) (real and equal), \(D < 0\) (complex roots).

1. The formula for the mid-point of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[ Mid-point = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
2. Substituting \(A(-2, 8)\) and \(B(-6, 4)\):
\[ Mid-point = \left(\frac{-2 + (-6)}{2}, \frac{8 + 4}{2}\right) = \left(\frac{-8}{2}, \frac{12}{2}\right) = (-4, 6) \] Quick Tip: For mid-point calculations, take the average of the \(x\)-coordinates and the \(y\)-coordinates separately.

Step 1: Draw the figure and note that \(\triangle OAP\) is a right triangle. Here:
- \(OA\) is the hypotenuse,
- \(OP\) is the radius, and
- \(AP = AQ = 2\) cm is the tangent.


Step 2: Apply the Pythagoras theorem: \[ OA^2 = OP^2 + AP^2 \]
Substitute \(AP = 2\): \[ OA^2 = OP^2 + 2^2 \]

Step 3: Express \(OA\) in terms of \(OP\). Since \(OA = OP + 2\), let \(OP = r\). Then: \[ (r + 2)^2 = r^2 + 4 \]

Step 4: Expand and simplify: \[ r^2 + 4r + 4 = r^2 + 4 \quad \Rightarrow \quad 4r = 0 \quad \Rightarrow \quad r = 2 \]

Step 5: Therefore, the radius is \(2\) cm. Quick Tip: For tangents meeting at a \(90^\circ\) angle, use the Pythagoras theorem with radius and tangent length.

Step 1: In \(\triangle ABC\), the sum of angles is \(180^\circ\). Using \(\triangle ABC \sim \triangle DEF\), corresponding angles are equal.


Step 2: Write the angle sum property: \[ \angle A + \angle B + \angle C = 180^\circ \]

Step 3: Substitute \(\angle A = 55^\circ\) and \(\angle B = \angle E = 45^\circ\): \[ 55^\circ + 45^\circ + \angle C = 180^\circ \]

Step 4: Simplify for \(\angle C\): \[ \angle C = 180^\circ - (55^\circ + 45^\circ) = 80^\circ \] Quick Tip: For similar triangles, use the angle sum property and corresponding angle relationships.

Step 1: Since \(DE \parallel BC\), apply the Basic Proportionality Theorem (BPT): \[ \frac{AD}{DB} = \frac{AE}{EC} \]

Step 2: Substitute the given values: \[ \frac{2x}{4} = \frac{x + 2}{3} \]

Step 3: Cross-multiply: \[ 3(2x) = 4(x + 2) \]

Step 4: Expand and simplify: \[ 6x = 4x + 8 \quad \Rightarrow \quad 2x = 8 \quad \Rightarrow \quad x = 4 \] Quick Tip: When parallel lines divide triangles, apply the Basic Proportionality Theorem (BPT).

Step 1: Rolling a die produces outcomes \(1, 2, 3, 4, 5, 6\). Since \(8\) is not a possible outcome, the probability is zero.

Step 2: The Reason (R) correctly explains why the Assertion (A) is true: The probability of an impossible event is zero. Quick Tip: The probability of an impossible event is always zero.

Step 1: The common difference is given by: \[ d = a_2 - a_1 = 1 - 5 = -4 \]
Hence, Assertion (A) is false.

Step 2: The formula \(d = a_n - a_{n-1}\) is correct, making Reason (R) true. Quick Tip: Verify assertions using the definitions or formulas directly to validate statements.

Step 1: Use the standard trigonometric values: \[ \sin 30^\circ = \frac{1}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 0^\circ = 1, \quad \tan 45^\circ = 1 \]

Step 2: Substitute these values into the given expression: \[ \sin^2 30^\circ + \cos^2 45^\circ - \cos 0^\circ \cdot \tan 45^\circ \] \[ = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 - 1 \times 1 \]

Step 3: Simplify each term: \[ \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}, \quad \cos 0^\circ \cdot \tan 45^\circ = 1 \]

Step 4: Add and subtract: \[ \sin^2 30^\circ + \cos^2 45^\circ - \cos 0^\circ \cdot \tan 45^\circ = \frac{1}{4} + \frac{1}{2} - 1 \] \[ = \frac{1}{4} + \frac{2}{4} - \frac{4}{4} = \frac{3}{4} - \frac{4}{4} = -\frac{1}{4} \]

Final Answer: \(-\frac{1}{4}\)
Quick Tip: Always substitute the standard values for trigonometric functions before simplifying. Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) when needed.

Step 1: Let the point \((3, y)\) divide the line segment in the ratio \(k : 1\). Using the section formula: \[ x = \frac{kx_2 + x_1}{k + 1}, \quad y = \frac{ky_2 + y_1}{k + 1} \]
Here, \((x_1, y_1) = (-2, -5)\) and \((x_2, y_2) = (6, 3)\).

Step 2: Substitute \(x = 3\) in the formula for \(x\): \[ 3 = \frac{k(6) + (-2)}{k + 1} \]
Simplify: \[ 3(k + 1) = 6k - 2 \quad \Rightarrow \quad 3k + 3 = 6k - 2 \quad \Rightarrow \quad 3k = 5 \quad \Rightarrow \quad k = \frac{5}{3} \]

Step 3: Substitute \(k = \frac{5}{3}\) in the formula for \(y\): \[ y = \frac{k(3) + (-5)}{k + 1} = \frac{\frac{5}{3}(3) - 5}{\frac{5}{3} + 1} \]
Simplify the numerator and denominator: \[ y = \frac{5 - 5}{\frac{5}{3} + 1} = \frac{0}{\frac{8}{3}} = 0 \]

Final Answer: The ratio is \(5 : 3\) and \(y = 0\).
Quick Tip: Use the section formula: \[ \left(x, y\right) = \left(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\right) \] to divide line segments in a given ratio.

Step 1: Let the point on the \(y\)-axis be \(P(0, y)\). Since \(P\) is equidistant from \(A\) and \(B\), we have: \[ PA = PB \] \[ PA^2 = PB^2 \]

Step 2: Write the distance formulas: \[ PA^2 = (0 - 6)^2 + (y - 5)^2, \quad PB^2 = (0 - (-4))^2 + (y - 3)^2 \]
Simplify: \[ PA^2 = 6^2 + (y - 5)^2 = 36 + (y^2 - 10y + 25) \] \[ PB^2 = 4^2 + (y - 3)^2 = 16 + (y^2 - 6y + 9) \]

Step 3: Set \(PA^2 = PB^2\): \[ 36 + y^2 - 10y + 25 = 16 + y^2 - 6y + 9 \]

Step 4: Simplify and solve for \(y\): \[ 61 - 10y = 25 - 6y \quad \Rightarrow \quad 61 - 25 = 10y - 6y \quad \Rightarrow \quad 36 = 4y \quad \Rightarrow \quad y = 9 \]

Final Answer: The point is \(P(0, 9)\).
Quick Tip: To find equidistant points, equate the square of the distances from the point to each given point to eliminate square roots.

Step 1: Substitute \(x = \frac{2}{3}\) into the quadratic equation \(kx^2 - x - 2 = 0\): \[ k\left(\frac{2}{3}\right)^2 - \frac{2}{3} - 2 = 0 \]

Step 2: Simplify each term: \[ k\left(\frac{4}{9}\right) - \frac{2}{3} - 2 = 0 \quad \Rightarrow \quad \frac{4k}{9} - \frac{2}{3} - 2 = 0 \]

Step 3: Eliminate fractions by multiplying through by 9: \[ 4k - 6 - 18 = 0 \quad \Rightarrow \quad 4k = 24 \quad \Rightarrow \quad k = 6 \]

Final Answer: \(k = 6\)
Quick Tip: Substitute the root into the quadratic equation and simplify to find the unknown coefficient.

Step 1: Use the Pythagoras theorem for the right triangle formed by the radius, tangent, and distance from the centre to the external point: \[ OA^2 = OP^2 + AP^2 \]
Here, \(OA\) is the distance from the centre to the external point (\(OA = 10\) cm), \(AP\) is the tangent length (\(AP = 6\) cm), and \(OP\) is the radius.

Step 2: Substitute the known values: \[ 10^2 = OP^2 + 6^2 \]

Step 3: Simplify: \[ 100 = OP^2 + 36 \quad \Rightarrow \quad OP^2 = 100 - 36 = 64 \quad \Rightarrow \quad OP = \sqrt{64} = 8 \]

Final Answer: Radius = \(8\) cm
Quick Tip: For tangents drawn from an external point, apply the Pythagoras theorem: \((Distance from centre)^2 = (Radius)^2 + (Tangent length)^2\).

Step 1: From Vieta’s formulas, for the polynomial \(2x^2 + 7x + 5\): \[ \alpha + \beta = -\frac{coefficient of x}{coefficient of x^2} = -\frac{7}{2}, \quad \alpha\beta = \frac{constant term}{coefficient of x^2} = \frac{5}{2} \]

Step 2: Use the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Step 3: Substitute the values of \(\alpha + \beta\) and \(\alpha\beta\): \[ \alpha^2 + \beta^2 = \left(-\frac{7}{2}\right)^2 - 2\left(\frac{5}{2}\right) \]
Simplify: \[ \alpha^2 + \beta^2 = \frac{49}{4} - \frac{10}{2} = \frac{49}{4} - \frac{20}{4} = \frac{29}{4} \]

Step 4: Add \(\alpha\beta\) to the result: \[ \alpha^2 + \beta^2 + \alpha\beta = \frac{29}{4} + \frac{5}{2} = \frac{29}{4} + \frac{10}{4} = \frac{39}{4} \]

Final Answer: \(\frac{39}{4}\)
Quick Tip: Use the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) to simplify expressions involving zeroes of a polynomial.

Step 1: Let the zeroes of the polynomial be \(\alpha\) and \(\frac{1}{\alpha}\). From Vieta’s formulas: \[ \alpha + \frac{1}{\alpha} = -\frac{coefficient of x}{coefficient of x^2} = -\frac{37}{6}, \quad \alpha \cdot \frac{1}{\alpha} = \frac{constant term}{coefficient of x^2} = \frac{p - 2}{6} \]

Step 2: Simplify \(\alpha \cdot \frac{1}{\alpha}\): \[ \alpha \cdot \frac{1}{\alpha} = 1 \quad \Rightarrow \quad \frac{p - 2}{6} = 1 \]

Step 3: Solve for \(p\): \[ p - 2 = 6 \quad \Rightarrow \quad p = 8 \]

Final Answer: \(p = 8\)
Quick Tip: For zeroes that are reciprocals, use the property \(\alpha \cdot \frac{1}{\alpha} = 1\) and Vieta’s formulas to solve for unknown coefficients.

Step 1: Let \(p(x) = 6x^2 + 37x - (p - 2)\), where \(a = 6\), \(b = 37\), and \(c = -(p - 2)\).

Let the zeroes of \(p(x)\) be \(\alpha\) and \(\frac{1}{\alpha}\).


Step 2: From Vieta’s formulas: \[ \alpha \cdot \frac{1}{\alpha} = \frac{c}{a} \]
Substitute \(a = 6\) and \(c = -(p - 2)\): \[ \alpha \cdot \frac{1}{\alpha} = \frac{-(p - 2)}{6} \]

Step 3: Since \(\alpha \cdot \frac{1}{\alpha} = 1\), equate: \[ 1 = \frac{-(p - 2)}{6} \]

Step 4: Solve for \(p\): \[ 6 \cdot 1 = -(p - 2) \quad \Rightarrow \quad 6 = -p + 2 \quad \Rightarrow \quad p = 2 - 6 \quad \Rightarrow \quad p = -4 \]

Final Answer: \(p = -4\)
Quick Tip: When zeroes are reciprocals, use the property \(\alpha \cdot \frac{1}{\alpha} = 1\) and apply Vieta’s formulas to relate coefficients.

Step 1: Assume that \(6 + 3\sqrt{2}\) is a rational number. Let: \[ 6 + 3\sqrt{2} = x \quad where \(x\) is a rational number. \]

Step 2: Rearrange to isolate \(\sqrt{2}\): \[ 3\sqrt{2} = x - 6 \] \[ \sqrt{2} = \frac{x - 6}{3} \]

Step 3: Analyze the result:
- Since \(x\) is a rational number, and \(6\) and \(3\) are also rational numbers, \(\frac{x - 6}{3}\) must be a rational number.
- Thus, \(\sqrt{2}\) is a rational number.

Step 4: Contradiction:
- This contradicts the given fact that \(\sqrt{2}\) is an irrational number.
- Therefore, our initial assumption that \(6 + 3\sqrt{2}\) is rational must be incorrect.

Step 5: Conclusion: \[ 6 + 3\sqrt{2} \quad is an irrational number. \] Quick Tip: To prove a number is irrational, assume it is rational and derive a contradiction using the properties of rational and irrational numbers.

Step 1: Write the quadratic polynomial: \[ p(x) = 6x^2 - 7x - 3 \]

Step 2: Factorize the quadratic polynomial: \[ p(x) = 6x^2 - 7x - 3 = (2x - 3)(3x + 1) \]

Step 3: Find the zeroes:
- Set \(2x - 3 = 0\): \[ x = \frac{3}{2} \]
- Set \(3x + 1 = 0\): \[ x = -\frac{1}{3} \]
Thus, the zeroes are \(\frac{3}{2}\) and \(-\frac{1}{3}\).

Step 4: Verify the sum of the zeroes:
- The sum of the zeroes is: \[ \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9}{6} - \frac{2}{6} = \frac{7}{6} \]
- The coefficient of \(x\) is \(-7\), and the coefficient of \(x^2\) is \(6\). By Vieta’s formula: \[ Sum of zeroes = -\frac{Coefficient of x}{Coefficient of x^2} = -\frac{-7}{6} = \frac{7}{6} \]
- The sum is verified.

Step 5: Verify the product of the zeroes:
- The product of the zeroes is: \[ \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \]
- The constant term is \(-3\), and the coefficient of \(x^2\) is \(6\). By Vieta’s formula: \[ Product of zeroes = \frac{Constant term}{Coefficient of x^2} = \frac{-3}{6} = -\frac{1}{2} \]
- The product is verified.

Step 6: Conclusion:
- The sum and product of the zeroes match the relationships given by Vieta’s formulas. Hence, the verification is complete. Quick Tip: To verify zeroes of a quadratic polynomial, use Vieta’s formulas: \[ Sum of zeroes = -\frac{Coefficient of x}{Coefficient of x^2}, \quad Product of zeroes = \frac{Constant term}{Coefficient of x^2}. \]

Step 1: Expand the left-hand side (LHS): \[ (\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 \] \[ = \sin^2 \theta + \csc^2 \theta + 2\sin \theta \csc \theta + \cos^2 \theta + \sec^2 \theta + 2\cos \theta \sec \theta \]

Step 2: Use trigonometric identities: \[ \sin \theta \csc \theta = 1, \quad \cos \theta \sec \theta = 1, \quad \sin^2 \theta + \cos^2 \theta = 1 \]
Substitute these into the equation: \[ LHS = (\sin^2 \theta + \cos^2 \theta) + \csc^2 \theta + \sec^2 \theta + 2(1) + \csc^2 \theta + \sec^2 \theta \] \[ LHS = 1 + \csc^2 \theta + \sec^2 \theta + 2 \]

Step 3: Use the definitions of \(\csc^2 \theta\) and \(\sec^2 \theta\) in terms of \(\tan^2 \theta\) and \(\cot^2 \theta\): \[ \csc^2 \theta = 1 + \cot^2 \theta, \quad \sec^2 \theta = 1 + \tan^2 \theta \]
Substitute: \[ LHS = 1 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) + 2 \]

Step 4: Simplify: \[ LHS = 1 + 1 + \cot^2 \theta + 1 + \tan^2 \theta + 2 = 7 + \tan^2 \theta + \cot^2 \theta \]

Conclusion: \[ LHS = RHS \]
The given identity is proved.
Quick Tip: Use standard trigonometric identities like \(\sin^2 \theta + \cos^2 \theta = 1\), \(\csc^2 \theta = 1 + \cot^2 \theta\), and \(\sec^2 \theta = 1 + \tan^2 \theta\) to simplify such expressions step by step.

Step 1: Recall the property of a circumscribed parallelogram:
- A parallelogram circumscribes a circle if the sum of the lengths of opposite sides is equal.

Step 2: Let the parallelogram have sides \(AB\), \(BC\), \(CD\), and \(DA\). Then: \[ AB + CD = BC + DA \]

Step 3: In a parallelogram:
- Opposite sides are equal: \(AB = CD\) and \(BC = DA\).
- Substituting this property: \[ AB + AB = BC + BC \quad \Rightarrow \quad 2AB = 2BC \quad \Rightarrow \quad AB = BC \]

Step 4: Conclusion:
- Since all sides are equal, the parallelogram is a rhombus. Quick Tip: A parallelogram circumscribing a circle is always a rhombus because the tangential property ensures all sides are equal.

Step 1: Consider a circle with centre \(O\) and an external point \(P\). Let \(PA\) and \(PB\) be the tangents drawn from \(P\) to the circle, touching the circle at \(A\) and \(B\).

Step 2: Join \(O\) to \(P\), \(O\) to \(A\), and \(O\) to \(B\):
- \(OA\) and \(OB\) are radii of the circle.
- \(PA\) and \(PB\) are tangents.

Step 3: Use the tangential property:
- The radius is perpendicular to the tangent at the point of contact. Hence: \[ \angle OAP = \angle OBP = 90^\circ \]

Step 4: In \(\triangle OAP\) and \(\triangle OBP\):
- \(OP\) is common,
- \(OA = OB\) (radii),
- \(\angle OAP = \angle OBP = 90^\circ\).

Step 5: By RHS (Right angle-Hypotenuse-Side) congruence: \[ \triangle OAP \cong \triangle OBP \]

Step 6: Conclude that corresponding sides are equal: \[ PA = PB \]

Conclusion:
The lengths of the tangents drawn from an external point to a circle are equal.
Quick Tip: For tangents drawn from an external point, use the RHS congruence criterion to prove equality of tangent lengths.

Step 1: Given:
- Length of the rope (radius of the grazing area), \(r = 14 \, m\),
- Side of the equilateral triangle, \(a = 20 \, m\),
- The angle subtended by each vertex in the equilateral triangle, \(\theta = 60^\circ\).


Step 2: Area grazed by the horse:
The horse grazes a sector of the circle. The area of the sector is given by: \[ Area of sector = \frac{\theta}{360^\circ} \cdot \pi r^2 \]
Substitute the values: \[ Area grazed = \frac{60^\circ}{360^\circ} \cdot \frac{22}{7} \cdot 14 \cdot 14 = \frac{1}{6} \cdot \frac{22}{7} \cdot 196 = \frac{308}{3} \, m^2 \, or 102.67 \, m^2. \]

Step 3: Total area of the equilateral triangle:
The area of an equilateral triangle is: \[ Area = \frac{\sqrt{3}}{4} a^2 \]
Substitute \(a = 20\): \[ Area of triangle = \frac{\sqrt{3}}{4} \cdot 20^2 = 100\sqrt{3} \, m^2 \, or 173 \, m^2. \]

Step 4: Area where the horse cannot graze: \[ Required area = Total area of triangle - Area grazed \] \[ Required area = 100\sqrt{3} - 102.67 \, m^2 \, or approximately 70.33 \, m^2. \]

Final Answer: \(70.33 \, m^2\).
Quick Tip: The area grazed by the horse is a sector of a circle. Use the sector area formula and subtract it from the total area of the triangle.

Step 1: Given:
- Length of the minute hand, \(r = 14 \, cm\),
- Angle swept by the minute hand in 60 minutes is \(360^\circ\).


Step 2: Calculate the angle swept in 5 minutes: \[ Angle swept in 5 minutes = \frac{5}{60} \cdot 360^\circ = 30^\circ \]

Step 3: Area swept by the minute hand:
The area swept is a sector of a circle, given by: \[ Area of sector = \frac{\theta}{360^\circ} \cdot \pi r^2 \]
Substitute \(\theta = 30^\circ\) and \(r = 14\): \[ Area swept = \frac{30^\circ}{360^\circ} \cdot \frac{22}{7} \cdot 14 \cdot 14 = \frac{1}{12} \cdot \frac{22}{7} \cdot 196 = \frac{154}{3} \, cm^2 \, or approximately 51.33 \, cm^2. \]

Final Answer: \(51.33 \, cm^2\).
Quick Tip: To find the area swept by the clock’s hand, first calculate the angle swept for the given time interval and use the sector area formula.

Step 1: Let the points of trisection be \(P\) and \(Q\), such that \(AP : PB = 1 : 2\) and \(AQ : QB = 2 : 1\).


Step 2: Use the section formula: \[ (x, y) = \left(\frac{m_2 x_1 + m_1 x_2}{m_1 + m_2}, \frac{m_2 y_1 + m_1 y_2}{m_1 + m_2}\right) \]
Here, \((x_1, y_1) = (5, -3)\) and \((x_2, y_2) = (-4, 3)\).


Step 3: Find the coordinates of \(P\):
- For \(P\), \(AP : PB = 1 : 2\), so \(m_1 = 1\), \(m_2 = 2\): \[ P = \left(\frac{2(5) + 1(-4)}{1 + 2}, \frac{2(-3) + 1(3)}{1 + 2}\right) \]
Simplify: \[ P = \left(\frac{10 - 4}{3}, \frac{-6 + 3}{3}\right) = \left(\frac{6}{3}, \frac{-3}{3}\right) = (2, -1) \]

Step 4: Find the coordinates of \(Q\):
- For \(Q\), \(AQ : QB = 2 : 1\), so \(m_1 = 2\), \(m_2 = 1\): \[ Q = \left(\frac{1(5) + 2(-4)}{2 + 1}, \frac{1(-3) + 2(3)}{2 + 1}\right) \]
Simplify: \[ Q = \left(\frac{5 - 8}{3}, \frac{-3 + 6}{3}\right) = \left(\frac{-3}{3}, \frac{3}{3}\right) = (-1, 1) \]

Final Answer: The points of trisection are \(P(2, -1)\) and \(Q(-1, 1)\).
Quick Tip: For dividing a line segment in a ratio, use the section formula: \[ (x, y) = \left(\frac{m_2 x_1 + m_1 x_2}{m_1 + m_2}, \frac{m_2 y_1 + m_1 y_2}{m_1 + m_2}\right) \]

Step 1: Given:

- Radius of the cone and hemisphere, \(r = 7 \, cm\),

- Total height of the toy, \(h_{total} = 31 \, cm\).

The height of the cone is: \[ h_{cone} = h_{total} - r = 31 - 7 = 24 \, cm. \]

Step 2: Find the slant height (\(l\)) of the cone:
The slant height is given by: \[ l = \sqrt{r^2 + h_{cone}^2} \]
Substitute \(r = 7\) and \(h_{cone} = 24\): \[ l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, cm. \]

Step 3: Surface area of the toy:
- Curved surface area of the cone: \[ CSA_{cone} = \pi r l = \frac{22}{7} \cdot 7 \cdot 25 = 550 \, cm^2 \]
- Curved surface area of the hemisphere: \[ CSA_{hemisphere} = 2 \pi r^2 = 2 \cdot \frac{22}{7} \cdot 7 \cdot 7 = 308 \, cm^2 \]
- Total surface area: \[ Total surface area = CSA_{cone} + CSA_{hemisphere} = 550 + 308 = 858 \, cm^2. \]

Final Answer: Total surface area = \(858 \, cm^2\).
Quick Tip: For combined solids, calculate the individual surface areas and add them appropriately, avoiding double-counting of shared surfaces.

Step 1: Given:

- Radius of the cylinder and hemisphere, \(r = 4.2 \, cm\)

- Height of the cylinder, \(h = 15 \, cm\).


Step 2: Surface area of the article:
- Curved surface area of the cylinder: \[ CSA_{cylinder} = 2 \pi r h \]
Substitute \(r = 4.2\) and \(h = 15\): \[ CSA_{cylinder} = 2 \cdot \frac{22}{7} \cdot 4.2 \cdot 15 = 2 \cdot 22 \cdot 15 = 660 \, cm^2. \]

- Surface area of the two hemispheres: \[ CSA_{hemispheres} = 2 \cdot 2 \pi r^2 = 4 \pi r^2 \]
Substitute \(r = 4.2\): \[ CSA_{hemispheres} = 4 \cdot \frac{22}{7} \cdot 4.2 \cdot 4.2 = 4 \cdot 22 \cdot 2.52 = 221.76 \, cm^2. \]

- Total surface area: \[ Total surface area = CSA_{cylinder} + CSA_{hemispheres} = 660 + 221.76 = 881.76 \, cm^2. \]

Final Answer: Total surface area = \(881.76 \, cm^2\).
Quick Tip: For objects with scooped-out portions, subtract overlapping areas and add only the visible curved surfaces.

Step 1: Let the monthly incomes of the two persons be \(9x\) and \(7x\), respectively.

Let their monthly expenditures be \(4y\) and \(3y\), respectively.


Step 2: Use the savings relation: \[ Income - Expenditure = Savings \]
For the first person: \[ 9x - 4y = 2000 \quad (i) \]
For the second person: \[ 7x - 3y = 2000 \quad (ii) \]

Step 3: Solve the two equations:
From equation (i): \[ 4y = 9x - 2000 \quad \Rightarrow \quad y = \frac{9x - 2000}{4} \]
Substitute this into equation (ii): \[ 7x - 3\left(\frac{9x - 2000}{4}\right) = 2000 \]
Simplify: \[ 7x - \frac{27x - 6000}{4} = 2000 \] \[ \frac{28x - 27x + 6000}{4} = 2000 \] \[ x = 2000 \]

Step 4: Calculate the incomes:
- For the first person: \[ Income = 9x = 9 \cdot 2000 = Rs.~18,000 \]
- For the second person: \[ Income = 7x = 7 \cdot 2000 = Rs.~14,000 \]

Final Answer: The monthly incomes are Rs.~18,000 and Rs.~14,000, respectively.
Quick Tip: Use ratios and savings relations to form linear equations, then solve step-by-step for the variables.

Step 1: Given:
- \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}\), where \(AD\) and \(PM\) are medians.

Step 2: By the property of medians:
The medians of two triangles divide them into two smaller triangles of equal area. Hence, the corresponding sides of \(\triangle ABC\) and \(\triangle PQR\) are proportional.

Step 3: Use the Side-Side-Side (SSS) similarity criterion:
Since: \[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}, \] \(\triangle ABC \sim \triangle PQR\) by the SSS similarity criterion.

Final Answer: \(\triangle ABC \sim \triangle PQR\).
Quick Tip: To prove similarity using SSS, ensure that all corresponding sides of the two triangles are proportional.

Step 1: Consider \(\triangle ABC\) with a line \(DE \parallel BC\), intersecting \(AB\) at \(D\) and \(AC\) at \(E\).


Step 2: By the Basic Proportionality Theorem (BPT):
If a line is parallel to one side of a triangle, it divides the other two sides in the same ratio: \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Step 3: Proof:
- In \(\triangle ABC\), since \(DE \parallel BC\), \(\triangle ADE \sim \triangle ABC\) (by AA similarity criterion).
- From the similarity: \[ \frac{AD}{AB} = \frac{AE}{AC} \quad \Rightarrow \quad \frac{AD}{DB} = \frac{AE}{EC}. \]

Conclusion: The line \(DE\) divides the two sides \(AB\) and \(AC\) in the same ratio. Quick Tip: Use the Basic Proportionality Theorem (BPT) whenever a line is drawn parallel to one side of a triangle.

Step 1: Identify the total number of consumers (\(N\)) and the median class.

- Total number of consumers: \[ N = 70 \]
- Median class corresponds to the cumulative frequency just greater than \(\frac{N}{2} = \frac{70}{2} = 35\). From the table, the median class is \(125 - 145\).


Step 2: Note down the values:
- Lower boundary of the median class (\(l\)): \(125\),
- Frequency of the median class (\(f\)): \(20\),
- Cumulative frequency of the class before the median class (\(CF\)): \(22\),
- Class width (\(h\)): \(20\).

Step 3: Use the median formula: \[ Median = l + \left(\frac{\frac{N}{2} - CF}{f}\right) \cdot h \]

Step 4: Substitute the values: \[ Median = 125 + \left(\frac{35 - 22}{20}\right) \cdot 20 \]
Simplify: \[ Median = 125 + \left(\frac{13}{20}\right) \cdot 20 = 125 + 13 = 138 \]

Final Answer: The median of the data is \(138\) units.
Quick Tip: To find the median in grouped data, always identify the class interval where the cumulative frequency exceeds \(\frac{N}{2}\), then apply the median formula step by step.

Step 1: Given probabilities:

Probability of no children: \(P(no children) = 51% = \frac{51}{100} = 0.51\),
Probability of one child: \(P(one child) = 20% = \frac{20}{100} = 0.20\),
Probability of two children: \(P(two children) = 19% = \frac{19}{100} = 0.19\),
Probability of three children: \(P(three children) = 7% = \frac{7}{100} = 0.07\),
Probability of four or more children: \(P(four or more children) = 3% = \frac{3}{100} = 0.03\).


Step 2: Solve each sub-question:


(i) Probability of two or three children:
\[ P(two or three children) = P(two children) + P(three children) \]
Substitute values: \[ P(two or three children) = 0.19 + 0.07 = 0.26 \]

Final Answer: \(P(two or three children) = 0.26\).


(ii) Probability of more than one child:

Families with more than one child have either two children, three children, or four or more children. Thus: \[ P(more than one child) = P(two children) + P(three children) + P(four or more children) \]
Substitute values: \[ P(more than one child) = 0.19 + 0.07 + 0.03 = 0.29 \]

Final Answer: \(P(more than one child) = 0.29\).



(iii)(a) Probability of less than three children:

Families with less than three children have either no children, one child, or two children. Thus: \[ P(less than three children) = P(no children) + P(one child) + P(two children) \]
Substitute values: \[ P(less than three children) = 0.51 + 0.20 + 0.19 = 0.90 \]

Final Answer: \(P(less than three children) = 0.90\).


(iii)(b) Probability of more than two children:

Families with more than two children have either three children or four or more children. Thus: \[ P(more than two children) = P(three children) + P(four or more children) \]
Substitute values: \[ P(more than two children) = 0.07 + 0.03 = 0.10 \]

Final Answer: \(P(more than two children) = 0.10\).
Quick Tip: To calculate probabilities, add the probabilities of all relevant events. Ensure that the total probability of all events sums to 1.

Step 1: Observe the pattern: The number of pairs of shoes in each row forms an arithmetic progression (AP): \[ 3, 5, 7, 9, \dots \]
Here, the first term (\(a\)) is \(3\), and the common difference (\(d\)) is \(5 - 3 = 2\).


(i) Number of pairs of shoes in the 6th row:

The general term of an AP is given by: \[ a_n = a + (n - 1)d \]
Substitute \(a = 3\), \(d = 2\), and \(n = 6\): \[ a_6 = 3 + (6 - 1) \cdot 2 = 3 + 10 = 13 \]

Final Answer: \(13\) pairs of shoes are displayed in the 6th row.



(ii) Difference of pairs of shoes in the 1st row and the 6th row:

From part (i), we know:
- Number of pairs in the 1st row (\(a_1\)) = \(3\),
- Number of pairs in the 6th row (\(a_6\)) = \(13\).

The difference is: \[ a_6 - a_1 = 13 - 3 = 10 \]

Final Answer: The difference is \(10\) pairs.



(iii)(a) Total number of pairs of shoes displayed in the first 15 rows:

The sum of the first \(n\) terms of an AP is given by: \[ S_n = \frac{n}{2} \cdot [2a + (n - 1)d] \]
Substitute \(n = 15\), \(a = 3\), and \(d = 2\): \[ S_{15} = \frac{15}{2} \cdot [2(3) + (15 - 1) \cdot 2] \]
Simplify: \[ S_{15} = \frac{15}{2} \cdot [6 + 28] = \frac{15}{2} \cdot 34 = 15 \cdot 17 = 255 \]

Final Answer: The total number of pairs displayed in the first 15 rows is \(255\).



(iii)(b) Total amount earned if all shoes in the 4th row are sold:

From the AP, the number of pairs of shoes in the 4th row (\(a_4\)) is: \[ a_4 = a + (4 - 1)d = 3 + 3 \cdot 2 = 3 + 6 = 9 \]
Price of each pair = Rs.~500. Total amount earned: \[ Total amount = a_4 \cdot price per pair = 9 \cdot 500 = Rs.~4500 \]

Final Answer: Total amount earned is Rs.~4500.
Quick Tip: For AP problems: 1. Use the general term formula \(a_n = a + (n - 1)d\) to find individual terms. 2. Use the sum formula \(S_n = \frac{n}{2}[2a + (n - 1)d]\) to calculate the total of the first \(n\) terms.

Step 1: Set up the known values:
- Distance between poles \(AB\) and \(PQ\) = 40 m,
- Height of pole \(AB\) = \(h_1\),
- Height of pole \(PQ\) = \(h_2\).



(i) Find the height of pole \(AB\):

From the geometry of the problem and using trigonometric ratios: \[ \tan 60^\circ = \frac{h_1}{40} \quad \Rightarrow \quad h_1 = 40 \cdot \sqrt{3} \quad \Rightarrow \quad h_1 = 40 \cdot 1.732 = 69.28 \, m. \]

Final Answer: The height of pole \(AB\) is \(69.28\) m.



(ii) Find the height of pole \(PQ\):

Using the same approach: \[ \tan 30^\circ = \frac{h_2}{40} \quad \Rightarrow \quad h_2 = 40 \cdot \frac{1}{\sqrt{3}} \quad \Rightarrow \quad h_2 = \frac{40}{1.732} = 23.09 \, m. \]

Final Answer: The height of pole \(PQ\) is \(23.09\) m.




(iii)(a) Find the distance \(BQ\):

The height difference between the poles is: \[ h_1 - h_2 = 69.28 - 23.09 = 46.19 \, m. \]
Using the given angle of elevation (\(30^\circ\)): \[ \tan 30^\circ = \frac{h_1 - h_2}{BQ} \quad \Rightarrow \quad BQ = \frac{h_1 - h_2}{\tan 30^\circ} \] \[ BQ = \frac{46.19}{\frac{1}{\sqrt{3}}} = 46.19 \cdot \sqrt{3} = 46.19 \cdot 1.732 = 80.01 \, m. \]

Final Answer: The distance \(BQ\) is \(80.01\) m.



(iii)(b) How far down the wire from pole \(AB\) is the coupling?

The coupling is at a height of 20 m from the ground. The height of pole \(AB\) is \(69.28\) m. The length of the wire from \(A\) to \(C\) forms part of the triangle with base 40 m.

Using similar triangles: \[ Fraction of wire down = \frac{Height from the top of \(AB\) to the coupling}{Total height of \(AB\)} = \frac{69.28 - 20}{69.28} = \frac{49.28}{69.28}. \]

The length of the wire from \(A\) to \(C\) is: \[ Wire length = \sqrt{40^2 + 69.28^2} = \sqrt{1600 + 4799.84} = \sqrt{6399.84} = 80 \, m. \]

The distance from \(A\) to the coupling is: \[ Distance down = \frac{49.28}{69.28} \cdot 80 = 56.91 \, m. \]

Final Answer: The coupling is \(56.91\) m down the wire from \(A\).
Quick Tip: For problems involving poles and angles of elevation, always use trigonometric ratios like \(\tan \theta = \frac{height}{base}\). For coupling points, use proportionality with similar triangles.