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

Step 1: Understanding the concept.

From an external point, we can draw tangents to a circle such that each tangent touches the circle at exactly one point.


Step 2: Visualize the geometry.

If we take any external point \( P \) outside a circle with center \( O \), two tangents can be drawn from \( P \), touching the circle at points \( A \) and \( B \). Both tangents are equal in length, i.e., \( PA = PB \).


Step 3: Conclusion.

Hence, the maximum number of tangents that can be drawn from an external point to a circle is two.
Quick Tip: From an external point to a circle, exactly two tangents can be drawn, and both are equal in length.

Step 1: Recall the formula.

The distance of any point \((x, y)\) from the y-axis is given by the absolute value of its x-coordinate. That is, \[ Distance = |x| \]

Step 2: Substitute the given point.

For the point \((7, 3)\), the x-coordinate is \(7\). Hence, \[ Distance = |7| = 7 \]

Step 3: Conclusion.

Therefore, the distance of the point \((7, 3)\) from the y-axis is 7 units.
Quick Tip: The distance of a point from the y-axis is always the absolute value of its x-coordinate.

Step 1: Given relation.

We have \( p \sin \theta = q \cos \theta \).


Step 2: Divide both sides by \(\cos \theta\).
\[ \tan \theta = \dfrac{q}{p} \]

Step 3: Express \(\sin \theta\) and \(\cos \theta\) in terms of \(p\) and \(q\).

Let us assume the hypotenuse to be \(\sqrt{p^2 + q^2}\).
Hence, \[ \sin \theta = \dfrac{q}{\sqrt{p^2 + q^2}}, \quad \cos \theta = \dfrac{p}{\sqrt{p^2 + q^2}} \]

Step 4: Find \(\csc \theta\).
\[ \csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt{p^2 + q^2}}{q} \]

Step 5: Conclusion.

The value of \(\csc \theta\) is \(\dfrac{\sqrt{p^2 + q^2}}{q}\).
Quick Tip: Whenever \( p \sin \theta = q \cos \theta \), divide both sides by \(\cos \theta\) to get \(\tan \theta = \dfrac{q}{p}\), then derive other trigonometric ratios easily.

Step 1: Pairing of angles.

Notice that \(\tan(89^\circ) = \cot(1^\circ)\), \(\tan(88^\circ) = \cot(2^\circ)\), and so on.


Step 2: Simplify the product.
\[ \tan 1^\circ \times \tan 89^\circ = \tan 1^\circ \times \cot 1^\circ = 1 \]
Similarly, each pair multiplies to 1: \[ (\tan 1^\circ \tan 89^\circ)(\tan 2^\circ \tan 88^\circ) \dots (\tan 44^\circ \tan 46^\circ) \tan 45^\circ = 1 \]

Step 3: Conclusion.

Hence, the total product = 1.
Quick Tip: Remember, \(\tan(90^\circ - \theta) = \cot \theta\). Pairing complementary angles often simplifies trigonometric products.

Step 1: Recall the condition for equal roots.

For a quadratic \(ax^2 + bx + c = 0\), roots are equal when \[ D = b^2 - 4ac = 0 \]

Step 2: Substitute values.

Here, \(a = 3, b = 5, c = -q\). \[ b^2 - 4ac = 0 \implies 25 - 4(3)(-q) = 0 \]

Step 3: Simplify.
\[ 25 + 12q = 0 \Rightarrow q = -\dfrac{25}{12} \]

Step 4: Verify given options.

Correct answer matches option (A) \(-\dfrac{25}{12}\).
(Note: If the printed options differ, the logic confirms this value.)
Quick Tip: For equal roots, always apply the discriminant condition \(D = 0\).

Step 1: Identify the first term and common difference.

First term \(a = -62\), second term = \(-59\).
Hence, common difference \(d = -59 - (-62) = 3\).


Step 2: Use the formula for the nth term.
\[ a_n = a + (n - 1)d \]
For the 11th term (\(n = 11\)): \[ a_{11} = -62 + (11 - 1)(3) = -62 + 30 = -32 \]

Step 3: Conclusion.

Hence, the 11th term = \(-32\).
Quick Tip: In an arithmetic progression, the nth term is found by \(a_n = a + (n - 1)d\).

Step 1: Recall the basic probability rule.
\[ P(E) + P(\overline{E}) = 1 \]

Step 2: Substitute the value of \(P(E)\).
\[ 0.05 + P(\overline{E}) = 1 \implies P(\overline{E}) = 1 - 0.05 = 0.95 \]

Step 3: Conclusion.

Therefore, the value of \(P(\overline{E})\) is 0.95.
Quick Tip: The probability of the complement of an event is always \(1 - P(E)\).

Step 1: Identify the total number of balls.

The bag contains 3 red balls and 5 black balls. \[ Total balls = 3 + 5 = 8 \]

Step 2: Determine the favorable outcomes.

The favorable outcomes (drawing a red ball) = 3.


Step 3: Apply the probability formula.
\[ P(Red ball) = \dfrac{Number of favorable outcomes}{Total outcomes} = \dfrac{3}{8} \]

Step 4: Conclusion.

Hence, the probability of drawing a red ball is \(\dfrac{3}{8}\).
Quick Tip: In probability, always divide the number of favorable outcomes by the total number of possible outcomes.

Step 1: Recall the Basic Proportionality Theorem (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.

Thus, \[ \dfrac{AD}{DB} = \dfrac{AE}{EC} \]

Step 2: Substitute the given values.

From the figure, \[ AD = 1.3 cm, \quad DB = 3.9 cm, \quad AE = 1.5 cm \]
Let \(EC = x\). Then, \[ \dfrac{1.3}{3.9} = \dfrac{1.5}{x} \]

Step 3: Simplify to find \(x\).
\[ x = \dfrac{1.5 \times 3.9}{1.3} = \dfrac{5.85}{1.3} \approx 4.5 cm \]

Step 4: Conclusion.

Therefore, the measure of \(CE\) is approximately \(4.5 \, cm\).
Quick Tip: Use the Basic Proportionality Theorem whenever a line parallel to one side of a triangle divides the other two sides proportionally.

Step 1: Given data.

For \(\triangle MNL\): \(MN = 3\) cm, \(ML = 4.5\) cm, and \(\angle M = 70^\circ\).

For \(\triangle PQR\): \(PQ = 2\) cm, \(QR = 3\) cm, and \(\angle Q = 70^\circ\).


Step 2: Compare sides including the equal angles.
\[ \dfrac{MN}{QR} = \dfrac{3}{3} = 1, \quad \dfrac{ML}{PQ} = \dfrac{4.5}{2} = 2.25 \]
These are not equal, but let’s check other possible corresponding sides.

If we consider \(\triangle NML\) and \(\triangle QPR\): \[ \dfrac{MN}{QP} = \dfrac{3}{2} = 1.5, \quad \dfrac{ML}{QR} = \dfrac{4.5}{3} = 1.5 \]
The sides are in the same ratio and included angles are equal (\(70^\circ\)).

Step 3: Conclusion.

Hence, by the SAS similarity criterion, \[ \triangle NML \sim \triangle QPR \] Quick Tip: When two triangles have one equal angle and the sides including these angles are in proportion, they are similar by the SAS criterion.

Step 1: Formula for surface area of a sphere.
\[ Surface Area = 4\pi r^2 \]

Step 2: Find the radius.

Given diameter = \(\dfrac{1}{2}\) cm \[ r = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} cm \]

Step 3: Substitute in the formula.
\[ Surface Area = 4\pi \left(\dfrac{1}{4}\right)^2 = 4\pi \times \dfrac{1}{16} = \dfrac{\pi}{4} cm^2 \]

Step 4: Conclusion.

Therefore, the surface area of the sphere is \(\dfrac{\pi}{4}\) cm\(^2\).
Quick Tip: Always remember, Surface Area of a Sphere = \(4\pi r^2\) and Volume = \(\dfrac{4}{3}\pi r^3\).

Step 1: Formula for length of an arc.
\[ Length of arc = \dfrac{\theta}{360^\circ} \times 2\pi r \]

Step 2: Substitute the given values.
\[ \theta = 30^\circ, \quad r = 6 cm \] \[ Arc length = \dfrac{30}{360} \times 2\pi \times 6 \]

Step 3: Simplify.
\[ = \dfrac{1}{12} \times 12\pi = \dfrac{\pi}{1} = \pi \]
Wait — we simplify carefully: \[ \dfrac{30}{360} = \dfrac{1}{12}, \quad 2\pi \times 6 = 12\pi \] \[ Arc length = \dfrac{1}{12} \times 12\pi = \pi cm \]

Step 4: Correct the simplification (angle check).

Oops — on rechecking, angle \(30^\circ\) gives: \[ Arc length = \dfrac{30}{360} \times 2\pi \times 6 = \dfrac{1}{12} \times 12\pi = \pi cm \]
So the correct answer is actually (D) \pi cm.

Step 5: Conclusion.

The measure of the arc = \(\pi\) cm.
Quick Tip: Arc length depends on the central angle: \(\dfrac{\theta}{360^\circ} \times 2\pi r\). Always convert the angle into a fraction of 360°.

Step 1: Identify the given data.

Radius \( r = 5 \) cm, \( OQ = 12 \) cm. We have to find \( PQ \).

Step 2: Use the property of tangent and radius.

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

Step 3: Apply the Pythagoras theorem.
\[ OQ^2 = OP^2 + PQ^2 \] \[ PQ^2 = OQ^2 - OP^2 = 12^2 - 5^2 = 144 - 25 = 119 \] \[ PQ = \sqrt{119} cm \]

Step 4: Conclusion.

Hence, the length of the tangent \( PQ = \sqrt{119} \) cm.
Quick Tip: When a tangent and radius meet, they form a right angle. Use the Pythagoras theorem in such tangent problems.

Step 1: Find the prime factors.

182 = 2 × 7 × 13

78 = 2 × 3 × 13


Step 2: Find the common factors.

Common prime factors = 2 and 13.

Step 3: Multiply the common factors.
\[ HCF = 2 \times 13 = 26 \]

Step 4: Conclusion.

Hence, the HCF of 182 and 78 is 26.
Quick Tip: To find HCF, multiply the smallest powers of all common prime factors.

Step 1: Recall the formula for curved surface area (CSA) of a cylinder.
\[ CSA = 2\pi r h \]

Step 2: Substitute the given values.
\( r = 3.5 cm, \, h = 8.4 cm \) \[ CSA = 2\pi \times 3.5 \times 8.4 \]

Step 3: Simplify.
\[ CSA = 2\pi \times 29.4 = 58.8\pi cm^2 \]

Step 4: Conclusion.

Hence, the curved surface area of the cylinder is \( 58.8\pi \, cm^2 \).
Quick Tip: Always remember: Curved Surface Area of a Cylinder = \(2\pi r h\), and Total Surface Area = \(2\pi r (r + h)\).

Step 1: Formula for the area of a sector.
\[ Area of a sector = \dfrac{\theta}{360^\circ} \times \pi r^2 \]

Step 2: Substitute the given values.
\[ r = 4 cm, \, \theta = 60^\circ \] \[ Area = \dfrac{60}{360} \times \pi \times (4)^2 = \dfrac{1}{6} \times 16\pi = \dfrac{8}{3}\pi \, cm^2 \]

Step 3: Conclusion.

Hence, the area of the sector is \(\dfrac{8}{3}\pi\) cm\(^2\).
Quick Tip: For finding the area of a sector, always use the fraction of the total circle: \(\dfrac{\theta}{360^\circ} \times \pi r^2\).

Step 1: Recall the formula for discriminant.

For any quadratic equation \(ax^2 + bx + c = 0\), \[ D = b^2 - 4ac \]

Step 2: Substitute the values.

Here, \(a = 1, b = 1, c = -1\). \[ D = (1)^2 - 4(1)(-1) = 1 + 4 = 5 \]

Step 3: Conclusion.

Therefore, the discriminant is \(D = 5\).
Quick Tip: Discriminant helps determine the nature of roots: - \(D > 0\): two distinct real roots - \(D = 0\): equal real roots - \(D < 0\): imaginary roots

Step 1: Write in standard form.
\[ 4x^2 - 4x + 1 = 0 \]
So, \(a = 4, b = -4, c = 1\).

Step 2: Formula for sum of roots.
\[ Sum of roots = -\dfrac{b}{a} = -\dfrac{-4}{4} = 1 \]

Wait, recheck original equation: \(1 - 4x + 4x^2 = 0\).
This rearranges to \(4x^2 - 4x + 1 = 0\), giving sum \(=\dfrac{4}{4}=1\).
So the correct answer is (C) 1.

Step 3: Conclusion.

Hence, the sum of the roots is \(1\).
Quick Tip: For a quadratic equation \(ax^2 + bx + c = 0\): Sum of roots = \(-\dfrac{b}{a}\), Product of roots = \(\dfrac{c}{a}\).

Step 1: Find class marks (midpoints).

For each class, \[ x_i = \dfrac{Upper limit + Lower limit}{2} \]
\[ \begin{array}{c|c|c} Class & f_i & x_i
\hline 0-10 & 4 & 5
10-20 & 7 & 15
20-30 & 5 & 25
30-40 & 8 & 35
40-50 & 6 & 45
\end{array} \]

Step 2: Find \(f_i x_i\) and total.
\[ \begin{array}{c|c|c} f_i & x_i & f_i x_i
4 & 5 & 20
7 & 15 & 105
5 & 25 & 125
8 & 35 & 280
6 & 45 & 270
\end{array} \]
\[ \Sigma f_i = 30, \quad \Sigma f_i x_i = 800 \]

Step 3: Apply the formula for mean.
\[ \bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{800}{30} = 26.66 \]

Step 4: Conclusion.

Hence, the mean = 26.66.
Quick Tip: Always multiply each class frequency by its class mark and divide by the total frequency to get the mean.

Step 1: Find cumulative frequencies.
\[ \begin{array}{c|c|c} Class & f & Cumulative frequency (CF)
0-10 & 8 & 8
10-20 & 6 & 14
20-30 & 11 & 25
30-40 & 18 & 43
40-50 & 6 & 49
\end{array} \]

Step 2: Determine \(N/2\).
\[ N = 49 \Rightarrow N/2 = 24.5 \]

Step 3: Identify the median class.

The class in which the cumulative frequency first exceeds 24.5 is 30–40.

Step 4: Conclusion.

Hence, the median class is 30–40.
Quick Tip: To find the median class, locate where the cumulative frequency first becomes greater than \(\dfrac{N}{2}\).

Step 1: Use the distance formula.

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

Step 2: Substitute the given values.

Here, \((x_1, y_1) = (x, 5)\) and \((x_2, y_2) = (2, -3)\), and \(d = 17\). \[ 17 = \sqrt{(2 - x)^2 + (-3 - 5)^2} \] \[ 17 = \sqrt{(2 - x)^2 + 64} \]

Step 3: Square both sides.
\[ 289 = (2 - x)^2 + 64 \] \[ 225 = (2 - x)^2 \]

Step 4: Solve for \(x\).
\[ 2 - x = \pm 15 \Rightarrow \begin{cases} x = -13, & if 2 - x = 15
x = 17, & if 2 - x = -15 \end{cases} \]

Step 5: Conclusion.

Hence, the values of \(x\) are \(-13\) and \(17\).
Quick Tip: Always use the distance formula \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find unknown coordinates when the distance is given.

Step 1: Recall the condition for collinearity.

Three points are collinear if the slopes between them are equal. \[ Slope of AB = Slope of BC \]

Step 2: Compute slopes.

Let \(A(1, 4)\), \(B(a, -2)\), and \(C(-3, 16)\). \[ Slope of AB = \dfrac{-2 - 4}{a - 1} = \dfrac{-6}{a - 1} \] \[ Slope of BC = \dfrac{16 - (-2)}{-3 - a} = \dfrac{18}{-3 - a} = \dfrac{-18}{a + 3} \]

Step 3: Equate the slopes.
\[ \dfrac{-6}{a - 1} = \dfrac{-18}{a + 3} \Rightarrow 6(a + 3) = 18(a - 1) \] \[ 6a + 18 = 18a - 18 \Rightarrow 12a = 36 \Rightarrow a = 3 \]

Step 4: Conclusion.

Hence, \(a = 3\).
Quick Tip: For three points to be collinear, their slopes must be equal. Equate the slopes and solve for the variable.

Step 1: Take the LHS.
\[ LHS = \dfrac{1 + \sin \theta}{\cos \theta} + \dfrac{\cos \theta}{1 + \sin \theta} \]

Step 2: Take a common denominator.
\[ LHS = \dfrac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos \theta (1 + \sin \theta)} \]

Step 3: Expand the numerator.
\[ (1 + \sin \theta)^2 + \cos^2 \theta = 1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta \] \[ = 2(1 + \sin \theta) \]

Step 4: Simplify.
\[ LHS = \dfrac{2(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)} = \dfrac{2}{\cos \theta} = 2 \sec \theta \]

Step 5: Conclusion.

Hence proved, \[ \dfrac{1 + \sin \theta}{\cos \theta} + \dfrac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta \] Quick Tip: Use \(\sin^2 \theta + \cos^2 \theta = 1\) whenever both \(\sin \theta\) and \(\cos \theta\) appear together.

Step 1: Compute cumulative frequencies.
\[ \begin{array}{c|c|c} Class & f & Cumulative Frequency (CF)
0–10 & 6 & 6
10–20 & 9 & 15
20–30 & 20 & 35
30–40 & 15 & 50
40–50 & 9 & 59
\end{array} \]

Step 2: Find \(N/2\).
\[ N = 59, \quad N/2 = 29.5 \]

Step 3: Locate the median class.

The class with cumulative frequency just greater than 29.5 is 20–30.

Step 4: Apply the median formula.
\[ Median = L + \left(\dfrac{\dfrac{N}{2} - CF}{f}\right) \times h \]
Substitute: \(L = 20, \, CF = 15, \, f = 20, \, h = 10\) \[ Median = 20 + \left(\dfrac{29.5 - 15}{20}\right) \times 10 = 20 + 7.25 = 27.25 \]

Step 5: Conclusion.

Hence, the median = 27.25.
Quick Tip: For the median in a frequency table, always find the class where cumulative frequency first exceeds \(N/2\).

Step 1: Prime factorization.
\[ 92 = 2^2 \times 23 \] \[ 510 = 2 \times 3 \times 5 \times 17 \]

Step 2: Take all unique prime factors with highest powers.
\[ LCM = 2^2 \times 3 \times 5 \times 17 \times 23 \]

Step 3: Multiply.
\[ 4 \times 3 \times 5 \times 17 \times 23 = 23460 \]

Step 4: Conclusion.

Hence, the LCM of 92 and 510 is 23460.
Quick Tip: Take the highest powers of all prime factors from both numbers to find the LCM.

Step 1: Assume the opposite.

Let \(\sqrt{3}\) be rational.
Then it can be expressed as: \[ \sqrt{3} = \dfrac{p}{q}, \quad where p, q are integers and \gcd(p, q) = 1 \]

Step 2: Square both sides.
\[ 3 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 3q^2 \]

Step 3: Analyze divisibility.

Since \(p^2\) is divisible by 3, \(p\) must also be divisible by 3.
Let \(p = 3k\).

Step 4: Substitute back.
\[ p^2 = (3k)^2 = 9k^2 \Rightarrow 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2 \]
Thus, \(q^2\) is also divisible by 3, so \(q\) is divisible by 3.

Step 5: Contradiction.

This contradicts the assumption that \(p\) and \(q\) have no common factor other than 1.

Step 6: Conclusion.

Therefore, \(\sqrt{3}\) is irrational.
Quick Tip: Use the method of contradiction to prove irrationality — assume rationality and reach a contradiction.

Step 1: Identify the modal class.

The class with the highest frequency is \(30–40\), so it is the modal class.

Step 2: Write the given data.
\[ L = 30, \quad f_1 = 23, \quad f_0 = 21, \quad f_2 = 14, \quad h = 10 \]

Step 3: Apply the formula for mode.
\[ Mode = L + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \]

Step 4: Substitute the values.
\[ Mode = 30 + \left(\dfrac{23 - 21}{2(23) - 21 - 14}\right) \times 10 \] \[ = 30 + \left(\dfrac{2}{46 - 35}\right) \times 10 = 30 + \dfrac{20}{11} = 31.82 \]

Step 5: Conclusion.

Hence, the mode = 31.82 (approx.).
Quick Tip: The modal class is the class interval with the highest frequency. Always apply the mode formula using \(f_1, f_0, f_2\) correctly.

Step 1: Construction.

Let \(P\) be a point outside the circle with center \(O\).
From \(P\), draw two tangents \(PA\) and \(PB\) to the circle, touching it at \(A\) and \(B\) respectively.

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

These are radii of the circle. Therefore, \[ OA = OB \]

Step 3: Observe right angles.

Since \(PA\) and \(PB\) are tangents, they are perpendicular to the radii at the point of contact. \[ \angle OAP = \angle OBP = 90^\circ \]

Step 4: Consider triangles \(\triangle OAP\) and \(\triangle OBP\).

In both triangles, \[ OA = OB \quad (radii)
OP = OP \quad (common side)
\angle OAP = \angle OBP = 90^\circ \]
Hence, \[ \triangle OAP \cong \triangle OBP \quad (by RHS congruence) \]

Step 5: Conclusion.

By congruence, \[ PA = PB \]
Therefore, the lengths of tangents drawn from an external point to a circle are equal.
Quick Tip: When proving tangents equal, always use the congruence of right triangles formed by radii and tangents (RHS rule).

Step 1: Given condition.

In \(\triangle ABC\), \(D\) lies on \(BC\) such that \(\angle ADC = \angle BAC\).

Step 2: Draw auxiliary lines.

Join \(AD\) and \(CA\).

Step 3: Observe similar triangles.

In \(\triangle CAD\) and \(\triangle CBA\): \[ \angle ADC = \angle BAC \quad (given)
\angle ACD = \angle ACB \quad (common) \]
Hence, \[ \triangle CAD \sim \triangle CBA \quad (by AA similarity) \]

Step 4: Write the ratio of corresponding sides.
\[ \dfrac{CA}{CB} = \dfrac{CD}{CA} \]

Step 5: Cross-multiply.
\[ CA^2 = CB \times CD \]

Step 6: Conclusion.

Hence proved that \(CA^2 = CB \cdot CD\).
Quick Tip: In geometry proofs involving equal angles, check for AA similarity — it often helps establish proportional sides.

Step 1: Let the digits of the number be.

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

Step 2: Write the given conditions.

The sum of the digits is 9: \[ x + y = 9 \quad (i) \]
9 times the number is equal to twice the number formed by reversing the digits: \[ 9(10x + y) = 2(10y + x) \]

Step 3: Simplify the equation.
\[ 90x + 9y = 20y + 2x \] \[ 88x = 11y \Rightarrow 8x = y \quad (ii) \]

Step 4: Substitute (ii) into (i).
\[ x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1 \] \[ y = 8x = 8 \]

Step 5: Find the number.
\[ Number = 10x + y = 10(1) + 8 = 18 \]

Step 6: Check the condition.
\[ 9 \times 18 = 162, \quad 2 \times 81 = 162 \]
Hence, the condition is satisfied.

Step 7: Conclusion.

The required number is 18.
Quick Tip: For two-digit number problems, always assume digits as \(x\) and \(y\) and form equations using their sum and reversal relations.

Step 1: Identify coefficients.

Here, \(a = 2\), \(b = -5\), and \(c = 3\).

Step 2: Use the quadratic formula.
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Substitute the values.
\[ x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)} \] \[ x = \dfrac{5 \pm \sqrt{25 - 24}}{4} = \dfrac{5 \pm 1}{4} \]

Step 4: Simplify.
\[ x = \dfrac{5 + 1}{4} = \dfrac{6}{4} = \dfrac{3}{2} \quad and \quad x = \dfrac{5 - 1}{4} = 1 \]

Step 5: Conclusion.

Hence, the roots of the equation are \(x = 1\) and \(x = \dfrac{3}{2}\).
Quick Tip: Always check the discriminant \(b^2 - 4ac\) before solving; it helps determine the nature of roots.

Step 1: Let the breadth be \(x\) m.

Then, the length = \(2x\) m (given).

Step 2: Write the formula for perimeter.
\[ Perimeter = 2(l + b) \]
Substitute: \[ 120 = 2(2x + x) \Rightarrow 120 = 6x \Rightarrow x = 20 \]

Step 3: Find the length.
\[ l = 2x = 2 \times 20 = 40 \]

Step 4: Verify using the area condition.
\[ Area = l \times b = 40 \times 20 = 800 \, m^2 \]
The condition is satisfied.

Step 5: Conclusion.

Hence, the length = 40 m and breadth = 20 m.
Quick Tip: For rectangle problems, always form two equations — one from perimeter and one from area — to find unknown dimensions.

Step 1: Eliminate the fractions by taking the LCM of denominators.

For the first equation: \[ \dfrac{3}{2}x - \dfrac{5}{3}y = -2 \]
Multiply both sides by 6 (LCM of 2 and 3): \[ 9x - 10y = -12 \quad (i) \]

For the second equation: \[ \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \]
Multiply both sides by 6: \[ 2x + 3y = 13 \quad (ii) \]

Step 2: Solve the two linear equations.

Equation (i): \(9x - 10y = -12\)
Equation (ii): \(2x + 3y = 13\)

Step 3: Multiply (ii) by 3 to align coefficients of \(y\).
\[ 6x + 9y = 39 \quad (iii) \]

Step 4: Eliminate \(y\).

Multiply (i) by 3: \[ 27x - 30y = -36 \quad (iv) \]
Multiply (ii) by 10: \[ 20x + 30y = 130 \quad (v) \]

Add (iv) and (v): \[ 47x = 94 \Rightarrow x = 2 \]

Step 5: Substitute \(x = 2\) in equation (ii).
\[ 2(2) + 3y = 13 \Rightarrow 4 + 3y = 13 \Rightarrow 3y = 9 \Rightarrow y = 3 \]

Step 6: Conclusion.

Hence, \(x = 2\) and \(y = 3\).
Quick Tip: To eliminate fractions, always multiply through by the LCM of denominators before applying elimination or substitution.

Step 1: Let the first term and common difference be \(a\) and \(d\).

The general term of an A.P. is \(a_n = a + (n - 1)d\).

Step 2: Use the given conditions.
\[ a_2 = a + d = 14 \quad (i) \] \[ a_3 = a + 2d = 18 \quad (ii) \]

Step 3: Subtract (i) from (ii).
\[ (a + 2d) - (a + d) = 18 - 14 \Rightarrow d = 4 \]

Step 4: Substitute \(d = 4\) in (i).
\[ a + 4 = 14 \Rightarrow a = 10 \]

Step 5: Use the sum formula of an A.P.
\[ S_n = \dfrac{n}{2}[2a + (n - 1)d] \]
For \(n = 51\): \[ S_{51} = \dfrac{51}{2}[2(10) + 50(4)] \] \[ = \dfrac{51}{2}[20 + 200] = \dfrac{51}{2} \times 220 = 51 \times 110 = 5610 \]

Step 6: Conclusion.

Hence, the sum of 51 terms of the A.P. is 5610.
Quick Tip: For A.P. problems, always identify the first term and common difference from the given terms before applying the sum formula.

Step 1: Let the height of the tower be \( h \) m and the height of the flagstaff be \( x \) m.

Total height of tower + flagstaff = \( h + x \).
Distance from the tower = \( 10 \, m \).

Step 2: Use trigonometric ratios for each angle of elevation.


For the top of the tower: \[ \tan 45^\circ = \dfrac{h}{10} \Rightarrow 1 = \dfrac{h}{10} \Rightarrow h = 10 \, m \]

For the top of the flagstaff: \[ \tan 60^\circ = \dfrac{h + x}{10} \Rightarrow \sqrt{3} = \dfrac{h + x}{10} \Rightarrow h + x = 10\sqrt{3} \]

Step 3: Substitute the value of \( h \).
\[ 10 + x = 10\sqrt{3} \Rightarrow x = 10(\sqrt{3} - 1) \] \[ x = 10(1.732 - 1) = 7.32 \, m \]

Step 4: Conclusion.

Hence, the length of the flagstaff is approximately 7.32 m.
Quick Tip: In height and distance problems, always draw a right triangle diagram and use \(\tan \theta = \frac{opposite}{adjacent}\) for elevation or depression.

Step 1: Let the height of the pillar be \( h \) m.

Let the horizontal distance between the tower and pillar be \( x \) m.

Step 2: Use trigonometric ratios for the angles of depression.


For the top of the pillar (\(45^\circ\)): \[ \tan 45^\circ = \dfrac{50 - h}{x} \Rightarrow 1 = \dfrac{50 - h}{x} \Rightarrow x = 50 - h \]

For the bottom of the pillar (\(60^\circ\)): \[ \tan 60^\circ = \dfrac{50}{x} \Rightarrow \sqrt{3} = \dfrac{50}{x} \Rightarrow x = \dfrac{50}{\sqrt{3}} \]

Step 3: Equate both expressions for \(x\).
\[ 50 - h = \dfrac{50}{\sqrt{3}} \Rightarrow h = 50 - \dfrac{50}{\sqrt{3}} \]

Step 4: Simplify.
\[ h = 50\left(1 - \dfrac{1}{\sqrt{3}}\right) = 50\left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = \dfrac{50(\sqrt{3} - 1)}{1.732} \] \[ h \approx 50(0.577) = 28.85 \, m \]

Step 5: Conclusion.

Hence, the height of the pillar is approximately 28.87 m.
Quick Tip: Angles of depression are measured from the horizontal line of sight; their corresponding angle of elevation at the other end is equal.

Step 1: Write the formula for the area of a sector.
\[ Area of sector = \dfrac{\theta}{360^\circ} \times \pi r^2 \]

Step 2: Substitute the given values for the minor sector.
\[ \theta = 60^\circ, \quad r = 15 \, cm \] \[ Area of minor sector = \dfrac{60}{360} \times 3.14 \times 15^2 \] \[ = \dfrac{1}{6} \times 3.14 \times 225 = 117.75 \, cm^2 \]

Step 3: Find the area of the major sector.

Since the total area of the circle is: \[ \pi r^2 = 3.14 \times 15^2 = 706.5 \, cm^2 \] \[ Area of major sector = 706.5 - 117.75 = 588.75 \, cm^2 \]

Step 4: Conclusion.
\[ \boxed{Minor sector area = 117.75 \, cm^2} \] \[ \boxed{Major sector area = 588.75 \, cm^2} \] Quick Tip: Always use \(\dfrac{\theta}{360^\circ}\) for finding the fractional part of the circle corresponding to the sector’s angle.

Step 1: Given data.

Radius, \( r = 3.5 \, cm \)

Height of the cylinder, \( h = 10 \, cm \)

Step 2: Total surface area (TSA) of the item.

Since hemispheres are taken out from both ends, there are no circular bases.
Thus, \[ TSA = Curved surface area of cylinder + 2 \times Curved surface area of hemisphere \]

Step 3: Write the formulas.
\[ CSA of cylinder = 2\pi rh \] \[ CSA of one hemisphere = 2\pi r^2 \]
Therefore, \[ TSA = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) \]

Step 4: Substitute the given values.
\[ TSA = 2 \times 3.14 \times 3.5 (10 + 2 \times 3.5) \] \[ = 6.28 \times 3.5 \times 17 = 6.28 \times 59.5 = 373.66 \, cm^2 \]

Step 5: Conclusion.

Hence, the total surface area of the item is approximately 373.66 cm\(^2\).
Quick Tip: Always exclude the base area when hemispheres are removed from a cylinder. Add the curved surface area of the cylinder and two hemispheres only.