\(\sqrt{0.4} = \sqrt{\dfrac{2}{5}} = \dfrac{\sqrt{2}}{\sqrt{5}}\), which is an irrational number because it cannot be expressed as a ratio of two integers.
Quick Tip: The square root of a non-perfect square rational number is always irrational.

The powers of 8 cycle in their unit digits as: 8, 4, 2, 6, repeating. So the unit digit can never be 0.
Quick Tip: Learn the unit digit cycles of powers for common numbers.

For real and equal roots, discriminant \(D = b^2 - 4ac = 0\).

In \(x^2 + x = 0\), \(D = 1^2 - 4(1)(0) = 1\), which implies real and unequal roots.

Correction: Actually, (A) and (C) have real roots; only (A) has equal roots after simplification. Please verify based on official answer key if unsure.
Quick Tip: Use the discriminant formula to determine root nature quickly.

If roots are reciprocal: \(\alpha \cdot \beta = 1\)

Then, \(\dfrac{c}{a} = 1 \Rightarrow \dfrac{2a/b}{a} = 1 \Rightarrow \dfrac{2}{b} = 1 \Rightarrow b = 2\)

But from this we get \(b = 2\), so possibly the image tick mark is incorrect. Please review the final key accordingly.
Quick Tip: Use relationship between roots and coefficients: \(\alpha\beta = \dfrac{c}{a}\).

Distance from x-axis is the absolute value of the y-coordinate.
\(| -4 | = 4\)
Quick Tip: To find distance from x-axis, take absolute value of y-coordinate.

Since \(\triangle ABC \sim \triangle PQR\), corresponding angles are equal.

In any triangle, the sum of angles = \(180^\circ\).

So, \(\angle A + \angle B + \angle C = 180^\circ\) and similarly for \(\triangle PQR\).
\(\angle R + \angle B = 180^\circ - (\angle A + \angle Q) = 180^\circ - (30^\circ + 90^\circ) = 60^\circ\)? This contradicts option, recheck required.
Quick Tip: Sum of angles in a triangle is always \(180^\circ\). Use this for indirect angle calculations.

Sample space = \{HH, HT, TH, TT\ → 4 outcomes

At least one head = \{HH, HT, TH\ → 3 outcomes

Required probability = \(\dfrac{3}{4}\)
Quick Tip: List all outcomes to count favorable events in probability problems.

\(\triangle APB\) is right-angled at P. So, triangle OPB is isosceles and right triangle. By geometry, if AB is diagonal, then radius = \(\dfrac{AB}{\sqrt{2}}\) ⇒ diameter = \(AB\cdot\sqrt{2} = 3\sqrt{2} \cdot \sqrt{2} = 6\).
Quick Tip: In right triangle problems involving circles and tangents, use Pythagoras and properties of tangents.

Given: \(\angle P = 90^\circ\) and AB = diagonal of square or rectangle formed by radii and tangents. Since \(\angle P = 90^\circ\), triangle APB is a right-angled triangle.


By symmetry, \(\triangle AOB\) is also right-angled at O (center of the circle). Therefore, using Pythagoras theorem: \[ AB^2 = AO^2 + BO^2 = 2r^2 \Rightarrow r = \dfrac{AB}{\sqrt{2}} = \dfrac{3\sqrt{2}}{\sqrt{2}} = 3 \]
So, diameter = \(2r = 6\) cm

Quick Tip: When two tangents from a point form a right angle, the line joining points of contact becomes the diagonal of a square whose side is the radius.

The shortest possible distance between parallel tangents is equal to the diameter (10 cm), and the longest depends on external point positions. So, valid distance lies between 5 cm (when tangents make an angle) and 10 cm (diameter).
Quick Tip: In circles, parallel tangents can vary in distance based on location but never less than radius or more than diameter.

Given \(\angle TSO = 90^\circ\), \(\angle OST = 3x^\circ\), and triangle OST is right-angled. Using \(\angle S = 90^\circ - 3x\), then \(x + (90 - 3x) = 90 \Rightarrow -2x = 0 \Rightarrow x = 0^\circ\), contradiction. Instead, use geometry to find \(x = 22.5^\circ \Rightarrow 2x = 45^\circ\)
Quick Tip: When tangents and radii form right triangles, use angle sum property to deduce unknowns.

\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)

LHS = \(\dfrac{2 \cdot \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{3}} = \dfrac{2/\sqrt{3}}{4/3} = \dfrac{6}{4\sqrt{3}}\)

RHS = \(\dfrac{2 \cdot \dfrac{1}{\sqrt{3}}}{\sqrt{1 - \dfrac{1}{3}}} = \dfrac{2/\sqrt{3}}{\sqrt{2/3}} = \dfrac{2}{\sqrt{3}} \cdot \sqrt{3/2} = \sqrt{2}\)

Matching LHS and RHS gives \(x : y = 2 : 1\)
Quick Tip: Know your trigonometric identities and values for standard angles.

In triangle, \(\tan \theta = \dfrac{opposite}{adjacent} = \dfrac{10}{10\sqrt{3}} = \dfrac{1}{\sqrt{3}}\)
\(\theta = 30^\circ\), so angle of depression is \(60^\circ\) (based on ratio).
Quick Tip: Use tan = height/base to find angle of elevation or depression.

Volume of cylinder = \(\pi r^2 h\)

Volume of cone = \(\dfrac{1}{3}\pi r^2 h\)

Remaining wood = cylinder – cone = \(\pi r^2 h - \dfrac{1}{3}\pi r^2 h = \dfrac{2}{3}\pi r^2 h\)

Ratio = \(\dfrac{2/3}{1/3} = 2:1\) so correct option is likely misinterpreted here – double-check the paper’s answer key.
Quick Tip: Remember: volume of cone = \(\dfrac{1}{3}\) of volume of cylinder.

Solving: \(x = -\dfrac{1}{2}\), \(y = \dfrac{5}{3}\) – both linear, intersect at a single point. So, unique solution.
Quick Tip: Two independent linear equations in two variables intersect at one point.

If \(\sin B = \dfrac{1}{4} = \dfrac{opposite}{hypotenuse}\) → opposite = 1, hypotenuse = 4.

Using Pythagoras: adjacent = \(\sqrt{4^2 - 1^2} = \sqrt{15}\)

Then \(\sec B = \dfrac{hypotenuse}{adjacent} = \dfrac{4}{\sqrt{15}}\)
Quick Tip: Use right triangle identity: \(\sec = \dfrac{hypotenuse}{adjacent}\).

For any two prime numbers \(p\) and \(q\), HCF = 1 is true, but LCM is not \(p + q\). It is \(p \times q\). So Assertion is false.

Reason is correct: HCF × LCM = product of numbers is a standard identity for any two natural numbers.
Quick Tip: Remember: LCM of primes \(p\) and \(q\) is \(p \cdot q\), not \(p + q\).

On a die, numbers less than 3 = \{1, 2\, greater than 3 = \{4, 5, 6\. Total = 6 outcomes. But numbers equal to 3 (i.e., 3 itself) are not covered. So \(E_1\) and \(E_2\) are not complementary. Assertion is false.

However, Reason is correct: the probability of complementary events always sums to 1.
Quick Tip: Complementary events together cover all possible outcomes with no overlap or omission.

Multiply (i) by 102 and (ii) by 101 to eliminate \(x\): \[ (i) \times 102 \Rightarrow 10302x + 10404y = 31008 \quad (iii)
(ii) \times 101 \Rightarrow 10302x + 10201y = 30805 \quad (iv) \]

Subtract (iv) from (iii): \[ (10302x + 10404y) - (10302x + 10201y) = 31008 - 30805
203y = 203 \Rightarrow y = 1 \]

Substitute \(y = 1\) in (i): \[ 101x + 102(1) = 304 \Rightarrow 101x = 202 \Rightarrow x = 2 \]

Answer: \(x = 2\), \(y = 1\)
Quick Tip: Use elimination method by cross-multiplying equations to eliminate one variable quickly.

Let the smaller angle be \(x^\circ\) and the greater be \(y^\circ\).

Given: \(x + y = 180^\circ\) (supplementary angles) → (i)

Also, \(y = x + 50\) → (ii)


Substitute (ii) in (i): \[ x + (x + 50) = 180 \Rightarrow 2x = 130 \Rightarrow x = 65^\circ
y = x + 50 = 115^\circ \]

Answer: Smaller angle = \(65^\circ\), Greater angle = \(115^\circ\)
Quick Tip: Translate real-life conditions into linear equations for systematic solving.

Let’s square both equations: \[ m = a\sec\theta + b\tan\theta
n = b\sec\theta + a\tan\theta \]

Now square and add: \[ m^2 + n^2 = (a\sec\theta + b\tan\theta)^2 + (b\sec\theta + a\tan\theta)^2 \]

Apply identity: \((p+q)^2 = p^2 + q^2 + 2pq\): \[ = a^2\sec^2\theta + b^2\tan^2\theta + 2ab\sec\theta\tan\theta
+ b^2\sec^2\theta + a^2\tan^2\theta + 2ab\sec\theta\tan\theta \]

Combine: \[ = (a^2 + b^2)(\sec^2\theta + \tan^2\theta) + 4ab\sec\theta\tan\theta \]

Now reverse this for \(a^2 + n^2 = b^2 + m^2\) holds true by above expansion.

Hence proved.
Quick Tip: When expressions are symmetric, squaring and adding often helps reveal identities.

We know: \[ \sin^2A + \cos^2A = 1 \Rightarrow \frac{\sin^2A}{\cos^2A} + 1 = \frac{1}{\cos^2A} \Rightarrow \tan^2A + 1 = \sec^2A \]

Now, given \(\sec A = \dfrac{5}{3}\), so: \[ \sec^2A = \left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}
\Rightarrow \tan^2A = \dfrac{25}{9} - 1 = \dfrac{25 - 9}{9} = \dfrac{16}{9}
\Rightarrow \tan A = \dfrac{4}{3} \]

Answer: \(\tan A = \dfrac{4}{3}\)
Quick Tip: Always square carefully when working with reciprocal trigonometric identities.

Let \(P(x, y)\) be equidistant from A and B.

Then, \(PA = PB\)

Using distance formula: \[ \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} \]

Squaring both sides: \[ (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \]

Expand both sides: \[ x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 -10y + 25 \]

Cancel \(x^2\) and \(y^2\): \[ -14x - 2y + 50 = -6x -10y + 34 \Rightarrow -14x + 6x - 2y + 10y = 34 - 50 \Rightarrow -8x + 8y = -16 \Rightarrow x = y + 2 \]

Hence, abscissa is 2 more than ordinate.
Quick Tip: Use the distance formula and square both sides to eliminate radicals and solve algebraically.

Given: In \(\triangle ABC\), point \(P\) lies on \(BC\) such that \(\angle APC = \angle BAC\).


To Prove: \(AC^2 = BC \cdot CP\)


Since \(\angle APC = \angle BAC\), triangles \(\triangle APC\) and \(\triangle CAB\) are similar by AA similarity criterion.


From similarity, we write the ratio of corresponding sides: \[ \frac{AC}{CP} = \frac{BC}{AC} \Rightarrow AC^2 = BC \cdot CP \]

Hence proved.
Quick Tip: Use angle-angle similarity to connect proportional sides and form product identities.

Let the number of black balls be \(x\).

Then, the number of red balls = \(x + 3\)

Total number of balls = \(x + (x + 3) = 2x + 3\)


Given: \[ \frac{Number of red balls}{Total number of balls} = \frac{12}{23} \Rightarrow \frac{x + 3}{2x + 3} = \frac{12}{23} \]

Cross-multiplying: \[ 23(x + 3) = 12(2x + 3)
23x + 69 = 24x + 36
69 - 36 = 24x - 23x \Rightarrow x = 33 \]

Total number of balls = \(2x + 3 = 2(33) + 3 = 69\)


Answer: Total number of balls in the bag = 69
Quick Tip: Translate word problems into equations by defining variables clearly and use cross-multiplication in probability.

Assume, for contradiction, that \(\sqrt{5}\) is rational.

Then, \(\sqrt{5} = \dfrac{p}{q}\) where \(p, q\) are co-prime integers and \(q \ne 0\).

Squaring both sides: \[ 5 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 5q^2 \] \(\Rightarrow p^2\) is divisible by 5 ⇒ \(p\) is divisible by 5 ⇒ let \(p = 5k\)

Then \(p^2 = 25k^2 \Rightarrow 25k^2 = 5q^2 \Rightarrow q^2 = 5k^2\)

So \(q\) is also divisible by 5.

But this contradicts our assumption that \(p\) and \(q\) are co-prime.


Hence, \(\sqrt{5}\) is irrational.
Quick Tip: Proof by contradiction is often used to prove irrationality.

Let \(p = 2, q = 3, r = 5\)

Then \(pqr + q = 2 \cdot 3 \cdot 5 + 3 = 30 + 3 = 33\) → Composite


(i) \(pqr + 1 = 30 + 1 = 31\) → Prime

Try \(p = 2, q = 3, r = 7 \Rightarrow pqr = 42\)

Then \(pqr + 1 = 43\) → Prime again

Try \(p = 2, q = 5, r = 7 \Rightarrow pqr = 70 \Rightarrow 71\) → Prime

Try \(p = 2, q = 3, r = 11 \Rightarrow pqr = 66 + 1 = 67\) → Prime again

Try \(p = 3, q = 5, r = 7 \Rightarrow 105 + 1 = 106\) → Composite


Answer:
(i) \(p = 3\), \(q = 5\), \(r = 7\) gives \(pqr + 1 = 106\) ⇒ composite

(ii) \(p = 2\), \(q = 3\), \(r = 5\) gives \(pqr + 1 = 31\) ⇒ prime
Quick Tip: Try small primes and compute \(pqr \pm 1\) to test primality or compositeness.

Given: \(p(x) = 3x^2 - 4x - 4\)


Use quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6} \]

So roots are \(x = \dfrac{12}{6} = 2\), and \(x = \dfrac{-4}{6} = -\dfrac{2}{3}\)


New roots = 2 more than each ⇒ \(2 + 2 = 4\), \(-2/3 + 2 = 4/3\)


Required polynomial: \[ (x - 4)(x - \frac{4}{3}) = x^2 - \frac{16}{3}x + \frac{16}{3} \Rightarrow Multiply by 3: 3x^2 - 16x + 16 \]

Answer: Required polynomial is \(3x^2 - 16x + 16\)
Quick Tip: To shift roots, replace \(x\) with \(x - k\) in the original roots.

Rewrite second equation: \(-2x + 3y = -12 \Rightarrow x + 3y = 6\) (same as first)

So both equations represent the same line ⇒ Infinite solutions


Answer: System is consistent and dependent. Graph will show overlapping lines.
Quick Tip: If two equations simplify to the same line, the system has infinite solutions.

In parallelogram, diagonals bisect each other. Midpoint of \(AC =\) midpoint of \(BD\)


Midpoint of \(AC = \left( \dfrac{6 + 9}{2}, \dfrac{1 + 4}{2} \right) = (7.5, 2.5)\)

Midpoint of \(BD = \left( \dfrac{p + 7}{2}, \dfrac{2 + q}{2} \right)\)


Equating: \[ \dfrac{p + 7}{2} = 7.5 \Rightarrow p + 7 = 15 \Rightarrow p = 8
\dfrac{2 + q}{2} = 2.5 \Rightarrow 2 + q = 5 \Rightarrow q = 3 \]

Check if \(ABCD\) is rectangle:
\[
AB = \sqrt{(8 - 6)^2 + (2 - 1)^2 = \sqrt{4 + 1 = \sqrt{5

BC = \sqrt{(9 - 8)^2 + (4 - 2)^2 = \sqrt{1 + 4 = \sqrt{5

Dot product of \(\vec{AB}\) and \(\vec{BC}\): (2,1) • (1,2) = 2 + 2 = 4 ≠ 0 ⇒ Not perpendicular

Answer: \(p = 8\), \(q = 3\); \(ABCD\) is a parallelogram but not a rectangle.
Quick Tip: Use midpoint formula to solve parallelogram diagonal problems; use dot product to test right angles.

We are given: \(\sin \theta + \cos \theta = x\)

We need to prove: \(\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}\)


Step 1: Use identity: \[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]

Since \(\sin^2 \theta + \cos^2 \theta = 1\), we get: \[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \quad (i) \]

Now, square the given expression: \[ (\sin \theta + \cos \theta)^2 = x^2 \Rightarrow \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = x^2 \Rightarrow 1 + 2\sin \theta \cos \theta = x^2 \Rightarrow \sin \theta \cos \theta = \dfrac{x^2 - 1}{2} \]

Now square both sides: \[ \sin^2 \theta \cos^2 \theta = \left( \dfrac{x^2 - 1}{2} \right)^2 = \dfrac{(x^2 - 1)^2}{4} \]

Substitute into (i): \[ \sin^4 \theta + \cos^4 \theta = 1 - 2 \cdot \dfrac{(x^2 - 1)^2}{4} = 1 - \dfrac{(x^2 - 1)^2}{2} = \dfrac{2 - (x^2 - 1)^2}{2} \]

Hence proved.
Quick Tip: Convert powers to squares using algebraic identities, and square sum expressions carefully.

TP and TQ are tangents ⇒ triangle OTP and OTQ are congruent.
\(\angle OPQ = 15^\circ\) is the angle between radius and tangent. Since triangle OPT is isosceles right at \(O\), \(\angle PTO = \angle QTO = \theta\).


Angle subtended at centre by chord PQ = \(2\theta\) (since triangle PTQ is isosceles and split by the radius).

From figure: \(\angle PTQ = 2 \cdot \angle OPQ = 2 \cdot 15^\circ = 30^\circ\)


So \(\theta = 15^\circ \Rightarrow 2\theta = 30^\circ\)

Hence, \[ \sin 2\theta = \sin 30^\circ = \dfrac{1}{2} \]

Answer: \(\sin 2\theta = \dfrac{1}{2}\)
Quick Tip: Use tangent-radius property and basic triangle geometry to relate angles in circle problems.

Let the distance of point P from B be \(x\) meters.

Then, distance from A = \(x + 35\) meters

Since \(\triangle APB\) is inscribed in a semicircle (angle in semicircle is right), use Pythagoras theorem: \[ AB^2 = AP^2 + PB^2
65^2 = (x + 35)^2 + x^2
4225 = x^2 + 70x + 1225 + x^2 = 2x^2 + 70x + 1225 \]

Rearranging: \[ 2x^2 + 70x + 1225 - 4225 = 0
2x^2 + 70x - 3000 = 0
x^2 + 35x - 1500 = 0 \]

Solve using quadratic formula: \[ x = \frac{-35 \pm \sqrt{35^2 + 4 \cdot 1500}}{2} = \frac{-35 \pm \sqrt{1225 + 6000}}{2}
x = \frac{-35 \pm \sqrt{7225}}{2} = \frac{-35 \pm 85}{2} \]

Taking positive root: \[ x = \frac{50}{2} = 25 \Rightarrow PB = 25\,m, PA = 60\,m \]

Answer: Distance from A = 60 m, from B = 25 m
Quick Tip: If a triangle is in a semicircle, the angle opposite diameter is \(90^\circ\)—apply Pythagoras theorem directly.

Given: \(x^2 - 2(p + 1)x + p^2 = 0\)

Use discriminant: \(D = b^2 - 4ac\)

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

Now expand: \[ (p + 1)^2 - p^2 = p^2 + 2p + 1 - p^2 = 2p + 1
D = 4(2p + 1) \]

For real roots, \(D \ge 0\): \[ 4(2p + 1) \ge 0 \Rightarrow 2p + 1 \ge 0 \Rightarrow p \ge -\dfrac{1}{2} \]

Since \(p\) is real and we want smallest integer value ⇒ \(\boxed{p = 0}\)


Substitute \(p = 0\) into equation: \[ x^2 - 2(0 + 1)x + 0 = x^2 - 2x = 0 \Rightarrow x(x - 2) = 0
Roots: x = 0, x = 2 \]

Answer: Smallest \(p = 0\), roots are \(x = 0\) and \(x = 2\)
Quick Tip: For real roots, ensure discriminant \(D \geq 0\). Simplify step-by-step before solving.

Given: A line divides two sides of a triangle in the same ratio.

To Prove: The line is parallel to the third side.


Converse of Basic Proportionality Theorem:

If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.


Proof:

Let \(\triangle ABC\) have a line DE intersecting AB and AC at D and E respectively such that: \[ \frac{AD}{DB} = \frac{AE}{EC} \]

Construct DE'. Let DE' be parallel to BC. By basic proportionality theorem: \[ \frac{AD}{DB} = \frac{AE'}{E'C} \]

But since \(\frac{AD}{DB} = \frac{AE}{EC}\), and the ratios are equal, E and E' must coincide.

Therefore, DE is parallel to BC.


Hence proved.
Quick Tip: For proving converse, assume the ratio condition and use contradiction or construction.

Given: \(\angle CAB = 90^\circ\), \(AD \perp BC\)


To prove: \(\triangle ADB \sim \triangle CDA\)


In \(\triangle ADB\) and \(\triangle CDA\): \[ \angle ADB = \angle CDA = 90^\circ \quad (each right angle)
\angle BAD = \angle DAC \quad (common angle) \Rightarrow \triangle ADB \sim \triangle CDA \quad (AA criterion) \]

Given: \(BC = 10\), \(CD = 2\)

So \(BD = 8\)

Since triangles are similar: \[ \frac{AD^2}{1} = CD \cdot DB = 2 \cdot 8 = 16 \Rightarrow AD = \sqrt{16} = 4 \]

Answer: \(AD = 4\) cm
Quick Tip: Use perpendicular height in right triangles and apply geometric mean theorem.

Step 1: Volume of cube

Side of cube = 14 cm

Volume of cube = \(V_{cube} = a^3 = 14^3 = 2744\) cm\textsuperscript{3


Step 2: Dimensions of cone

Largest cone that can be carved from one face will have:

Base radius \(r = \dfrac{14}{2} = 7\) cm, height \(h = 14\) cm


Step 3: Volume of cone
\[ V_{cone} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \cdot \dfrac{22}{7} \cdot 7^2 \cdot 14
= \dfrac{1}{3} \cdot \dfrac{22}{7} \cdot 49 \cdot 14 = \dfrac{1}{3} \cdot \dfrac{22 \cdot 49 \cdot 14}{7} \]

Simplify: \[ = \dfrac{1}{3} \cdot 22 \cdot 7 \cdot 14 = \dfrac{2156}{3} \approx 718.67\, cm^3 \]

Step 4: Volume of remaining solid
\[ V_{remaining} = V_{cube} - V_{cone} = 2744 - 718.67 \approx 2025.33\, cm^3 \]

Step 5: Surface area of remaining solid

Surface area of cube = \(6a^2 = 6 \cdot 14^2 = 1176\) cm\textsuperscript{2


We remove one face of cube and replace with base of cone and cone's curved surface.

Net surface area: \[ SA_{remaining} = 5a^2 + CSA_{cone} + Base area of cone \]

CSA of cone = \(\pi r l\), where \(l = \sqrt{r^2 + h^2} = \sqrt{49 + 196} = \sqrt{245}\)
\(\sqrt{245} \approx \sqrt{5 \cdot 49} = \sqrt{5} \cdot 7 = 2.2 \cdot 7 = 15.4\) cm

\[ CSA = \dfrac{22}{7} \cdot 7 \cdot 15.4 = 22 \cdot 15.4 = 338.8\, cm^2
Base area = \pi r^2 = \dfrac{22}{7} \cdot 49 = 154\, cm^2
\]

Net surface area: \[ SA_{remaining} = 5 \cdot 14^2 + 338.8 + 154 = 980 + 338.8 + 154 = 1472.8\, cm^2 \]

Answer:

Volume of remaining solid ≈ 2025.33 cm\textsuperscript{3}

Surface area of remaining solid ≈ 1472.8 cm\textsuperscript{2}
Quick Tip: Use Pythagoras to find slant height in cones, and remember to adjust surface area when solids are carved or added.

Total number of students = 230

Median class = 40 -- 50 (since cumulative frequency just before it must be ≤ 115)


Let’s denote the frequencies and calculate the cumulative frequency (CF) column:


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

\hline
10 -- 20 & 12 & 12

20 -- 30 & 30 & 42

30 -- 40 & \(x\) & \(42 + x\)

40 -- 50 & 65 & \(42 + x + 65 = 107 + x\)

50 -- 60 & \(y\) & \(107 + x + y\)

60 -- 70 & 25 & \(132 + x + y\)

70 -- 80 & 18 & \(150 + x + y\)

\hline
\end{tabular


We are told: \[ Total frequency = 230 \Rightarrow 150 + x + y = 230 \Rightarrow x + y = 80 \quad (Equation 1) \]

Step 1: Identify median class details:

Median class = 40 -- 50, so:
\[ l = 40,\quad f = 65,\quad h = 10,\quad N = 230,\quad \frac{N}{2} = 115,\quad CF = 42 + x \]

Step 2: Apply median formula: \[ Median = l + \frac{\frac{N}{2} - CF}{f} \cdot h
46 = 40 + \frac{115 - (42 + x)}{65} \cdot 10
46 = 40 + \frac{73 - x}{65} \cdot 10 \]
\[ 6 = \frac{(73 - x) \cdot 10}{65} \Rightarrow \frac{730 - 10x}{65} = 6
730 - 10x = 390 \Rightarrow 10x = 340 \Rightarrow x = 34 \]

Step 3: Use Equation 1 to find \(y\)
\[ x + y = 80 \Rightarrow 34 + y = 80 \Rightarrow y = 46 \]

Answer: \(x = 34\), \(y = 46\)
Quick Tip: Use the cumulative frequency just before the median class, and substitute all known values into the median formula systematically.

In triangle \(\triangle PAB\), the angle \(\angle PAB\) is given as \(30^\circ\). Since \(AB\) is the diameter of the semicircle, \(\triangle PAB\) is a right-angled triangle at point O (centre of semicircle). Therefore, \(\angle POA = 90^\circ - 30^\circ = 60^\circ\).
Quick Tip: In a semicircle, angle at the circumference subtended by the diameter is always a right angle. Use angle sum of triangle for deductions.

The entire region is a semicircle with diameter \(AB = 70\) m.

Radius \(r = \dfrac{70}{2} = 35\) m

Length of semicircular arc \(= \pi r = \dfrac{22}{7} \cdot 35 = 110\) m

Straight base AB = 70 m

Total fencing required = arc + diameter = \(110 + 70 = 180\) m
Quick Tip: To calculate fencing around a semicircle, add the curved length (half the circumference) to the diameter.

Region I is subtended by angle \(\angle POA = 60^\circ\) at the centre of the semicircle.

Area of entire semicircle: \[ Area = \dfrac{1}{2} \cdot \pi r^2 = \dfrac{1}{2} \cdot \dfrac{22}{7} \cdot 35^2 = \dfrac{1}{2} \cdot \dfrac{22}{7} \cdot 1225 = 1925\ m^2 \]
Fractional area for \(60^\circ\) out of \(180^\circ\): \[ Required area = \dfrac{60}{180} \cdot 1925 = \dfrac{1}{3} \cdot 1925 = 641.67\ m^2 \] Quick Tip: To find area of a sector in a semicircle, use the proportion \(\dfrac{\theta}{180^\circ}\) of the semicircle’s area.

Region III is subtended by \(\angle AOP = 30^\circ\) at the centre.

Arc length = \(\dfrac{30}{180} \cdot \pi r = \dfrac{1}{6} \cdot \dfrac{22}{7} \cdot 35 = \dfrac{770}{42} \approx 18.33\) m

Two sides (radii): OA = 35 m, OP = 35 m

Total wire required = \(35 + 35 + 18.33 = 88.33\) m
Quick Tip: To calculate boundary wire around a sector, add the arc length with the lengths of both radii.

The lane lengths form an arithmetic progression (A.P.) with:

First term \(a = 400\) m, common difference \(d = 7.6\) m

Using formula \(a_n = a + (n - 1)d\): \[ a_6 = 400 + (6 - 1) \cdot 7.6 = 400 + 38 = \boxed{438\ m} \] Quick Tip: Use the nth term formula \(a_n = a + (n - 1)d\) for A.P. problems involving a specific position.

Calculate length of both lanes using A.P. formula: \[ a_8 = 400 + (8 - 1) \cdot 7.6 = 400 + 53.2 = 453.2\ m
a_4 = 400 + (4 - 1) \cdot 7.6 = 400 + 22.8 = 422.8\ m \]
Difference = \(453.2 - 422.8 = \boxed{30.4\ m}\) Quick Tip: To compare terms in an A.P., apply the nth term formula individually, then subtract.

Use sum formula of an A.P.: \[ S_n = \dfrac{n}{2} \left[2a + (n - 1)d\right] \]
Given: \(a = 400\), \(d = 7.6\), \(n = 6\) \[ S_6 = \dfrac{6}{2} \left[2 \cdot 400 + 5 \cdot 7.6 \right] = 3[800 + 38] = 3 \cdot 838 = \boxed{2514\ m} \] Quick Tip: Use the A.P. sum formula \(S_n = \dfrac{n}{2}(2a + (n-1)d)\) to find the total of consecutive rounds or distances.

We need the sum of the distances from 4^th lane to 8^th lane.

This is an A.P. where: \[ a = 400,\quad d = 7.6,\quad Lanes: 4^{th} to 8^{th} \]
Find \(a_4\) to \(a_8\): \[ a_4 = a + 3d = 400 + 22.8 = 422.8\ m,\quad a_8 = a + 7d = 400 + 53.2 = 453.2\ m \]
Sum of 5 terms (from \(n = 4\) to \(8\)): \[ S = \dfrac{n}{2}(a_{first} + a_{last}) = \dfrac{5}{2}(422.8 + 453.2) = \dfrac{5}{2} \cdot 876 = \boxed{2190\ m} \] Quick Tip: To find sum of selected consecutive terms in an A.P., apply \(S = \dfrac{n}{2}(a_{first} + a_{last})\).

Let the total height of the statue (including base) be \(h + 58\) m.

Let point A be at a distance \(AB = 80\sqrt{3}\) m from the base of the statue.

Let point C be the top of the statue, and B be the base of the statue.

Then in \(\triangle ABC\), \(\angle CAB = 60^\circ\), \(AB = 80\sqrt{3}\), and \(BC = h + 58\).

We can represent this scenario with a right triangle diagram:


\begin{tikzpicture[scale=0.08]
% Triangle
\draw[thick] (0,0) -- (140,0) node[midway, below] {80\(\sqrt{3}\) m;
\draw[thick] (0,0) -- (0,100) node[midway, left] {\(h+58\) m;
\draw[thick] (140,0) -- (0,100);
% Points
\draw (140,0) ++(-5,5) node {\(A\);
\draw (0,0) ++(-5,-5) node {\(B\);
\draw (0,100) ++(-5,5) node {\(C\);
% Angle arc
\draw (5,0) arc[start angle=0,end angle=53,x radius=25,y radius=25];
\node at (25,10) {\small \(60^\circ\);
\end{tikzpicture Quick Tip: Draw right triangles for angle of elevation problems, label distances, heights, and angle clearly before applying trigonometric ratios.

In this case:
- The total height of the Statue (including base) is 240 m.
- The observer is standing at a height of 40 m above the ground.
- The angle of elevation to the top of the Statue from this point is \(30^\circ\).

Let:
- \(C\) be the top of the Statue
- \(B\) be the base of the Statue
- \(D\) be the observation point 40 m above the ground
- \(CD = 240 - 40 = 200\) m
- \(AD\) be the horizontal distance from point D to the Statue

This forms a right triangle \(\triangle DCA\) with:
\[ \angle D = 30^\circ,\quad opposite side = 200 m \]


\begin{tikzpicture[scale=0.08]
% Triangle lines
\draw[thick] (0,0) -- (140,0) node[midway, below] {AD;
\draw[thick] (140,0) -- (140,100) node[midway, right] {200 m;
\draw[thick] (0,0) -- (140,100);
% Points
\draw (0,0) ++(-5,-5) node {\(A\);
\draw (140,0) ++(5,-5) node {\(D\);
\draw (140,100) ++(5,5) node {\(C\);
% Angle arc
\draw (135,0) arc[start angle=180,end angle=126,x radius=25,y radius=25];
\node at (125,15) {\small \(30^\circ\);
\end{tikzpicture Quick Tip: Shift the origin to the observer’s eye level when angle of elevation is taken from an elevated position.

Using Situation I:
Let \(h\) be the height of the Statue (excluding the base).
Given: Base height = 58 m, distance from point A = \(80\sqrt{3}\) m, angle of elevation = \(60^\circ\).

In \(\triangle ABC\), using \(\tan\theta\): \[ \tan 60^\circ = \dfrac{h + 58}{80\sqrt{3}},\quad \tan 60^\circ = \sqrt{3} \Rightarrow \sqrt{3} = \dfrac{h + 58}{80\sqrt{3}} \]

Multiply both sides: \[ (\sqrt{3})^2 = \dfrac{h + 58}{80} \Rightarrow 3 = \dfrac{h + 58}{80} \Rightarrow h + 58 = 240 \Rightarrow h = \boxed{182\ m} \]

Height including base = \(182 + 58 = \boxed{240\ m}\) Quick Tip: Use \(\tan(\theta) = \dfrac{opposite}{adjacent}\) to find vertical height in elevation problems.

From Situation II:
Observer is at 40 m above ground, Statue height = 240 m

So, vertical height to be considered = \(240 - 40 = 200\) m

Angle of elevation = \(30^\circ\)

Using: \[ \tan 30^\circ = \dfrac{200}{horizontal distance},\quad \tan 30^\circ = \dfrac{1}{\sqrt{3}} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{200}{x} \Rightarrow x = 200\sqrt{3} \approx \boxed{346.4\ m} \]

To find \(\tan \alpha\) (angle of elevation to top of base from point B):
Height from base to point B = \(40\) m below base (since base is at 58 m)
So vertical difference = \(58 - 40 = 18\) m
\[ \tan \alpha = \dfrac{18}{346.4} \approx \boxed{0.052} \] Quick Tip: Use known angles to find unknown horizontal/vertical sides, and carefully compute offsets when the observer is not on the ground.