Step 1: Recall the formula relating HCF and LCM.

We know that: \[ HCF \times LCM = Product of the two numbers \]

Step 2: Substitute the known values.

Here, the numbers are 96 and 404. \[ Product = 96 \times 404 \] \[ HCF = 4 \]

Step 3: Compute the product.
\[ 96 \times 404 = 38784 \]

Step 4: Calculate the LCM.
\[ LCM = \frac{Product}{HCF} = \frac{38784}{4} = 9696 \]


Final Answer: \[ \boxed{9696} \] Quick Tip: Use the formula \(HCF \times LCM = Product of the numbers\) to easily solve LCM/HCF problems.

Step 1: Recall the distance formula.

The distance from the origin \((0, 0)\) to any point \((x, y)\) is given by: \[ d = \sqrt{x^2 + y^2} \]

Step 2: Substitute the coordinates.

For the point \((2, 5)\): \[ d = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} \]

Step 3: Simplify the expression.
\[ d = \sqrt{29} \]


Final Answer: \[ \boxed{\sqrt{29}} \] Quick Tip: Use the formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) to calculate distances. For the distance from the origin, it simplifies to \(\sqrt{x^2 + y^2}\).

Step 1: Identify the value of the trigonometric function.
\[ \tan 45^\circ = 1 \]

Step 2: Substitute this value into the expression.
\[ \sqrt{2} + \tan 45^\circ = \sqrt{2} + 1 \]

Step 3: Conclude the result.

This matches option (C).


Final Answer: \[ \boxed{\sqrt{2} + 1} \] Quick Tip: Keep in mind that \(\tan 45^\circ = 1\). This simple substitution simplifies the problem.

Step 1: Recall the standard trigonometric identity.
\[ \sec^2 A = 1 + \tan^2 A \]

Step 2: Substitute into the expression.
\[ \sec^2 A - \tan^2 A = (1 + \tan^2 A) - \tan^2 A \]

Step 3: Simplify the expression.
\[ \sec^2 A - \tan^2 A = 1 \]


Final Answer: \[ \boxed{1} \] Quick Tip: Remember the Pythagorean identity: \(\sec^2 A = 1 + \tan^2 A\). This identity helps solve many similar problems.

Step 1: Substitute \(x = 1\) into the equation.
\[ 2(1) + 3y = 11 \]

Step 2: Simplify.
\[ 2 + 3y = 11 \] \[ 3y = 11 - 2 = 9 \]

Step 3: Solve for \(y\).
\[ y = \frac{9}{3} = 3 \]


Final Answer: \[ \boxed{3} \] Quick Tip: Always substitute carefully and simplify step by step in linear equations.

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

Step 2: Identify the values.

Here, \(a = 21\), \(d = 18 - 21 = -3\), and \(a_n = -81\).

Step 3: Substitute into the formula.
\[ -81 = 21 + (n-1)(-3) \]

Step 4: Simplify.
\[ -81 = 21 - 3n + 3 \] \[ -81 = 24 - 3n \] \[ -81 - 24 = -3n \] \[ -105 = -3n \] \[ n = \frac{105}{3} = 35 \]

Oops — correction: let’s carefully recompute.

Recheck Step 4: \[ -81 = 21 + (n-1)(-3) \] \[ -81 = 21 - 3n + 3 \] \[ -81 = 24 - 3n \] \[ -81 - 24 = -3n \] \[ -105 = -3n \] \[ n = 35 \]

So the correct term is the 35th, not 33rd.


Final Answer: \[ \boxed{35^{th}} \] Quick Tip: Always double-check the arithmetic in A.P. problems. Using the \(a_n = a + (n-1)d\) formula systematically avoids errors.

Step 1: Count the total balls.

Total balls = \(2 + 3 + 4 = 9\).

Step 2: Count the favorable outcomes.

Balls that are not red = Blue + White = \(2 + 3 = 5\).

Step 3: Apply probability formula.
\[ P(Not Red) = \frac{Favorable outcomes}{Total outcomes} = \frac{5}{9} \]


Final Answer: \[ \boxed{\tfrac{5}{9}} \] Quick Tip: When asked for "not" probability, simply subtract the unfavorable cases from the total, or use complement rule: \(P(Not A) = 1 - P(A)\).

Step 1: Recall probability basics.

The probability of any event lies between 0 and 1.

Step 2: Certain event.

If an event is certain (sure to happen), its probability is the maximum possible value.

Step 3: Conclusion.

Therefore, the probability of a sure event is: \[ P(Sure Event) = 1 \]


Final Answer: \[ \boxed{1} \] Quick Tip: Remember: Impossible event probability = 0, Sure event probability = 1.

Step 1: Recall Basic Proportionality Theorem (Thales' theorem).

If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

So, \[ \frac{AD}{DB} = \frac{AE}{EC} \]

Step 2: Substitute the values from the figure.

Here, \(AD = 3\), \(DB = 6\), \(AE = 2\), and \(EC = 4\).
\[ \frac{AD}{DB} = \frac{3}{6} = \frac{1}{2}, \quad \frac{AE}{EC} = \frac{2}{4} = \frac{1}{2} \]

So, triangles \(ADE\) and \(ABC\) are similar.

Step 3: Ratio of corresponding sides.

Since \(\triangle ADE \sim \triangle ABC\), \[ \frac{DE}{BC} = \frac{AD}{AB} = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3} \]

Wait — let’s recheck carefully.

Step 3 (Revised):

For similar triangles: \[ \frac{DE}{BC} = \frac{AD}{AB} \]

Now, \(AD = 3\), \(AB = AD + DB = 3 + 6 = 9\).
\[ \frac{DE}{BC} = \frac{3}{9} = \frac{1}{3} \]


Final Answer: \[ \boxed{\tfrac{1}{3}} \] Quick Tip: In problems with parallel lines inside triangles, always apply the Basic Proportionality Theorem: \(\triangle ADE \sim \triangle ABC\). Use side ratios carefully to avoid mistakes.

Step 1: Apply Pythagoras theorem.

In a right-angled triangle, \[ OQ^2 = O P^2 + P Q^2 \]
where \(O\) = centre, \(P\) = point of tangency, \(Q\) = external point.

Step 2: Substitute values.
\[ 25^2 = r^2 + 24^2 \]

Step 3: Simplify.
\[ 625 = r^2 + 576 \] \[ r^2 = 625 - 576 = 49 \] \[ r = 7 \, cm \]

Oops — correction: I mistakenly selected option (C). The correct radius is **7 cm**.


Final Answer: \[ \boxed{7 \, cm} \] Quick Tip: Use the Pythagoras theorem: \((distance from centre)^2 = (radius)^2 + (tangent length)^2\).

Step 1: Recall arc length formula.
\[ l = \frac{\theta}{360^\circ} \times 2\pi r \]

Step 2: Substitute values.
\[ l = \frac{30}{360} \times 2\pi \times 6 \]

Step 3: Simplify.
\[ l = \frac{1}{12} \times 12\pi = \pi \]


Final Answer: \[ \boxed{\pi \, cm} \] Quick Tip: Arc length is always proportional to the central angle: \(l = \frac{\theta}{360^\circ} \times circumference\).

Step 1: Surface area formula.

Total Surface Area = CSA of cylinder + CSA of hemisphere + base area of cylinder

Step 2: Cylinder CSA.
\[ CSA of cylinder = 2\pi r h = 2\pi (3)(5) = 30\pi \]

Step 3: Hemisphere CSA.
\[ CSA of hemisphere = 2\pi r^2 = 2\pi (3^2) = 18\pi \]

Step 4: Base area of cylinder.
\[ \pi r^2 = \pi (3^2) = 9\pi \]

Step 5: Total.
\[ 30\pi + 18\pi + 9\pi = 57\pi \]

So the correct answer is (D) \(57\pi \, cm^2\), not (B).


Final Answer: \[ \boxed{57\pi \, cm^2} \] Quick Tip: Be careful: when a hemisphere is placed on a cylinder, we exclude the common base but include the cylinder base.

Step 1: Recall tangent property.

At the point of contact, the tangent is always perpendicular to the radius.

Step 2: Conclusion.
\[ \angle(radius, tangent) = 90^\circ \]


Final Answer: \[ \boxed{90^\circ} \] Quick Tip: Always remember: A tangent to a circle makes a right angle with the radius at the point of contact.

Step 1: Recall identities.
\[ 1 + \tan^2 A = \sec^2 A, \quad 1 + \cot^2 A = \csc^2 A \]

Step 2: Rewrite expression.
\[ \dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \dfrac{\sec^2 A}{\csc^2 A} \]

Step 3: Simplify.
\[ \dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A \]


Final Answer: \[ \boxed{\tan^2 A} \] Quick Tip: Use Pythagorean identities \(\sec^2 A = 1 + \tan^2 A\) and \(\csc^2 A = 1 + \cot^2 A\) to simplify ratios easily.

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

Step 2: Substitute values.
\[ = \frac{60}{360} \times \pi \times (7^2) \] \[ = \frac{1}{6} \times \pi \times 49 \]

Step 3: Simplify.
\[ = \frac{49\pi}{6} \]

Taking \(\pi = \tfrac{22}{7}\): \[ = \frac{49 \times 22}{6 \times 7} = \frac{154}{6} = \frac{77}{3} \, cm^2 \]

Correction — the correct answer is (B) \(\tfrac{77}{3}\).


Final Answer: \[ \boxed{\tfrac{77}{3} \, cm^2} \] Quick Tip: Always check if \(\pi\) needs to be taken as \(\frac{22}{7}\). For exact forms, leave answers in terms of \(\pi\).

Step 1: Recall surface area formula.
\[ Surface Area = 4\pi r^2 \]

Step 2: Convert mixed fraction.
\[ 452 \tfrac{4}{7} = \frac{3168}{7} \, cm^2 \]

Step 3: Solve for \(r\).
\[ 4\pi r^2 = \frac{3168}{7} \]

Take \(\pi = \tfrac{22}{7}\): \[ 4 \times \frac{22}{7} \times r^2 = \frac{3168}{7} \] \[ \frac{88}{7} r^2 = \frac{3168}{7} \] \[ 88 r^2 = 3168 \] \[ r^2 = 36 \] \[ r = 6 \, cm \]

Correction — the correct answer is (A) 6 cm, not 9 cm.


Final Answer: \[ \boxed{6 \, cm} \] Quick Tip: Always first convert mixed fractions to improper fractions before substituting.

Step 1: Recall the product of roots formula.

For a quadratic \(ax^2 + bx + c\), product of zeroes = \(\dfrac{c}{a}\).

Step 2: Identify coefficients.

Here, \(a = 4\), \(b = -4\), \(c = -1\).

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


Final Answer: \[ \boxed{-\tfrac{1}{4}} \] Quick Tip: Always remember: Sum of roots = \(-b/a\), Product of roots = \(c/a\).

Step 1: Recall discriminant formula.
\[ D = b^2 - 4ac \]

Step 2: Identify coefficients.

Here, \(a = 3\), \(b = -4\), \(c = \tfrac{4}{3}\).

Step 3: Substitute values.
\[ D = (-4)^2 - 4(3)\left(\tfrac{4}{3}\right) \] \[ D = 16 - 16 = 0 \]


Final Answer: \[ \boxed{0} \] Quick Tip: If \(D = 0\), the quadratic has equal real roots.

Step 1: Find midpoints of each class.
\[ Midpoints = 2, \; 6, \; 10, \; 14, \; 18 \]

Step 2: Multiply frequency and midpoint.
\[ f \times x = 4 \times 2 + 6 \times 6 + 6 \times 10 + 5 \times 14 + 4 \times 18 \] \[ = 8 + 36 + 60 + 70 + 72 = 246 \]

Step 3: Sum of frequencies.
\[ \Sigma f = 4 + 6 + 6 + 5 + 4 = 25 \]

Step 4: Calculate mean.
\[ \bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{246}{25} = 9.84 \]

Correction: The correct answer is (C) 9.84, not (D).


Final Answer: \[ \boxed{9.84} \] Quick Tip: For grouped data, mean is always \(\dfrac{\Sigma f x}{\Sigma f}\), where \(x\) is the class midpoint.

Step 1: Total frequency.
\[ N = 6 + 8 + 10 + 9 + 12 = 45 \]

Step 2: Find median class.

Median class is the class containing \(\dfrac{N}{2} = \dfrac{45}{2} = 22.5\).

Step 3: Cumulative frequencies.
\[ Cumulative frequencies = 6, \; 14, \; 24, \; 33, \; 45 \]

The \(22.5^{th}\) value lies in the class \(20-30\).


Final Answer: \[ \boxed{20-30} \] Quick Tip: The median class is found by locating the \(\dfrac{N}{2}\)-th item using cumulative frequencies.

Using distance formula: \[ \sqrt{(0-5)^2+(1+3)^2} = \sqrt{(0-x)^2+(1-6)^2} \] \[ \sqrt{25+16} = \sqrt{x^2+25} \] \[ \sqrt{41} = \sqrt{x^2+25} \] \[ x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \]


Final Answer: \[ \boxed{x = 4 \; or \; x = -4} \] Quick Tip: Equidistant condition always gives equality of distances from the given point. Apply distance formula carefully.

The modal class is the class with highest frequency = \(30-40\).

Formula: \[ Mode = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]

Here, \(L = 30, f_1=21, f_0=18, f_2=15, h=10\).
\[ Mode = 30 + \left( \frac{21-18}{2(21)-18-15} \right)\times 10 = 30 + \frac{3}{9} \times 10 = 30 + 3.33 = 33.33 \]


Final Answer: \[ \boxed{33.33} \] Quick Tip: The mode lies in the class with highest frequency. Use the formula with \(f_0, f_1, f_2\).

Step 1: Express \(y\) from first equation.
\[ y=3-5x \]

Step 2: Substitute into second equation.
\[ 6x-5(3-5x)=\tfrac{1}{2} \] \[ 6x-15+25x=\tfrac{1}{2} \] \[ 31x-15=\tfrac{1}{2} \]

Step 3: Simplify.
\[ 31x=\tfrac{1}{2}+15 = \tfrac{1}{2}+ \tfrac{30}{2} = \tfrac{31}{2} \] \[ x=\tfrac{31}{62} = \tfrac{1}{2} \]

Step 4: Find \(y\).
\[ y=3-5\left(\tfrac{1}{2}\right)=3-\tfrac{5}{2}=\tfrac{1}{2} \]


Final Answer: \[ \boxed{x=\tfrac{1}{2}, \; y=\tfrac{1}{2}} \] Quick Tip: For solving pairs of linear equations, substitution method is effective when one equation can be easily expressed in terms of a variable.

Step 1: Let length = \(l\), breadth = \(b\).

Area = \(lb\).

Condition 1: \[ (l+2)(b-2) = lb-28 \] \[ lb-2l+2b-4 = lb-28 \] \[ -2l+2b= -24 \quad \Rightarrow \quad l-b=12 \quad ...(1) \]

Condition 2: \[ (l-1)(b+2) = lb+33 \] \[ lb+2l-b-2=lb+33 \] \[ 2l-b=35 \quad ...(2) \]

Step 2: Solve equations.

From (1): \(l=b+12\).

Substitute in (2): \[ 2(b+12)-b=35 \] \[ 2b+24-b=35 \] \[ b+24=35 \quad \Rightarrow \quad b=11 \]
\[ l=b+12=23 \]

Step 3: Find area.
\[ Area = l \times b = 23 \times 11 = 253 \]


Final Answer: \[ \boxed{253 \, m^2} \] Quick Tip: Translate word problems into equations systematically. Rectangle problems often reduce to simultaneous equations.

Step 1: Let distance from \(O\) to foot of building = \(x\), height of flagstaff = \(h\).


For building of height 20 m, \[ \tan 30^\circ = \frac{20}{x} \quad \Rightarrow \quad x=\frac{20}{\tfrac{1}{\sqrt{3}}}=20\sqrt{3} \]

Step 2: Apply to total height (building + flagstaff).

Total height = \(20+h\).
\[ \tan 45^\circ = \frac{20+h}{x} \] \[ 1 = \frac{20+h}{20\sqrt{3}} \quad \Rightarrow \quad 20+h=20\sqrt{3} \] \[ h=20(\sqrt{3}-1) \]

Step 3: Approximate value.
\[ h=20(1.732-1)=20(0.732)=14.64 \, m \]


Final Answer: \[ \boxed{Height of flagstaff = 20(\sqrt{3}-1) \approx 14.64 \, m}, \quad \boxed{Distance of building from O = 20\sqrt{3} \approx 34.64 \, m} \] Quick Tip: In such problems, always first find the horizontal distance using one triangle, then use it again in the larger triangle with flagstaff.

Step 1: Let breadth of river = \(AB\).

Let point on bridge = \(P\), banks = \(A, B\), with \(PA \perp AB\), height \(PA=4.5\).

Step 2: For angle of depression \(30^\circ\).
\[ \tan 30^\circ = \frac{PA}{AD} = \frac{4.5}{AD} \quad \Rightarrow \quad AD=\frac{4.5}{1/\sqrt{3}}=4.5\sqrt{3} \]

Step 3: For angle of depression \(45^\circ\).
\[ \tan 45^\circ = \frac{PA}{DB} = \frac{4.5}{DB} \quad \Rightarrow \quad DB=4.5 \]

Step 4: Total breadth.
\[ AB=AD+DB=4.5\sqrt{3}+4.5=4.5(\sqrt{3}+1) \] \[ =4.5(1.732+1)=4.5 \times 2.732=12.29 \, m \]


Final Answer: \[ \boxed{AB=4.5(\sqrt{3}+1) \approx 12.29 \, m} \] Quick Tip: Angles of depression are measured from the horizontal, so they form right triangles with the vertical height of the bridge.

Step 1: Volume of cube.
\[ V_{cube} = a^3 = 14^3 = 2744 \, cm^3 \]

Step 2: Volume of largest sphere inscribed in cube.

Radius of sphere = half of cube’s side = \(\tfrac{14}{2} = 7\) cm.
\[ V_{sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (7^3) = \frac{4}{3}\pi (343) = \frac{1372}{3}\pi \]

Using \(\pi=\tfrac{22}{7}\): \[ V_{sphere} = \frac{1372}{3}\times \frac{22}{7} = \frac{30184}{21} \approx 1437.33 \, cm^3 \]

Step 3: Volume of peeled wood.
\[ V_{peeled} = V_{cube} - V_{sphere} = 2744 - 1437.33 = 1306.67 \, cm^3 \]


Final Answer: \[ \boxed{1306.67 \, cm^3 \; (approx)} \] Quick Tip: The largest sphere that can fit inside a cube has diameter equal to the side of the cube.

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

Step 2: Outer sector (radius 14).
\[ Outer area = \frac{30}{360}\pi (14^2) = \frac{1}{12}\pi (196) = \frac{196}{12}\pi = \frac{49}{3}\pi \]

Step 3: Inner sector (radius 7).
\[ Inner area = \frac{30}{360}\pi (7^2) = \frac{1}{12}\pi (49) = \frac{49}{12}\pi \]

Step 4: Shaded area (difference).
\[ Shaded area = \frac{49}{3}\pi - \frac{49}{12}\pi \] \[ = \left(\frac{196-49}{12}\right)\pi = \frac{147}{12}\pi = \frac{49}{4}\pi \]

Using \(\pi=\tfrac{22}{7}\): \[ Shaded area = \frac{49}{4}\times \frac{22}{7} = \frac{154}{4} = 38.5 \, cm^2 \]


Final Answer: \[ \boxed{38.5 \, cm^2} \] Quick Tip: For ring-shaped sectors, subtract the area of smaller sector from the larger one.