To find the sides, we factorize the volume polynomial \(V(x) = x^3 - 7x + 6\).


Substitute \(x=1\) into the polynomial: \(1^3 - 7(1) + 6 = 1 - 7 + 6 = 0\).


Since \(V(1) = 0\), \((x-1)\) is a factor.


Perform division: \((x^3 - 7x + 6) \div (x-1) = x^2 + x - 6\).


Now factor the quadratic quotient: \(x^2 + x - 6 = (x+3)(x-2)\).


Thus, the dimensions are \((x-1)\), \((x-2)\), and \((x+3)\).


Comparing the three linear factors, \(x+3\) is the largest for any valid \(x\).
Quick Tip: Use the Rational Root Theorem to find the first integer root (factors of the constant term). Once one factor is found, reducing the cubic to a quadratic makes the rest easy.

Let the roots be \(\alpha = \frac{4+\sqrt{7}}{2}\) and \(\beta = \frac{4-\sqrt{7}}{2}\).


Sum of roots \(S = \alpha + \beta = \frac{4+\sqrt{7} + 4-\sqrt{7}}{2} = \frac{8}{2} = 4\).


Product of roots \(P = \alpha \beta = \frac{(4+\sqrt{7})(4-\sqrt{7})}{4} = \frac{16-7}{4} = \frac{9}{4}\).


The equation is \(x^2 - Sx + P = 0\).


Substituting values: \(x^2 - 4x + \frac{9}{4} = 0\).


Multiplying by 4 to clear the fraction: \(4x^2 - 16x + 9 = 0\).
Quick Tip: For conjugate roots \(\frac{a \pm \sqrt{b}}{c}\), the sum is simply \(\frac{2a}{c}\) and the product is \(\frac{a^2 - b}{c^2}\).

Make the coefficients of \(x\) and \(y\) same for both lines. Multiply the first equation by 2.


Eq 1: \(6x + 8y - 12 = 0\). (\(C_1 = -12\))


Eq 2: \(6x + 8y + 7 = 0\). (\(C_2 = 7\))


Distance \(d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} = \frac{|-12 - 7|}{\sqrt{6^2 + 8^2}}\).

\(d = \frac{|-19|}{\sqrt{36 + 64}} = \frac{19}{\sqrt{100}} = \frac{19}{10}\).
Quick Tip: Never apply the distance formula between parallel lines until the \(x\) and \(y\) coefficients are identical.

Let sides be \(3x, 4x, 5x\). Perimeter \(= 12x = 144 \Rightarrow x = 12\).


Sides are \(36, 48, 60\).


Since \(3^2 + 4^2 = 5^2\), this is a right-angled triangle.


Area \(= \frac{1}{2} \times base \times height = \frac{1}{2} \times 36 \times 48\).


Area \(= 18 \times 48 = 864 cm^2\).
Quick Tip: The ratio 3:4:5 always indicates a right-angled triangle. You can skip Heron's formula and directly use Area = \(0.5 \times Leg_1 \times Leg_2\).

Volume \(V = \pi r^2 h = 448 \pi\).

\(\pi r^2 (7) = 448 \pi \Rightarrow 7r^2 = 448 \Rightarrow r^2 = 64 \Rightarrow r = 8\).


Lateral Surface Area \(A = 2 \pi r h = 2 \pi (8)(7) = 112 \pi\).


Using \(\pi \approx 22/7\), \(A = 112 \times \frac{22}{7} = 16 \times 22 = 352 cm^2\).
Quick Tip: Check options first; if they are integers, substitute \(\pi = 22/7\) or check for divisibility by 11.

Given, \[ A = 4x + \frac{1}{x} \]

Taking LCM, \[ A = \frac{4x^2 + 1}{x} \]

Thus, the simplified value of \(A\) is \[ \frac{4x^2 + 1}{x} \]

Hence, the correct option is (C). Quick Tip: In competitive exams, if an option perfectly matches the simplified form of a variable defined in the question, check if the question text might have a typo asking for that variable.

Total rotation in 24 hours = \(360^\circ\).


Angular speed = \(\frac{360}{24} = 15^\circ\) per hour.


Time duration = 3 hours 20 minutes = \(3 + \frac{20}{60}\) hours = \(3 + \frac{1}{3} = \frac{10}{3}\) hours.


Angle turned = Speed \(\times\) Time = \(15 \times \frac{10}{3}\).


Angle = \(5 \times 10 = 50^\circ\).
Quick Tip: Convert mixed time units (hours and minutes) into a single fraction of hours before multiplying by the rate.

The formula for the main diagonal of a cube with side \(a\) is \(d = a\sqrt{3}\).


Given side \(a = 6\) cm.


Diagonal = \(6\sqrt{3}\) cm.
Quick Tip: Face diagonal = \(a\sqrt{2}\). Main diagonal (space diagonal) = \(a\sqrt{3}\).

Let \(x = y^{\frac{1}{3}}\). Then \(y^{\frac{2}{3}} = x^2\).


The equation becomes \(x^2 - 2x - 15 = 0\).


Factorize: \((x - 5)(x + 3) = 0\).


So, \(x = 5\) or \(x = -3\).


Case 1: \(y^{\frac{1}{3}} = 5 \Rightarrow y = 5^3 = 125\).


Case 2: \(y^{\frac{1}{3}} = -3 \Rightarrow y = (-3)^3 = -27\).


Roots are \(125, -27\).
Quick Tip: Transform equations with rational exponents into standard quadratic forms by substitution (e.g., \(t = y^{1/n}\)).

The figure shows a cyclic quadrilateral ABCD inscribed in a circle.


Property of cyclic quadrilateral: Opposite angles sum to \(180^\circ\).

\(\angle B + \angle D = 180^\circ \Rightarrow y^\circ + 120^\circ = 180^\circ \Rightarrow y = 60\).

\(\angle A + \angle C = 180^\circ \Rightarrow x^\circ + 130^\circ = 180^\circ \Rightarrow x = 50\).


Thus, \(x = 50, y = 60\).
Quick Tip: In any cyclic quadrilateral (all four vertices on a circle), opposite angles are always supplementary (add up to \(180^\circ\)).

Rearrange the equation: \(\sqrt{3}x = 2\sqrt{3} + 4 + 2\).

\(\sqrt{3}x = 2\sqrt{3} + 6\).


Divide by \(\sqrt{3}\): \(x = \frac{2\sqrt{3}}{\sqrt{3}} + \frac{6}{\sqrt{3}}\).

\(x = 2 + 2\sqrt{3}\).


Factor out 2: \(x = 2(1 + \sqrt{3})\).
Quick Tip: Rationalizing the denominator (multiplying top and bottom by the root) is often the key step in simplifying expressions with surds. Here, \(6/\sqrt{3} = 2\sqrt{3}\).

Formula for centroid \(G(x, y)\) of triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is \((\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})\).

\(x = \frac{4 + 2 + 0}{3} = \frac{6}{3} = 2\).

\(y = \frac{6 + (-2) + 2}{3} = \frac{6}{3} = 2\).


Centroid is \((2, 2)\).
Quick Tip: The centroid is simply the average of the x-coordinates and the average of the y-coordinates.

Let the angles be \(x, 5x, 12x\).


Sum of angles in a triangle is \(180^\circ\).

\(x + 5x + 12x = 180\).

\(18x = 180 \Rightarrow x = 10\).


The biggest angle is \(12x = 12(10) = 120^\circ\).
Quick Tip: Sum of ratio parts often maps directly to the total sum (180 for triangles). Always find the value of '1 unit' first.

Let \(a = x\) and \(b = \frac{2}{x}\).


The expression is in the form \((a - b)(a^2 + a \cdot b + b^2)\).


Check the middle term: \(a \cdot b = x \cdot \frac{2}{x} = 2\). This matches the middle term '2' in the second bracket.


This is the expansion for the identity \(a^3 - b^3\).


Result \(= (x)^3 - (\frac{2}{x})^3 = x^3 - \frac{8}{x^3}\).
Quick Tip: Memorize algebraic identities like \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\) to spot patterns immediately without expanding everything.

\((A+B)\)'s 1 day work = \(1/72\).

\((B+C)\)'s 1 day work = \(1/120\).

\((A+C)\)'s 1 day work = \(1/90\).


Summing all three: \(2(A+B+C) = \frac{1}{72} + \frac{1}{120} + \frac{1}{90}\).


LCM of 72, 120, 90 is 360.

\(2(A+B+C) = \frac{5 + 3 + 4}{360} = \frac{12}{360} = \frac{1}{30}\).

\((A+B+C) = \frac{1}{60}\).


To find A alone, subtract \((B+C)\) from \((A+B+C)\):

\(A = \frac{1}{60} - \frac{1}{120} = \frac{2-1}{120} = \frac{1}{120}\).


A takes 120 days.
Quick Tip: Work rate problems are easier when you sum the combined rates. Remember the sum gives \(2(A+B+C)\), so always divide by 2.

For collinear points, the slope between any two pairs must be equal.


Slope \(m_1\) between \((5,5)\) and \((10,k)\) is \(\frac{k-5}{10-5} = \frac{k-5}{5}\).


Slope \(m_2\) between \((5,5)\) and \((-5,1)\) is \(\frac{1-5}{-5-5} = \frac{-4}{-10} = \frac{2}{5}\).


Equate slopes: \(\frac{k-5}{5} = \frac{2}{5}\).

\(k - 5 = 2 \Rightarrow k = 7\).
Quick Tip: Collinear points lie on the same line, so slopes \(m_{AB} = m_{BC} = m_{AC}\). Alternatively, the area of the triangle formed by them is zero.

Given that \(AP\) is a tangent to the circle at point \(A\) and \(AB\) is a chord.

By the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the opposite arc.
\[ \therefore \angle ABC = \angle QAC = 30^\circ \]

Given, \[ \angle BAP = 80^\circ \]

Since \(P\), \(A\), and \(Q\) lie on a straight line, \[ \angle QAB = 180^\circ - 80^\circ = 100^\circ \]

Now, in \(\triangle AQC\), \[ \angle AQC = \angle QAB - \angle QAC \] \[ \angle AQC = 100^\circ - 30^\circ = 50^\circ \]
\[ \therefore \angle AQC = 50^\circ \]

Hence, the correct option is (B). Quick Tip: In secant-tangent problems, if you see values like \(x\) and \(y\), the exterior intersection angle is often the difference of associated arc angles or segment angles. Here \(80 - 30 = 50\).

Given \(\sin x + \sin^2 x = 1\).

\(\sin x = 1 - \sin^2 x\).


We know \(1 - \sin^2 x = \cos^2 x\). So, \(\sin x = \cos^2 x\).


We need to find \(\cos^2 x + \cos^4 x\).


Substitute \(\cos^2 x = \sin x\):


Expression \(= \sin x + (\sin x)^2 = \sin x + \sin^2 x\).


But the value of this is given as 1.


Therefore, the value is 1.
Quick Tip: This is a standard identity substitution problem. Always express the target equation in terms of the given relation.

Let \(2^x = 5^y = 10^{-z} = k\).


Then \(2 = k^{1/x}\), \(5 = k^{1/y}\), and \(10 = k^{-1/z}\).


We know that \(2 \times 5 = 10\).


Substitute the \(k\) terms: \(k^{1/x} \times k^{1/y} = k^{-1/z}\).


Using exponent laws: \(k^{(1/x + 1/y)} = k^{-1/z}\).


Equating exponents: \(\frac{1}{x} + \frac{1}{y} = -\frac{1}{z}\).

\(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0\).
Quick Tip: For equations of type \(a^x = b^y = c^z\), set them equal to a constant \(k\) and express bases \(a, b, c\) in terms of \(k\). Then use the multiplicative relationship between bases (\(a \cdot b = c\)).

Let the expression be \(E\). Take log of the expression (assuming base 10 or \(e\)).

\(\log E = (\log y - \log z)\log X + (\log z - \log x)\log Y + (\log x - \log y)\log Z\).


Assuming bases \(X, Y, Z\) correspond to variables \(x, y, z\) in the exponents (or simply cyclic symmetry with \(X=x, Y=y, Z=z\)):

\(\log E = (\log y - \log z)\log x + (\log z - \log x)\log y + (\log x - \log y)\log z\).


Expand: \(\log y \log x - \log z \log x + \log z \log y - \log x \log y + \log x \log z - \log y \log z\).


All terms cancel out. \(\log E = 0\).

\(E = 10^0 = 1\).
Quick Tip: Cyclic expressions of the form \(\sum a(b-c)\) or products like \(\prod x^{y-z}\) often simplify to 0 (for sums) or 1 (for products) due to term cancellation.

Given speed of train \(v = 36 km/h\).


Convert speed to m/s: \(36 \times \frac{5}{18} = 10 m/s\).


Time taken to pass a post \(t = 40 seconds\).


Distance traveled (Length of train) \(d = speed \times time\).

\(d = 10 \times 40 = 400 meters\).
Quick Tip: To convert km/h to m/s, multiply by \(5/18\). To convert m/s to km/h, multiply by \(18/5\).

Rationalize the denominator inside the square root by multiplying numerator and denominator by \((1+\sin x)\).


Expression \(= \sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}} = \sqrt{\frac{(1+\sin x)^2}{1-\sin^2 x}}\).


Since \(1-\sin^2 x = \cos^2 x\), expression \(= \sqrt{\frac{(1+\sin x)^2}{\cos^2 x}}\).


Taking the square root gives \(\frac{1+\sin x}{\cos x}\).


Split the fraction: \(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\).


Result \(= \sec x + \tan x\).
Quick Tip: Rationalizing the denominator using the conjugate (e.g., multiplying by \(1+\sin x\) for \(1-\sin x\)) is a standard technique for simplifying trigonometric fractions under radicals.

Let the side of the equilateral triangle be \(a\).


Area formula: \(\frac{\sqrt{3}}{4} a^2 = 4\sqrt{3}\).


Divide both sides by \(\sqrt{3}\): \(\frac{1}{4} a^2 = 4\).


Multiply by 4: \(a^2 = 16 \Rightarrow a = 4\).


Perimeter \(= 3a = 3(4) = 12 cm\).
Quick Tip: Memorize the formula for the area of an equilateral triangle: \(A = \frac{\sqrt{3}}{4} a^2\).

We can express \(\frac{1}{125}\) as \(\frac{1}{5^3} = 5^{-3}\).


The expression becomes \(\log_5 (5^{-3})\).


Using the property \(\log_b (b^x) = x\), the value is \(-3\).
Quick Tip: Express the argument of the logarithm as a power of the base to solve immediately.

Volume ratio \(\frac{V_1}{V_2} = \frac{1}{8}\).


Since volume is proportional to the cube of the radius (\(r^3\)), \(\left(\frac{r_1}{r_2}\right)^3 = \frac{1}{8}\).


Taking the cube root, \(\frac{r_1}{r_2} = \frac{1}{2}\).


Surface area is proportional to the square of the radius (\(r^2\)).


Area ratio \(\frac{A_1}{A_2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
Quick Tip: If the ratio of corresponding lengths is \(k\), the ratio of areas is \(k^2\) and the ratio of volumes is \(k^3\).

Convert all terms to powers of 5.

\(5\sqrt{5} = 5^1 \cdot 5^{1/2} = 5^{1.5}\).


The LHS expression is \(5^{1.5} \times 5^3 \div 5^{-1.5}\).


Using laws of exponents \(x^m \times x^n \div x^p = x^{m+n-p}\):


Exponent \(= 1.5 + 3 - (-1.5) = 1.5 + 3 + 1.5 = 6\).


So, LHS \(= 5^6\).


Equating to RHS: \(5^6 = 5^{a+2}\).

\(a + 2 = 6 \Rightarrow a = 4\).
Quick Tip: Remember that dividing by a number with a negative exponent is equivalent to multiplying by that number with a positive exponent.

The expression is a difference of two squares: \((a^2b^2)^2 - (4c^2)^2\).


Factor using \(x^2 - y^2 = (x+y)(x-y)\):

\(= (a^2b^2 + 4c^2)(a^2b^2 - 4c^2)\).


The second term \((a^2b^2 - 4c^2)\) is also a difference of squares: \((ab)^2 - (2c)^2\).


Factor further: \((ab + 2c)(ab - 2c)\).


Combined expression: \((a^2b^2 + 4c^2)(ab + 2c)(ab - 2c)\).
Quick Tip: Always check if factors obtained from \(a^2-b^2\) can be factored further using the same identity.

Time taken for \(\frac{3}{4}\) work = 12 days.


Time for full work = \(12 \times \frac{4}{3} = 16\) days.


Time for \(\frac{1}{2}\) work = \(\frac{1}{2} \times 16 = 8\) days.
Quick Tip: Use unitary method: Calculate time for 1 unit of work first, then multiply by the desired fraction of work.

First, arrange the data in ascending order: 20, 22, 23, 25, 26, 29, 31, 32, 34, 35.


Total number of observations \(n = 10\) (even number).


The median is the average of the two middle terms (\(\frac{n}{2}\)th and \((\frac{n}{2}+1)\)th).


5th term = 26, 6th term = 29.


Median = \(\frac{26 + 29}{2} = \frac{55}{2} = 27.5\).
Quick Tip: Don't forget to sort the data set before finding the median. For an even number of items, average the two middle values.

Using the identity \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\).


Here \(a=x\) and \(b=\frac{1}{x}\).

\(x^3 + \frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 - 3(x)(\frac{1}{x})\left(x+\frac{1}{x}\right)\).


Substitute \(x+\frac{1}{x} = \sqrt{3}\):

\(= (\sqrt{3})^3 - 3(1)(\sqrt{3})\).

\(= 3\sqrt{3} - 3\sqrt{3} = 0\).
Quick Tip: If \(x + 1/x = \sqrt{3}\), then \(x^3 + 1/x^3 = 0\) and \(x^6 = -1\). This is a useful standard result in algebra.

Perimeter of square = \(4 \times side = 4 \times 11 = 44 cm\).


Given Circumference of circle = Perimeter of square = \(44\).

\(2\pi r = 44 \Rightarrow 2 \times \frac{22}{7} \times r = 44\).

\(\frac{44}{7} r = 44 \Rightarrow r = 7 cm\).


Area of circle = \(\pi r^2 = \frac{22}{7} \times 7^2 = 22 \times 7 = 154 cm^2\).
Quick Tip: Common circle values: if \(r=7\), Area=154, Circumference=44. Memorizing this saves calculation time.

We know that \(3A = 2A + A\).

\(\tan 3A = \tan(2A + A)\).


Using expansion formula: \(\tan 3A = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}\).


Cross multiply: \(\tan 3A (1 - \tan 2A \tan A) = \tan 2A + \tan A\).

\(\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A\).


Rearrange terms: \(\tan 3A \tan 2A \tan A = \tan 3A - \tan 2A - \tan A\).
Quick Tip: The identity \(\tan(A+B+C) = \sum \tan A - \prod \tan A\) (when sum is multiple of \(\pi\)) or specifically \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\) is a standard trigonometric result.

\(\tan(A+B) = \sqrt{3} = \tan 60^\circ \Rightarrow A+B = 60^\circ\).

\(\cos(A-B) = \frac{\sqrt{3}}{2} = \cos 30^\circ \Rightarrow A-B = 30^\circ\).


Adding both equations: \(2A = 90^\circ \Rightarrow A = 45^\circ\).


Subtracting second from first: \(2B = 30^\circ \Rightarrow B = 15^\circ\).
Quick Tip: Know the standard trigonometric values for \(30^\circ, 45^\circ, 60^\circ\) by heart.

Calculate slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 5}{-4 - 3} = \frac{-3}{-7} = \frac{3}{7}\).


Use point-slope form with point \((3,5)\): \(y - 5 = \frac{3}{7}(x - 3)\).


Multiply by 7: \(7(y - 5) = 3(x - 3)\).

\(7y - 35 = 3x - 9\).


Rearrange to general form: \(3x - 7y - 9 + 35 = 0\).

\(3x - 7y + 26 = 0\).
Quick Tip: Formula for line through two points: \(\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}\).

Expression: \(\sqrt[3]{\frac{72.9}{0.4096}}\).


Multiply numerator and denominator by 10000 to clear decimals:

\(= \sqrt[3]{\frac{729000}{4096}}\).

\(\sqrt[3]{729000} = \sqrt[3]{729 \times 1000} = 9 \times 10 = 90\).

\(\sqrt[3]{4096} = 16\) (since \(16^3 = 4096\)).


Value \(= \frac{90}{16} = \frac{45}{8}\).

\(45 \div 8 = 5.625\).
Quick Tip: Recognize powers of small integers: \(9^3 = 729\) and \(2^{12} = 4096\) (so \(16^3 = 4096\)).

Let the width of the verandah be \(w\).


Dimensions of room: \(15 \times 12\). Area \(= 180 m^2\).


Outer dimensions including verandah: \((15+2w)\) and \((12+2w)\).


Area of verandah = Outer Area - Inner Area.

\(90 = (15+2w)(12+2w) - 180\).

\(270 = 180 + 30w + 24w + 4w^2\).

\(90 = 4w^2 + 54w\).


Divide by 2: \(2w^2 + 27w - 45 = 0\).


Factorize: \(2w^2 + 30w - 3w - 45 = 0\).

\(2w(w+15) - 3(w+15) = 0 \Rightarrow (2w-3)(w+15) = 0\).

\(w = 1.5\) or \(w = -15\). Width must be positive.

\(w = 1.5 m\).
Quick Tip: For a path of width \(w\) around a rectangle \(L \times B\), Area \(= (L+2w)(B+2w) - LB = 2w(L+B+2w)\).

Calculate \(m^2 - n^2 = (m+n)(m-n)\).

\(m+n = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = 2\tan \theta\).

\(m-n = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = 2\sin \theta\).


So, \(m^2 - n^2 = (2\tan \theta)(2\sin \theta) = 4\tan \theta \sin \theta\).


Now calculate \(mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta\).

\(= \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta (\frac{1}{\cos^2 \theta} - 1) = \sin^2 \theta \tan^2 \theta\).


So, \(\sqrt{mn} = \sin \theta \tan \theta\).


Substituting this back, \(m^2 - n^2 = 4\sqrt{mn}\).
Quick Tip: This is a standard trigonometric identity problem. Remember \(m^2-n^2 = 4\sqrt{mn}\) for \(m = \tan \theta + \sin \theta, n = \tan \theta - \sin \theta\).

Use the identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\).


Here \(A = 20^\circ\) and \(B = 70^\circ\).


Expression \(= \cos(20^\circ + 70^\circ) = \cos 90^\circ\).


Since \(\cos 90^\circ = 0\), the value is 0.
Quick Tip: Recognize the cosine addition formula pattern immediately: \(CC - SS = \cos(Sum)\).

Given, \[ p(x) = 4x^2(x^2 - 3x + 2) \] \[ p(x) = 4x^2(x - 1)(x - 2) \]

Also, \[ q(x) = 12x(x - 2)(x^2 - 4) \] \[ q(x) = 12x(x - 2)(x - 2)(x + 2) \] \[ q(x) = 12x(x - 2)^2(x + 2) \]

Now, \[ LCM of coefficients = LCM of 4 and 12 = 12 \]
\[ LCM of x^2 and x = x^2 \]

Distinct factors involved are: \[ (x - 1), (x - 2), (x + 2) \]

Taking highest powers: \[ (x - 1)^1, (x - 2)^2, (x + 2)^1 \]
\[ \therefore LCM = 12x^2(x - 1)(x - 2)^2(x + 2) \]

Now, \[ (x - 1)(x - 2) = x^2 - 3x + 2 \] \[ (x - 2)(x + 2) = x^2 - 4 \]
\[ \therefore LCM = 12x^2(x^2 - 3x + 2)(x^2 - 4) \]

Hence, the correct option is (B). Quick Tip: LCM involves taking the highest power of every prime factor and algebraic factor present in the expressions.

Complementary angles sum to \(90^\circ\).


Complement of \(75^\circ = 90^\circ - 75^\circ\).

\(= 15^\circ\).
Quick Tip: Complementary = Sum is 90. Supplementary = Sum is 180.

Simplify the terms in the denominator:

\(\sqrt{20} = 2\sqrt{5}\), \(\sqrt{40} = 2\sqrt{10}\), \(\sqrt{80} = 4\sqrt{5}\).


Denominator \(D = \sqrt{10} + 2\sqrt{5} + 2\sqrt{10} - \sqrt{5} - 4\sqrt{5}\).


Group like terms: \((\sqrt{10} + 2\sqrt{10}) + (2\sqrt{5} - \sqrt{5} - 4\sqrt{5})\).

\(D = 3\sqrt{10} - 3\sqrt{5} = 3(\sqrt{10} - \sqrt{5})\).


Expression \(= \frac{15}{3(\sqrt{10} - \sqrt{5})} = \frac{5}{\sqrt{10} - \sqrt{5}}\).


Rationalize: \(\frac{5(\sqrt{10} + \sqrt{5})}{(\sqrt{10} - \sqrt{5})(\sqrt{10} + \sqrt{5})} = \frac{5(\sqrt{10} + \sqrt{5})}{10 - 5} = \frac{5(\sqrt{10} + \sqrt{5})}{5}\).

\(= \sqrt{10} + \sqrt{5} = \sqrt{5}(\sqrt{2} + 1)\).
Quick Tip: Simplify surds (\(\sqrt{80} \to 4\sqrt{5}\)) before performing addition or subtraction.

The data set has \(n=7\) items. Mean \(= 7\).


Sum of items \(= Mean \times n = 7 \times 7 = 49\).


Sum given \(= 5 + 3 + 0.5 + 4.5 + a + 8.5 + 9.5\).


Sum \(= 8 + 5 + a + 18 = 31 + a\).


Equating sums: \(31 + a = 49\).

\(a = 49 - 31 = 18\).
Quick Tip: Mean = Sum of terms / Number of terms.

Based on the figure properties (Angles in the same segment of a circle are equal).


The angle subtended by arc DC (or similar arc depending on exact labeling) at point B is given as \(55^\circ\) (implied by the visual similarity in such problems).


If \(\angle DBC = 55^\circ\) (or \(\angle DAC\)), then \(\angle DEC\) subtended by the same arc at the circumference is equal.


Therefore, \(\angle DEC = 55^\circ\).
Quick Tip: Angles subtended by the same arc at the circumference are equal. This is the "Angles in the same segment" theorem.

Let numbers be \(x\) and \(y\).

\(x + y = 11\).

\(xy = 30\).


This corresponds to the quadratic equation \(t^2 - (x+y)t + xy = 0\).

\(t^2 - 11t + 30 = 0\).


Factorizing: \((t - 5)(t - 6) = 0\).


The numbers are 5 and 6.
Quick Tip: Check options directly: \(6+5=11\) and \(6 \times 5=30\). Done.

Slope of first line \(3x - 2y = 5\): \(m_1 = \frac{-coeff of x}{coeff of y} = \frac{-3}{-2} = \frac{3}{2}\).


Slope of second line \(2x + ky + 7 = 0\): \(m_2 = \frac{-2}{k}\).


For perpendicular lines, product of slopes \(m_1 \times m_2 = -1\).

\(\frac{3}{2} \times \left(\frac{-2}{k}\right) = -1\).

\(\frac{-3}{k} = -1\).

\(k = 3\).
Quick Tip: Lines \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) are perpendicular if \(a_1a_2 + b_1b_2 = 0\). Here: \(3(2) + (-2)(k) = 0 \Rightarrow 6 - 2k = 0 \Rightarrow k = 3\).

LCM of coefficients 12 and 18 is 36.


For variables, take the highest power present in the terms.

\(x\): max(\(x^2, x^4\)) = \(x^4\).

\(y\): max(\(y^3, y^2\)) = \(y^3\).

\(z\): max(\(z^2, z^3\)) = \(z^3\).


LCM is \(36x^4y^3z^3\).
Quick Tip: LCM involves the highest powers; HCF involves the lowest powers.

\(\tan 15^\circ = \tan(45^\circ - 30^\circ)\).


Formula: \(\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).

\(= \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}\).


Rationalize by multiplying numerator and denominator by \((\sqrt{3}-1)\).

\(= \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2}\).

\(= 2 - \sqrt{3}\).
Quick Tip: Memorize \(\tan 15^\circ = 2 - \sqrt{3}\) and \(\tan 75^\circ = 2 + \sqrt{3}\).

Simplify each term:


1. \(\sin \theta\)


2. \(\cos(90 + \theta) = -\sin \theta\) (2nd quadrant, cos is negative).


3. \(\sin(180 - \theta) = \sin \theta\) (2nd quadrant, sin is positive).


4. \(\sin(180 + \theta) = -\sin \theta\) (3rd quadrant, sin is negative).


Sum \(= \sin \theta - \sin \theta + \sin \theta - \sin \theta = 0\).
Quick Tip: Use ASTC rule (All, Sin, Tan, Cos) to determine signs of trig functions in different quadrants.

Use log laws: \(\log A - \log B - \log C = \log \frac{A}{B \times C}\).

\(= \log \left( \frac{14/15}{(3/25) \times (7/9)} \right)\).


Calculate denominator: \(\frac{3}{25} \times \frac{7}{9} = \frac{1}{25} \times \frac{7}{3} = \frac{7}{75}\).


Expression becomes \(\log \left( \frac{14/15}{7/75} \right)\).

\(= \log \left( \frac{14}{15} \times \frac{75}{7} \right)\).


Cancel terms: \(\frac{14}{7} = 2\), \(\frac{75}{15} = 5\).

\(= \log (2 \times 5) = \log 10\).


Assuming base 10, \(\log_{10} 10 = 1\).
Quick Tip: Combine log terms into a single fraction before calculating. \(\log x - \log y = \log(x/y)\).

Principal \(P = 24000\).


Rate \(R = 10%\) per annum \(\Rightarrow 5%\) per half-year.


Time \(T = 1.5\) years \(\Rightarrow n = 3\) half-years.


Amount \(A = P(1 + \frac{r}{100})^n = 24000(1 + \frac{5}{100})^3\).

\(A = 24000(\frac{21}{20})^3 = 24000 \times \frac{9261}{8000}\).

\(A = 3 \times 9261 = 27783\).


Compound Interest \(= A - P = 27783 - 24000 = 3783\).
Quick Tip: For semi-annual compounding, halve the rate and double the time periods.

Given focal length \(f = +12\) cm (convex lens).


Magnification \(m = \frac{size of image}{size of object}\). Since the image is real, it is inverted, so \(m = -\frac{1}{2}\).


We know \(m = \frac{v}{u}\), so \(\frac{v}{u} = -\frac{1}{2} \Rightarrow v = -\frac{u}{2}\).


Using the lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).


Substitute \(v\): \(\frac{1}{12} = \frac{1}{-u/2} - \frac{1}{u}\).

\(\frac{1}{12} = -\frac{2}{u} - \frac{1}{u} = -\frac{3}{u}\).

\(u = -36\) cm.


The distance is the magnitude \(|u| = 36\) cm.
Quick Tip: For real images formed by a single lens, magnification is negative. For virtual images, magnification is positive.

Newton's Law of Gravitation states \(F = G \frac{m_1 m_2}{r^2}\).


The gravitational constant \(G\) is a universal constant.


Gravitational force depends only on the masses and the distance between them.


It is independent of the intervening medium.


Therefore, the force in water (Q) is equal to the force in air (P).
Quick Tip: Unlike electrostatic force, gravitational force does not change with the medium.

The refractive index \(\mu\) of a medium is related to wavelength \(\lambda\) (Cauchy's relation): \(\mu\) decreases as \(\lambda\) increases.


Wavelength order: Red > Green > Violet (\(\lambda_R > \lambda_G > \lambda_V\)).


Refractive index order: \(\mu_V > \mu_G > \mu_R\).


Velocity in medium is \(v = c/\mu\).


Since velocity is inversely proportional to refractive index: \(V_R > V_G > V_V\).


Rearranging: \(V_V < V_G < V_R\).
Quick Tip: Red travels fastest in glass (least deviation), Violet travels slowest (most deviation). Remember "Red Racing car".

Radius \(r = 1\) m.


In half a revolution:


Distance covered = Length of semi-circular arc = \(\pi r = \pi(1) = \pi\).


Displacement = Shortest distance between start and end points (Diameter) = \(2r = 2(1) = 2\).


Ratio = Distance / Displacement = \(\pi / 2\).
Quick Tip: Displacement in circular motion is the chord length, while distance is the arc length. For half circle, chord is diameter.

According to the Cartesian sign convention:


For a mirror, a negative focal length (\(f < 0\)) indicates a Concave Mirror.


For a lens, a negative focal length (\(f < 0\)) indicates a Concave Lens.


Since both have \(f = -15\) cm, both are concave.
Quick Tip: Negative focal length always implies "Concave" for both mirrors and lenses. Positive implies "Convex".

Heat produced per second is Power (\(P\)).

\(P = 100\) J/s = 100 W.


Resistance \(R = 4 \, \Omega\).


Formula: \(P = \frac{V^2}{R}\).

\(100 = \frac{V^2}{4}\).

\(V^2 = 400\).

\(V = \sqrt{400} = 20\) V.
Quick Tip: Power formulas: \(P = VI = I^2R = V^2/R\). Choose the one connecting the given variables.

Measured value = \(9.92\). Standard value = \(9.80\).


Absolute Error = \(|9.92 - 9.80| = 0.12\).


Percentage Error = \(\frac{Error}{Standard Value} \times 100\).

\(= \frac{0.12}{9.80} \times 100\).

\(= \frac{12}{9.8} = \frac{120}{98} \approx 1.224\).


Rounding to two decimal places gives 1.22%.
Quick Tip: Percentage Error = (Difference / Actual Value) 100.

Using \(s = ut + \frac{1}{2}at^2\) with \(u=0\):


Total height \(h = \frac{1}{2} g T^2\).


Let distance fallen in time \(t = T/5\) be \(h'\).

\(h' = \frac{1}{2} g (T/5)^2 = \frac{1}{2} g \frac{T^2}{25}\).


Substituting \(h\): \(h' = \frac{h}{25}\).


Position above ground = Total Height - Distance Fallen.


Position = \(h - \frac{h}{25} = \frac{24h}{25}\).
Quick Tip: Distance fallen is proportional to \(t^2\). If time is \(1/n\) of total, distance fallen is \(1/n^2\) of total. Remaining height is \(1 - 1/n^2\).

Mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\).


Concave mirror: \(f = -15\) cm, \(u = -25\) cm.

\(\frac{1}{v} + \frac{1}{-25} = \frac{1}{-15}\).

\(\frac{1}{v} = \frac{1}{25} - \frac{1}{15}\).

\(\frac{1}{v} = \frac{3 - 5}{75} = \frac{-2}{75}\).

\(v = -37.5\) cm.


The screen must be placed 37.5 cm in front of the mirror.
Quick Tip: When solving mirror/lens problems, always write down u, v, f with their proper signs according to Cartesian convention before calculating.

Mass \(m = 2000\) kg.


Velocity \(v = 18\) km/h \(= 18 \times \frac{5}{18} = 5\) m/s.


Work done = Change in Kinetic Energy.

\(W = \frac{1}{2} m v^2 - 0\) (magnitude).

\(W = \frac{1}{2} \times 2000 \times (5)^2\).

\(W = 1000 \times 25 = 25000\) J.

\(W = 2.5 \times 10^4\) J.
Quick Tip: Always convert velocity to SI units (m/s) before calculating energy.

Weight loss in water = Weight in Air - Weight in Water = \(75 - 67 = 8\) gm.


Weight loss in liquid = Weight in Air - Weight in Liquid = \(75 - 51 = 24\) gm.


Relative Density of Liquid = \(\frac{Loss in Liquid}{Loss in Water}\).


RD \(= \frac{24}{8} = 3\).


Density \(= 3 \times \rho_{water}\). Assuming \(\rho_{water} = 1\) g/cc.


Density \(= 3\) g/cm\(^3\).
Quick Tip: Buoyant force is proportional to the density of the fluid. The ratio of weight loss equals the ratio of fluid densities.

Let the forces be \(F_1 = 16\) N and \(F_2 = 2\) N acting at ends A and B respectively. Rod length \(d = 21\) cm.


Since forces are unlike (opposite directions), the resultant \(R\) acts outside the rod on the side of the larger force (\(F_1\)).


Let the resultant act at distance \(x\) from the larger force (16 N). The distance from the smaller force is \((21 + x)\).


Taking moments about the resultant point (torque must balance):

\(16 \times x = 2 \times (21 + x)\).

\(16x = 42 + 2x\).

\(14x = 42\).

\(x = 3\) cm.
Quick Tip: For unlike parallel forces, the resultant lies outside the segment, closer to the larger force. Formula: \(x = \frac{F_{small} \cdot L}{F_{large} - F_{small}}\).

Electric Field \(E = 30\) N/C.


Charge \(q = 20\) C.


Force \(F = qE\).

\(F = 20 \times 30 = 600\) N.
Quick Tip: Simple definition: Electric field is force per unit charge (\(E = F/q\)).

Mass \(m = 6\) kg. Acceleration \(a = 6\) m/s\(^2\).


Net Force \(F_{net} = ma = 6 \times 6 = 36\) N.


Applied Force \(F_{app} = 54\) N.


Equation of motion: \(F_{app} - Friction = F_{net}\).

\(54 - F = 36\).

\(F = 54 - 36 = 18\) N.
Quick Tip: Newton's Second Law applies to the Net Force. Always subtract opposing forces like friction from the applied force.

Acceleration due to gravity \(g = \frac{GM}{R^2}\).


Since \(M\) is constant, \(g \propto R^{-2}\).


For small percentage changes, we can use the derivative approximation:

\(\frac{\Delta g}{g} \approx -2 \frac{\Delta R}{R}\).


Given \(\frac{\Delta R}{R} = -4%\) (shrinks implies negative).

\(\frac{\Delta g}{g} \approx -2 (-4%) = +8%\).


The value increases by 8%.
Quick Tip: For a relation \(Y = k X^n\), the percentage change is \(%Y \approx n \times %X\). Here \(n=-2\).

Electric Potential \(V\) is Work done per unit charge.

\(V = \frac{W}{Q}\).


Given \(W = 200\) J and \(Q = 10\) C.

\(V = \frac{200}{10} = 20\) Volts.
Quick Tip: Potential is simply Work divided by Charge (\(V=W/q\)).

According to Faraday's Law of Induction: \(emf = -N \frac{\Delta \phi}{\Delta t}\).


Number of turns \(N = 20\).


Change in flux \(\Delta \phi = Final - Initial = 0 - 0.3 = -0.3\) Wb.


Time \(\Delta t = 1\) s.

\(emf = -20 \times \frac{-0.3}{1} = 6\) V.
Quick Tip: EMF is proportional to the rate of change of magnetic flux multiplied by the number of turns.

Frequency \(f = 500\) Hz.


Wavelength \(\lambda = 80\) cm \(= 0.8\) m.


Wave velocity \(v = f \lambda = 500 \times 0.8 = 400\) m/s.


Distance \(d = 1\) km \(= 1000\) m.


Time \(t = \frac{d}{v} = \frac{1000}{400}\).

\(t = 2.5\) seconds.
Quick Tip: Ensure units are consistent (meters for wavelength and distance) before calculating velocity and time.

Impedance \(Z\) in a series LCR circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).


Given \(R = 5\), \(X_L = 20\), \(X_C = 20\).

\(Z = \sqrt{5^2 + (20 - 20)^2}\).

\(Z = \sqrt{25 + 0} = 5 \, \Omega\).


(This is the condition of resonance where \(Z = R\)).
Quick Tip: When \(X_L = X_C\), the circuit is in resonance and the impedance is minimum and equal to the Resistance \(R\).

Let resistances be \(R_1\) and \(R_2\).


Series: \(R_1 + R_2 = 50\).


Parallel: \(\frac{R_1 R_2}{R_1 + R_2} = 8\).


Substitute the sum: \(\frac{R_1 R_2}{50} = 8 \Rightarrow R_1 R_2 = 400\).


We need two numbers that add to 50 and multiply to 400.


Let's test options:


(B) \(10 + 40 = 50\) and \(10 \times 40 = 400\).


So the resistances are 10 and 40.
Quick Tip: Instead of solving the quadratic equation \(x^2 - Sx + P = 0\), simply check which option satisfies both Sum and Product conditions.

Initial velocity \(u = 0\) m/s (dropped).


Acceleration \(a = g = 10 m/s^2\).


Height \(s = 20\) m.


Using the third equation of motion: \(v^2 = u^2 + 2as\).

\(v^2 = 0^2 + 2(10)(20) = 400\).

\(v = \sqrt{400} = 20 m/s\).


Using the first equation of motion: \(v = u + at\).

\(20 = 0 + 10t\).

\(10t = 20 \Rightarrow t = 2 s\).
Quick Tip: When an object is dropped, initial velocity is zero. Use \(v^2 = 2gh\) to find velocity and \(t = \sqrt{2h/g}\) to find time quickly.

The force on a moving charge in a magnetic field is given by Lorentz force formula: \(F = qvB \sin \theta\).


Where \(\theta\) is the angle between the velocity vector and the magnetic field vector.


The problem states the proton moves "in parallel of field", which means \(\theta = 0^\circ\).


Since \(\sin 0^\circ = 0\), the force \(F = qvB(0) = 0\).
Quick Tip: Magnetic force is zero if the charged particle moves parallel or anti-parallel to the magnetic field lines.

Let the final equilibrium temperature be \(T\).


Heat gained by colder water = Heat lost by hotter water.

\(m_1 c (T - T_1) = m_2 c (T_2 - T)\).


The specific heat \(c\) is the same for both and cancels out.

\(100(T - 60) = 180(95 - T)\).


Divide both sides by 20: \(5(T - 60) = 9(95 - T)\).

\(5T - 300 = 855 - 9T\).

\(5T + 9T = 855 + 300\).

\(14T = 1155\).

\(T = \frac{1155}{14} = 82.5^\circ C\).
Quick Tip: For mixing same substances, weighted average formula works: \(T_{mix} = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}\).

Given Capacitance \(C = 3 \mu F = 3 \times 10^{-6} F\).


Charge \(Q = 108 \mu C = 108 \times 10^{-6} C\).


The relationship between charge, capacitance, and potential difference is \(Q = CV\).


So, \(V = \frac{Q}{C}\).

\(V = \frac{108 \times 10^{-6}}{3 \times 10^{-6}}\).

\(V = \frac{108}{3} = 36 V\).
Quick Tip: Remember the triangle relation Q = CV. If units are consistent (\(\mu F\) and \(\mu C\)), you can divide the numbers directly without converting powers of 10.

Heat required for Copper \(Q_{Cu} = m_1 s_1 \Delta T_1\).

\(m_1 = 6 kg, s_1 = 0.09, \Delta T_1 = 20 - 10 = 10\).

\(Q_{Cu} = 6 \times 0.09 \times 10 = 5.4\) units.


Heat required for Lead \(Q_{Pb} = m_2 s_2 \Delta T_2\).

\(m_2 = 3 kg, s_2 = ?, \Delta T_2 = 100 - 20 = 80\).

\(Q_{Pb} = 3 \times s_2 \times 80 = 240 s_2\).


Given \(Q_{Cu} = Q_{Pb}\).

\(5.4 = 240 s_2\).

\(s_2 = \frac{5.4}{240} = \frac{0.54}{24} = 0.0225\).


Rounding to three decimal places matches option (A).
Quick Tip: Always ensure \(\Delta T\) is positive and correctly calculated. Equate \(m_1 s_1 \Delta T_1 = m_2 s_2 \Delta T_2\).

Using Faraday's First Law of Electrolysis: \(m = ZIt\).

\(I = 2\) Amperes.

\(t = 30\) minutes \(= 30 \times 60 = 1800\) seconds.


For \(Cu^{2+}\), valency \(n = 2\). Equivalent mass \(E = \frac{Atomic Mass}{n} = \frac{63.5}{2} = 31.75\).


Electrochemical equivalent \(Z = \frac{E}{96500}\).

\(m = \frac{31.75}{96500} \times 2 \times 1800\).

\(m = \frac{31.75 \times 3600}{96500} = \frac{31.75 \times 36}{965}\).

\(m \approx \frac{1143}{965} \approx 1.1844\) gm.
Quick Tip: Remember to convert time into seconds. Formula: \(m = \frac{Atomic Mass}{Valency \times 96500} \times I \times t\).

(A) \(K_2O\): Oxygen is bonded to Metal (K). O is -2.


(B) \(O_3\): Elemental form. O is 0.


(C) \(Na_2O_2\): Peroxide. O is -1.


(D) \(F_2O\) (Oxygen difluoride): Fluorine is more electronegative (-1) than Oxygen.


Let ox. no. of O be \(x\).

\(2(-1) + x = 0 \Rightarrow x = +2\).
Quick Tip: Oxygen generally has -2 oxidation state, except in peroxides (-1), superoxides (-1/2), and when bonded to Fluorine (+1 or +2).

Nitrogen gas exists as diatomic molecules \(N_2\).


Molar mass of \(N_2 = 14 \times 2 = 28\) g/mol.


Number of moles \(n = \frac{Given Mass}{Molar Mass} = \frac{2.8}{28} = 0.1\) mol.


Number of molecules = moles \(\times\) Avogadro's Number (\(N_A\)).

\(= 0.1 \times 6.023 \times 10^{23} = 6.023 \times 10^{22}\).
Quick Tip: Always assume nitrogen means nitrogen gas (\(N_2\)) unless specified as nitrogen atoms.

Check the oxidation states of Sulphur in reactants:


In \(SO_2\), O is -2, so S is +4.


In \(H_2S\), H is +1, so S is -2.


In product S (elemental sulphur), oxidation state is 0.


Sulphur from \(SO_2\) goes from +4 to 0 (Reduction).


Sulphur from \(H_2S\) goes from -2 to 0 (Oxidation).


Thus, Sulphur is both oxidised and reduced in this reaction (Comproportionation).
Quick Tip: When an element in two different oxidation states reacts to form a single product with an intermediate oxidation state, it is a comproportionation reaction.

An amalgam is an alloy of mercury with another metal.


For example, Dental amalgam (Ag-Hg), Sodium amalgam (Na-Hg).


Mercury (Hg) is the mandatory component.
Quick Tip: Iron, Platinum, Tungsten, and Tantalum do not form amalgams with mercury.

Given: Initial amount \(N_0 = 1\) gm.


Half-life \(T_{1/2} = 150\) days.


Total time \(t = 600\) days.


Number of half-lives \(n = \frac{t}{T_{1/2}} = \frac{600}{150} = 4\).


Amount remaining \(N = N_0 \left(\frac{1}{2}\right)^n\).

\(N = 1 \times \left(\frac{1}{2}\right)^4 = \frac{1}{16}\) gm.
Quick Tip: The formula for remaining amount is \(N = N_0(1/2)^n\). Always calculate 'n' first.

Hardness in water is caused by dissolved cations, primarily Calcium (\(Ca^{2+}\)) and Magnesium (\(Mg^{2+}\)).


These ions are typically present as bicarbonates, chlorides, or sulfates.


Sodium and Potassium salts do not cause hardness (soft water).
Quick Tip: Temporary hardness is due to bicarbonates of Ca/Mg; Permanent hardness is due to chlorides/sulfates of Ca/Mg.

Calculate percentage of Oxygen: \(100 - (38.71 + 9.67) = 100 - 48.38 = 51.62%\).


Calculate moles of each element:


Moles of C = \(38.71 / 12 \approx 3.22\).


Moles of H = \(9.67 / 1 = 9.67\).


Moles of O = \(51.62 / 16 \approx 3.22\).


Find simple ratio by dividing by the smallest value (3.22):


C : \(3.22 / 3.22 = 1\).


H : \(9.67 / 3.22 \approx 3\).


O : \(3.22 / 3.22 = 1\).


Empirical Formula is \(CH_3O\).
Quick Tip: Empirical formula represents the simplest whole-number ratio of atoms in a compound.

Thermosetting plastics are polymers that irreversibly harden upon heating and cannot be remolded.


Bakelite (Phenol-formaldehyde resin) is a classic example of a thermosetting polymer.


Polyethylene and PVC are thermoplastics (can be remolded).
Quick Tip: Thermoplastics soften on heating; Thermosets harden permanently on heating (due to cross-linking).

The atomic number of Copper (Cu) is 29.


Expected configuration: \([Ar] 4s^2 3d^9\).


Actual configuration: \([Ar] 4s^1 3d^{10}\).


This exception occurs because a fully filled d-orbital (\(3d^{10}\)) is more stable than a partially filled one (\(3d^9\)). One electron jumps from 4s to 3d.
Quick Tip: Chromium (Cr, 24) and Copper (Cu, 29) are the two most common exceptions to standard Aufbau filling rules in the first transition series.

Ionic (or saline) hydrides are formed by s-block elements (Group 1 and 2 metals) with hydrogen.

\(BH_3\), \(SiH_4\), and \(PH_3\) are covalent hydrides formed by p-block non-metals.

\(MgH_2\) is a hydride of an alkaline earth metal. While it has significant covalent polymeric character, it is classified with the ionic hydrides in this context relative to the others.
Quick Tip: Alkali metals and heavier Alkaline Earth metals form Ionic Hydrides. Non-metals form Covalent Hydrides.

The Ostwald process is used for the manufacture of Nitric Acid (\(HNO_3\)).


The first step is the catalytic oxidation of Ammonia (\(NH_3\)) to Nitric Oxide (NO).

\(4NH_3 + 5O_2 \xrightarrow{Pt/Rh gauge, 500 K} 4NO + 6H_2O\).


Platinum (Pt) or Platinum-Rhodium alloy is used as the catalyst.
Quick Tip: Haber's process (\(NH_3\)): Fe catalyst. Contact process (\(H_2SO_4\)): \(V_2O_5\). Ostwald's process (\(HNO_3\)): Pt.

Let's count electrons for each species:


(A) \(O^{2-}\) (\(8+2=10\)), \(F^-\) (\(9+1=10\)). Same.


(B) \(P^{3-}\) (\(15+3=18\)), Ar (18). Same.


(C) \(Na^+\) (\(11-1=10\)), \(Al^{3+}\) (\(13-3=10\)). Same.


(D) \(Mg^{2+}\) (\(12-2=10\)), Ar (18). Different.
Quick Tip: Isoelectronic species have the same number of electrons. Calculate electrons = Atomic Number - Charge.

Relationship: Molecular Weight = Equivalent Weight \(\times\) Basicity.


For a dibasic acid, Basicity = 2.


Given Equivalent Weight = 12.


Molecular Weight \(= 12 \times 2 = 24\).
Quick Tip: For acids: Eq. Wt = Mol. Wt / Basicity (\(H^+\) ions). For bases: Eq. Wt = Mol. Wt / Acidity (\(OH^-\) ions).

A mixture of ethyne (acetylene) and oxygen is burnt to produce a very hot flame known as the Oxy-acetylene flame.


This flame is used for welding and cutting metals.


Formula: \(C_2H_2\) (Ethyne).
Quick Tip: Common name for Ethyne is Acetylene.

For isoelectronic species (all have 10 electrons here: \(Al^{3+}\), \(Mg^{2+}\), \(N^{3-}\)), the radius decreases as the nuclear charge (atomic number) increases.

\(Al^{3+}\) (Z=13): Strongest attraction, smallest radius.

\(Mg^{2+}\) (Z=12): Intermediate.

\(N^{3-}\) (Z=7): Weakest attraction, largest radius.


Correct order: \(Al^{3+} < Mg^{2+} < N^{3-}\).
Quick Tip: In an isoelectronic series, Radius \(\propto 1/Atomic Number\). Anions are larger than cations.

Mass of Reactant (\(CaCO_3\)) = 10.0 gm.


Mass of Products (\(CaO + CO_2\)) = 5.6 gm + 4.4 gm = 10.0 gm.


Since Total Mass of Reactants = Total Mass of Products, the data supports the Law of Conservation of Mass.
Quick Tip: "Mass can neither be created nor destroyed." Check if Sum of Reactants = Sum of Products.

Cracking (pyrolysis) is the process of breaking down large, complex hydrocarbon molecules (higher alkanes) into smaller, simpler molecules (lower alkanes and alkenes) by the action of heat and pressure (and often catalysts).


This is used in petroleum refining to produce petrol/gasoline.
Quick Tip: Cracking breaks big molecules into small ones (like cracking a nut).

Antipyretics are chemical substances used to lower body temperature in high fever (e.g., Paracetamol, Aspirin).


Analgesics reduce pain. Antibiotics kill bacteria. Tranquilizers treat anxiety/mental stress.
Quick Tip: "Anti" = against, "Pyretic" = related to fire/heat (fever).

The monomer shown is Tetrafluoroethylene (\(CF_2=CF_2\)).


Polymerization of tetrafluoroethylene yields Polytetrafluoroethylene (PTFE).


The trade name for PTFE is Teflon.
Quick Tip: Teflon is the non-stick coating on cookware. Monomer is Tetra-Fluoro-Ethylene.

Structure analysis: \(HC \equiv C - CH(CH_3) - CH_2CH_3\). (Note: The figure shows an ethyl group upwards and methyl rightwards on the 3rd carbon).


1. Identify the longest carbon chain containing the triple bond. The chain is 5 carbons long (Pentyne).


2. Number the chain starting from the end closer to the triple bond.

\(C^1 \equiv C^2 - C^3 - C^4 - C^5\).


3. The triple bond is at C-1. So, "1-Pentyne".


4. There is a methyl substituent at C-3.


Full name: 3-Methyl-1-pentyne.
Quick Tip: Always give the double/triple bond the lowest possible number, taking precedence over alkyl substituents.

Synthetic detergents are cleansing agents which have all the properties of soap but do not contain any soap.


They are typically sodium salts of long chain alkyl sulphonic acids or long chain alkyl hydrogen sulphates.


Soaps, on the other hand, are salts of long chain carboxylic acids (fatty acids).
Quick Tip: Soap = Carboxylic salt (\(R-COO^-Na^+\)). Detergent = Sulphonic/Sulphate salt (\(R-SO_3^-Na^+\)).

Structure of 2-Butanone: \(CH_3 - CO - CH_2CH_3\).


The carbonyl group (\(CO\)) is attached to a Methyl group (\(CH_3\)) and an Ethyl group (\(CH_2CH_3\)).


In common nomenclature for ketones, we name the alkyl groups attached to the carbonyl carbon.


Name: Ethyl Methyl Ketone (EMK).
Quick Tip: Acetone is Dimethyl Ketone. 2-Butanone is Ethyl Methyl Ketone.

A real gas obeys the ideal gas equation (\(PV=nRT\)) only under specific conditions where intermolecular forces are negligible.


This occurs at High Temperature (molecules move too fast for interaction) and Low Pressure (molecules are far apart).


Among the given options, High temperature is the correct condition.
Quick Tip: Real gases approach Ideality at High T and Low P. (Remember: High-Low).

According to Graham's Law of Diffusion, the rate of diffusion (\(r\)) of a gas is inversely proportional to the square root of its density (\(d\)) or molar mass (\(M\)).


Formula: \(r \propto \frac{1}{\sqrt{d}}\).


Therefore, option (B) is the correct mathematical representation.
Quick Tip: Heavier gases diffuse slower. Rate \(\propto 1/\sqrt{Mass}\).