Comprehensive Advanced Mathematics and Quantitative Aptitude Study Guide

Course Curriculum and Module Overview

• Advanced quantitative aptitude and applied mathematical topics for competitive civil examinations include:

  • Higher arithmetic and commercial mathematics.

  • Classical 2D and 3D mensuration and Euclidean geometry.

  • Circular and analytical trigonometry, including heights and distances.

  • Coordinate geometry and Diophantine equations.

  • Discrete mathematics, including permutations, combinations, and probability theory.

  • Descriptive statistics and measures of central tendency and dispersion.

  • Classical number systems and algebraic structures.

Mensuration and Euclidean Geometry

• Surface Area Variations in Divided Cylinders:

  • Initial cylinder specifications:

  • Base radius: $r = 7\,\text{cm}$.

  • Height: $h = 21\,\text{cm}$.

  • Initial Total Surface Area ($TSA_1$):

  • Formula: $TSA_1 = 2\pi r(r + h)$.

  • Calculation: $TSA_1 = 2 \times \frac{22}{7} \times 7 \times (7 + 21) = 44 \times 28 = 1232\,\text{cm}^2$.

  • The cylinder is divided into four sections by two distinct cuts:

  • Cut 1 is parallel to the base, bisecting the cylinder into two cylinders of height $\frac{h}{2}$. This adds two circular cross-sections: $\Delta A_1 = 2(\pi r^2) = 2 \times \frac{22}{7} \times 49 = 308\,\text{cm}^2$.

  • Cut 2 is made along the axis from top to bottom base through the center. This bisects both halves along the diameter $2r$ across total height $h$, adding two complete rectangular faces of dimensions $2r \times h$: $\Delta A_2 = 2(2r \times h) = 4rh = 4 \times 7 \times 21 = 588\,\text{cm}^2$.

  • Total increase in surface area:

  • $\Delta A = \Delta A_1 + \Delta A_2 = 308\,\text{cm}^2 + 588\,\text{cm}^2 = 896\,\text{cm}^2$.

  • Percentage increase in total surface area:

  • $\text{Percentage Increase} = \frac{\Delta A}{TSA_1} \times 100\% = \frac{896}{1232} \times 100\% = \frac{8}{11} \times 100\% = 72.72\%$.

• Exterior Angles and Regular Polygons:

  • Let the number of sides of two regular polygons be $n_1$ and $n_2$, with the ratio $n_1 : n_2 = 5 : 7$.

  • Let $n_1 = 5x$ and $n_2 = 7x$.

  • The measure of an exterior angle of a regular polygon with $n$ sides is given by $\frac{360^\circ}{n}$.

  • The difference between the exterior angles is given as $20\frac{4}{7}^\circ = \frac{144^\circ}{7}$.

  • Equation:

  • $\frac{360^\circ}{5x} - \frac{360^\circ}{7x} = \frac{144^\circ}{7}$

  • $\frac{360^\circ}{x} \left(\frac{7 - 5}{35}\right) = \frac{144^\circ}{7}$

  • $\frac{360^\circ}{x} \times \frac{2}{35} = \frac{144^\circ}{7}$

  • $\frac{720^\circ}{35x} = \frac{144^\circ}{7}$

  • $\frac{144^\circ}{7x} = \frac{144^\circ}{7} \implies x = 2$

  • Evaluating the sides:

  • $n_1 = 5(2) = 10$

  • $n_2 = 7(2) = 14$

• Intersecting Chords and Subtended Central Angles:

  • Chords $AB$ and $CD$ of a circle intersect internally at point $P$.

  • Interior angle at the intersection point: $\angle APC = 50^\circ$.

  • Minor arc $AC$ subtends a central angle of $60^\circ$.

  • Theorem: The measure of the interior angle formed by two chords intersecting within a circle is equal to half the sum of the central angles subtended by the intercepted arcs:

  • $\angle APC = \frac{1}{2}(\text{Arc } AC + \text{Arc } BD)$

  • Substituting the given values:

  • $50^\circ = \frac{1}{2}(60^\circ + \text{Arc } BD)$

  • $100^\circ = 60^\circ + \text{Arc } BD$

  • $\text{Arc } BD = 40^\circ$

• Largest Inscribed Sphere in a Right Square Pyramid:

  • Pyramid specifications:

  • Base is a square with side length $a = 14\,\text{cm}$.

  • Height from apex to base center: $h = 24\,\text{cm}$.

  • Cross-section along the vertical plane bisecting opposite base edges:

  • Forms an isosceles triangle with base $b = a = 14\,\text{cm}$ and vertical height $h = 24\,\text{cm}$.

  • Distance from base center to midpoint of a base side: $\frac{a}{2} = 7\,\text{cm}$.

  • Slant height of the pyramid ($l$):

    • $l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\,\text{cm}$.

  • The inscribed sphere corresponds to the incircle of this triangular cross-section:

  • Cross-sectional area ($\Delta$):

    • $\Delta = \frac{1}{2} \times a \times h = \frac{1}{2} \times 14 \times 24 = 168\,\text{cm}^2$.

  • Semi-perimeter of triangular cross-section ($s$):

    • $s = \frac{a + 2l}{2} = \frac{14 + 2(25)}{2} = \frac{64}{2} = 32\,\text{cm}$.

  • Inradius of sphere ($r$):

    • $r = \frac{\Delta}{s} = \frac{168}{32} = \frac{21}{4} = 5.25\,\text{cm}$.

  • Volume of the inscribed sphere ($V$):

  • Formula: $V = \frac{4}{3}\pi r^3$.

  • Calculation:

    • $V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{4}\right)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{9261}{64} = \frac{88}{21} \times \frac{9261}{64} = \frac{11 \times 441}{8} = \frac{4851}{8} = 606.375\,\text{cm}^3$.

    • Using standard decimal expansion: $V \approx 606.13\,\text{cm}^3$.

• Volume of Material in a Hollow Pyramid Spire:

  • Geometry:

  • Regular right square pyramid with outer base edge $A = 20\,\text{m}$ and outer vertical height $H = 15\,\text{m}$.

  • Wall thickness along the base: $t = 1\,\text{m}$.

  • Outer pyramid volume ($V_{\text{outer}}$):

  • $V_{\text{outer}} = \frac{1}{3} \times A^2 \times H = \frac{1}{3} \times (20)^2 \times 15 = 400 \times 5 = 2000\,\text{m}^3$.

  • Inner hollow pyramid dimensions:

  • Inner base edge: $a = A - 2t = 20 - 2(1) = 18\,\text{m}$.

  • By geometric similitude, linear scaling factor $k = \frac{a}{A} = \frac{18}{20} = 0.9$.

  • Inner height: $h = k \times H = 0.9 \times 15 = 13.5\,\text{m}$.

  • Inner hollow volume ($V_{\text{inner}}$):

    • $V_{\text{inner}} = \frac{1}{3} \times (18)^2 \times 13.5 = \frac{1}{3} \times 324 \times 13.5 = 108 \times 13.5 = 1458\,\text{m}^3$.

    • Alternatively, using the cube of the scale factor: $V_{\text{inner}} = V_{\text{outer}} \times k^3 = 2000 \times (0.9)^3 = 2000 \times 0.729 = 1458\,\text{m}^3$.

  • Volume of construction material:

  • $V_{\text{material}} = V_{\text{outer}} - V_{\text{inner}} = 2000\,\text{m}^3 - 1458\,\text{m}^3 = 542\,\text{m}^3$.

• Total Surface Area of a Trapezoidal Prism:

  • Prism base is an isosceles trapezium:

  • Parallel sides: $a = 20\,\text{cm}$, $b = 10\,\text{cm}$.

  • Non-parallel sides: $c = 13\,\text{cm}$, $d = 13\,\text{cm}$.

  • Altitude of trapezium ($h_t$):

  • Horizontal projection of each slant side: $\frac{a - b}{2} = \frac{20 - 10}{2} = 5\,\text{cm}$.

  • Using the Pythagorean theorem: $h_t = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\,\text{cm}$.

  • Base area ($A_{\text{base}}$):

  • $A_{\text{base}} = \frac{1}{2}(a + b) \times h_t = \frac{1}{2}(20 + 10) \times 12 = 15 \times 12 = 180\,\text{cm}^2$.

  • Base perimeter ($P_{\text{base}}$):

  • $P_{\text{base}} = a + b + c + d = 20 + 10 + 13 + 13 = 56\,\text{cm}$.

  • Height of the prism ($H$):

  • Given volume $V = 3240\,\text{cm}^3$.

  • $V = A_{\text{base}} \times H \implies 3240 = 180 \times H \implies H = \frac{3240}{180} = 18\,\text{cm}$.

  • Total Surface Area ($TSA$):

  • $TSA = 2(A_{\text{base}}) + P_{\text{base}} \times H$

  • $TSA = 2(180) + 56 \times 18 = 360 + 1008 = 1368\,\text{cm}^2$.

• Common Tangents and Radii of Two Circles:

  • Length of Direct Common Tangent (external tangent): $L_{\text{ext}} = 9$.

  • Formula: $L_{\text{ext}}^2 = d^2 - (r_1 - r_2)^2 \implies d^2 - (r_1 - r_2)^2 = 81$.

  • Length of Transverse Common Tangent (internal tangent): $L_{\text{int}} = 5$.

  • Formula: $L_{\text{int}}^2 = d^2 - (r_1 + r_2)^2 \implies d^2 - (r_1 + r_2)^2 = 25$.

  • Subtracting the second equation from the first:

  • $(d^2 - (r_1 - r_2)^2) - (d^2 - (r_1 + r_2)^2) = 81 - 25$

  • $(r_1 + r_2)^2 - (r_1 - r_2)^2 = 56$

  • $4r_1 r_2 = 56$

  • $r_1 r_2 = \frac{56}{4} = 14$.

• Triangle Area via Coordinate Geometry:

  • Given vertex coordinates: $A(2, 3)$, $B(7, 1)$, and $C(4, 6)$.

  • Area formula:

  • $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

  • Substitution:

  • $\text{Area} = \frac{1}{2} |2(1 - 6) + 7(6 - 3) + 4(3 - 1)|$

  • $\text{Area} = \frac{1}{2} |2(-5) + 7(3) + 4(2)|$

  • $\text{Area} = \frac{1}{2} |-10 + 21 + 8| = \frac{1}{2} |19| = 9.5\,\text{square units}$.

Trigonometry and Heights & Distances

• Height of an Aerial Balloon Between Observation Points:

  • Two observation points $A$ and $B$ lie on flat ground separated by distance $AB = 1200\,\text{m}$.

  • The balloon is situated at an unknown height $h$ above a ground point $C$ located between $A$ and $B$.

  • Angles of elevation: $\angle PAC = 60^\circ$ and $\angle PBC = 30^\circ$.

  • Relating distances:

  • In $\triangle PAC$: $\tan(60^\circ) = \frac{h}{AC} \implies AC = \frac{h}{\sqrt{3}}$.

  • In $\triangle PBC$: $\tan(30^\circ) = \frac{h}{BC} \implies BC = h\sqrt{3}$.

  • Total ground distance:

  • $AC + BC = AB = 1200\,\text{m}$

  • $\frac{h}{\sqrt{3}} + h\sqrt{3} = 1200$

  • $h\left(\frac{1 + 3}{\sqrt{3}}\right) = 1200$

  • $h\left(\frac{4}{\sqrt{3}}\right) = 1200$

  • $h = \frac{1200\sqrt{3}}{4} = 300\sqrt{3}\,\text{m}$.

• Evaluation of Higher-Power Trigonometric Expressions:

  • Given relation: $\tan^2(\theta) + \cot^2(\theta) = 50$.

  • Fundamental identities:

  • $\sec^2(\theta) = 1 + \tan^2(\theta)$

  • $\csc^2(\theta) = 1 + \cot^2(\theta)$

  • Target expression: $\sec^4(\theta) + \csc^4(\theta)$.

  • $\sec^4(\theta) + \csc^4(\theta) = (1 + \tan^2(\theta))^2 + (1 + \cot^2(\theta))^2$

  • $= (1 + 2\tan^2(\theta) + \tan^4(\theta)) + (1 + 2\cot^2(\theta) + \cot^4(\theta))$

  • $= 2 + 2(\tan^2(\theta) + \cot^2(\theta)) + (\tan^4(\theta) + \cot^4(\theta))$

  • Evaluating $\tan^4(\theta) + \cot^4(\theta)$:

  • $(\tan^2(\theta) + \cot^2(\theta))^2 = \tan^4(\theta) + \cot^4(\theta) + 2\tan^2(\theta)\cot^2(\theta)$

  • $50^2 = \tan^4(\theta) + \cot^4(\theta) + 2(1)$

  • $\tan^4(\theta) + \cot^4(\theta) = 2500 - 2 = 2498$

  • Substitution into target expression:

  • $\sec^4(\theta) + \csc^4(\theta) = 2 + 2(50) + 2498 = 2 + 100 + 2498 = 2600$.

• Continuous Product of Sines:

  • Expression to evaluate: $P = \sin(186^\circ) \sin(222^\circ) \sin(246^\circ) \sin(258^\circ)$.

  • Reduction to first quadrant using $\sin(180^\circ + \alpha) = -\sin(\alpha)$:

  • $\sin(186^\circ) = -\sin(6^\circ)$

  • $\sin(222^\circ) = -\sin(42^\circ)$

  • $\sin(246^\circ) = -\sin(66^\circ)$

  • $\sin(258^\circ) = -\sin(78^\circ)$

  • The product of four negative terms is positive:

  • $P = \sin(6^\circ) \sin(42^\circ) \sin(66^\circ) \sin(78^\circ)$

  • Product identity: $\sin(\alpha) \sin(60^\circ - \alpha) \sin(60^\circ + \alpha) = \frac{1}{4} \sin(3\alpha)$.

  • Pairing $\alpha = 6^\circ$:

    • $\sin(6^\circ) \sin(66^\circ) = \frac{\frac{1}{4}\sin(18^\circ)}{\sin(54^\circ)}$

  • Pairing $\alpha = 18^\circ$ with $\sin(42^\circ) \sin(78^\circ)$, noting $60^\circ - 18^\circ = 42^\circ$ and $60^\circ + 18^\circ = 78^\circ$:

    • $\sin(42^\circ) \sin(78^\circ) = \frac{\frac{1}{4}\sin(54^\circ)}{\sin(18^\circ)}$

  • Combining factors:

  • $P = \left(\frac{\frac{1}{4}\sin(18^\circ)}{\sin(54^\circ)}\right) \times \left(\frac{\frac{1}{4}\sin(54^\circ)}{\sin(18^\circ)}\right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

Arithmetic and Commercial Mathematics

• Motion Analysis and Speed Reductions After Rail Accidents:

  • A train journeys from point $P$ to point $Q$.

  • Scenario 1: The train has an accident after $90\,\text{km}$. Speed decreases to $\frac{2}{3}s$, causing an arrival delay of $30\,\text{minutes} = \frac{1}{2}\,\text{hour}$.

  • Delay on the remaining distance $(D - 90)$:

    • $\frac{D - 90}{\frac{2}{3}s} - \frac{D - 90}{s} = \frac{1}{2}$

    • $\frac{D - 90}{s} \left(\frac{3}{2} - 1\right) = \frac{1}{2}$

    • $\frac{D - 90}{2s} = \frac{1}{2} \implies \frac{D - 90}{s} = 1\,\text{hour}$

  • Scenario 2: The accident occurs $60\,\text{km}$ farther ahead (at $150\,\text{km}$ from $P$). The delay is $10\,\text{minutes} = \frac{1}{6}\,\text{hour}$.

  • Delay on distance $(D - 150)$:

    • $\frac{D - 150}{2s} = \frac{1}{6} \implies \frac{D - 150}{s} = \frac{1}{3}\,\text{hour}$

  • Solving for normal speed $s$ and total distance $D$:

  • Subtracting Scenario 2 from Scenario 1:

    • $\frac{(D - 90) - (D - 150)}{s} = 1 - \frac{1}{3}$

    • $\frac{60}{s} = \frac{2}{3} \implies s = \frac{60 \times 3}{2} = 90\,\text{km/h}$

  • Substituting $s = 90\,\text{km/h}$:

    • $\frac{D - 90}{90} = 1 \implies D - 90 = 90 \implies D = 180\,\text{km}$.

• Linear Races and Time Headstarts:

  • Total race length: $1000\,\text{m}$.

  • Performance metrics:

  • Runner $A$ beats runner $B$ by $35\,\text{seconds}$: $t_B = t_A + 35$.

  • In a direct match, $B$ beats $C$ by $35\,\text{seconds}$: $t_C = t_B + 35 = t_A + 70$.

  • $A$ beats $C$ by $200\,\text{m}$: When $A$ finishes $1000\,\text{m}$ (at time $t_A$), $C$ has covered $800\,\text{m}$.

  • Speed of runner $C$:

  • $v_C = \frac{800}{t_A} = \frac{1000}{t_C}$

  • $\frac{t_C}{t_A} = \frac{1000}{800} = \frac{5}{4}$

  • Substituting $t_C = t_A + 70$:

  • $\frac{t_A + 70}{t_A} = \frac{5}{4}$

  • $1 + \frac{70}{t_A} = \frac{5}{4} \implies \frac{70}{t_A} = \frac{1}{4} \implies t_A = 280\,\text{seconds}$.

• Composite Discount on Multiple Articles:

  • Article 1 (Shirt): Marked Price (MP) = ₹$1200$, Discount = $15\%$.

  • Single shirt discount = $0.15 \times 1200 = ₹180$.

  • Article 2 (Pants): Marked Price (MP) = ₹$2000$, Discount = $25\%$.

  • Single pant discount = $0.25 \times 2000 = ₹500$.

  • Customer transaction: 3 shirts and 2 pants:

  • Total Marked Price: $3(1200) + 2(2000) = 3600 + 4000 = ₹7600$.

  • Total Discount Value: $3(180) + 2(500) = 540 + 1000 = ₹1540$.

  • Overall effective discount percentage:

    • $\text{Discount \%} = \frac{1540}{7600} \times 100\% = \frac{154}{76} \times 10\% = \frac{77}{38} \times 10\% \approx 20.26\%$.

• Parameter Equivalence in Simple Interest:

  • Given ratio of Principal ($P$) to Total Amount ($A$) after $t_1 = 5\,\text{years}$ is $4 : 7$.

  • Simple interest accumulated in 5 years: $SI_1 = A - P = 7 - 4 = 3\,\text{units}$.

  • Annual interest rate equation:

    • $SI = \frac{P \times R \times T}{100} \implies 3 = \frac{4 \times R \times 5}{100} = \frac{20R}{100} = \frac{R}{5} \implies R = 15\%\,\text{per annum}$.

  • Target: Determine duration $t_2$ such that the ratio of Principal to Simple Interest is $5 : 6$.

  • Set $P = 5$ and $SI_2 = 6$:

    • $6 = \frac{5 \times 15 \times t_2}{100} = \frac{75 t_2}{100} = \frac{3}{4} t_2$

    • $t_2 = \frac{6 \times 4}{3} = 8\,\text{years}$.

• Successive Liquid Dilution and Replacement:

  • Initial vessel volume = $100\,\text{litres}$.

  • Initial Milk-to-Water ratio = $4 : 1$.

  • Initial Milk = $80\,\text{litres}$, Initial Water = $20\,\text{litres}$.

  • Process 1: $20\,\text{litres}$ of mixture removed and replaced with $20\,\text{litres}$ of pure water:

  • Fraction of mixture retained = $\frac{100 - 20}{100} = \frac{4}{5}$.

  • Remaining Milk: $80 \times \frac{4}{5} = 64\,\text{litres}$.

  • Resulting Water: $100 - 64 = 36\,\text{litres}$.

  • Process 2: $25\,\text{litres}$ of new mixture removed and replaced with $25\,\text{litres}$ of pure milk:

  • Pure milk is added; thus the dilution factor applies directly to the existing water content.

  • Fraction of mixture retained = $\frac{100 - 25}{100} = \frac{75}{100} = \frac{3}{4}$.

  • Remaining Water: $36 \times \frac{3}{4} = 27\,\text{litres}$.

  • Final Milk content: $100 - 27 = 73\,\text{litres}$.

  • Final ratio of Milk to Water: $73 : 27$.

• Electoral Distribution and Voter Margin:

  • Total registered voters on electoral list = $N$.

  • Voter turnout and validity structure:

  • $10\%$ of voters did not cast votes: participating voters = $0.90N$.

  • 60 voters cast blank/invalid ballots: total valid votes cast = $0.90N - 60$.

  • Candidate voting tallies:

  • Winning candidate received $47\%$ of all registered voters: $V_{\text{winner}} = 0.47N$.

  • Rival candidate's share: $V_{\text{rival}} = (0.90N - 60) - 0.47N = 0.43N - 60$.

  • Given victory margin: $V_{\text{winner}} - V_{\text{rival}} = 308$.

  • $0.47N - (0.43N - 60) = 308$

  • $0.04N + 60 = 308$

  • $0.04N = 248$

  • $N = \frac{248}{0.04} = 6200\,\text{voters}$.

• Compound Interest Multi-Year Compounding:

  • Principal $P = ₹4500$, annual interest rate $r = 10\%$, time period $t = 5\,\text{years}$, compounded annually.

  • Formula: $A = P\left(1 + \frac{r}{100}\right)^t$.

  • Calculation:

  • $A = 4500 \times (1.10)^5$

  • $(1.10)^5 = 1.61051$

  • $A = 4500 \times 1.61051 = ₹7247.295 \approx ₹7247.3$.

• Partnership Allocations for Working and Capital Partners:

  • Capital investments: $A = ₹24000$, $B = ₹36000$, $C = ₹12000$.

  • Investment ratio:

  • $A : B : C = 24000 : 36000 : 12000 = 2 : 3 : 1$.

  • Sum of capital parts: $2 + 3 + 1 = 6$.

  • Managerial compensation:

  • Working partner $A$ receives $26.67\% = 26\frac{2}{3}\% = \frac{80}{300} = \frac{4}{15}$ of total profit $P$.

  • Residual profit for capital allocation: $P - \frac{4}{15}P = \frac{11}{15}P$.

  • Individual earnings from capital share:

  • $A_{\text{cap}} = \frac{2}{6} \times \frac{11}{15}P = \frac{11}{45}P$

  • $B_{\text{cap}} = \frac{3}{6} \times \frac{11}{15}P = \frac{11}{30}P$

  • $C_{\text{cap}} = \frac{1}{6} \times \frac{11}{15}P = \frac{11}{90}P$

  • Combined share of $B$ and $C$: $(B + C) = \frac{4}{6} \times \frac{11}{15}P = \frac{22}{45}P$.

  • Total earnings of $A$:

  • $A_{\text{total}} = \frac{4}{15}P + \frac{11}{45}P = \frac{12 + 11}{45}P = \frac{23}{45}P$.

  • Profit differential condition:

  • $A_{\text{total}} - (B + C) = ₹300$

  • $\frac{23}{45}P - \frac{22}{45}P = 300 \implies \frac{1}{45}P = 300 \implies P = ₹13500$

  • Total amount received by $A$:

  • $A_{\text{total}} = \frac{23}{45} \times 13500 = 23 \times 300 = ₹6900$.

• Profit Expansion via Selling Price Multipliers:

  • Initial Cost Price: $CP$, initial Selling Price: $SP_1$, initial Profit: $P_1 = SP_1 - CP$.

  • Modified Selling Price: $SP_2 = 2 \times SP_1$.

  • Modified Profit: $P_2 = SP_2 - CP = 2SP_1 - CP$.

  • Given condition: $P_2 = 7P_1$.

  • $2SP_1 - CP = 7(SP_1 - CP)$

  • $2SP_1 - CP = 7SP_1 - 7CP$

  • $6CP = 5SP_1 \implies \frac{SP_1}{CP} = \frac{6}{5}$

  • Initial profit percentage:

  • $\text{Profit \%} = \frac{SP_1 - CP}{CP} \times 100\% = \frac{6 - 5}{5} \times 100\% = \frac{1}{5} \times 100\% = 20\%$.

• Linear Diophantine Item Acquisition:

  • Let $t$ be the number of plastic toys at ₹$7$ each, and $s$ be the number of plastic spoons at ₹$5$ each.

  • Non-negative integer constraints: $t, s \in \mathbb{Z}_{\ge 0}$.

  • Total expenditure equation: $7t + 5s = 38$.

  • Rearranging for integer solutions:

  • $5s = 38 - 7t$

  • Testing integer values of $t$ where $38 - 7t \ge 0$ and ends in 0 or 5:

    • If $t = 1$: $38 - 7 = 31$ (not divisible by 5).

    • If $t = 2$: $38 - 14 = 24$ (not divisible by 5).

    • If $t = 3$: $38 - 21 = 17$ (not divisible by 5).

    • If $t = 4$: $38 - 28 = 10 \implies 5s = 10 \implies s = 2$ (valid integer pair).

    • If $t = 5$: $38 - 35 = 3$ (not divisible by 5).

  • Number of plastic toys purchased: $t = 4$.

• Time and Work with Collective Inefficiencies:

  • Individual construction durations for three men: $12\,\text{hours}$, $15\,\text{hours}$, and $20\,\text{hours}$.

  • Let total capacity of the wall be $W = \operatorname{LCM}(12, 15, 20) = 60\,\text{units}$.

  • Theoretical individual hourly production rates:

  • $r_A = \frac{60}{12} = 5\,\text{units/h}$

  • $r_B = \frac{60}{15} = 4\,\text{units/h}$

  • $r_C = \frac{60}{20} = 3\,\text{units/h}$

  • Theoretical combined rate: $5 + 4 + 3 = 12\,\text{units/h}$.

  • Actual collective completion duration: $15\,\text{hours}$.

  • Actual combined rate: $\frac{60}{15} = 4\,\text{units/h}$.

  • Rate shortfall:

  • $\Delta r = 12 - 4 = 8\,\text{units/h}$.

  • Given that this deficit equals $340\,\text{bricks/hour}$:

  • $8\,\text{units/h} = 340\,\text{bricks/h} \implies 1\,\text{unit} = \frac{340}{8} = 42.5\,\text{bricks}$.

  • Total wall capacity in bricks:

  • $W = 60 \times 42.5 = 2550\,\text{bricks}$.

• Average Shifts Across Groups:

  • Total population: 14 persons.

  • The average weight of the first 13 persons is $78\,\text{kg}$.

  • The weight of the 14th person ($W_{14}$) is $39\,\text{kg}$ higher than the average weight ($A$) of all 14 persons:

  • $W_{14} = A + 39$.

  • Total weight equation:

  • $14A = 13(78) + W_{14}$

  • $14A = 1014 + (A + 39)$

  • $13A = 1053 \implies A = \frac{1053}{13} = 81\,\text{kg}$.

  • Weight of the 14th individual:

  • $W_{14} = 81 + 39 = 120\,\text{kg}$.

  • Alternative deviation approach:

  • The $39\,\text{kg}$ excess carried by the 14th person over the overall mean must balance the shortfall of the remaining 13 members.

  • Shift in average: $\Delta = \frac{39}{13} = 3\,\text{kg}$.

  • Overall mean $A = 78 + 3 = 81\,\text{kg}$.

  • $W_{14} = 81 + 39 = 120\,\text{kg}$.

Algebra and Number Systems

• Classification of Real Roots and Discriminant Analysis:

• For a general quadratic equation $ax^2 + bx + c = 0$, the discriminant is defined as $D = b^2 - 4ac$.

• Summary of root classifications based on $D$:

  • If $D > 0$: Roots are real and distinct (वास्तविक व भिन्न).

  • If $D = 0$: Roots are real and equal (वास्तविक व समान).

  • If $D < 0$: Roots are complex/not real (अवास्तविक).

  • If $D$ is a perfect square ($D = m^2, m \in \mathbb{Q}$): Roots are rational (परिमेय).

  • If $D$ is not a perfect square: Roots are irrational conjugate pairs (अपरिमेय).

  • Parameter Constraints for Integer Roots:

• Equation: $x^2 - 5x + k = 0$.

• For the quadratic to possess strictly integer roots, two conditions must be met:

  • The discriminant $D = (-5)^2 - 4(1)(k) = 25 - 4k$ must be the square of an integer: $25 - 4k = m^2$, where $m \in \mathbb{Z}_{\ge 0}$.

  • By the quadratic formula, the roots are $x = \frac{5 \pm m}{2}$. For $x$ to be an integer, $5 \pm m$ must be even, requiring $m$ to be odd.

• Non-negativity constraint on $k$: $k \ge 0 \implies 25 - 4k \le 25 \implies m^2 \le 25$.

• Permissible odd values for $m$ within $[0, 5]$ are $m \in \{1, 3, 5\}$:

  • Case 1 ($m = 5$): $25 - 4k = 25 \implies 4k = 0 \implies k = 0$.

    • Roots: $x = 0, 5$ (integers).

  • Case 2 ($m = 3$): $25 - 4k = 9 \implies 4k = 16 \implies k = 4$.

    • Roots: $x = 1, 4$ (integers).

  • Case 3 ($m = 1$): $25 - 4k = 1 \implies 4k = 24 \implies k = 6$.

    • Roots: $x = 2, 3$ (integers).

• There are exactly 3 non-negative integer values of $k$ ($k \in \{0, 4, 6\}$).

  • Set Properties of the Real Number Continuum:

• Rational numbers ($\mathbb{Q}$) encompass all numbers that can be expressed as $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \ne 0$.

• Irrational numbers ($\mathbb{Q}'$ or $\mathbb{R} \setminus \mathbb{Q}$) comprise non-terminating, non-repeating decimals that cannot be represented as an integer ratio.

• The union of the set of rational numbers and the set of irrational numbers forms the complete set of Real Numbers ($\mathbb{R}$):

  • $\mathbb{Q} \cup \mathbb{Q}' = \mathbb{R}$.

Probability, Combinatorics, and Statistics

• Three-Set Cardinality and Disjoint Event Probability:

  • Total cohort in Grade 11: $N = 60$.

  • Subject enrollments:

  • Physics ($P$): $n(P) = 40$.

  • Biology ($B$): $n(B) = 17$.

  • Chemistry ($C$): $n(C) = 20$.

  • Universal participation condition: $n(P \cup B \cup C) = 60$.

  • Triple intersection: $n(P \cap B \cap C) = 5$.

  • General principle of inclusion-exclusion for three sets:

  • $n(P \cup B \cup C) = n(P) + n(B) + n(C) - [n(P \cap B) + n(B \cap C) + n(C \cap P)] + n(P \cap B \cap C)$

  • $60 = 40 + 17 + 20 - \sum n(\text{two}) + 5$

  • $60 = 82 - \sum n(\text{two}) \implies \sum n(\text{two}) = 22$, where $\sum n(\text{two}) = n(P \cap B) + n(B \cap C) + n(C \cap P)$.

  • Partitioning Venn regions:

  • Number of students enrolled in exactly two subjects ($E_2$):

    • $E_2 = \sum n(\text{two}) - 3 \times n(P \cap B \cap C) = 22 - 3(5) = 22 - 15 = 7$.

  • Number of students enrolled in all three subjects ($E_3$):

    • $E_3 = 5$.

  • Number of students enrolled in exactly one subject ($E_1$):

    • $E_1 = N - (E_2 + E_3) = 60 - (7 + 5) = 60 - 12 = 48$.

  • Probability that a randomly selected student chose exactly one subject:

  • $P(E_1) = \frac{E_1}{N} = \frac{48}{60} = \frac{4}{5} = 0.80$.

• Mathematical Theorems Governing Measures of Dispersion:

  • Statement 1: Adding a constant $c$ to every observation $x_i$:

  • New data point $y_i = x_i + c$.

  • Mean shifts by $c$: $\mu_y = \mu_x + c$.

  • Variance remains unchanged: $\operatorname{Var}(Y) = \frac{1}{n}\sum (y_i - \mu_y)^2 = \frac{1}{n}\sum (x_i + c - (\mu_x + c))^2 = \operatorname{Var}(X)$. (True).

  • Statement 2: Multiplying each data point by $-1$:

  • New data point $y_i = -x_i$.

  • Variance property: $\operatorname{Var}(aX) = a^2 \operatorname{Var}(X)$.

  • For $a = -1$: $\operatorname{Var}(-X) = (-1)^2 \operatorname{Var}(X) = \operatorname{Var}(X)$. Variance does not change. (False; the statement incorrectly asserted that variance changes).

  • Statement 3: Properties of the Coefficient of Variation ($CV$):

  • Defined as $CV = \frac{\sigma}{\mu} \times 100\%$.

  • Since standard deviation ($\sigma$) and mean ($\mu$) share identical physical dimensions, their quotient is a dimensionless scalar invariant under linear transformations of scale/measurement units. (True).

  • Statement 4: Inequality between Mean Deviation and Standard Deviation:

  • Mean deviation about the median minimizes absolute error: $\frac{1}{n}\sum |x_i - M| \le \frac{1}{n}\sum |x_i - \mu|$.

  • By Lyapunov's / Cauchy-Schwarz inequality, $MD_{\text{mean}} \le \sigma$.

  • Consequently, $MD_{\text{median}} \le MD_{\text{mean}} \le \sigma$ holds universally. (True).

  • Correct assertions: Statements 1, 3, and 4 are valid.

• Binomial Probability in Multi-Sensor Missions:

  • Independent detection probability per sensor: $p = 0.75$, failure probability: $q = 1 - p = 0.25$.

  • Total deployed sensors: $n = 4$.

  • Mission success criteria: at least 3 sensors must detect water ($X \ge 3$):

  • $P(X \ge 3) = P(X = 3) + P(X = 4)$.

  • Binomial probabilities:

  • $P(X = 3) = \binom{4}{3} p^3 q^1 = 4 \times (0.75)^3 \times (0.25) = 4 \times \frac{27}{64} \times \frac{1}{4} = \frac{27}{64} = 0.421875$.

  • $P(X = 4) = \binom{4}{4} p^4 q^0 = 1 \times (0.75)^4 = \frac{81}{256} = 0.31640625$.

  • Summation:

  • $P(X \ge 3) = \frac{27}{64} + \frac{81}{256} = \frac{108 + 81}{256} = \frac{189}{256} \approx 0.73828 \approx 0.738$.

• Computation of Population Standard Deviation:

  • Discrete age series: $25, 30, 35, 40, 45, 50$ ($n = 6$).

  • Arithmetic mean ($\mu$):

  • $\mu = \frac{25 + 30 + 35 + 40 + 45 + 50}{6} = \frac{225}{6} = 37.5\,\text{years}$.

  • Squared deviations from the mean ($(x_i - \mu)^2$):

  • $(25 - 37.5)^2 = (-12.5)^2 = 156.25$

  • $(30 - 37.5)^2 = (-7.5)^2 = 56.25$

  • $(35 - 37.5)^2 = (-2.5)^2 = 6.25$

  • $(40 - 37.5)^2 = (2.5)^2 = 6.25$

  • $(45 - 37.5)^2 = (7.5)^2 = 56.25$

  • $(50 - 37.5)^2 = (12.5)^2 = 156.25$

  • Sum of squared deviations:

  • $\sum (x_i - \mu)^2 = 2(156.25 + 56.25 + 6.25) = 2(218.75) = 437.5$.

  • Variance ($\sigma^2$) and standard deviation ($\sigma$):

  • $\sigma^2 = \frac{437.5}{6} \approx 72.9167$

  • $\sigma = \sqrt{72.9167} \approx 8.539 \approx 8.54\,\text{years}$.

• Grouped Continuous Distribution and Median Formulation: