Math Fundamentals Cheatsheet

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Ratio and Proportion Ratio: A comparison of two quantities of the same kind by division. It can be written as $a:b$ or $\frac{a}{b}$. Example: If there are 3 apples and 2 oranges, the ratio of apples to oranges is $3:2$. Proportion: An equality of two ratios. If $a:b = c:d$, then $ad = bc$. Example: $2:3 = 4:6$ is a proportion because $2 \times 6 = 3 \times 4$. Comparing Ratios: To compare ratios like $12:54$ and $14:63$, simplify them: $12:54 = \frac{12}{54} = \frac{2}{9}$ $14:63 = \frac{14}{63} = \frac{2}{9}$ Since both simplify to $\frac{2}{9}$, the ratios are equal. Permutations and Combinations Permutation ($P(n, k)$ or $_nP_k$): The number of ways to arrange $k$ items from a set of $n$ distinct items, where the order matters. Formula: $P(n, k) = \frac{n!}{(n-k)!}$ Example: How many ways to arrange 3 letters from ABC? $P(3, 3) = \frac{3!}{(3-3)!} = \frac{6}{1} = 6$ (ABC, ACB, BAC, BCA, CAB, CBA). Combination ($C(n, k)$ or $_nC_k$ or $\binom{n}{k}$): The number of ways to choose $k$ items from a set of $n$ distinct items, where the order does NOT matter. Formula: $C(n, k) = \frac{n!}{k!(n-k)!}$ Example: How many ways to choose 2 letters from ABC? $C(3, 2) = \frac{3!}{2!(3-2)!} = \frac{6}{2 \times 1} = 3$ (AB, AC, BC). Difference: Permutation considers order, Combination does not. Applications of Combinations Committee Selection: A committee of 5 can be selected from 8 persons in $C(8, 5)$ ways. $C(8, 5) = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ ways. Committee with Conditions (Americans/Englishmen): From 7 Englishmen and 4 Americans, a committee of 6: Exactly 3 Americans: Choose 3 Americans from 4 ($C(4,3)$) AND 3 Englishmen from 7 ($C(7,3)$). $C(4,3) \times C(7,3) = \frac{4!}{3!1!} \times \frac{7!}{3!4!} = 4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 4 \times 35 = 140$ ways. At least 2 Americans: Sum of (2 Americans, 4 Englishmen) + (3 Americans, 3 Englishmen) + (4 Americans, 2 Englishmen) $C(4,2)C(7,4) + C(4,3)C(7,3) + C(4,4)C(7,2)$ $6 \times 35 + 4 \times 35 + 1 \times 21 = 210 + 140 + 21 = 371$ ways. Permutations of Words with Repeated Letters The number of permutations of $n$ objects where $p_1$ are alike of one kind, $p_2$ are alike of another kind, etc., is $\frac{n!}{p_1! p_2! ...}$. 'EXAMINATION': $n=11$. E(1), X(1), A(2), M(1), I(2), N(2), T(1), O(1). Permutations = $\frac{11!}{2! \times 2! \times 2!} = \frac{39,916,800}{8} = 4,989,600$. 'BUSINESS': $n=8$. B(1), U(1), S(3), I(1), N(1), E(1). Permutations = $\frac{8!}{3!} = \frac{40,320}{6} = 6,720$. 'INDIA': $n=5$. I(2), N(1), D(1), A(1). Permutations = $\frac{5!}{2!} = \frac{120}{2} = 60$. 'MATHEMATICS': $n=11$. M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1). Permutations = $\frac{11!}{2! \times 2! \times 2!} = \frac{39,916,800}{8} = 4,989,600$. 'FAILURE' with vowels together: Vowels (A, I, U, E) must stay together. Treat 'AIUE' as one unit. Letters: F, L, R, (AIUE). Total 4 units. Arrangement of 4 units: $4!$ ways. Arrangement of vowels within their unit: $4!$ ways. Total permutations = $4! \times 4! = 24 \times 24 = 576$. Interest Calculations Simple Interest (SI): Interest calculated only on the principal amount. Formula: $SI = \frac{P \times R \times T}{100}$ Total Amount ($A$) = $P + SI$ Example: Rs 1200 becomes Rs 1560 at 10% p.a. SI. Find time. $SI = 1560 - 1200 = 360$ $360 = \frac{1200 \times 10 \times T}{100} \implies T = \frac{360 \times 100}{1200 \times 10} = 3$ years. Find Principal: What principal becomes Rs 2232 at 8% p.a. in 3 years (same time as above)? $A = P(1 + \frac{RT}{100}) \implies 2232 = P(1 + \frac{8 \times 3}{100}) = P(1 + \frac{24}{100}) = P(1.24)$ $P = \frac{2232}{1.24} = 1800$ Rs. Compound Interest (CI): Interest calculated on the principal and accumulated interest from previous periods. Formula for amount: $A = P(1 + \frac{R}{100n})^{nt}$, where $n$ is compounding frequency per year. $CI = A - P$ Example: CI on Rs 6,950 for 3 years, half-yearly. 6% p.a. for first 2 years, 9% p.a. for third year. Year 1 Amount ($A_1$): $P_0 = 6950$, $R=6\%$, $n=2$, $t=1$. $A_1 = 6950(1 + \frac{0.06}{2})^{2 \times 1} = 6950(1.03)^2 = 6950 \times 1.0609 = 7374.235$ Year 2 Amount ($A_2$): $P_1 = A_1$, $R=6\%$, $n=2$, $t=1$. $A_2 = 7374.235(1.03)^2 = 7374.235 \times 1.0609 = 7822.40$ (approx) Year 3 Amount ($A_3$): $P_2 = A_2$, $R=9\%$, $n=2$, $t=1$. $A_3 = 7822.40(1 + \frac{0.09}{2})^{2 \times 1} = 7822.40(1.045)^2 = 7822.40 \times 1.092025 = 8541.27$ (approx) Compounded Amount = Rs 8,541.27 Loan Distribution Example: Mahajan lent Rs 10,000 to two persons. First at 5% p.a., second at 6% p.a. After 4 years, total amount Rs 12,240. Let $x$ be the amount borrowed by the first person. Then $10000-x$ by the second. Amount from first: $x(1 + \frac{5 \times 4}{100}) = x(1.2)$ Amount from second: $(10000-x)(1 + \frac{6 \times 4}{100}) = (10000-x)(1.24)$ Total amount: $1.2x + 1.24(10000-x) = 12240$ $1.2x + 12400 - 1.24x = 12240$ $-0.04x = 12240 - 12400 = -160$ $x = \frac{-160}{-0.04} = 4000$ First person borrowed Rs 4,000; second person borrowed Rs 6,000. Mixed Compounding Frequencies: Deposit $P = 300$. First 5 years at 6% p.a. compounded quarterly ($n=4$). $A_5 = 300(1 + \frac{0.06}{4})^{4 \times 5} = 300(1.015)^{20} \approx 300 \times 1.346855 = 404.0565$ Next 10 years at 8% p.a. compounded semi-annually ($n=2$). $A_{15} = A_5(1 + \frac{0.08}{2})^{2 \times 10} = 404.0565(1.04)^{20} \approx 404.0565 \times 2.191123 = 885.34$ Compounded amount at end of 15 years $\approx$ Rs 885.34. Linear Equations Solving Systems of Linear Equations Substitution Method: Solve one equation for one variable, then substitute into the other equation. Elimination Method: Multiply equations by constants to make coefficients of one variable opposites, then add equations. Example 1: $6x - 7y = -120$ and $7x - 4y = 11$ Multiply first by 4, second by 7: $24x - 28y = -480$ $49x - 28y = 77$ Subtract first from second: $(49x - 28y) - (24x - 28y) = 77 - (-480) \implies 25x = 557 \implies x = \frac{557}{25}$ Substitute $x$ into an equation to find $y$. Example 2: $2x + 10y = 21$ and $6x - 5y = -8$ Multiply second by 2: $12x - 10y = -16$ Add to first: $(2x + 10y) + (12x - 10y) = 21 + (-16) \implies 14x = 5 \implies x = \frac{5}{14}$ Substitute $x$ into an equation: $2(\frac{5}{14}) + 10y = 21 \implies \frac{5}{7} + 10y = 21 \implies 10y = 21 - \frac{5}{7} = \frac{147-5}{7} = \frac{142}{7}$ $y = \frac{142}{70} = \frac{71}{35}$ Age Problems Example: Ratio of present age of father to daughter is $5:3$. After 10 years, ratio is $3:2$. Let father's present age be $5x$, daughter's be $3x$. After 10 years: Father $(5x+10)$, Daughter $(3x+10)$. Ratio: $\frac{5x+10}{3x+10} = \frac{3}{2}$ $2(5x+10) = 3(3x+10)$ $10x + 20 = 9x + 30$ $x = 10$ Present ages: Father $5 \times 10 = 50$ years, Daughter $3 \times 10 = 30$ years. Graphical Representation of Functions Linear Function: $y = mx + c$ Graph is a straight line. $m$: slope (rate of change). $c$: y-intercept (where the line crosses the y-axis). Example: $y = 2x + 1$ When $x=0, y=1$. When $y=0, x=-0.5$. Plot these points and draw a straight line. Quadratic Function: $y = ax^2 + bx + c$ ($a \neq 0$) Graph is a parabola (U-shaped curve). If $a > 0$, parabola opens upwards (minimum point). If $a Vertex: $(-\frac{b}{2a}, f(-\frac{b}{2a}))$ Roots (x-intercepts): Where $y=0$, found by quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Example: $y = x^2 - 4$ Opens upwards. Vertex at $(0, -4)$. Roots at $x = \pm 2$. Matrices Matrix: A rectangular array of numbers, symbols, or expressions arranged in rows and columns. Order of a Matrix: The number of rows by the number of columns (e.g., $m \times n$ matrix has $m$ rows and $n$ columns). Types of Matrices: Row Matrix: Has only one row (e.g., $[1 \quad 2 \quad 3]$). Column Matrix: Has only one column (e.g., $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$). Square Matrix: Number of rows equals number of columns (e.g., $2 \times 2$, $3 \times 3$). Diagonal Matrix: A square matrix where all non-diagonal elements are zero. Scalar Matrix: A diagonal matrix where all diagonal elements are equal. Identity Matrix ($I$): A diagonal matrix where all diagonal elements are 1. (e.g., $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$). Zero/Null Matrix: All elements are zero. Symmetric Matrix: $A = A^T$ (transpose of $A$). Skew-Symmetric Matrix: $A = -A^T$. Matrix Operations Scalar Multiplication: For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and scalar $k$, $kA = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix}$. Matrix Addition/Subtraction: Only possible for matrices of the same order. Add/subtract corresponding elements. If $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$, then $2A + 2B = 2\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} + 2\begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} + \begin{pmatrix} 10 & 12 \\ 14 & 16 \end{pmatrix} = \begin{pmatrix} 12 & 16 \\ 20 & 24 \end{pmatrix}$. Determinant of a Matrix: $2 \times 2$ Matrix: For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $\det(A) = ad - bc$. Example: $\det \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = (2)(4) - (3)(1) = 8 - 3 = 5$. $3 \times 3$ Matrix: For $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, $\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$. Example: $\det \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} = 1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5)$ $= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)$ $= -24 - 2(-20) + 3(-5) = -24 + 40 - 15 = 1$. Inverse of a $3 \times 3$ Matrix ($A^{-1}$): $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, where $\text{adj}(A)$ is the adjugate (or adjoint) matrix. $\text{adj}(A) = (\text{cofactor matrix of } A)^T$. Steps: Calculate $\det(A)$. If $\det(A)=0$, inverse does not exist. Find the cofactor matrix $C$ where $C_{ij} = (-1)^{i+j} M_{ij}$ ($M_{ij}$ is the minor). Find the adjugate matrix $\text{adj}(A) = C^T$. Multiply $\text{adj}(A)$ by $\frac{1}{\det(A)}$. Example: For $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, we found $\det(A) = 1$. Cofactor matrix: $C_{11} = -24$, $C_{12} = 20$, $C_{13} = -5$ $C_{21} = 18$, $C_{22} = -15$, $C_{23} = 4$ $C_{31} = 5$, $C_{32} = -4$, $C_{33} = 1$ So, $C = \begin{pmatrix} -24 & 20 & -5 \\ 18 & -15 & 4 \\ 5 & -4 & 1 \end{pmatrix}$ $\text{adj}(A) = C^T = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$ $A^{-1} = \frac{1}{1} \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix} = \begin{pmatrix} -24 & 18 & 5 \\ 20 & -15 & -4 \\ -5 & 4 & 1 \end{pmatrix}$ Word Problems & Ratios Rounaq's Weight: Weighs 56.7 kg. Reduces weight in ratio $7:6$. New weight = $56.7 \times \frac{6}{7} = 8.1 \times 6 = 48.6$ kg. School Students Ratio: 720 students, boys to girls $3:5$. Total parts = $3+5=8$. Number of boys = $\frac{3}{8} \times 720 = 270$. Number of girls = $\frac{5}{8} \times 720 = 450$. If 18 new girls admitted: Girls = $450 + 18 = 468$. New ratio of girls to total students is not specified, assuming `boys:girls` ratio changes to $2:3$. Let $x$ be new boys. New ratio $\frac{270+x}{468} = \frac{2}{3}$ $3(270+x) = 2(468)$ $810 + 3x = 936$ $3x = 126 \implies x = 42$. 42 new boys may be admitted. Percentage Problem: 60% of students are girls. Total girls = 690. Let total students be $S$. $0.60 \times S = 690 \implies S = \frac{690}{0.60} = 1150$. Total students = 1150. Number of boys = $1150 - 690 = 460$. Number Formation Numbers between 10 and 100 (2-digit numbers) from {3,0,4,5,6} unique digits: First digit cannot be 0. So 4 choices for the first digit ({3,4,5,6}). Second digit can be any of the remaining 4 digits (including 0). Total numbers = $4 \times 4 = 16$. Combinations with Conditions Group of 3 astronauts from 5 Indian and 4 American, at least one Indian: Total ways to choose 3 from 9: $C(9,3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. Ways with no Indian (all 3 Americans): $C(4,3) = 4$. Ways with at least one Indian = Total ways - Ways with no Indian = $84 - 4 = 80$. Alternatively: 1 Indian, 2 American: $C(5,1)C(4,2) = 5 \times 6 = 30$ 2 Indian, 1 American: $C(5,2)C(4,1) = 10 \times 4 = 40$ 3 Indian, 0 American: $C(5,3)C(4,0) = 10 \times 1 = 10$ Total = $30 + 40 + 10 = 80$. Committee of 6 from 7 Indians and 4 Pakistanis, at least two Pakistanis: Total members to choose from = 11. Committee size = 6. Possible compositions with at least 2 Pakistanis: 2 Pakistanis, 4 Indians: $C(4,2) \times C(7,4) = 6 \times 35 = 210$ 3 Pakistanis, 3 Indians: $C(4,3) \times C(7,3) = 4 \times 35 = 140$ 4 Pakistanis, 2 Indians: $C(4,4) \times C(7,2) = 1 \times 21 = 21$ Total ways = $210 + 140 + 21 = 371$.