MATHEMATICS

\displaystyle \text{(Maximum Marks: 100)}

\displaystyle \text{(Time Allowed: Three Hours)}

\displaystyle \text{(Candidates are allowed additional 15 minutes for only reading the paper.}
\displaystyle \text{They must NOT start writing during this time)}

\displaystyle \text{The Question Paper consists of three sections A, B and C.}
\displaystyle \text{Candidates are required to attempt all questions from Section A and all questions}
\displaystyle \text{EITHER from Section B OR Section C.}

\displaystyle \text{Section A: Internal choice has been provided in three questions of four marks each}
\displaystyle \text{and two questions of six marks each.}

\displaystyle \text{Section B: Internal choice has been provided in two questions of four marks each.}

\displaystyle \text{Section C: Internal choice has been provided in two questions of four marks each.}

\displaystyle \text{All working, including rough work, should be done on the same sheet as, and adjacent}
\displaystyle \text{to, the rest of the answer.}

\displaystyle \text{The intended marks for questions or parts of questions are given in brackets [ ].}

\displaystyle \text{Mathematical tables and graph papers are provided.}

\displaystyle \textbf{Section - A (80 Marks)}

\displaystyle \textbf{Question 1: } \hspace{10.0cm} [10 \text{ times } 3]

\displaystyle \text{(i) Solve for } x \text{ and } y \text{ if } \begin{bmatrix} x^2 \\ y^2 \end{bmatrix}+2\begin{bmatrix} 2x \\ 3y \end{bmatrix}=3\begin{bmatrix} 7 \\ -3 \end{bmatrix}.
\displaystyle \text{Answer:} 
\displaystyle \begin{bmatrix} x^2 \\ y^2 \end{bmatrix}+\begin{bmatrix} 4x \\ 6y \end{bmatrix}=\begin{bmatrix} 21 \\ -9 \end{bmatrix} 
\displaystyle \therefore \begin{bmatrix} x^2+4x \\ y^2+6y \end{bmatrix}=\begin{bmatrix} 21 \\ -9 \end{bmatrix} 
\displaystyle \therefore x^2+4x=21 \text{ and } y^2+6y=-9 
\displaystyle x^2+4x-21=0 
\displaystyle x^2+7x-3x-21=0 
\displaystyle x(x+7)-3(x+7)=0 
\displaystyle (x-3)(x+7)=0 
\displaystyle \therefore x=3 \text{ or } x=-7 
\displaystyle y^2+6y+9=0 
\displaystyle (y+3)^2=0 
\displaystyle \therefore y=-3 
\displaystyle \therefore (x,y)=(3,-3) \text{ or } (-7,-3) 
\\

\displaystyle \text{(ii) Prove that } \sec^2(\tan^{-1}2)+\mathrm{cosec}^2(\cot^{-1}3)=15.
\displaystyle \text{Answer:} 
\displaystyle \text{Let } \theta=\tan^{-1}2 
\displaystyle \therefore \tan\theta=2 
\displaystyle \sec^2\theta=1+\tan^2\theta=1+2^2=5 
\displaystyle \therefore \sec^2(\tan^{-1}2)=5 
\displaystyle \text{Let } \phi=\cot^{-1}3 
\displaystyle \therefore \cot\phi=3 
\displaystyle \mathrm{cosec}^2\phi=1+\cot^2\phi=1+3^2=10 
\displaystyle \therefore \mathrm{cosec}^2(\cot^{-1}3)=10 
\displaystyle \therefore \sec^2(\tan^{-1}2)+\mathrm{cosec}^2(\cot^{-1}3)=5+10=15 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(iii) Find the equation of the hyperbola whose transverse and conjugate axes are the } x \text{ and } y
\displaystyle \text{axes respectively, given that the length of the conjugate axis is } 5 \text{ and distance between the foci is } 13.
\displaystyle \text{Answer:} 
\displaystyle \text{Equation: } \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 
\displaystyle 2b=5 \Rightarrow b=\frac{5}{2},\ b^2=\frac{25}{4} 
\displaystyle 2c=13 \Rightarrow c=\frac{13}{2},\ c^2=\frac{169}{4} 
\displaystyle c^2=a^2+b^2 \Rightarrow a^2=\frac{169}{4}-\frac{25}{4}=36 
\displaystyle \therefore \frac{x^2}{36}-\frac{y^2}{\frac{25}{4}}=1 \Rightarrow \frac{x^2}{36}-\frac{4y^2}{25}=1 
\\

\displaystyle \text{(iv) From the equation of two regression lines, } 4x+3y+7=0 \text{ and } 3x+4y+8=0, \text{ find:}
\displaystyle \text{(a) Mean of } x \text{ and } y \quad \text{(b) Regression coefficient} \quad \text{(c) Coefficient of correlation}
\displaystyle \text{Answer:} 
\displaystyle \text{Means are point of intersection: solve }4x+3y+7=0,\ 3x+4y+8=0 
\displaystyle x=-\frac{4}{7},\ y=-\frac{11}{7} 
\displaystyle \text{Thus } \overline{x}=-\frac{4}{7},\ \overline{y}=-\frac{11}{7} 
\displaystyle \text{From }4x+3y+7=0 \Rightarrow x=-\frac{3}{4}y-\frac{7}{4} \Rightarrow b_{xy}=-\frac{3}{4} 
\displaystyle \text{From }3x+4y+8=0 \Rightarrow y=-\frac{3}{4}x-2 \Rightarrow b_{yx}=-\frac{3}{4} 
\displaystyle r=\pm\sqrt{b_{xy}b_{yx}}=\pm\frac{3}{4} \Rightarrow r=-\frac{3}{4} \text{ (both negative)} 
\\

\displaystyle \text{(v) Evaluate: } \int e^x(\tan x+\log \sec x)\ dx.
\displaystyle \text{Answer:} 
\displaystyle \text{Let }u=\log \sec x \Rightarrow \frac{du}{dx}=\tan x 
\displaystyle \int e^x(\tan x+\log \sec x)\ dx=\int e^x\left(\frac{du}{dx}+u\right)dx 
\displaystyle =\int \frac{d}{dx}(e^x u)\ dx=e^x u+C 
\displaystyle =e^x \log \sec x+C 
\\

\displaystyle \text{(vi) Evaluate: } \lim\limits_{x\to \frac{\pi}{2}}(\cos x\cdot \log \tan x).
\displaystyle \text{Answer:} 
\displaystyle L=\lim\limits_{x\to \frac{\pi}{2}}\frac{\log \tan x}{\sec x} 
\displaystyle \text{Using L'Hospital's rule, }L=\lim\limits_{x\to \frac{\pi}{2}}\frac{\frac{\sec^2 x}{\tan x}}{\sec x\tan x} 
\displaystyle =\lim\limits_{x\to \frac{\pi}{2}}\frac{\sec x}{\tan^2 x}=\lim\limits_{x\to \frac{\pi}{2}}\frac{\cos x}{\sin^2 x}=0 
\\

\displaystyle \text{(vii) Find the locus of the complex number } Z=x+iy \text{ given } \left|\frac{x+iy-2i}{x+iy+2i}\right|=\sqrt{2}.
\displaystyle \text{Answer:} 
\displaystyle \left|\frac{x+i(y-2)}{x+i(y+2)}\right|=\sqrt{2} 
\displaystyle \frac{\sqrt{x^2+(y-2)^2}}{\sqrt{x^2+(y+2)^2}}=\sqrt{2} 
\displaystyle x^2+(y-2)^2=2\{x^2+(y+2)^2\} 
\displaystyle x^2+y^2-4y+4=2x^2+2y^2+8y+8 
\displaystyle x^2+y^2+12y+4=0 
\displaystyle \therefore x^2+(y+6)^2=32 
\\

\displaystyle \text{(viii) Evaluate: } \int\limits_{1}^{2}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\ dx.
\displaystyle \text{Answer:} 
\displaystyle I=\int\limits_{1}^{2}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\ dx 
\displaystyle \text{Using }x\mapsto 3-x,\ I=\int\limits_{1}^{2}\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\ dx 
\displaystyle 2I=\int\limits_{1}^{2}\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}\ dx=\int\limits_{1}^{2}1\ dx=1 
\displaystyle \therefore I=\frac{1}{2} 
\\

\displaystyle \text{(ix) Three persons } A,B \text{ and } C \text{ shoot to hit a target. If in trials, } A \text{ hits the target } 4
\displaystyle \text{times out of } 5 \text{ shots, } B \text{ hits } 3 \text{ times in } 4 \text{ shots and } C \text{ hits } 2 \text{ times in } 3 \text{ trials. Find:}
\displaystyle \text{(a) Exactly two persons hit the target} \quad \text{(b) At least two persons hit the target}
\displaystyle \text{Answer:} 
\displaystyle P(A)=\frac{4}{5},\ P(B)=\frac{3}{4},\ P(C)=\frac{2}{3} 
\displaystyle P(A')=\frac{1}{5},\ P(B')=\frac{1}{4},\ P(C')=\frac{1}{3} 
\displaystyle \text{(a) }P(\text{exactly two hit})=P(ABC')+P(AB'C)+P(A'BC) 
\displaystyle =\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}+\frac{4}{5}\cdot\frac{1}{4}\cdot\frac{2}{3}+\frac{1}{5}\cdot\frac{3}{4}\cdot\frac{2}{3} 
\displaystyle =\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{13}{30} 
\displaystyle \text{(b) }P(\text{at least two hit})=P(\text{exactly two hit})+P(ABC) 
\displaystyle =\frac{13}{30}+\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{13}{30}+\frac{2}{5}=\frac{5}{6} 
\\

\displaystyle \text{(x) Solve the differential equation: } (xy^2+x)\ dx+(x^2y+y)\ dy=0.
\displaystyle \text{Answer:} 
\displaystyle x(y^2+1)\ dx+y(x^2+1)\ dy=0 
\displaystyle \frac{x}{x^2+1}\ dx+\frac{y}{y^2+1}\ dy=0 
\displaystyle \int \frac{x}{x^2+1}\ dx+\int \frac{y}{y^2+1}\ dy=C 
\displaystyle \frac{1}{2}\log(x^2+1)+\frac{1}{2}\log(y^2+1)=C 
\displaystyle \log\{(x^2+1)(y^2+1)\}=C 
\displaystyle \therefore (x^2+1)(y^2+1)=C 
\\

\displaystyle \textbf{Question 2:}

\displaystyle \text{(a) Using properties of determinants, prove that:}
\displaystyle \left| \begin{array}{ccc} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{array} \right|=a^3 \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \Delta=\left| \begin{array}{ccc} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{array} \right| 
\displaystyle \text{Applying }C_3\to C_3-C_2,\ C_2\to C_2-C_1 
\displaystyle \Delta=\left| \begin{array}{ccc} a & b & c \\ 2a & a+2b & a+b+2c \\ 3a & 3a+3b & 4a+3b+3c \end{array} \right| 
\displaystyle \text{Applying }C_3\to C_3-C_2,\ C_2\to C_2-C_1 
\displaystyle \Delta=\left| \begin{array}{ccc} a & b-a & c-b \\ 2a & -a+2b & c-a-b \\ 3a & 3b & a+c \end{array} \right| 
\displaystyle \text{Applying }C_3\to C_3-C_2,\ C_2\to C_2-C_1 
\displaystyle \Delta=\left| \begin{array}{ccc} a & b-2a & c-2b+a \\ 2a & 2b-3a & c-3b \\ 3a & 3b-3a & a+c-3b \end{array} \right| 
\displaystyle \text{A simpler way is to use row operations on the original determinant.} 
\displaystyle \text{Applying }R_3\to R_3-3R_1,\ R_2\to R_2-2R_1 
\displaystyle \Delta=\left| \begin{array}{ccc} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{array} \right| 
\displaystyle \text{Applying }R_3\to R_3-3R_2 
\displaystyle \Delta=\left| \begin{array}{ccc} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 0 & a \end{array} \right| 
\displaystyle \therefore \Delta=a\cdot a\cdot a=a^3 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) Find the product of matrices } A \text{ and } B \text{ where:}
\displaystyle A=\begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix},\quad B=\begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix}.
\displaystyle \text{Hence, solve the following equations by matrix method:}
\displaystyle x+y+2z=1,\quad 3x+2y+z=7,\quad 2x+y+3z=2 \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle AB=\begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} 
\displaystyle =\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=4I 
\displaystyle \therefore A B=4I\Rightarrow B^{-1}=\frac{1}{4}A 
\displaystyle \text{The given equations can be written as }BX=K 
\displaystyle \text{where }X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad K=\begin{bmatrix}1\\7\\2\end{bmatrix} 
\displaystyle X=B^{-1}K=\frac{1}{4}AK 
\displaystyle X=\frac{1}{4}\begin{bmatrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{bmatrix}\begin{bmatrix}1\\7\\2\end{bmatrix} 
\displaystyle =\frac{1}{4}\begin{bmatrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\4\\-4\end{bmatrix} 
\displaystyle =\begin{bmatrix}2\\1\\-1\end{bmatrix} 
\displaystyle \therefore x=2,\quad y=1,\quad z=-1 
\\

\displaystyle \textbf{Question 3:}

\displaystyle \text{(a) Prove that: } \cos^{-1}\frac{63}{65}+2\tan^{-1}\frac{1}{5}=\sin^{-1}\frac{3}{5}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 

\displaystyle \text{(a) Prove that: } \cos^{-1}\frac{63}{65}+2\tan^{-1}\frac{1}{5}=\sin^{-1}\frac{3}{5}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let } \alpha=\cos^{-1}\frac{63}{65} \text{ and } \beta=\tan^{-1}\frac{1}{5} 
\displaystyle \therefore \cos\alpha=\frac{63}{65},\quad \tan\beta=\frac{1}{5} 
\displaystyle \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{3969}{4225}}=\frac{16}{65} 
\displaystyle \tan 2\beta=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\cdot\frac{1}{5}}{1-\frac{1}{25}}=\frac{5}{12} 
\displaystyle \therefore \sin 2\beta=\frac{5}{13},\quad \cos 2\beta=\frac{12}{13} 
\displaystyle \sin(\alpha+2\beta)=\sin\alpha\cos 2\beta+\cos\alpha\sin 2\beta 
\displaystyle =\frac{16}{65}\cdot\frac{12}{13}+\frac{63}{65}\cdot\frac{5}{13} 
\displaystyle =\frac{192+315}{845}=\frac{507}{845}=\frac{3}{5} 
\displaystyle \therefore \alpha+2\beta=\sin^{-1}\frac{3}{5} 
\displaystyle \therefore \cos^{-1}\frac{63}{65}+2\tan^{-1}\frac{1}{5}=\sin^{-1}\frac{3}{5} 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) (i) Write the Boolean expression corresponding to the circuit below:}
\displaystyle \text{(ii) Simplify the expression using laws of Boolean algebra and construct simplified circuit.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 

\\

\displaystyle \textbf{Question 4:}

\displaystyle \text{(a) Verify Rolle's theorem for the function:}
\displaystyle f(x)=\log \left\{\frac{x^2+ab}{(a+b)x}\right\} \text{ in the interval } [a,b] \text{ where } 0\notin [a,b]. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle f(a)=\log\left\{\frac{a^2+ab}{(a+b)a}\right\}=\log 1=0 
\displaystyle f(b)=\log\left\{\frac{b^2+ab}{(a+b)b}\right\}=\log 1=0 
\displaystyle \therefore f(a)=f(b) 
\displaystyle f'(x)=\frac{2x}{x^2+ab}-\frac{1}{x}=\frac{x^2-ab}{x(x^2+ab)} 
\displaystyle f'(x)=0\Rightarrow x^2-ab=0\Rightarrow x=\pm\sqrt{ab} 
\displaystyle \text{Since }0\notin[a,b],\ a\text{ and }b\text{ have the same sign.} 
\displaystyle \text{If }a,b>0,\text{ then }c=\sqrt{ab}\in(a,b). 
\displaystyle \text{If }a,b<0,\text{ then }c=-\sqrt{ab}\in(a,b). 
\displaystyle \therefore \text{there exists }c\in(a,b)\text{ such that }f'(c)=0. 
\displaystyle \text{Hence, Rolle's theorem is verified.} 
\\

\displaystyle \text{(b) Find the equation of the ellipse with its center at } (4,-1), \text{ focus at } (1,-1)
\displaystyle \text{and given that it passes through } (8,0). \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Centre }=(4,-1),\text{ focus }=(1,-1) 
\displaystyle \therefore \text{major axis is parallel to }x\text{-axis and }c=3 
\displaystyle \text{Equation is } \frac{(x-4)^2}{a^2}+\frac{(y+1)^2}{b^2}=1,\quad a^2=b^2+c^2 
\displaystyle \therefore a^2=b^2+9 
\displaystyle \text{Since it passes through }(8,0),\ \frac{16}{a^2}+\frac{1}{b^2}=1 
\displaystyle \frac{16}{b^2+9}+\frac{1}{b^2}=1 
\displaystyle 16b^2+b^2+9=b^2(b^2+9) 
\displaystyle b^4-8b^2-9=0\Rightarrow b^2=9 
\displaystyle \therefore a^2=9+9=18 
\displaystyle \therefore \text{Required equation is } \frac{(x-4)^2}{18}+\frac{(y+1)^2}{9}=1 
\\

\displaystyle \textbf{Question 5:}

\displaystyle \text{(a) If } e^y(x+1)=1, \text{ then show that } \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle e^y(x+1)=1 \Rightarrow e^y=\frac{1}{x+1} 
\displaystyle \therefore y=\log\left(\frac{1}{x+1}\right)=-\log(x+1) 
\displaystyle \frac{dy}{dx}=-\frac{1}{x+1} 
\displaystyle \frac{d^2y}{dx^2}=\frac{1}{(x+1)^2} 
\displaystyle \left(\frac{dy}{dx}\right)^2=\left(-\frac{1}{x+1}\right)^2=\frac{1}{(x+1)^2} 
\displaystyle \therefore \frac{d^2y}{dx^2}=\left(\frac{dy}{dx}\right)^2 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) A printed page is to have a total area of } 80 \text{ sq. cm with a margin of } 1 \text{ cm at the top}
\displaystyle \text{and on each side and a margin of } 1.5 \text{ cm at the bottom. What should be the dimensions of the page}
\displaystyle \text{so that the printed area will be maximum?} \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let the length and breadth of the page be }x\text{ cm and }y\text{ cm.} 
\displaystyle xy=80 \Rightarrow y=\frac{80}{x} 
\displaystyle \text{Printed area }A=(x-2)\left(y-\frac{5}{2}\right) 
\displaystyle A=(x-2)\left(\frac{80}{x}-\frac{5}{2}\right)=85-\frac{5x}{2}-\frac{160}{x} 
\displaystyle \frac{dA}{dx}=-\frac{5}{2}+\frac{160}{x^2} 
\displaystyle \frac{dA}{dx}=0 \Rightarrow \frac{160}{x^2}=\frac{5}{2}\Rightarrow x^2=64\Rightarrow x=8 
\displaystyle y=\frac{80}{8}=10 
\displaystyle \frac{d^2A}{dx^2}=-\frac{320}{x^3}<0\text{ at }x=8,\text{ hence printed area is maximum.} 
\displaystyle \therefore \text{dimensions of the page are }8\text{ cm}\times10\text{ cm.} 
\\

\displaystyle \textbf{Question 6:}

\displaystyle \text{(a) Evaluate: } \int \frac{dx}{x\{6\log x^2+7\log x+2\}}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let }t=\log x\Rightarrow \frac{dx}{x}=dt 
\displaystyle I=\int \frac{dt}{6t^2+7t+2}=\int \frac{dt}{(3t+2)(2t+1)} 
\displaystyle \frac{1}{(3t+2)(2t+1)}=\frac{3}{3t+2}-\frac{2}{2t+1} 
\displaystyle I=\int\left(\frac{3}{3t+2}-\frac{2}{2t+1}\right)dt 
\displaystyle I=\log(3t+2)-\log(2t+1)+C 
\displaystyle \therefore I=\log\left(\frac{3\log x+2}{2\log x+1}\right)+C 
\\

\displaystyle \text{(b) Find the area of the region bounded by the curves } x=4y-y^2 \text{ and the } y\text{-axis}. \hspace{.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{The curve meets the }y\text{-axis when }x=0 
\displaystyle 4y-y^2=0\Rightarrow y(4-y)=0\Rightarrow y=0,4 
\displaystyle \text{Required area}=\int\limits_{0}^{4}(4y-y^2)\ dy 
\displaystyle =\left[2y^2-\frac{y^3}{3}\right]_{0}^{4}=32-\frac{64}{3}=\frac{32}{3} 
\displaystyle \therefore \text{Area}=\frac{32}{3}\text{ square units} 
\\

\displaystyle \textbf{Question 7:}

\displaystyle \text{(a) } 10 \text{ candidates received percentage marks in two subjects as follows:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Candidate} & A & B & C & D & E & F & G & H & I & J \\ \hline \text{Mathematics Marks} & 80 & 88 & 76 & 74 & 68 & 65 & 40 & 43 & 40 & 80 \\ \hline \text{Statistics Marks} & 72 & 84 & 90 & 66 & 54 & 50 & 54 & 38 & 30 & 43 \\ \hline \end{array}
\displaystyle \text{Calculate Spearman's rank correlation coefficient and interpret your results.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Candidate} & A & B & C & D & E & F & G & H & I & J \\ \hline M & 80 & 88 & 76 & 74 & 68 & 65 & 40 & 43 & 40 & 80 \\ \hline R_1 & 2.5 & 1 & 4 & 5 & 6 & 7 & 9.5 & 8 & 9.5 & 2.5 \\ \hline S & 72 & 84 & 90 & 66 & 54 & 50 & 54 & 38 & 30 & 43 \\ \hline R_2 & 3 & 2 & 1 & 4 & 5.5 & 7 & 5.5 & 9 & 10 & 8 \\ \hline d = R_1 - R_2 & -0.5 & -1 & 3 & 1 & 0.5 & 0 & 4 & -1 & -0.5 & -5.5 \\ \hline d^2 & 0.25 & 1 & 9 & 1 & 0.25 & 0 & 16 & 1 & 0.25 & 30.25 \\ \hline \end{array}

\displaystyle \therefore \Sigma d^2=59 
\displaystyle \text{There are two ties in Mathematics and one tie in Statistics, each of length }2. 
\displaystyle \text{Correction}=\frac{2^3-2}{12}+\frac{2^3-2}{12}+\frac{2^3-2}{12}=\frac{18}{12}=\frac{3}{2} 
\displaystyle \rho=1-\frac{6\left(\Sigma d^2+\text{Correction}\right)}{n(n^2-1)} 
\displaystyle =1-\frac{6\left(59+\frac{3}{2}\right)}{10(100-1)}=1-\frac{363}{990}=\frac{627}{990} 
\displaystyle \therefore \rho=\frac{19}{30}=0.6333 
\displaystyle \text{Thus, there is a moderate positive correlation between Mathematics and Statistics marks.} 
\\

\displaystyle \text{(b) The following results were obtained with respect to two variables } x \text{ and } y:
\displaystyle \Sigma x=30,\quad \Sigma y=42,\quad \Sigma xy=199,\quad \Sigma x^2=184,\quad \Sigma y^2=318,\quad n=6
\displaystyle \text{Find the following:}
\displaystyle \text{(i) The regression coefficient}
\displaystyle \text{(ii) Correlation coefficient between } x \text{ and } y
\displaystyle \text{(iii) Regression equation of } y \text{ on } x
\displaystyle \text{(iv) The likely value of } y \text{ when } x=10 \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \overline{x}=\frac{\Sigma x}{n}=\frac{30}{6}=5,\quad \overline{y}=\frac{\Sigma y}{n}=\frac{42}{6}=7 
\displaystyle S_{xy}=\Sigma xy-\frac{\Sigma x\Sigma y}{n}=199-\frac{30\times42}{6}=-11 
\displaystyle S_{xx}=\Sigma x^2-\frac{(\Sigma x)^2}{n}=184-\frac{900}{6}=34 
\displaystyle S_{yy}=\Sigma y^2-\frac{(\Sigma y)^2}{n}=318-\frac{1764}{6}=24 
\displaystyle \text{(i) }b_{yx}=\frac{S_{xy}}{S_{xx}}=-\frac{11}{34},\quad b_{xy}=\frac{S_{xy}}{S_{yy}}=-\frac{11}{24} 
\displaystyle \text{(ii) }r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-11}{\sqrt{34\times24}} 
\displaystyle r=\frac{-11}{4\sqrt{51}}=-0.3852\text{ approximately} 
\displaystyle \text{(iii) Regression equation of }y\text{ on }x\text{ is }y-\overline{y}=b_{yx}(x-\overline{x}) 
\displaystyle y-7=-\frac{11}{34}(x-5) 
\displaystyle 34y-238=-11x+55 
\displaystyle \therefore 11x+34y=293 
\displaystyle \text{(iv) When }x=10,\quad 11(10)+34y=293 
\displaystyle 34y=183\Rightarrow y=\frac{183}{34}=5.382\text{ approximately} 
\\

\displaystyle \textbf{Question 8:}

\displaystyle \text{(a) A bag contains } 8 \text{ red and } 5 \text{ white balls. Two successive draws of } 3 \text{ balls are made at random}
\displaystyle \text{from the bag without replacement. Find the probability that the first draw yields } 3 \text{ white balls}
\displaystyle \text{and the second draw yields } 3 \text{ red balls.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle P(\text{first 3 white})=\frac{{}^{5}C_{3}}{{}^{13}C_{3}}=\frac{10}{286}=\frac{5}{143} 
\displaystyle \text{After removing 3 white balls, remaining balls: }8R,2W 
\displaystyle P(\text{next 3 red})=\frac{{}^{8}C_{3}}{{}^{10}C_{3}}=\frac{56}{120}=\frac{7}{15} 
\displaystyle \therefore P=\frac{5}{143}\cdot\frac{7}{15}=\frac{35}{2145}=\frac{7}{429} 
\\

\displaystyle \text{(b) A box contains } 30 \text{ bolts and } 40 \text{ nuts. Half of the bolts and half of the nuts are rusted.}
\displaystyle \text{If two items are drawn at random from the box, what is the probability that either both are rusted or both are bolts?} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Total items }=70,\ \text{rusted items }=15+20=35 
\displaystyle P(A=\text{both rusted})=\frac{{}^{35}C_{2}}{{}^{70}C_{2}}=\frac{595}{2415}=\frac{17}{69} 
\displaystyle P(B=\text{both bolts})=\frac{{}^{30}C_{2}}{{}^{70}C_{2}}=\frac{435}{2415}=\frac{29}{161} 
\displaystyle \text{Common cases (both rusted bolts)}=15 
\displaystyle P(A\cap B)=\frac{{}^{15}C_{2}}{{}^{70}C_{2}}=\frac{105}{2415}=\frac{1}{23} 
\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B) 
\displaystyle =\frac{17}{69}+\frac{29}{161}-\frac{1}{23}=\frac{103}{253} 
\\

\displaystyle \textbf{Question 9:}

\displaystyle \text{(a) Using De Moivre's theorem, prove that:}
\displaystyle \left(\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}\right)^n=\cos n\theta+i\sin n\theta,\text{ where } i=\sqrt{-1}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}=\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}\cdot\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta+i\sin\theta} 
\displaystyle =\frac{(1+\cos\theta+i\sin\theta)^2}{(1+\cos\theta)^2+\sin^2\theta} 
\displaystyle =\frac{(1+\cos\theta+i\sin\theta)^2}{2(1+\cos\theta)} 
\displaystyle =\frac{(1+\cos\theta)^2-\sin^2\theta+2i\sin\theta(1+\cos\theta)}{2(1+\cos\theta)} 
\displaystyle =\frac{2\cos\theta(1+\cos\theta)+2i\sin\theta(1+\cos\theta)}{2(1+\cos\theta)} 
\displaystyle =\cos\theta+i\sin\theta 
\displaystyle \therefore \left(\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}\right)^n=(\cos\theta+i\sin\theta)^n 
\displaystyle =\cos n\theta+i\sin n\theta \quad \text{by De Moivre's theorem.} 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) Solve the differential equation: } \frac{dy}{dx}-3y\cot x=\sin 2x,\text{ given } y=2,\text{ when } x=\frac{\pi}{2}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \frac{dy}{dx}-3y\cot x=\sin 2x 
\displaystyle \text{This is a linear differential equation with }P=-3\cot x 
\displaystyle \text{I.F.}=e^{\int -3\cot x\ dx}=e^{-3\log\sin x}=\frac{1}{\sin^3x} 
\displaystyle \therefore \frac{d}{dx}\left(\frac{y}{\sin^3x}\right)=\sin 2x\cdot\frac{1}{\sin^3x} 
\displaystyle =\frac{2\sin x\cos x}{\sin^3x}=2\cot x\ \mathrm{cosec}\ x 
\displaystyle \frac{y}{\sin^3x}=\int 2\cot x\ \mathrm{cosec}\ x\ dx=-2\mathrm{cosec}\ x+C 
\displaystyle y=\sin^3x\left(C-2\mathrm{cosec}\ x\right)=C\sin^3x-2\sin^2x 
\displaystyle \text{Given }y=2\text{ when }x=\frac{\pi}{2} 
\displaystyle 2=C(1)^3-2(1)^2\Rightarrow C=4 
\displaystyle \therefore y=4\sin^3x-2\sin^2x 
\\

\displaystyle \textbf{Question 10:}

\displaystyle \text{(a) For any of the vectors } \overrightarrow{a},\overrightarrow{b} \text{ and } \overrightarrow{c}, \text{ prove:}
\displaystyle [\overrightarrow{a}-\overrightarrow{b}\quad \overrightarrow{b}-\overrightarrow{c}\quad \overrightarrow{c}-\overrightarrow{a}]=0 \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let } \overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b},\quad \overrightarrow{v}=\overrightarrow{b}-\overrightarrow{c},\quad \overrightarrow{w}=\overrightarrow{c}-\overrightarrow{a} 
\displaystyle \therefore \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=(\overrightarrow{a}-\overrightarrow{b})+(\overrightarrow{b}-\overrightarrow{c})+(\overrightarrow{c}-\overrightarrow{a})=\overrightarrow{0} 
\displaystyle \therefore \overrightarrow{w}=-(\overrightarrow{u}+\overrightarrow{v}) 
\displaystyle [\overrightarrow{u}\ \overrightarrow{v}\ \overrightarrow{w}]=[\overrightarrow{u}\ \overrightarrow{v}\ -(\overrightarrow{u}+\overrightarrow{v})] 
\displaystyle =-[\overrightarrow{u}\ \overrightarrow{v}\ \overrightarrow{u}]-[\overrightarrow{u}\ \overrightarrow{v}\ \overrightarrow{v}]=0 
\displaystyle \therefore [\overrightarrow{a}-\overrightarrow{b}\quad \overrightarrow{b}-\overrightarrow{c}\quad \overrightarrow{c}-\overrightarrow{a}]=0 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \text{(b) In any triangle } ABC, \text{ prove by vector method: } \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let } \overrightarrow{AB}=\overrightarrow{c},\quad \overrightarrow{AC}=\overrightarrow{b},\quad \overrightarrow{BC}=\overrightarrow{a} 
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\frac{1}{2}bc\sin A 
\displaystyle \text{Also, area}=\frac{1}{2}ca\sin B=\frac{1}{2}ab\sin C 
\displaystyle \therefore bc\sin A=ca\sin B=ab\sin C 
\displaystyle \text{Dividing throughout by }abc,\quad \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} 
\displaystyle \therefore \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \textbf{Question 11:}

\displaystyle \text{(a) Find the shortest distance between the lines:}
\displaystyle \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7} \text{ and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{For the first line, } \overrightarrow{a_1}=8\widehat{i}-9\widehat{j}+10\widehat{k},\quad \overrightarrow{b_1}=3\widehat{i}-16\widehat{j}+7\widehat{k} 
\displaystyle \text{For the second line, } \overrightarrow{a_2}=15\widehat{i}+29\widehat{j}+5\widehat{k},\quad \overrightarrow{b_2}=3\widehat{i}+8\widehat{j}-5\widehat{k} 
\displaystyle \overrightarrow{a_2}-\overrightarrow{a_1}=7\widehat{i}+38\widehat{j}-5\widehat{k} 
\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\3&-16&7\\3&8&-5\end{vmatrix}=24\widehat{i}+36\widehat{j}+72\widehat{k} 
\displaystyle \text{Shortest distance}=\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})|}{|\overrightarrow{b_1}\times\overrightarrow{b_2}|} 
\displaystyle =\frac{|7(24)+38(36)-5(72)|}{\sqrt{24^2+36^2+72^2}} 
\displaystyle =\frac{1176}{84}=14 
\displaystyle \therefore \text{Shortest distance}=14\text{ units} 
\\

\displaystyle \text{(b) Find the equation of the plane passing through the line of intersection of the planes } x+2y+3z-5=0
\displaystyle \text{and } 3x-2y-z+1=0 \text{ and cutting off equal intercepts on the } x \text{ and } z \text{ axes.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Required plane is }(x+2y+3z-5)+\lambda(3x-2y-z+1)=0 
\displaystyle (1+3\lambda)x+(2-2\lambda)y+(3-\lambda)z+(\lambda-5)=0 
\displaystyle \text{Since intercepts on }x\text{-axis and }z\text{-axis are equal, coefficients of }x\text{ and }z\text{ are equal.} 
\displaystyle 1+3\lambda=3-\lambda\Rightarrow 4\lambda=2\Rightarrow \lambda=\frac{1}{2} 
\displaystyle \therefore \left(1+\frac{3}{2}\right)x+\left(2-1\right)y+\left(3-\frac{1}{2}\right)z+\left(\frac{1}{2}-5\right)=0 
\displaystyle \frac{5}{2}x+y+\frac{5}{2}z-\frac{9}{2}=0 
\displaystyle \therefore 5x+2y+5z-9=0 
\\

\displaystyle \textbf{Question 12:}

\displaystyle \text{(a) In a class of } 75 \text{ students, } 15 \text{ are above average, } 45 \text{ are average and the rest are below average achievers.}
\displaystyle \text{The probability that an above average achieving student fails is } 0.005, \text{ that an average achieving student fails is } 0.05
\displaystyle \text{and the probability of a below average achieving student failing is } 0.15. \text{ If a student is known to have passed,}
\displaystyle \text{what is the probability that he is a below average achiever?} \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle P(A)=\frac{15}{75}=\frac{1}{5},\quad P(B)=\frac{45}{75}=\frac{3}{5},\quad P(C)=\frac{15}{75}=\frac{1}{5} 
\displaystyle P(Pass|A)=1-0.005=0.995,\quad P(Pass|B)=1-0.05=0.95 
\displaystyle P(Pass|C)=1-0.15=0.85 
\displaystyle P(C|Pass)=\frac{P(C)P(Pass|C)}{P(A)P(Pass|A)+P(B)P(Pass|B)+P(C)P(Pass|C)} 
\displaystyle =\frac{\frac{1}{5}\times0.85}{\frac{1}{5}\times0.995+\frac{3}{5}\times0.95+\frac{1}{5}\times0.85} 
\displaystyle =\frac{0.17}{0.199+0.57+0.17}=\frac{0.17}{0.939}=\frac{170}{939} 
\displaystyle \therefore P(C|Pass)=0.181\text{ approximately} 
\\

\displaystyle \text{(b) The probability that a bulb produced by a factory will fuse in } 100 \text{ days of use is } 0.05.
\displaystyle \text{Find the probability that out of } 5 \text{ such bulbs, after } 100 \text{ days of use:}
\displaystyle \text{(i) None fuse.}
\displaystyle \text{(ii) Not more than one fuses.}
\displaystyle \text{(iii) More than one fuses.}
\displaystyle \text{(iv) At least one fuses.} \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Here }n=5,\quad p=0.05,\quad q=0.95 
\displaystyle \text{Let }X\text{ be the number of bulbs that fuse. Then }X\sim B(5,0.05) 
\displaystyle P(X=r)={}^{5}C_r(0.05)^r(0.95)^{5-r} 
\displaystyle \text{(i) }P(X=0)=(0.95)^5=0.7738 
\displaystyle \text{(ii) }P(X\leq1)=P(X=0)+P(X=1) 
\displaystyle =(0.95)^5+{}^{5}C_1(0.05)(0.95)^4=0.7738+0.2036=0.9774 
\displaystyle \text{(iii) }P(X>1)=1-P(X\leq1)=1-0.9774=0.0226 
\displaystyle \text{(iv) }P(X\geq1)=1-P(X=0)=1-0.7738=0.2262 
\\

\displaystyle \textbf{Question 13:}

\displaystyle \text{(a) Two tailors } P \text{ and } Q \text{ earn Rs. } 150 \text{ and Rs. } 200 \text{ per day respectively. } P \text{ can stitch}
\displaystyle 6 \text{ shirts and } 4 \text{ trousers a day, while } Q \text{ can stitch } 10 \text{ shirts and } 4 \text{ trousers per day.}
\displaystyle \text{How many days should each work to produce at least } 60 \text{ shirts and } 32 \text{ trousers at minimum labour cost?} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Let }P\text{ work for }x\text{ days and }Q\text{ work for }y\text{ days.} 
\displaystyle \text{Minimize }Z=150x+200y 
\displaystyle \text{Subject to }6x+10y\geq60,\quad 4x+4y\geq32,\quad x\geq0,\ y\geq0 
\displaystyle \text{i.e. }3x+5y\geq30,\quad x+y\geq8,\quad x\geq0,\ y\geq0 
\displaystyle \text{Corner points of feasible region are }(10,0),(5,3),(0,8) 
\displaystyle \begin{array}{|c|c|} \hline \text{Corner point} & Z=150x+200y \\ \hline (10,0) & 1500 \\ \hline (5,3) & 1350 \\ \hline (0,8) & 1600 \\ \hline \end{array} 
\displaystyle \text{Minimum cost is Rs. }1350\text{ at }(x,y)=(5,3). 
\displaystyle \therefore P\text{ should work for }5\text{ days and }Q\text{ should work for }3\text{ days.} 
\\

\displaystyle \text{(b) A machine costs Rs. } 97000 \text{ and its effective life is estimated to be } 12 \text{ years. If scrap realizes}
\displaystyle \text{Rs. } 2000 \text{ only, what amount should be retained out of the profits at the end of each year to accumulate}
\displaystyle \text{at compound interest of } 5\% \text{ per annum in order to buy a new machine after } 12 \text{ years?}
\displaystyle \text{Use } 1.05^{12}=1.769. \hspace{5.0cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Amount required after }12\text{ years}=97000-2000=95000 
\displaystyle \text{Let Rs. }A\text{ be retained at the end of each year.} 
\displaystyle 95000=A\left[\frac{(1.05)^{12}-1}{0.05}\right] 
\displaystyle 95000=A\left[\frac{1.769-1}{0.05}\right]=A\left(\frac{0.769}{0.05}\right) 
\displaystyle 95000=15.38A 
\displaystyle A=\frac{95000}{15.38}=6176.85 
\displaystyle \therefore \text{Rs. }6176.85\text{ approximately should be retained at the end of each year.} 
\\

\displaystyle \textbf{Question 14:}

\displaystyle \text{(a) A bill of Rs. } 1000 \text{ drawn on 7th May 2011 for six months was discounted on 29th August 2011}
\displaystyle \text{for cash payment of Rs. } 988. \text{ Find the rate of interest charged by the bank.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle \text{Nominal due date}=7\text{th November 2011} 
\displaystyle \text{Legal due date}=10\text{th November 2011} 
\displaystyle \text{Unexpired period from 29th August 2011 to 10th November 2011}=73\text{ days} 
\displaystyle \text{Banker's discount}=1000-988=\text{Rs. }12 
\displaystyle \text{Banker's discount}=\frac{\text{Face value}\times r\times t}{100} 
\displaystyle 12=\frac{1000\times r\times 73}{100\times365} 
\displaystyle r=\frac{12\times100\times365}{1000\times73}=6 
\displaystyle \therefore \text{Rate of interest charged by the bank}=6\%\text{ per annum} 
\\

\displaystyle \text{(b) If total cost function is given by } C=a+bx+cx^2, \text{ where } x \text{ is the quantity of output, show that:}
\displaystyle \frac{d}{dx}(AC)=\frac{1}{x}(MC-AC), \text{ where } MC \text{ is the marginal cost and } AC \text{ is the average cost.} \hspace{0.2cm} [5]
\displaystyle \text{Answer:} 
\displaystyle C=a+bx+cx^2 
\displaystyle AC=\frac{C}{x}=\frac{a}{x}+b+cx 
\displaystyle MC=\frac{dC}{dx}=b+2cx 
\displaystyle \frac{d}{dx}(AC)=\frac{d}{dx}\left(\frac{a}{x}+b+cx\right)=-\frac{a}{x^2}+c 
\displaystyle \frac{1}{x}(MC-AC)=\frac{1}{x}\left(b+2cx-\frac{a}{x}-b-cx\right) 
\displaystyle =\frac{1}{x}\left(cx-\frac{a}{x}\right)=c-\frac{a}{x^2} 
\displaystyle \therefore \frac{d}{dx}(AC)=\frac{1}{x}(MC-AC) 
\displaystyle \text{Hence proved.} 
\\

\displaystyle \textbf{Question 15:}

\displaystyle \text{(a) Find the consumer price index number for the year } 2010 \text{ using year } 2000 \text{ as the base year}
\displaystyle \text{by using method of weighted aggregates.} \hspace{5.0cm} [5]
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Commodity} & A & B & C & D & E \\ \hline \text{Year 2000 price (in Rs. per unit)} & 16 & 40 & 0.50 & 5.12 & 2.00 \\ \hline \text{Year 2010 price (in Rs. per unit)} & 20 & 60 & 0.50 & 6.25 & 1.50 \\ \hline \text{Weights} & 40 & 25 & 5.00 & 20.00 & 10.00 \\ \hline \end{array}
\displaystyle \text{Answer:} 
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Commodity} & p_0 & p_1 & w & p_0w \quad p_1w \\ \hline A & 16 & 20 & 40 & 640 \quad 800 \\ \hline B & 40 & 60 & 25 & 1000 \quad 1500 \\ \hline C & 0.50 & 0.50 & 5 & 2.5 \quad 2.5 \\ \hline D & 5.12 & 6.25 & 20 & 102.4 \quad 125 \\ \hline E & 2.00 & 1.50 & 10 & 20 \quad 15 \\ \hline \end{array} 
\displaystyle \Sigma p_0w=1764.9,\quad \Sigma p_1w=2442.5 
\displaystyle \text{Consumer Price Index}=\frac{\Sigma p_1w}{\Sigma p_0w}\times100 
\displaystyle =\frac{2442.5}{1764.9}\times100=138.39 
\displaystyle \therefore \text{Consumer price index number for }2010=138.39 
\\

\displaystyle \text{(b) Calculate the } 5 \text{ yearly moving averages of the number of students in a college from the following data}
\displaystyle \text{and plot them on the graph paper.} \hspace{5.0cm} [5]
\displaystyle \begin{array}{|c|cccccccccc|} \hline \text{Year} & 1981 & 1982 & 1983 & 1984 & 1985 & 1986 & 1987 & 1988 & 1989 & 1990 \\ \hline \text{Number} & 332 & 317 & 357 & 392 & 402 & 405 & 510 & 427 & 405 & 438 \\ \hline \end{array}
\displaystyle \text{Answer:} 
\displaystyle \begin{array}{|c|c|c|} \hline \text{Years} & \text{Total of 5 years} & \text{5-yearly moving average} \\ \hline 1981-1985 & 332+317+357+392+402=1800 & 360.0 \\ \hline 1982-1986 & 317+357+392+402+405=1873 & 374.6 \\ \hline 1983-1987 & 357+392+402+405+510=2066 & 413.2 \\ \hline 1984-1988 & 392+402+405+510+427=2136 & 427.2 \\ \hline 1985-1989 & 402+405+510+427+405=2149 & 429.8 \\ \hline 1986-1990 & 405+510+427+405+438=2185 & 437.0 \\ \hline \end{array} 
\displaystyle \text{Since period is }5,\text{ moving averages are placed against the middle years.} 
\displaystyle \begin{array}{|c|c|} \hline \text{Year} & \text{Moving Average} \\ \hline 1983 & 360.0 \\ \hline 1984 & 374.6 \\ \hline 1985 & 413.2 \\ \hline 1986 & 427.2 \\ \hline 1987 & 429.8 \\ \hline 1988 & 437.0 \\ \hline \end{array} 
\displaystyle \text{Plot the points }(1983,360.0),(1984,374.6),(1985,413.2),(1986,427.2), 
\displaystyle (1987,429.8),(1988,437.0)\text{ on graph paper and join them smoothly.} 
\\

 

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