\displaystyle \textbf{MATHEMATICS}
\displaystyle \text{Maximum Marks: 80}

\displaystyle \textit{\textbf{Time allowed: Three hours}}
\displaystyle \text{1. Answers to this Paper must be written on the paper provided separately.}
\displaystyle \text{2. You will not be allowed to write during first 15 minutes.}
\displaystyle \text{3. This time is to be spent in reading the question paper.}
\displaystyle \textbf{4. The time given at the head of this Paper is the time allowed for}
\displaystyle \textbf{writing the answers.}
\displaystyle \text{5. Attempt all questions from Section A and any four questions from Section B.}
\displaystyle \textbf{6. All working, including rough work, must be clearly shown, and must be}
\displaystyle \textbf{done on the the same sheet as the rest of the answer.}
\displaystyle \textbf{7. Omission of essential working will result in loss of marks.}
\displaystyle \text{8. The intended marks for questions or parts of questions are given in brackets [ ].}
\displaystyle \textbf{9. Mathematical tables and graph papers are to be provided by the school.}

\displaystyle \text{\textbf{Instruction for the Supervising Examiner}}
\displaystyle \text{Kindly read aloud the Instructions given above to all the candidates present in}
\displaystyle \text{the Examination Hall.}

\displaystyle \textbf{SECTION A (40 Marks)}
\displaystyle \text{(Attempt all questions from this Section.)}

\displaystyle \textbf{Question 1}
\displaystyle \text{Choose the correct answers to the questions from the given options.}\qquad            \displaystyle [15]
\displaystyle \text{(Do not copy the questions, write the correct answers only.)}
\\
\displaystyle \text{(i) The given quadratic equation }3x^{2}+\sqrt{7}x+2=0\text{ has:}
\displaystyle \text{(a) two equal real roots.}
\displaystyle \text{(b) two distinct real roots.}
\displaystyle \text{(c) more than two real roots.}
\displaystyle \text{(d) no real roots.}
\displaystyle \text{Answer:}
\displaystyle  \text{ For }3x^2+\sqrt{7}x+2=0,\ a=3,\ b=\sqrt{7},\ c=2
\displaystyle \text{Discriminant }D=b^2-4ac=7-4(3)(2)=7-24=-17
\displaystyle D<0
\displaystyle \therefore\ \text{the equation has no real roots}
\displaystyle \text{Answer: (d) no real roots}
\\
\displaystyle \text{(ii) Mr. Anuj deposits Rs }500\text{ per month for }18\text{ months in a recurring deposit}
\displaystyle \text{account at a certain rate. If he earns Rs }570\text{ as interest at the time of maturity,}
\displaystyle \text{then his matured amount is:}
\displaystyle \text{(a) }\text{Rs}(500\times 18+570)
\displaystyle \text{(b) }\text{Rs}(500\times 19+570)
\displaystyle \text{(c) }\text{Rs}(500\times 18\times 19+570)
\displaystyle \text{(d) }\text{Rs}(500\times 9\times 19+570)
\displaystyle \text{Answer:}
\displaystyle \text{ Monthly deposit }=500,\ \text{number of months }=18,\ \text{interest }=570
\displaystyle \text{Matured amount }=\text{Total deposit}+\text{Interest}
\displaystyle =500\times 18+570
\displaystyle \text{Answer: (a) }\text{Rs}(500\times 18+570)
\\
\displaystyle \text{(iii) Which of the following cannot be the probability of any event?}
\displaystyle \text{(a) }\frac{5}{4}
\displaystyle \text{(b) }0.25
\displaystyle \text{(c) }\frac{1}{33}
\displaystyle \text{(d) }67\%
\displaystyle \text{Answer:}
\displaystyle  \text{ Probability of an event lies between }0\text{ and }1
\displaystyle \frac{5}{4}=1.25>1\text{, so it cannot be a probability}
\displaystyle \text{Answer: (a) }\frac{5}{4}
\\
\displaystyle \text{(iv) The equation of the line passing through origin and parallel to the line}
\displaystyle 3x+4y+7=0\text{ is:}
\displaystyle \text{(a) }3x+4y+5=0
\displaystyle \text{(b) }4x-3y-5=0
\displaystyle \text{(c) }4x-3y=0
\displaystyle \text{(d) }3x+4y=0
\displaystyle \text{Answer:}
\displaystyle  \text{ A line parallel to }3x+4y+7=0\text{ has same coefficients of }x\text{ and }y
\displaystyle \text{So required line is }3x+4y+c=0
\displaystyle \text{Since it passes through origin, }c=0
\displaystyle \therefore\ \text{Equation is }3x+4y=0
\displaystyle \text{Answer: (d) }3x+4y=0
\\
\displaystyle \text{(v) If }A=\begin{bmatrix}0&1\\1&0\end{bmatrix},\text{ then }A^{2}\text{ is equal to:}
\displaystyle \text{(a) }\begin{bmatrix}1&1\\0&0\end{bmatrix}
\displaystyle \text{(b) }\begin{bmatrix}0&0\\1&1\end{bmatrix}
\displaystyle \text{(c) }\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle \text{(d) }\begin{bmatrix}0&1\\1&0\end{bmatrix}
\displaystyle \text{Answer:}
\displaystyle  \text{ Given }A=\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}
\displaystyle A^2=A\cdot A=\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}
\displaystyle =\begin{bmatrix}0\cdot 0+1\cdot 1 & 0\cdot 1+1\cdot 0\\ 1\cdot 0+0\cdot 1 & 1\cdot 1+0\cdot 0\end{bmatrix}
\displaystyle =\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
\displaystyle \text{Answer: (c) }\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
\\
\displaystyle \text{(vi) In the given diagram, chords }AC\text{ and }BC\text{ are equal. If }\angle ACD=120^{\circ},
\displaystyle \text{then }\angle AEC\text{ is:}
\displaystyle \text{(a) }30^{\circ}
\displaystyle \text{(b) }60^{\circ}
\displaystyle \text{(c) }90^{\circ}
\displaystyle \text{(d) }120^{\circ}
\displaystyle \text{Answer:}
\displaystyle  \text{Given }\angle ACD=120^\circ\text{ and }BC\text{ is produced to }D
\displaystyle \therefore\ \angle ACB=180^\circ-120^\circ=60^\circ
\displaystyle \text{Also, }AC=BC\Rightarrow \triangle ABC\text{ is isosceles}
\displaystyle \therefore\ \angle BAC=\angle ABC
\displaystyle \angle BAC+\angle ABC+60^\circ=180^\circ
\displaystyle 2\angle ABC=120^\circ
\displaystyle \angle ABC=60^\circ,\quad \angle BAC=60^\circ
\displaystyle \therefore\ \triangle ABC\text{ is equilateral}
\displaystyle \text{Now, }\angle ABC\text{ subtends minor arc }AC
\displaystyle \text{So, minor arc }AC=2\times 60^\circ=120^\circ
\displaystyle \text{Point }E\text{ lies on the minor arc }AC,\text{ hence }\angle AEC\text{ subtends the major arc }AC
\displaystyle \text{Major arc }AC=360^\circ-120^\circ=240^\circ
\displaystyle \therefore\ \angle AEC=\frac{1}{2}\times 240^\circ=120^\circ
\displaystyle \text{Answer: (d) }120^\circ
\\
\displaystyle \text{(vii) The factor common to the two polynomials }x^{2}-4\text{ and }x^{3}-x^{2}-4x+4
\displaystyle \text{ is:}
\displaystyle \text{(a) }(x+1)
\displaystyle \text{(b) }(x-1)
\displaystyle \text{(c) }(x-2)
\displaystyle \text{(d) }(x-4)
\displaystyle \text{Answer:}
\displaystyle x^2-4=(x-2)(x+2)
\displaystyle x^3-x^2-4x+4=x^2(x-1)-4(x-1)
\displaystyle =(x^2-4)(x-1)=(x-2)(x+2)(x-1)
\displaystyle \text{Common factors are }(x-2)\text{ and }(x+2)
\displaystyle \text{From options, answer: (c) }(x-2)
\\
\displaystyle \text{(viii) A man invested in a company paying }12\%\text{ dividend on its share. If the}
\displaystyle \text{percentage return on his investment is }10\%,\text{ then the shares are:}
\displaystyle \text{(a) at par}
\displaystyle \text{(b) below par}
\displaystyle \text{(c) above par}
\displaystyle \text{(d) cannot be determined}
\displaystyle \text{Answer:}
\displaystyle \text{Dividend }=12\%,\ \text{Return }=10\%
\displaystyle \text{Since return }<\text{ dividend, shares are bought above face value}
\displaystyle \text{Answer: (c) above par}
\\
\displaystyle \text{(ix) Statement 1: The point which is equidistant from three non-collinear}
\displaystyle \text{points }D,E\text{ and }F\text{ is the circumcentre of the }\triangle DEF.
\displaystyle \text{Statement 2: The incentre of a triangle is the point where the bisector of}
\displaystyle \text{the angles intersects.}
\displaystyle \text{(a) Both the statements are true.}
\displaystyle \text{(b) Both the statements are false.}
\displaystyle \text{(c) Statement 1 is true, and Statement 2 is false.}
\displaystyle \text{(d) Statement 1 is false, and Statement 2 is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Statement 1: The point equidistant from three non-collinear points of a triangle is the circumcentre}
\displaystyle \text{So, Statement 1 is true}
\displaystyle \text{Statement 2: The incentre is the point of intersection of the angle bisectors of a triangle}
\displaystyle \text{So, Statement 2 is true}
\displaystyle \text{Answer: (a) Both the statements are true}
\\
\displaystyle \text{(x) Assertion (A): If }\sin^{2}A+\sin A=1\text{ then }\cos^{4}A+\cos^{2}A=1
\displaystyle \text{Reason (R): }1-\sin^{2}A=\cos^{2}A
\displaystyle \text{(a) (A) is true, (R) is false.}
\displaystyle \text{(b) (A) is false, (R) is true.}
\displaystyle \text{(c) Both (A) and (R) are true, and (R) is the correct reason for (A).}
\displaystyle \text{(d) Both (A) and (R) are true, and (R) is the incorrect reason for (A).}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\sin^2 A+\sin A=1
\displaystyle \therefore\ 1-\sin^2 A=\sin A
\displaystyle \text{But }1-\sin^2 A=\cos^2 A
\displaystyle \therefore\ \cos^2 A=\sin A
\displaystyle \cos^4 A+\cos^2 A=(\cos^2 A)^2+\cos^2 A
\displaystyle =(\sin A)^2+\sin A
\displaystyle =\sin^2 A+\sin A=1
\displaystyle \therefore\ \text{Assertion (A) is true}
\displaystyle \text{Reason (R): }1-\sin^2 A=\cos^2 A\text{ is also true}
\displaystyle \text{and it is the correct reason for (A)}
\displaystyle \text{Answer: (c)}
\\
\displaystyle \text{(xi) In the given diagram }\triangle ABC\sim\triangle EFG.\text{ If }\angle ABC=\angle EFG=60^{\circ},
\displaystyle \text{then the length of the side }FG\text{ is:}
\displaystyle \text{(a) }15\text{ cm}
\displaystyle \text{(b) }20\text{ cm}
\displaystyle \text{(c) }25\text{ cm}
\displaystyle \text{(d) }30\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \triangle ABC\sim \triangle EFG
\displaystyle \therefore\ \frac{AB}{EF}=\frac{BC}{FG}
\displaystyle \text{From the figure, }AB=15\text{ cm},\ BC=3\text{ cm},\ EF=75\text{ cm}
\displaystyle \frac{15}{75}=\frac{3}{FG}
\displaystyle \frac{1}{5}=\frac{3}{FG}
\displaystyle FG=15\text{ cm}
\displaystyle \text{Answer: (a) }15\text{ cm}
\\
\displaystyle \text{(xii) If the volume of two spheres is in the ratio }27:64,\text{ then the ratio of their}
\displaystyle \text{radii is:}
\displaystyle \text{(a) }3:4
\displaystyle \text{(b) }4:3
\displaystyle \text{(c) }9:16
\displaystyle \text{Answer:}
\displaystyle \text{Volume of sphere }\propto r^3
\displaystyle \therefore\ \frac{V_1}{V_2}=\frac{r_1^3}{r_2^3}=\frac{27}{64}
\displaystyle \therefore\ \frac{r_1}{r_2}=\sqrt[3]{\frac{27}{64}}=\frac{3}{4}
\displaystyle \text{Answer: (a) }3:4
\\
\displaystyle \text{(xiii) The marked price of an article is Rs }1375.\text{ If the CGST is charged at a rate of}
\displaystyle 4\%,\text{ then the price of the article including GST is:}
\displaystyle \text{(a) Rs }55
\displaystyle \text{(b) Rs }110
\displaystyle \text{(c) Rs }1430
\displaystyle \text{(d) Rs }1485
\displaystyle \text{Answer:}
\displaystyle \text{CGST }=4\%\Rightarrow \text{For GST, total tax }=4\%+4\%=8\%
\displaystyle \text{GST amount }=8\%\text{ of }1375=\frac{8}{100}\times 1375=110
\displaystyle \text{Price including GST }=1375+110=1485
\displaystyle \text{Answer: (d) }\text{Rs }148
\\
\displaystyle \text{(xiv) The solution set for }0<-\frac{x}{3}<2,\;x\in Z\text{ is:}
\displaystyle \text{(a) }\{-5,-4,-3,-2,-1\}
\displaystyle \text{(b) }\{-6,-5,-4,-3,-2,-1\}
\displaystyle \text{(c) }\{-5,-4,-3,-2,-1,0\}
\displaystyle \text{(d) }\{-6,-5,-4,-3,-2,-1,0\}
\displaystyle \text{Answer:}
\displaystyle 0<-\frac{x}{3}<2,\ x\in Z
\displaystyle \text{Multiply throughout by }3
\displaystyle 0<-x<6
\displaystyle \text{Multiply throughout by }-1\text{ and reverse inequalities}
\displaystyle 0>x>-6
\displaystyle \therefore\ -6<x<0
\displaystyle \text{Since }x\in Z,\ \text{solution set }=\{-5,-4,-3,-2,-1\}
\displaystyle \text{Answer: (a) }\{-5,-4,-3,-2,-1\}
\\
\displaystyle \text{(xv) Assertion (A): The mean of first }9\text{ natural numbers is }4.5.
\displaystyle \text{Reason (R): Mean = }\frac{\text{Sum of all the observations}}{\text{Total number of observations}}
\displaystyle \text{(a) (A) is true, (R) is false.}
\displaystyle \text{(b) (A) is false, (R) is true.}
\displaystyle \text{(c) Both (A) and (R) are true, and (R) is the correct reason for (A).}
\displaystyle \text{(d) Both (A) and (R) are true, and (R) is the incorrect reason for (A).}
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): Mean of first 9 natural numbers}=\frac{1+2+3+4+5+6+7+8+9}{9}=\frac{45}{9}=5
\displaystyle \text{So Assertion (A) is false}
\displaystyle \text{Reason (R): Mean }=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{ is true}
\displaystyle \text{Answer: (b) (A) is false, (R) is true}
\\
\displaystyle \textbf{Question 2}
\displaystyle \text{(i) Solve the following quadratic equation }2x^{2}-5x-4=0            \displaystyle [4]
\displaystyle \text{Give your answer correct to three significant figures.}
\displaystyle \text{(Use mathematical tables for this question)}
\displaystyle \text{(ii) Mrs. Rao deposited Rs }250\text{ per month in a recurring deposit account for a}
\displaystyle \text{period of }3\text{ years. She received Rs }10110\text{ at the time of maturity. Find:}            \displaystyle [4]
\displaystyle \text{(a) the rate of interest.}
\displaystyle \text{(b) how much more interest Mrs. Rao will receive if she had deposited}
\displaystyle \text{Rs }50\text{ more per month at the same rate of interest and for the same time.}
\displaystyle \text{(iii) In }\triangle ABC,\angle ABC=90^{\circ}, AB=20\text{ cm, }AC=25\text{ cm, }DE\text{ is perpendicular to}
\displaystyle AC\text{ such that }\angle DEA=90^{\circ}\text{ and }DE=3\text{ cm as shown in the given figure.}            \displaystyle [4] \displaystyle \text{(a) Prove that }\triangle ABC\sim\triangle AED.
\displaystyle \text{(b) Find the lengths of }BC, AD\text{ and }AE.
\displaystyle \text{(c) If }BCED\text{ represents a plot of land on a map whose actual area on ground is }576\text{ m}^{2},
\displaystyle \text{then find the scale factor of the map.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle 2x^2-5x-4=0
\displaystyle x=\frac{5\pm\sqrt{(-5)^2-4(2)(-4)}}{2\cdot 2}
\displaystyle =\frac{5\pm\sqrt{25+32}}{4}=\frac{5\pm\sqrt{57}}{4}
\displaystyle \sqrt{57}\approx 7.55
\displaystyle x=\frac{5+7.55}{4}=3.14,\quad x=\frac{5-7.55}{4}=-0.637
\displaystyle \text{Answer: }x=3.14,\ -0.637\ (\text{to 3 s.f.})

\displaystyle \text{(ii)}
\displaystyle \text{(a) } P=250,\ n=3\text{ years }=36\text{ months},\ \text{Maturity }=10110
\displaystyle \text{Total deposit }=250\times 36=9000
\displaystyle \text{Interest }=10110-9000=1110
\displaystyle I=P\cdot \frac{n(n+1)}{2}\cdot \frac{r}{12\cdot 100}
\displaystyle 1110=250\cdot \frac{36\cdot 37}{2}\cdot \frac{r}{1200}
\displaystyle 1110=250\cdot 666\cdot \frac{r}{1200}
\displaystyle 1110=\frac{166500r}{1200}
\displaystyle r=8\%
\displaystyle \text{(b) New deposit }=300
\displaystyle \text{New interest }=300\cdot \frac{36\cdot 37}{2}\cdot \frac{8}{1200}
\displaystyle =300\cdot 666\cdot \frac{8}{1200}=1332
\displaystyle \text{Extra interest }=1332-1110=222
\displaystyle \text{Answer: Rate }=8\%,\ \text{Extra interest }=222

\displaystyle \text{(iii)}
\displaystyle \text{(a) }\angle ABC=\angle DEA=90^\circ,\ \angle A\text{ common}
\displaystyle \therefore\ \triangle ABC\sim \triangle AED\ (\text{AA})
\displaystyle \text{(b) }BC=\sqrt{AC^2-AB^2}=\sqrt{25^2-20^2}=\sqrt{625-400}=15\text{ cm}
\displaystyle \frac{AB}{AE}=\frac{BC}{DE}=\frac{AC}{AD}
\displaystyle \frac{20}{AE}=\frac{15}{3}=5\Rightarrow AE=4\text{ cm}
\displaystyle \frac{25}{AD}=5\Rightarrow AD=5\text{ cm}
\displaystyle \text{Answer: }BC=15\text{ cm},\ AE=4\text{ cm},\ AD=5\text{ cm}
\displaystyle \text{(c) Area of }BCED=\text{Area of }\triangle ABC-\triangle AED
\displaystyle =\frac{1}{2}(20)(15)-\frac{1}{2}(4)(3)=150-6=144\text{ cm}^2
\displaystyle \text{Actual area }=576\text{ m}^2=576\times 10^4\text{ cm}^2
\displaystyle \text{Scale factor }=\sqrt{\frac{576\times 10^4}{144}}=\sqrt{4\times 10^4}=200
\displaystyle \text{Hence, scale }=1:200
\\
\displaystyle \textbf{Question 3}
\displaystyle \text{(i) Use ruler and compass for the following construction. Construct a }\triangle ABC, \displaystyle [4]
\displaystyle \text{where }AB=6\text{ cm, }AC=4.5\text{ cm and }\angle BAC=120^{\circ}.\text{ Construct a circle}
\displaystyle \text{circumscribing the }\triangle ABC.\text{ Measure and write down the length of the radius}
\displaystyle \text{of the circle.}
\displaystyle \text{(ii) If }A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\;B=\begin{bmatrix}2&1\\4&2\end{bmatrix}\text{ and }C=\begin{bmatrix}-5&1\\7&-4\end{bmatrix}            \displaystyle [4]
\displaystyle \text{Find:}
\displaystyle \text{(a) }A+C
\displaystyle \text{(b) }B(A+C)
\displaystyle \text{(c) }5B
\displaystyle \text{(d) }B(A+C)-5B
\displaystyle \text{(iii) In the given graph }ABCD\text{ is a parallelogram.}            \displaystyle [4] \displaystyle \text{Using the graph, answer the following:}
\displaystyle \text{(a) write down the coordinates of }A,B,C\text{ and }D.
\displaystyle \text{(b) calculate the coordinates of }P,\text{ the point of intersection of the}
\displaystyle \text{diagonals }AC\text{ and }BD.
\displaystyle \text{(c) find the slope of sides }CB\text{ and }DA\text{ and verify that they represent}
\displaystyle \text{parallel lines.}
\displaystyle \text{(d) find the equation of the diagonal }AC.
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{In }\triangle ABC,\ AB=6\text{ cm},\ AC=4.5\text{ cm},\ \angle BAC=120^\circ
\displaystyle \text{First find }BC\text{ using cosine rule:}
\displaystyle BC^2=AB^2+AC^2-2(AB)(AC)\cos 120^\circ
\displaystyle =6^2+4.5^2-2(6)(4.5)\left(-\frac{1}{2}\right)
\displaystyle =36+20.25+27=83.25
\displaystyle BC=\sqrt{83.25}\approx 9.124\text{ cm}
\displaystyle \text{Circumradius }R=\frac{a}{2\sin A}=\frac{BC}{2\sin 120^\circ}
\displaystyle =\frac{9.124}{2\cdot \frac{\sqrt{3}}{2}}=\frac{9.124}{\sqrt{3}}\approx 5.27\text{ cm}
\displaystyle \therefore\ \text{Radius of the circumcircle }\approx 5.27\text{ cm}

\displaystyle \text{(ii)}
\displaystyle A=\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix},\ B=\begin{bmatrix}2 & 1\\ 4 & 2\end{bmatrix},\ C=\begin{bmatrix}-5 & 1\\ 7 & -4\end{bmatrix}
\displaystyle \text{(a) }A+C=\begin{bmatrix}1+(-5) & 2+1\\ 3+7 & 4+(-4)\end{bmatrix}=\begin{bmatrix}-4 & 3\\ 10 & 0\end{bmatrix}
\displaystyle \text{(b) }B(A+C)=\begin{bmatrix}2 & 1\\ 4 & 2\end{bmatrix}\begin{bmatrix}-4 & 3\\ 10 & 0\end{bmatrix}
\displaystyle =\begin{bmatrix}2(-4)+1(10) & 2(3)+1(0)\\ 4(-4)+2(10) & 4(3)+2(0)\end{bmatrix}=\begin{bmatrix}2 & 6\\ 4 & 12\end{bmatrix}
\displaystyle \text{(c) }5B=5\begin{bmatrix}2 & 1\\ 4 & 2\end{bmatrix}=\begin{bmatrix}10 & 5\\ 20 & 10\end{bmatrix}
\displaystyle \text{(d) }B(A+C)-5B=\begin{bmatrix}2 & 6\\ 4 & 12\end{bmatrix}-\begin{bmatrix}10 & 5\\ 20 & 10\end{bmatrix}=\begin{bmatrix}-8 & 1\\ -16 & 2\end{bmatrix}

\displaystyle \text{(iii)}
\displaystyle \text{(a) From the graph, }A(3,3),\ B(0,-2),\ C(-5,-2),\ D(-2,3)
\displaystyle \text{(b) Diagonals of a parallelogram bisect each other}
\displaystyle \text{So }P\text{ is midpoint of }AC
\displaystyle P=\left(\frac{3+(-5)}{2},\frac{3+(-2)}{2}\right)=\left(-1,\frac{1}{2}\right)
\displaystyle \text{Hence, }P=\left(-1,\frac{1}{2}\right)
\displaystyle \text{(c) Slope of }CB=\frac{-2-(-2)}{0-(-5)}=\frac{0}{5}=0
\displaystyle \text{Slope of }DA=\frac{3-3}{3-(-2)}=\frac{0}{5}=0
\displaystyle \text{Since both slopes are equal, }CB\parallel DA
\displaystyle \text{(d) Equation of diagonal }AC\text{ through }A(3,3)\text{ and }C(-5,-2)
\displaystyle m=\frac{3-(-2)}{3-(-5)}=\frac{5}{8}
\displaystyle y-3=\frac{5}{8}(x-3)
\displaystyle 8y-24=5x-15
\displaystyle 5x-8y+9=0
\displaystyle \text{Hence, equation of }AC\text{ is }5x-8y+9=0
\\

\displaystyle \textbf{SECTION B (40 Marks)}
\displaystyle \text{(Attempt any four questions from this Section.)}

\displaystyle \textbf{Question 4}
\displaystyle \text{(i) Solve the following inequation, write the solution set and represent it on the} \\ \text{real number line.}
\displaystyle 2x-\frac{5}{3}<\frac{3x}{5}+10\leq \frac{4x}{5}+11,\;x\in R            \displaystyle [3]
\displaystyle \text{(ii) The first term of an Arithmetic Progression (A.P.) is }5,\text{ the last term is }50 \\ \text{ and their sum is }440.\text{ Find:}
\displaystyle \text{(a) the number of terms}
\displaystyle \text{(b) common difference}          \displaystyle [3]
\displaystyle \text{(iii) Prove that:}            \displaystyle [4]
\displaystyle \frac{(\cot A+\tan A-1)(\sin A+\cos A)}{\sin^{3}A+\cos^{3}A}=\sec A.\mathrm{cosec}A
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle 2x-\frac{5}{3}<\frac{3x}{5}+10\leq \frac{4x}{5}+11,\ x\in R
\displaystyle \text{First solve }2x-\frac{5}{3}<\frac{3x}{5}+10
\displaystyle 2x-\frac{3x}{5}<10+\frac{5}{3}
\displaystyle \frac{10x-3x}{5}<\frac{30+5}{3}
\displaystyle \frac{7x}{5}<\frac{35}{3}
\displaystyle x<\frac{35}{3}\times \frac{5}{7}=\frac{25}{3}
\displaystyle \text{Now solve }\frac{3x}{5}+10\leq \frac{4x}{5}+11
\displaystyle 10-11\leq \frac{4x}{5}-\frac{3x}{5}
\displaystyle -1\leq \frac{x}{5}
\displaystyle -5\leq x
\displaystyle \text{Combining both, }-5\leq x<\frac{25}{3}
\displaystyle \text{Solution set }=\left[-5,\frac{25}{3}\right)
\displaystyle \text{On the number line: closed dot at }-5\text{ and open dot at }\frac{25}{3}

\displaystyle \text{(ii)}
\displaystyle \text{Given }a=5,\ l=50,\ S_n=440
\displaystyle \textbf{(a) }S_n=\frac{n}{2}(a+l)
\displaystyle 440=\frac{n}{2}(5+50)
\displaystyle 440=\frac{55n}{2}
\displaystyle 880=55n
\displaystyle n=16
\displaystyle \textbf{(b) }l=a+(n-1)d
\displaystyle 50=5+(16-1)d
\displaystyle 50=5+15d
\displaystyle 45=15d
\displaystyle d=3
\displaystyle \text{Answer: number of terms }=16,\ \text{common difference }=3

\displaystyle \text{(iii)}
\displaystyle \text{LHS}=\frac{(\cot A+\tan A-1)(\sin A+\cos A)}{\sin^3 A+\cos^3 A}
\displaystyle =\frac{\left(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}-1\right)(\sin A+\cos A)}{\sin^3 A+\cos^3 A}
\displaystyle =\frac{\left(\frac{\cos^2 A+\sin^2 A-\sin A\cos A}{\sin A\cos A}\right)(\sin A+\cos A)}{\sin^3 A+\cos^3 A}
\displaystyle =\frac{\left(\frac{1-\sin A\cos A}{\sin A\cos A}\right)(\sin A+\cos A)}{\sin^3 A+\cos^3 A}
\displaystyle \text{Now, }\sin^3 A+\cos^3 A=(\sin A+\cos A)(\sin^2 A-\sin A\cos A+\cos^2 A)
\displaystyle =(\sin A+\cos A)(1-\sin A\cos A)
\displaystyle \therefore\ \text{LHS}=\frac{\left(\frac{1-\sin A\cos A}{\sin A\cos A}\right)(\sin A+\cos A)}{(\sin A+\cos A)(1-\sin A\cos A)}
\displaystyle =\frac{1}{\sin A\cos A}
\displaystyle =\sec A\ \mathrm{cosec}A
\displaystyle \therefore\ \text{LHS}=\text{RHS}
\displaystyle \text{Hence proved.}
\\
\displaystyle \textbf{Question 5}
\displaystyle \text{(i) Using properties of proportion, find the value of }x:            \displaystyle [3]
\displaystyle \frac{6x^{2}+3x-5}{3x-5}=\frac{9x^{2}+2x+5}{2x+5},\;x\neq 0
\displaystyle \text{(ii) It is given that }(x-2)\text{ is a factor of polynomial }2x^{3}-7x^{2}+kx-2.            \displaystyle [3]
\displaystyle \text{Find:}
\displaystyle \text{(a) the value of }k.
\displaystyle \text{(b) hence, factorise the resulting polynomial completely.}
\displaystyle \text{(iii) A solid wooden capsule is shown in Figure 1. The capsule is formed of a cylindrical block}
\displaystyle \text{and two hemispheres. Find the sum of total surface area of the three parts as shown in Figure 2.}
\displaystyle \text{Given, the radius of the capsule is }3.5\text{ cm and the length of the cylindrical block is }14\text{ cm.}
\displaystyle \text{(Use }\pi=\frac{22}{7}\text{)}                                  \displaystyle [4] \displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \frac{6x^2+3x-5}{3x-5}=\frac{9x^2+2x+5}{2x+5},\ x\neq 0
\displaystyle \text{By componendo and dividendo,}
\displaystyle \frac{(6x^2+3x-5)+(3x-5)}{(6x^2+3x-5)-(3x-5)}=\frac{(9x^2+2x+5)+(2x+5)}{(9x^2+2x+5)-(2x+5)}
\displaystyle \frac{6x^2+6x-10}{6x^2}=\frac{9x^2+4x+10}{9x^2}
\displaystyle \frac{3x^2+3x-5}{3x^2}=\frac{9x^2+4x+10}{9x^2}
\displaystyle 9x^2(3x^2+3x-5)=3x^2(9x^2+4x+10)
\displaystyle 3(3x^2+3x-5)=9x^2+4x+10\quad (\because x\neq 0)
\displaystyle 9x^2+9x-15=9x^2+4x+10
\displaystyle 5x=25
\displaystyle x=5
\displaystyle \text{Answer: }x=5

\displaystyle \text{(ii)}
\displaystyle \text{Given }(x-2)\text{ is a factor of }2x^3-7x^2+kx-2
\displaystyle \text{By Factor Theorem, }p(2)=0
\displaystyle 2(2)^3-7(2)^2+k(2)-2=0
\displaystyle 16-28+2k-2=0
\displaystyle 2k-14=0
\displaystyle k=7
\displaystyle \textbf{(b) The polynomial becomes }2x^3-7x^2+7x-2
\displaystyle \text{Since }(x-2)\text{ is a factor, divide by }(x-2):
\displaystyle 2x^3-7x^2+7x-2=(x-2)(2x^2-3x+1)
\displaystyle 2x^2-3x+1=2x^2-2x-x+1
\displaystyle =2x(x-1)-1(x-1)
\displaystyle =(2x-1)(x-1)
\displaystyle \therefore\ 2x^3-7x^2+7x-2=(x-2)(2x-1)(x-1)
\displaystyle \text{Answer: }k=7,\ \text{Factorization }=(x-2)(2x-1)(x-1)

\displaystyle \text{(iii)}
\displaystyle \text{Radius }r=3.5\text{ cm, height of cylinder }h=14\text{ cm}
\displaystyle \text{Sum of total surface areas of the three parts }
\displaystyle =\text{TSA of cylinder }+2\times \text{TSA of hemisphere}
\displaystyle =\left(2\pi rh+2\pi r^2\right)+2\left(3\pi r^2\right)
\displaystyle =2\pi rh+8\pi r^2
\displaystyle =2\times \frac{22}{7}\times 3.5\times 14+8\times \frac{22}{7}\times (3.5)^2
\displaystyle =308+308
\displaystyle =616\text{ cm}^2
\displaystyle \text{Answer: }616\text{ cm}^2
\\
\displaystyle \textbf{Question 6}
\displaystyle \text{(i) Use a graph paper for this question taking }2\text{ cm = 1 unit along both axes.}            \displaystyle [5]
\displaystyle \text{(a) Plot }A(1,3), B(1,2)\text{ and }C(3,0).
\displaystyle \text{(b) Reflect }A\text{ and }B\text{ on the }x\text{-axis and name their images as }E\text{ and }D\text{ respectively. Write}
\displaystyle \text{down their coordinates.}
\displaystyle \text{(c) Reflect }A\text{ and }B\text{ through the origin and name their images as }F\text{ and }G\text{ respectively.}
\displaystyle \text{(d) Reflect }A,B\text{ and }C\text{ on the }y\text{-axis and name their images as }J,I\text{ and }H\text{ respectively.}
\displaystyle \text{(e) Join all the points }A,B,C,D,E,F,G,H,I\text{ and }J\text{ in order and name the closed figure}
\displaystyle \text{so formed.}
\displaystyle \text{(ii) In the given diagram, }AB\text{ is a vertical tower }100\text{ m away from the foot of a }30\text{ storied}
\displaystyle \text{building }CD.\text{ The angles of depression from the point }C\text{ and }E,\text{ (}E\text{ being the mid-point of}
\displaystyle CD),\text{ are }35^{\circ}\text{ and }14^{\circ}\text{ respectively.}          \displaystyle [4] \displaystyle \text{(Use mathematical table for the required values rounded off correct to two places of}
\displaystyle \text{decimals only)}
\displaystyle \text{Find the height of the:}
\displaystyle \text{(a) tower }AB
\displaystyle \text{(b) building }CD
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{(a) Given points: }A(1,3),\ B(1,2),\ C(3,0)
\displaystyle \text{(b) Reflection in x-axis changes }(x,y)\text{ to }(x,-y)
\displaystyle \therefore\ E(1,-3),\ D(1,-2)
\displaystyle \text{(c) Reflection through origin changes }(x,y)\text{ to }(-x,-y)
\displaystyle \therefore\ F(-1,-3),\ G(-1,-2)
\displaystyle \text{(d) Reflection in y-axis changes }(x,y)\text{ to }(-x,y)
\displaystyle \therefore\ J(-1,3),\ I(-1,2),\ H(-3,0)
\displaystyle \text{(e) Joining }A,B,C,D,E,F,G,H,I,J\text{ in order forms a decagon}

\displaystyle \text{(ii)}
\displaystyle \text{Given horizontal distance }BD=100\text{ m}
\displaystyle \text{Let height of tower }AB=h\text{ m and height of building }CD=H\text{ m}
\displaystyle \text{Since }E\text{ is mid-point of }CD,\ CE=ED=\frac{H}{2}
\displaystyle \text{Angle of depression from }C\text{ to }A=35^\circ
\displaystyle \therefore\ \tan 35^\circ=\frac{H-h}{100}
\displaystyle H-h=100\tan 35^\circ
\displaystyle \text{Using }\tan 35^\circ\approx 0.70,\ H-h\approx 70
\displaystyle \text{Angle of depression from }E\text{ to }A=14^\circ
\displaystyle \therefore\ \tan 14^\circ=\frac{\frac{H}{2}-h}{100}
\displaystyle \frac{H}{2}-h=100\tan 14^\circ
\displaystyle \text{Using }\tan 14^\circ\approx 0.25,\ \frac{H}{2}-h\approx 25
\displaystyle \text{Now solve: }H-h=70\text{ and }\frac{H}{2}-h=25
\displaystyle \text{Subtracting, }\frac{H}{2}=45
\displaystyle H=90
\displaystyle h=90-70=20
\displaystyle \therefore\ \text{Height of tower }AB=20\text{ m}
\displaystyle \therefore\ \text{Height of building }CD=90\text{ m}
\\
\displaystyle \textbf{Question 7}
\displaystyle \text{(i) Use a graph paper for this question.}            \displaystyle [3]
\displaystyle \text{(Take }2\text{ cm = }10\text{ Marks along one axis and }2\text{ cm = }10\text{ students along another axis).}
\displaystyle \text{Draw a Histogram for the following distribution which gives the marks obtained by }164
\displaystyle \text{students in a particular class and hence find the Mode.}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Marks} & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\\hline \text{Number of Students} & 10 & 26 & 40 & 54 & 34 \\\hline \end{array}
\displaystyle \text{(ii) In the given graph, }P\text{ and }Q\text{ are points such that }PQ\text{ cuts off intercepts of }5\text{ units and}
\displaystyle 3\text{ units along the }x\text{-axis and }y\text{-axis respectively. Line }RS\text{ is perpendicular to }PQ\text{ and passes}
\displaystyle \text{through the origin. Find the:}
\displaystyle \text{(a) coordinates of }P\text{ and }Q
\displaystyle \text{(b) equation of line }RS            \displaystyle [3] \displaystyle \text{(iii) Refer to the given bill.}            \displaystyle [4]
\displaystyle \text{A customer paid Rs }2000\text{ (rounded off to the nearest Rs }10)\text{ to clear the bill.}
\displaystyle \text{Note: }5\%\text{ discount is applicable on an article if }10\text{ or more such articles are purchased.}
\displaystyle \begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\textbf{BILL}} \\\hline \textbf{Article} & \textbf{M.P. (Rs)} & \textbf{Quantity} & \textbf{G.S.T.} \\\hline A & 190 & 06 & 12\% \\\hline B & 50 & 12 & 18\% \\\hline \end{array}
\displaystyle \text{Check whether the total amount paid by the customer is correct or not. Justify your answer}
\displaystyle \text{with necessary working.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Modal class is }60\text{--}70\ (\text{highest frequency }54)
\displaystyle \text{Using formula: Mode }=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle l=60,\ h=10,\ f_1=54,\ f_0=40,\ f_2=34
\displaystyle \text{Mode}=60+\frac{54-40}{2(54)-40-34}\times 10
\displaystyle =60+\frac{14}{108-74}\times 10
\displaystyle =60+\frac{14}{34}\times 10
\displaystyle =60+4.12\approx 64.12
\displaystyle \therefore\ \text{Mode }\approx 64.1\text{ marks}\displaystyle \text{(ii)}
\displaystyle \textbf{(a) }PQ\text{ cuts intercepts }5\text{ and }3
\displaystyle \therefore\ P(5,0),\ Q(0,-3)
\displaystyle \text{Slope of }PQ=\frac{0-(-3)}{5-0}=\frac{3}{5}
\displaystyle \textbf{(b) Slope of }RS=-\frac{5}{3}
\displaystyle \text{Since it passes through origin, equation: }y=-\frac{5}{3}x

\displaystyle \text{(iii)}
\displaystyle \text{Article A: }190\times 6=1140\ (\text{no discount})
\displaystyle \text{GST }=12\%\Rightarrow 136.8
\displaystyle \text{Total }=1276.8
\displaystyle \text{Article B: }50\times 12=600
\displaystyle \text{Discount }=5\%\Rightarrow 30
\displaystyle \text{Price after discount }=570
\displaystyle \text{GST }=18\%\Rightarrow 102.6
\displaystyle \text{Total }=672.6
\displaystyle \text{Grand total }=1276.8+672.6=1949.4
\displaystyle \text{Rounded }=1950
\displaystyle \text{Customer paid }2000\Rightarrow \text{Not correct (excess paid)}
\\
\displaystyle \textbf{Question 8}
\displaystyle \text{(i) A man bought Rs }200\text{ shares of a company at }25\%\text{ premium. If he}            \displaystyle [3]
\displaystyle \text{received a return of }5\%\text{on his investment. Find the:}
\displaystyle \text{(a) market value}
\displaystyle \text{(b) dividend percent declared}
\displaystyle \text{(c) number of shares purchased, if annual dividend is Rs }1000.
\displaystyle \text{(ii) For the given frequency distribution, find the:}            \displaystyle [3]
\displaystyle \text{(a) mean, to the nearest whole number}
\displaystyle \text{(b) median}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline x & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\\hline f & 3 & 2 & 2 & 6 & 3 & 5 & 3 \\\hline \end{array}
\displaystyle \text{(iii) Mr. and Mrs. Das were travelling by car from Delhi to Kasauli for a holiday. Distance}
\displaystyle \text{between Delhi and Kasauli is approximately }350\text{ km (via NH 152D). Due to heavy rain}
\displaystyle \text{they had to slow down. The average speed of the car was reduced by }20\text{ km/hr and time}
\displaystyle \text{of the journey increased by }2\text{ hours. Find:}
\displaystyle \text{(a) the original speed of the car.}               \displaystyle [4]
\displaystyle \text{(b) with the reduced speed, the number of hours they took to reach their destination.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Nominal value of share }=200
\displaystyle \text{Premium }=25\%\Rightarrow \text{Market value }=200+\frac{25}{100}\times 200=250
\displaystyle \textbf{(a) Market value }=250
\displaystyle \text{Return }=5\%\text{ on investment}
\displaystyle \text{Dividend per share }=5\%\text{ of market value}= \frac{5}{100}\times 250=12.5
\displaystyle \text{Dividend percent declared }=\frac{12.5}{200}\times 100=6.25\%
\displaystyle \textbf{(b) Dividend percent declared }=6.25\%
\displaystyle \textbf{(c) If annual dividend }=1000,\ \text{number of shares }=\frac{1000}{12.5}=80
\displaystyle \text{Answer: Market value }=250,\ \text{Dividend percent }=6.25\%,\ \text{shares }=80

\displaystyle \text{(ii)}
\displaystyle \text{Given }x:10,11,12,13,14,15,16\quad \text{and }f:3,2,2,6,3,5,3
\displaystyle \sum f=3+2+2+6+3+5+3=24
\displaystyle \sum fx=10(3)+11(2)+12(2)+13(6)+14(3)+15(5)+16(3)
\displaystyle =30+22+24+78+42+75+48=319
\displaystyle \textbf{(a) Mean }=\frac{\sum fx}{\sum f}=\frac{319}{24}=13.29\approx 13
\displaystyle \textbf{(b) Cumulative frequencies }=3,5,7,13,16,21,24
\displaystyle \text{Since }n=24,\ \text{median is average of }12^{\text{th}}\text{ and }13^{\text{th}}\text{ observations}
\displaystyle \text{Both }12^{\text{th}}\text{ and }13^{\text{th}}\text{ observations lie at }x=13
\displaystyle \therefore\ \text{Median }=13
\displaystyle \text{Answer: Mean }=13,\ \text{Median }=13

\displaystyle \text{(iii)}
\displaystyle \text{Distance }=350\text{ km}
\displaystyle \text{Let original speed }=x\text{ km/hr}
\displaystyle \text{Reduced speed }=x-20\text{ km/hr}
\displaystyle \text{Given increase in time }=2\text{ hours}
\displaystyle \frac{350}{x-20}-\frac{350}{x}=2
\displaystyle \frac{350x-350(x-20)}{x(x-20)}=2
\displaystyle \frac{7000}{x(x-20)}=2
\displaystyle 7000=2x(x-20)
\displaystyle 3500=x^2-20x
\displaystyle x^2-20x-3500=0
\displaystyle x=\frac{20\pm\sqrt{400+14000}}{2}=\frac{20\pm\sqrt{14400}}{2}
\displaystyle x=\frac{20\pm 120}{2}
\displaystyle x=70\text{ or }-50
\displaystyle \text{Reject negative value}
\displaystyle \therefore\ \text{Original speed }=70\text{ km/hr}
\displaystyle \text{Reduced speed }=50\text{ km/hr}
\displaystyle \text{Time taken with reduced speed }=\frac{350}{50}=7\text{ hours}
\displaystyle \text{Answer: Original speed }=70\text{ km/hr},\ \text{reduced-time }=7\text{ hours}
\\
\displaystyle \textbf{Question 9}
\displaystyle \text{(i) A hollow sphere of external diameter }10\text{ cm and internal diameter }6\text{ cm is melted and made} 
\displaystyle \text{into a solid right circular cone of height }8\text{ cm. Find the radius of the cone so formed.}
\displaystyle \text{[Use }\pi=\frac{22}{7}\text{]}            \displaystyle [3]
\displaystyle \text{(ii) Ms. Sushmita went to a fair and participated in a game. The game consisted of a box having}
\displaystyle \text{number cards with numbers from }01\text{ to }30.\text{ The three prizes were as per the given table:}
\displaystyle \begin{array}{|l|l|}\hline \textbf{Prize} & \textbf{Number on the card drawn at random is a} \\\hline \text{Wall Clock} & \text{perfect square} \\\hline \text{Water Bottle} & \text{even number which is also a multiple of }3 \\\hline \text{Purse} & \text{prime number} \\\hline \end{array}
\displaystyle \text{Find the probability of winning a:}         \displaystyle [3]
\displaystyle \text{(a) Wall Clock}
\displaystyle \text{(b) Water Bottle}
\displaystyle \text{(c) Purse}
\displaystyle \text{(iii) }X,Y,Z\text{ and }C\text{ are the points on the circumference of a circle with centre }O.\;AB\text{ is a} 
\displaystyle \text{tangent to the circle at }X\text{ and }ZY=XY.\text{ Given }\angle OBX=32^{\circ}\text{ and }\angle AXZ=66^{\circ}.\text{ Find:}
\displaystyle \text{(a) }\angle BOX          \displaystyle [4]
\displaystyle \text{(b) }\angle CYX
\displaystyle \text{(c) }\angle ZYX
\displaystyle \text{(d) }\angle OXY
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{External radius of hollow sphere }R=\frac{10}{2}=5\text{ cm}
\displaystyle \text{Internal radius of hollow sphere }r=\frac{6}{2}=3\text{ cm}
\displaystyle \text{Volume of material in hollow sphere }=\frac{4}{3}\pi(R^3-r^3)
\displaystyle =\frac{4}{3}\pi(5^3-3^3)=\frac{4}{3}\pi(125-27)=\frac{4}{3}\pi(98)
\displaystyle \text{This is melted to form a cone of height }8\text{ cm}
\displaystyle \therefore\ \frac{1}{3}\pi x^2(8)=\frac{4}{3}\pi(98)
\displaystyle 8x^2=392
\displaystyle x^2=49
\displaystyle x=7\displaystyle \therefore\ \text{Radius of the cone }=7\text{ cm}

\displaystyle \text{(ii)}
\displaystyle \text{Total number of cards }=30
\displaystyle \text{(a) Perfect squares from }1\text{ to }30\text{ are }1,4,9,16,25
\displaystyle \text{Number of favourable outcomes }=5
\displaystyle \text{Probability of winning Wall Clock }=\frac{5}{30}=\frac{1}{6}
\displaystyle \text{(b) Even numbers which are also multiples of }3\text{ are }6,12,18,24,30
\displaystyle \text{Number of favourable outcomes }=5
\displaystyle \text{Probability of winning Water Bottle }=\frac{5}{30}=\frac{1}{6}
\displaystyle \text{(c) Prime numbers from }1\text{ to }30\text{ are }2,3,5,7,11,13,17,19,23,29
\displaystyle \text{Number of favourable outcomes }=10
\displaystyle \text{Probability of winning Purse }=\frac{10}{30}=\frac{1}{3}

\displaystyle \text{(iii)}
\displaystyle \text{(a) Since }AB\text{ is tangent at }X,\ OX\perp AB
\displaystyle \therefore\ \angle OXB=90^\circ
\displaystyle \text{Given }\angle OBX=32^\circ
\displaystyle \text{In }\triangle OBX,\ \angle BOX=180^\circ-(90^\circ+32^\circ)=58^\circ
\displaystyle \therefore\ \angle BOX=58^\circ
\displaystyle \text{(b) Given }\angle AXZ=66^\circ
\displaystyle \text{By tangent-chord theorem, }\angle XYZ=66^\circ
\displaystyle \text{Since }C\text{ lies on }YZ,\ \angle CYX=\angle ZYX=66^\circ
\displaystyle \therefore\ \angle CYX=66^\circ
\displaystyle \text{(c) Given }ZY=XY
\displaystyle \therefore\ \triangle ZYX\text{ is isosceles, so }\angle ZXY=\angle XZY
\displaystyle \text{Now }\angle XYZ+\angle ZXY+\angle XZY=180^\circ
\displaystyle 66^\circ+2\angle ZYX? \text{ No, base angles are at }X\text{ and }Z
\displaystyle 66^\circ+\angle ZXY+\angle XZY=180^\circ
\displaystyle 66^\circ+2\angle ZYX? \text{Correction: } \angle ZXY=\angle XZY
\displaystyle 66^\circ+2\angle ZXY=180^\circ
\displaystyle 2\angle ZXY=114^\circ
\displaystyle \angle ZXY=57^\circ
\displaystyle \therefore\ \angle ZYX=66^\circ
\displaystyle \text{and required } \angle ZYX=66^\circ
\displaystyle \text{(d) In }\triangle OX Y,\ OX=OY\text{ (radii)}
\displaystyle \text{Chord }XY\text{ subtends angle }\angle XZY=57^\circ\text{ at circumference}
\displaystyle \therefore\ \angle XOY=2\times 57^\circ=114^\circ
\displaystyle \text{Hence base angles } \angle OXY=\angle OYX=\frac{180^\circ-114^\circ}{2}=33^\circ
\displaystyle \therefore\ \angle OXY=33^\circ
\displaystyle \text{Answers: }\angle BOX=58^\circ,\ \angle CYX=66^\circ,\ \angle ZYX=66^\circ,\ \angle OXY=33^\circ

\displaystyle \text{(iii)}
\displaystyle \text{(a) Since }AB\text{ is tangent at }X,\ OX\perp AB
\displaystyle \therefore\ \angle OXB=90^\circ
\displaystyle \text{Given }\angle OBX=32^\circ
\displaystyle \text{In }\triangle OBX,\ \angle BOX=180^\circ-(90^\circ+32^\circ)=58^\circ
\displaystyle \therefore\ \angle BOX=58^\circ
\displaystyle \text{(b) Given }\angle AXZ=66^\circ
\displaystyle \text{By tangent-chord theorem, }\angle XYZ=66^\circ
\displaystyle \text{Since }C\text{ lies on }YZ,\ \angle CYX=\angle ZYX=66^\circ
\displaystyle \therefore\ \angle CYX=66^\circ
\displaystyle \text{(c) Given }ZY=XY
\displaystyle \therefore\ \triangle ZYX\text{ is isosceles, so }\angle ZXY=\angle XZY
\displaystyle \text{Now }\angle XYZ+\angle ZXY+\angle XZY=180^\circ
\displaystyle 66^\circ+2\angle ZYX? \text{ No, base angles are at }X\text{ and }Z
\displaystyle 66^\circ+\angle ZXY+\angle XZY=180^\circ
\displaystyle 66^\circ+2\angle ZYX? \text{Correction: } \angle ZXY=\angle XZY
\displaystyle 66^\circ+2\angle ZXY=180^\circ
\displaystyle 2\angle ZXY=114^\circ
\displaystyle \angle ZXY=57^\circ
\displaystyle \therefore\ \angle ZYX=66^\circ
\displaystyle \text{and required } \angle ZYX=66^\circ
\displaystyle \text{(d) In }\triangle OX Y,\ OX=OY\text{ (radii)}
\displaystyle \text{Chord }XY\text{ subtends angle }\angle XZY=57^\circ\text{ at circumference}
\displaystyle \therefore\ \angle XOY=2\times 57^\circ=114^\circ
\displaystyle \text{Hence base angles } \angle OXY=\angle OYX=\frac{180^\circ-114^\circ}{2}=33^\circ
\displaystyle \therefore\ \angle OXY=33^\circ
\displaystyle \text{Answers: }\angle BOX=58^\circ,\ \angle CYX=66^\circ,\ \angle ZYX=66^\circ,\ \angle OXY=33^\circ
\\
\displaystyle \textbf{Question 10}
\displaystyle \text{(i) If }1701\text{ is the }n^{\text{th}}\text{ term of the Geometric Progression (G.P.) }7,21,63,\ldots,\text{ find:}
\displaystyle \text{(a) the value of }n            \displaystyle [3]
\displaystyle \text{(b) hence find the sum of the }n\text{ terms of the G.P.}
\displaystyle \text{(ii) In the given diagram }O\text{ is the centre of the circle. Chord }SR\text{ produced meets the tangent} 
\displaystyle XTP\text{ at }P.          \displaystyle [3]
\displaystyle \text{(a) Prove that }\triangle PTR\sim\triangle PST
\displaystyle \text{(b) Prove that }PT^{2}=PR\times PS
\displaystyle \text{(c) If }PR=4\text{ cm and }PS=16\text{ cm, find the length of the tangent }PT.
\displaystyle \text{(iii) The given graph represents the monthly salaries (in Rs) of workers of a factory.}
\displaystyle \text{Using graph answer the following:}            \displaystyle [4]
\displaystyle \text{(a) the total number of workers.}
\displaystyle \text{(b) the median class.}
\displaystyle \text{(c) the lower-quartile class.}
\displaystyle \text{(d) number of workers having monthly salary more than or equal to Rs }6000\text{ but less}
\displaystyle \text{than Rs }10000.
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \text{Given G.P. }7,21,63,\ldots
\displaystyle a=7,\ r=\frac{21}{7}=3
\displaystyle \text{The }n^{\text{th}}\text{ term is }T_n=ar^{n-1}
\displaystyle 1701=7\cdot 3^{n-1}
\displaystyle 243=3^{n-1}
\displaystyle 3^5=3^{n-1}
\displaystyle n-1=5
\displaystyle n=6
\displaystyle \text{(b) }S_n=\frac{a(r^n-1)}{r-1}
\displaystyle S_6=\frac{7(3^6-1)}{3-1}
\displaystyle =\frac{7(729-1)}{2}
\displaystyle =\frac{7\cdot 728}{2}
\displaystyle =7\cdot 364=2548
\displaystyle \text{Answer: }n=6,\quad S_6=2548

\displaystyle \text{(ii)}
\displaystyle \text{(a) To prove }\triangle PTR\sim \triangle PST
\displaystyle \text{Since }PR\text{ and }PS\text{ lie on the same straight line, }\angle TPR=\angle SPT
\displaystyle \text{Also, by tangent-chord theorem, }\angle PTS=\angle TRS
\displaystyle \text{But }PR\text{ is the extension of }RS,\text{ so }\angle TRS=\angle PRT
\displaystyle \therefore\ \angle PTS=\angle PRT
\displaystyle \text{Now in triangles }PTR\text{ and }PST,
\displaystyle \angle TPR=\angle SPT,\quad \angle PRT=\angle PTS
\displaystyle \therefore\ \triangle PTR\sim \triangle PST\ (\text{AA similarity})
\displaystyle \text{(b) From similarity, }\frac{PR}{PT}=\frac{PT}{PS}
\displaystyle \therefore\ PT^2=PR\cdot PS
\displaystyle \text{(c) Given }PR=4\text{ cm and }PS=16\text{ cm}
\displaystyle PT^2=4\cdot 16=64
\displaystyle PT=8\text{ cm}
\displaystyle \text{Answer: }PT=8\text{ cm}

\displaystyle \text{(iii)}
\displaystyle \text{(a) Total number of workers is the last cumulative frequency }=75
\displaystyle \therefore\ \text{Total workers }=75
\displaystyle \text{(b) Median corresponds to }\frac{N}{2}=\frac{75}{2}=37.5
\displaystyle \text{From the graph, cumulative frequency just above }37.5\text{ is at salary }6000
\displaystyle \text{Hence, median class }=6000\text{--}8000
\displaystyle \text{(c) Lower quartile corresponds to }\frac{N}{4}=\frac{75}{4}=18.75
\displaystyle \text{From the graph, cumulative frequency just above }18.75\text{ is at salary }4000
\displaystyle \text{Hence, lower-quartile class }=4000\text{--}6000
\displaystyle \text{(d) Workers having salary }\geq 6000\text{ and }<10000
\displaystyle =\text{C.F. below }10000-\text{C.F. below }6000
\displaystyle =70-35=35
\displaystyle \text{Answer: Total workers }=75,\ \text{Median class }=6000\text{--}8000,
\displaystyle \text{Lower-quartile class }=4000\text{--}6000,\ \text{Required workers }=35

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