\displaystyle \textbf{MATHEMATICS}

\displaystyle \text{Maximum Marks: 80}
\displaystyle \textbf{Time allowed: Two and half hours}
\displaystyle \text{Answers to this Paper must be written on the paper provided separately.}
\displaystyle \text{You will not be allowed to write during first 15 minutes.}
\displaystyle \text{This time is to be spent in reading the question paper.}
\displaystyle \textbf{The time given at the head of this Paper is the time allowed for writing the answers.}

\displaystyle \text{Attempt all questions from Section A and any four questions from Section B.}
\displaystyle \textbf{All working, including rough work, must be clearly shown, and must be}
\displaystyle \textbf{done on the same sheet as the rest of the answer.}
\displaystyle \textbf{Omission of essential working will result in loss of marks.}
\displaystyle \text{The intended marks for questions or parts of questions are given in brackets [ ].}
\displaystyle \textbf{Mathematical tables and graph papers are to be provided by the school.}

\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. }    \displaystyle \text{[15]}
\displaystyle \text{(Do not copy the questions, write the correct answers only.)}

\displaystyle \text{(i) For an Intra-state sale, the CGST paid by a dealer to the Central government is } \text{Rs }120.
\displaystyle \text{If the marked price of the article is } \text{Rs }2000,\text{ the rate of GST is:}
\displaystyle \text{(a) }6\% \quad \text{(b) }10\% \quad \text{(c) }12\% \quad \text{(d) }16.67\%
\displaystyle \text{Answer:}
\displaystyle \text{Given CGST }=120,\ \text{Marked price }=2000
\displaystyle \text{For intra-state sale, CGST = SGST}
\displaystyle \text{Total GST }=120+120=240
\displaystyle \text{GST rate }=\frac{240}{2000}\times 100=12\%
\displaystyle \text{Answer: (c) }12\%
\\
\displaystyle \text{(ii) What must be subtracted from the polynomial } x^{3}+x^{2}-2x+1,\text{ so that}
\displaystyle \text{the result is exactly divisible by }(x-3)?
\displaystyle \text{(a) }-31 \quad \text{(b) }-30 \quad \text{(c) }30 \quad \text{(d) }31
\displaystyle \text{Answer:}
\displaystyle \text{Let }p(x)=x^3+x^2-2x+1
\displaystyle \text{For divisibility by }(x-3),\ \text{remainder }=p(3)
\displaystyle p(3)=27+9-6+1=31
\displaystyle \text{So, subtract }31\text{ to make remainder zero}
\displaystyle \text{Answer: (d) }31
\\
\displaystyle \text{(iii) The roots of the quadratic equation } px^{2}-qx+r=0 \text{ are real and equal if:}
\displaystyle \text{(a) } p^{2}=4qr \quad \text{(b) } q^{2}=4pr \quad \text{(c) } -q^{2}=4pr \quad \text{(d) } p^{2}>4qr
\displaystyle \text{Answer:}
\displaystyle \text{Given quadratic }px^2-qx+r=0
\displaystyle \text{For real and equal roots, discriminant }D=0
\displaystyle D=b^2-4ac=(-q)^2-4(p)(r)
\displaystyle =q^2-4pr=0
\displaystyle \therefore\ q^2=4pr
\displaystyle \text{Answer: (b) }q^2=4pr
\\
\displaystyle \text{(iv) If matrix } A=\begin{bmatrix}2&2\\0&2\end{bmatrix} \text{ and } A^{2}=\begin{bmatrix}4&x\\0&4\end{bmatrix},   \text{then the value of }x\text{ is:}
\displaystyle \text{(a) }2 \quad \text{(b) }4 \quad \text{(c) }8 \quad \text{(d) }10
\displaystyle \text{Answer:}
\displaystyle A=\begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}
\displaystyle A^2=A\cdot A=\begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}\begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}
\displaystyle =\begin{bmatrix}2\cdot 2+2\cdot 0 & 2\cdot 2+2\cdot 2 \\ 0\cdot 2+2\cdot 0 & 0\cdot 2+2\cdot 2\end{bmatrix}
\displaystyle =\begin{bmatrix}4 & 4+4 \\ 0 & 4\end{bmatrix}=\begin{bmatrix}4 & 8 \\ 0 & 4\end{bmatrix}
\displaystyle \therefore\ x=8
\displaystyle \text{Answer: (c) }8
\\
\displaystyle \text{(v) The median of the following observations arranged in ascending order is }64.
\displaystyle \text{Find the value of }x: \quad 27,31,46,52,x,x+4,71,79,85,90
\displaystyle \text{(a) }60 \quad \text{(b) }61 \quad \text{(c) }62 \quad \text{(d) }66
\displaystyle \text{Answer:}
\displaystyle \text{Total observations }=9
\displaystyle \text{Median is the } \frac{9+1}{2}=5^{th}\text{ term}
\displaystyle \text{So, }5^{th}\text{ term }=x
\displaystyle \text{Given median }=64
\displaystyle \therefore\ x=64
\displaystyle \text{Answer: }64
\\
\displaystyle \text{(vi) Points } A(x,y), B(3,-2) \text{ and } C(4,-5) \text{ are collinear.}   \text{The value of }y\text{ in terms of }x\text{ is:}
\displaystyle \text{(a) }3x-11 \quad \text{(b) }11-3x \quad \text{(c) }3x-7 \quad \text{(d) }7-3x
\displaystyle \text{Answer:}
\displaystyle \text{Since points }A(x,y),\ B(3,-2),\ C(4,-5)\text{ are collinear, their slopes are equal}
\displaystyle \text{Slope }BC=\frac{-5-(-2)}{4-3}=\frac{-3}{1}=-3
\displaystyle \text{Slope }AB=\frac{y-(-2)}{x-3}=\frac{y+2}{x-3}
\displaystyle \text{Equate slopes: }\frac{y+2}{x-3}=-3
\displaystyle y+2=-3(x-3)
\displaystyle y+2=-3x+9
\displaystyle y=-3x+7
\displaystyle \text{Answer: (d) }7-3x
\\
\displaystyle \text{(vii) The given table shows the distance covered and the time taken by a train moving}
\displaystyle \text{at a uniform speed along a straight track.}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Distance (in m)} & 60 & 90 & y \\\hline \text{Time (in sec)} & 2 & x & 5 \\\hline \end{array}
\displaystyle \text{The values of }x\text{ and }y\text{ are:}
\displaystyle \text{(a) }x=4,\,y=150
\displaystyle \text{(b) }x=3,\,y=100
\displaystyle \text{(c) }x=4,\,y=100
\displaystyle \text{(d) }x=3,\,y=150
\displaystyle \text{Answer:}
\displaystyle \text{Since speed is uniform, }\frac{\text{distance}}{\text{time}}=\text{constant}
\displaystyle \text{From first pair: speed }=\frac{60}{2}=30\ \text{m/s}
\displaystyle \text{For second pair: }\frac{90}{x}=30\Rightarrow x=3
\displaystyle \text{For third pair: }\frac{y}{5}=30\Rightarrow y=150
\displaystyle \text{Answer: (d) }x=3,\ y=150
\\
\displaystyle \text{(viii) The }7^{\text{th}}\text{ term of the given Arithmetic Progression (A.P.):}
\displaystyle \frac{1}{a},\left(\frac{1}{a}+1\right),\left(\frac{1}{a}+2\right)\ldots\text{ is:}
\displaystyle \text{(a) }\left(\frac{1}{a}+6\right)
\displaystyle \text{(b) }\left(\frac{1}{a}+7\right)
\displaystyle \text{(c) }\left(\frac{1}{a}+8\right)
\displaystyle \text{(d) }\left(\frac{1}{a}+7\right)
\displaystyle \text{Answer:}
\displaystyle \text{Given A.P. } \frac{1}{a},\left(\frac{1}{a}+1\right),\left(\frac{1}{a}+2\right),\ldots
\displaystyle \text{First term }A=\frac{1}{a},\quad \text{common difference }d=1
\displaystyle \text{The }n^{\text{th}}\text{ term of an A.P. is }a_n=A+(n-1)d
\displaystyle \text{So, }7^{\text{th}}\text{ term }=\frac{1}{a}+(7-1)\cdot 1
\displaystyle =\frac{1}{a}+6
\displaystyle \text{Answer: (a) }\left(\frac{1}{a}+6\right)
\\
\displaystyle \text{(ix) The sum invested to purchase }15\text{ shares of a company of nominal value}
\displaystyle \text{Rs }75\text{ available at a discount of }20\%\text{ is:}
\displaystyle \text{(a) Rs }60
\displaystyle \text{(b) Rs }90
\displaystyle \text{(c) Rs }1350
\displaystyle \text{(d) Rs }900
\displaystyle \text{Answer:}
\displaystyle \text{Nominal value per share }=75
\displaystyle \text{Discount }=20\%\Rightarrow \text{Discount amount }=0.2\times 75=15
\displaystyle \text{Market price per share }=75-15=60
\displaystyle \text{Total cost for 15 shares }=15\times 60=900
\displaystyle \text{Answer: (d) }900
\\
\displaystyle \text{(x) The circumcentre of a triangle is the point which is:}
\displaystyle \text{(a) at equal distance from the three sides of the triangle.}
\displaystyle \text{(b) at equal distance from the three vertices of the triangle.}
\displaystyle \text{(c) the point of intersection of the three medians.}
\displaystyle \text{(d) the point of intersection of the three altitudes of the triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{The circumcentre of a triangle is the point equidistant from all three vertices}
\displaystyle \text{Answer: (b) at equal distance from the three vertices of the triangle}
\\
\displaystyle \text{(xi) Statement 1:} \qquad \sin^{2}\theta+\cos^{2}\theta=1
\displaystyle \text{Statement 2:} \qquad \mathrm{cosec}^{2}\theta+\cot^{2}\theta=1
\displaystyle \text{Which of the following is valid?}
\displaystyle \text{(a) only 1}
\displaystyle \text{(b) only 2}
\displaystyle \text{(c) both 1 and 2}
\displaystyle \text{(d) neither 1 nor 2}
\displaystyle \text{Answer:}
\displaystyle \text{Statement 1: }\sin^2\theta+\cos^2\theta=1\ \text{(true identity)}
\displaystyle \text{Statement 2: }\mathrm{cosec}^2\theta+\cot^2\theta=1
\displaystyle \text{But correct identity is }\mathrm{cosec}^2\theta=1+\cot^2\theta
\displaystyle \text{So Statement 2 is false}
\displaystyle \text{Answer: (a) only 1}
\\

\displaystyle \text{(xii) In the given diagram, PS and PT are the tangents to the circle.} \\  SQ \parallel \text{ PT and}   \angle SPT=80^{\circ}.\text{ The value of }\angle QST\text{ is:}
\displaystyle \text{(a) }140^{\circ}
\displaystyle \text{(b) }90^{\circ}
\displaystyle \text{(c) }80^{\circ}
\displaystyle \text{(d) }50^{\circ}
\displaystyle \text{Answer:}

\displaystyle \text{Given }PS\text{ and }PT\text{ are tangents, }SQ\parallel PT,\ \angle SPT=80^\circ
\displaystyle \text{Angle between two tangents }=\ 180^\circ-\angle SOT
\displaystyle \therefore\ \angle SOT=180^\circ-80^\circ=100^\circ
\displaystyle \text{Since }SQ\parallel PT,\ \angle QST=\angle \text{angle between }PT\text{ and }ST
\displaystyle \text{Now, angle between tangent }PT\text{ and chord }ST=\frac{1}{2}\angle SOT
\displaystyle \therefore\ \angle QST=\frac{1}{2}\times 100^\circ=50^\circ
\displaystyle \text{Answer: (d) }50^\circ
\\
\displaystyle \text{(xiii) Assertion (A): A die is thrown once and the probability of getting an even}
\displaystyle \text{number is }\frac{2}{3}\text{.}
\displaystyle \text{Reason (R): The sample space for even numbers on a die is }\{2,4,6\}
\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.}
\displaystyle \text{(d) Both A and R are false.}
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): Probability of getting an even number on a die} \\ =\frac{3}{6}=\frac{1}{2}\neq \frac{2}{3}
\displaystyle \text{So Assertion is false}
\displaystyle \text{Reason (R): Even numbers on a die are }\{2,4,6\}\text{ (true)}
\displaystyle \text{Answer: (b) A is false, R is true}
\\

\displaystyle \text{(xiv) A rectangular sheet of paper of size }11\text{ cm }\times 7\text{ cm is first rotated about the side}
\displaystyle 11\text{ cm and then about the side }7\text{ cm to form a cylinder, as shown in the diagram.}
\displaystyle \text{The ratio of their curved surface areas is:}
\displaystyle \text{(a) }1:1
\displaystyle \text{(b) }7:11
\displaystyle \text{(c) }11:7
\displaystyle \text{(d) }\frac{11\pi}{7}:\frac{7\pi}{11}
\displaystyle \text{Answer:}

\displaystyle \text{When the sheet is rotated about side }11\text{ cm,}
\displaystyle \text{height }h_1=11\text{ cm and circumference }=7\text{ cm}
\displaystyle \therefore\ \text{CSA}_1=2\pi r_1 h_1=(\text{circumference})\times (\text{height})=7\times 11=77
\displaystyle \text{When the sheet is rotated about side }7\text{ cm,}
\displaystyle \text{height }h_2=7\text{ cm and circumference }=11\text{ cm}
\displaystyle \therefore\ \text{CSA}_2=2\pi r_2 h_2=(\text{circumference})\times (\text{height})=11\times 7=77
\displaystyle \text{Hence, the ratio of their curved surface areas is }77:77=1:1
\displaystyle \text{Answer: (a) }1:1

\\
\displaystyle \text{(xv) In the given diagram, }\triangle ABC\sim\triangle PQR.\text{ If AD and PS are bisectors of }\angle BAC \\ \text{and }   \angle QPR\text{ respectively then:}
\displaystyle \text{(a) }\triangle ABC\sim\triangle PQS
\displaystyle \text{(b) }\triangle ABD\sim\triangle PQS
\displaystyle \text{(c) }\triangle ABD\sim\triangle PSR
\displaystyle \text{(d) }\triangle ABC\sim\triangle PSR
\displaystyle \text{Answer:}

\displaystyle \text{Given }\triangle ABC\sim \triangle PQR
\displaystyle \therefore\ \angle BAC=\angle QPR,\ \angle ABC=\angle PQR,\ \angle ACB=\angle PRQ
\displaystyle \text{Also, }AD\text{ bisects }\angle BAC\text{ and }PS\text{ bisects }\angle QPR
\displaystyle \therefore\ \angle BAD=\frac{1}{2}\angle BAC,\ \angle QPS=\frac{1}{2}\angle QPR
\displaystyle \text{Since }\angle BAC=\angle QPR,\ \angle BAD=\angle QPS
\displaystyle \text{Now }D\text{ lies on }BC\text{ and }S\text{ lies on }QR
\displaystyle \therefore\ \angle ABD=\angle ABC,\ \angle PQS=\angle PQR
\displaystyle \text{But }\angle ABC=\angle PQR
\displaystyle \therefore\ \angle ABD=\angle PQS
\displaystyle \text{Hence, }\triangle ABD\sim \triangle PQS\text{ (by AA similarity)}
\displaystyle \text{Answer: (b) }\triangle ABD\sim \triangle PQS
\\
\displaystyle \textbf{Question 2}
\displaystyle \text{(i)} \qquad A=\begin{bmatrix}x&0\\1&1\end{bmatrix}, B=\begin{bmatrix}4&0\\y&1\end{bmatrix} \text{ and } C=\begin{bmatrix}4&0\\x&1\end{bmatrix}                                                             \displaystyle \text{[4]}
\displaystyle \text{Find the values of }x\text{ and }y,\text{ if }AB=C.
\displaystyle \text{(ii) A solid metallic cylinder is cut into two identical halves along its height (as}       \displaystyle \text{[4]}
\displaystyle \text{shown in the diagram). The diameter of the cylinder is }7\text{ cm and the height is }10\text{ cm. Find:}
\displaystyle \text{(a) The total surface area (both the halves).}
\displaystyle \text{(b) The total cost of painting the two halves at the rate of Rs }30\text{ per cm}^{2}
\displaystyle \text{(Use }\pi=\frac{22}{7}\text{)} \displaystyle \text{(iii) }15,30,60,120\ldots \text{ are in G.P. (Geometric Progression).}                                           \displaystyle \text{[4]}

\displaystyle \text{(a) Find the }n^{\text{th}}\text{ term of this G.P. in terms of }n.
\displaystyle \text{(b) How many terms of the above G.P. will give the sum }945?
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle A=\begin{bmatrix}x & 0\\ 1 & 1\end{bmatrix},\ B=\begin{bmatrix}4 & 0\\ y & 1\end{bmatrix},\ C=\begin{bmatrix}4 & 0\\ x & 1\end{bmatrix}
\displaystyle AB=\begin{bmatrix}x & 0\\ 1 & 1\end{bmatrix}\begin{bmatrix}4 & 0\\ y & 1\end{bmatrix}
\displaystyle =\begin{bmatrix}4x & 0\\ 4+y & 1\end{bmatrix}
\displaystyle \text{Given }AB=C=\begin{bmatrix}4 & 0\\ x & 1\end{bmatrix}
\displaystyle 4x=4\Rightarrow x=1
\displaystyle 4+y=x\Rightarrow 4+y=1\Rightarrow y=-3
\displaystyle \text{Answer: }x=1,\ y=-3

\displaystyle \textbf{(ii)}
\displaystyle \text{Diameter }=7\text{ cm}\Rightarrow r=\frac{7}{2}=3.5\text{ cm},\ h=10\text{ cm}
\displaystyle \text{Each half-cylinder has:}
\displaystyle \text{Curved surface }=\frac{1}{2}(2\pi rh)=\pi rh
\displaystyle \text{Two semicircular ends }=\pi r^2
\displaystyle \text{Rectangular face from cut }=2rh
\displaystyle \text{So TSA of one half }=\pi rh+\pi r^2+2rh
\displaystyle =\pi(3.5)(10)+\pi(3.5)^2+2(3.5)(10)
\displaystyle =35\pi+12.25\pi+70=47.25\pi+70
\displaystyle \text{For two halves: }2(47.25\pi+70)=94.5\pi+140
\displaystyle =94.5\cdot \frac{22}{7}+140=297+140=437\text{ cm}^2
\displaystyle \text{(b) Cost }=437\times 30=13110
\displaystyle \text{Answer: TSA }=437\text{ cm}^2,\ \text{Cost }=13110

\displaystyle \textbf{(iii)}
\displaystyle \text{G.P.: }15,30,60,120,\ldots
\displaystyle a=15,\ r=2
\displaystyle \text{(a) }T_n=ar^{n-1}=15\cdot 2^{n-1}
\displaystyle \text{(b) }S_n=a\frac{r^n-1}{r-1}=15(2^n-1)
\displaystyle 15(2^n-1)=945
\displaystyle 2^n-1=63\Rightarrow 2^n=64\Rightarrow n=6
\displaystyle \text{Answer: }T_n=15\cdot 2^{n-1},\ n=6

\displaystyle \textbf{Question 3}
\displaystyle \text{(i) Factorize: } \sin^{3}\theta+\cos^{3}\theta                                                                                                    \displaystyle \text{[4]}
\displaystyle \text{Hence, prove the following identity: }   \frac{\sin^{3}\theta+\cos^{3}\theta}{\sin\theta+\cos\theta}+\sin\theta\cos\theta=1
\displaystyle \text{(ii) In the given diagram, }O\text{ is the centre of the circle. PR and PT are two}                 \displaystyle \text{[4]}
\displaystyle \text{tangents drawn from the external point }P\text{ and touching the circle at }Q\text{ and }S
\displaystyle \text{respectively. } MN\text{ is a diameter of the circle. Given }\angle PQM=42^{\circ}\text{ and }\angle PSM=25^{\circ}.
\displaystyle \text{Find:}
\displaystyle \text{(a) } \angle OQM
\displaystyle \text{(b) } \angle QNS
\displaystyle \text{(c) } \angle QOS
\displaystyle \text{(d) } \angle QMS
\displaystyle \text{(iii) Use graph sheet for this question. Take }2\text{ cm }=1\text{ unit along the axes.}                 \displaystyle \text{[5]}
\displaystyle \text{(a) Plot }A(0,3), B(2,1) \text{ and }C(4,-1).
\displaystyle \text{(b) Reflect point }B\text{ and }C\text{ in }y\text{-axis and name their images as }B'\text{ and }C'
\displaystyle \text{respectively. Plot and write coordinates of the points }B'\text{ and }C'.
\displaystyle \text{(c) Reflect point }A\text{ in the line }BB'\text{ and name its image as }A'.
\displaystyle \text{(d) Plot and write coordinates of point }A'.
\displaystyle \text{(e) Join the points }ABA'B'\text{ and give the geometrical name of the closed figure}
\displaystyle \text{so formed.}
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \sin^3\theta+\cos^3\theta=(\sin\theta+\cos\theta)(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta)
\displaystyle =(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)
\displaystyle \text{Hence, } \frac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}=1-\sin\theta\cos\theta
\displaystyle \text{LHS of identity }=(1-\sin\theta\cos\theta)+\sin\theta\cos\theta=1
\displaystyle \text{Hence proved.}

\displaystyle \textbf{(ii)}
\displaystyle \text{(a) }\angle PQM=42^\circ\Rightarrow \angle QNM=42^\circ\ (\text{alternate segment theorem})
\displaystyle \angle QOM=2\angle QNM=2\times 42^\circ=84^\circ
\displaystyle \text{Answer: }84^\circ
\displaystyle \text{(b) }\angle PSM=25^\circ\Rightarrow \angle QNS=25^\circ
\displaystyle \text{Answer: }25^\circ
\displaystyle \text{(c) }\angle QOS=2\angle QNS=2\times 25^\circ=50^\circ
\displaystyle \text{Answer: }50^\circ
\displaystyle \text{(d) }\angle QMS=\angle QNS+\angle QNM=25^\circ+42^\circ=67^\circ
\displaystyle \text{Answer: }67^\circ

\displaystyle \textbf{(iii)}
\displaystyle \text{(a) }A(0,3),\ B(2,1),\ C(4,-1)
\displaystyle \text{(b) Reflection in y-axis: }B'(-2,1),\ C'(-4,-1)
\displaystyle \text{(c) Line }BB'\text{ is }x=0\text{ (y-axis)}
\displaystyle \text{(d) Reflection of }A(0,3)\text{ in }x=0\Rightarrow A'(0,3)
\displaystyle \text{(e) Points }A,B,A',B'\text{ form a parallelogram}

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

\displaystyle \textbf{Question 4}
\displaystyle \text{(i) Suresh has a recurring deposit account in a bank. He deposits Rs }2000\text{ per month}
\displaystyle \text{and the bank pays interest at the rate of }8\%\text{ per annum. If he gets Rs }1040\text{ as interest}
\displaystyle \text{at the time of maturity, find in years total time for which the account was held.}       \displaystyle \text{[3]}
\displaystyle \text{(ii) The following table gives the duration of movies in minutes.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Duration} & 100-110 & 110-120 & 120-130 & 130-140 & 140-150 & 150-160 \\ \text{(in minutes)} & & & & & & \\\hline \text{No. of} & 5 & 10 & 17 & 8 & 6 & 4 \\ \text{movies} & & & & & & \\\hline \end{array}
\displaystyle \text{Using step - deviation method, find the mean duration of the movies.}                         \displaystyle \text{[3]}
\displaystyle \text{(iii) If } \frac{(a+b)^{3}}{(a-b)^{3}}=\frac{64}{27}                                                                                                               \displaystyle \text{[4]}
\displaystyle \text{(a) Find } \frac{a+b}{a-b}
\displaystyle \text{(b) Hence using properties of proportion, find } a:b.
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \text{Monthly deposit }P=2000,\ \text{Rate }r=8\%\text{ p.a.},\ \text{Interest }I=1040
\displaystyle \text{For a recurring deposit, }I=P\cdot \frac{n(n+1)}{2}\cdot \frac{r}{12\cdot 100}
\displaystyle 1040=2000\cdot \frac{n(n+1)}{2}\cdot \frac{8}{1200}
\displaystyle 1040=\frac{40}{3}n(n+1)
\displaystyle n(n+1)=78
\displaystyle n^2+n-78=0
\displaystyle (n-12)(n+13)=0
\displaystyle n=12\ \text{months}
\displaystyle \therefore\ \text{Total time }=1\text{ year}

\displaystyle \textbf{(ii)}
\displaystyle \text{Class intervals: }100\text{--}110,\ 110\text{--}120,\ 120\text{--}130,\ 130\text{--}140,\ 140\text{--}150,\ 150\text{--}160
\displaystyle \text{Frequencies }f_i: 5,\ 10,\ 17,\ 8,\ 6,\ 4
\displaystyle \text{Class marks }x_i: 105,\ 115,\ 125,\ 135,\ 145,\ 155
\displaystyle \text{Take assumed mean }A=125,\ \text{class size }h=10
\displaystyle u_i=\frac{x_i-A}{h}=-2,\ -1,\ 0,\ 1,\ 2,\ 3
\displaystyle f_iu_i=-10,\ -10,\ 0,\ 8,\ 12,\ 12
\displaystyle \sum f_i=5+10+17+8+6+4=50
\displaystyle \sum f_iu_i=-10-10+0+8+12+12=12
\displaystyle \overline{x}=A+h\cdot \frac{\sum f_iu_i}{\sum f_i}
\displaystyle \overline{x}=125+10\cdot \frac{12}{50}
\displaystyle \overline{x}=125+2.4=127.4
\displaystyle \therefore\ \text{Mean duration of movies }=127.4\text{ minutes}

\displaystyle \textbf{(iii)}
\displaystyle \frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}
\displaystyle \textbf{(a) Taking cube root on both sides,}
\displaystyle \frac{a+b}{a-b}=\frac{4}{3}
\displaystyle \text{Answer: }\frac{a+b}{a-b}=\frac{4}{3}
\displaystyle \textbf{(b) Using property of proportion, if }\frac{a+b}{a-b}=\frac{4}{3},\text{ then }
\displaystyle \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{4+3}{4-3}
\displaystyle \frac{2a}{2b}=\frac{7}{1}
\displaystyle \frac{a}{b}=\frac{7}{1}
\displaystyle \therefore\ a:b=7:1

\\
\displaystyle \textbf{Question 5}
\displaystyle \text{(i) The given graph with a histogram represents the number of plants of different} \displaystyle \text{[5]}
\displaystyle \text{heights grown in a school campus. Study the graph carefully and answer the} \\ \text{following questions:} \displaystyle \text{(a) Make a frequency table with respect to the class boundaries and their corresponding}
\displaystyle \text{frequencies.}
\displaystyle \text{(b) State the modal class.}
\displaystyle \text{(c) Identify and note down the mode of the distribution.}
\displaystyle \text{(d) Find the number of plants whose height range is between }80\text{ cm to }90\text{ cm.}
\displaystyle \text{(ii) The angle of elevation of the top of a }100\text{ m high tree from two points }                 \displaystyle \text{[5]}
\displaystyle A\text{ and }B \text{ on the opposite side of the tree are }52^{\circ}\text{ and }45^{\circ}\text{ respectively. Find the distance }
\displaystyle AB,\text{ to the nearest } \text{metre.} \displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \text{From the histogram, the frequency table is:}
\displaystyle \begin{array}{|c|c|}\hline \text{Height (in cm)} & \text{Number of plants} \\ \hline 30-40 & 4 \\ \hline 40-50 & 2 \\ \hline 50-60 & 8 \\ \hline 60-70 & 12 \\ \hline 70-80 & 6 \\ \hline 80-90 & 3 \\ \hline 90-100 & 4 \\ \hline \end{array}
\displaystyle \textbf{(a) }\text{The required frequency table is given above.}
\displaystyle \textbf{(b) }\text{The class with highest frequency is }60\text{--}70
\displaystyle \therefore\ \text{Modal class }=60\text{--}70
\displaystyle \textbf{(c) }\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle \text{Here }l=60,\ h=10,\ f_1=12,\ f_0=8,\ f_2=6
\displaystyle \text{Mode}=60+\frac{12-8}{2(12)-8-6}\times 10
\displaystyle =60+\frac{4}{24-14}\times 10
\displaystyle =60+\frac{4}{10}\times 10
\displaystyle =60+4=64
\displaystyle \therefore\ \text{Mode of the distribution }=64\text{ cm}
\displaystyle \textbf{(d) }\text{For height range }80\text{--}90\text{ cm, frequency }=3
\displaystyle \therefore\ \text{Number of plants }=3

\displaystyle \textbf{(ii)}
\displaystyle \text{Let }DC=100\text{ m be the height of the tree}
\displaystyle \text{In right }\triangle ADC,\ \tan 52^\circ=\frac{DC}{AC}
\displaystyle \tan 52^\circ=\frac{100}{AC}
\displaystyle AC=\frac{100}{\tan 52^\circ}
\displaystyle \text{In right }\triangle BDC,\ \tan 45^\circ=\frac{DC}{BC}
\displaystyle 1=\frac{100}{BC}
\displaystyle BC=100
\displaystyle \text{Therefore, }AB=AC+BC=\frac{100}{\tan 52^\circ}+100
\displaystyle \tan 52^\circ\approx 1.279
\displaystyle AC\approx \frac{100}{1.279}\approx 78.19
\displaystyle AB\approx 78.19+100=178.19
\displaystyle \therefore\ AB\approx 178\text{ m}
\displaystyle \text{Answer: }178\text{ m (nearest metre)}

\\
\displaystyle \textbf{Question 6}
\displaystyle \text{(i) Solve the following quadratic equation for }x\text{ and give your answer correct to three}
\displaystyle \text{significant figures: } \quad 2x^{2}-10x+5=0                                                                              \displaystyle \text{[3]}
\displaystyle \text{(Use mathematical tables if necessary)}
\displaystyle \text{(ii) The }n^{\text{th}}\text{ term of an Arithmetic Progression (A.P.) is given by the relation }           \displaystyle \text{[3]}
\displaystyle T_{n}=6(7-n).
\displaystyle \text{Find:}
\displaystyle \text{(a) its first term and common difference}
\displaystyle \text{(b) sum of its first }25\text{ terms}
\displaystyle \text{(iii) In the given diagram } \triangle ADB \text{ and } \triangle ACB \text{ are two right angled triangles with}   \displaystyle \text{[4]}
\displaystyle \angle ADB=\angle BCA=90^{\circ}.\text{ If }AB=10\text{ cm, }AD=6\text{ cm, }BC=2.4\text{ cm and } \\ DP=4.5\text{ cm}
\displaystyle \text{(a) Prove that } \triangle APL \sim \triangle BPC
\displaystyle \text{(b) Find the length of }BD\text{ and }PB
\displaystyle \text{(c) Hence, find the length of }PA
\displaystyle \text{(d) Find area } \triangle APD:\text{ area } \triangle BPC
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle 2x^2-10x+5=0
\displaystyle x=\frac{10\pm\sqrt{(-10)^2-4(2)(5)}}{2\cdot 2}
\displaystyle =\frac{10\pm\sqrt{100-40}}{4}=\frac{10\pm\sqrt{60}}{4}
\displaystyle =\frac{10\pm 7.746}{4}
\displaystyle x=\frac{17.746}{4}=4.44\ (\text{approx}),\quad x=\frac{2.254}{4}=0.564
\displaystyle \text{Answer: }x=4.44,\ 0.564\ (\text{to 3 s.f.})

\displaystyle \textbf{(ii)}
\displaystyle T_n=6(7-n)=42-6n
\displaystyle \textbf{(a) }T_1=42-6=36\Rightarrow a=36
\displaystyle d=T_2-T_1=(42-12)-36=30-36=-6
\displaystyle \text{Answer: }a=36,\ d=-6
\displaystyle \textbf{(b) }S_{25}=\frac{25}{2}[2a+(25-1)d]
\displaystyle =\frac{25}{2}[72+24(-6)]
\displaystyle =\frac{25}{2}[72-144]=\frac{25}{2}(-72)=-900
\displaystyle \text{Answer: }S_{25}=-900

\displaystyle \textbf{(iii)}
\displaystyle \textbf{(a) }\angle ADB=\angle BCA=90^\circ,\ \angle APD=\angle BPC
\displaystyle \therefore\ \triangle APD\sim \triangle BPC\ (\text{AA similarity})
\displaystyle \textbf{(b) In }\triangle ADB,\ AB^2=AD^2+BD^2
\displaystyle 10^2=6^2+BD^2\Rightarrow 100=36+BD^2\Rightarrow BD=8\text{ cm}
\displaystyle PB=BD-DP=8-4.5=3.5\text{ cm}
\displaystyle \textbf{(c) From similarity, }\frac{AP}{PB}=\frac{AD}{BC}=\frac{6}{2.4}=2.5
\displaystyle AP=2.5\times 3.5=8.75\text{ cm}
\displaystyle \textbf{(d) Area ratio }=\left(\frac{AP}{PB}\right)^2=(2.5)^2=6.25=25:4
\displaystyle \text{Answer: }BD=8\text{ cm},\ PB=3.5\text{ cm},\ PA=8.75\text{ cm},\ \text{Area ratio }=25:4

\displaystyle \textbf{Question 7}
\displaystyle \text{(i) In the given diagram, an isosceles } \triangle ABC \text{ is inscribed in a circle with centre }O. \displaystyle \text{[3]}
\displaystyle \text{PQ is a tangent to the circle at }C.\text{ OM is perpendicular to chord }AC\text{ and} \\ \angle COM=65^{\circ}.
\displaystyle \text{Find:}
\displaystyle \text{(a) } \angle ABC
\displaystyle \text{(b) } \angle BAC
\displaystyle \text{(c) } \angle BCQ
\displaystyle \text{(ii) Solve the following inequation, write down the solution set and represent it }        \displaystyle \text{[3]}
\displaystyle \text{on the real number line.}
\displaystyle -3+x\leq \frac{7x}{2}+2<8+2x,\;x\in I
\displaystyle \text{(iii) In the given diagram, }ABC\text{ is a triangle, where }B(4,-4)\text{ and }C(-4,-2). \\ \text{D is a point on }   AC.             \displaystyle \text{[4]}
\displaystyle \text{(a) Write down the coordinates of }A\text{ and }D.
\displaystyle \text{(b) Find the coordinates of the centroid of } \triangle ABC.
\displaystyle \text{(c) If }D\text{ divides }AC\text{ in the ratio }k:1,\text{ find the value of }k.
\displaystyle \text{(d) Find the equation of the line }BD. \displaystyle \text{Answer:}

\\
\displaystyle \textbf{Question 8}
\displaystyle \text{(i) The polynomial }3x^{3}+8x^{2}-15x+k\text{ has }(x-1)\text{ as a factor. Find the value of }k.     \displaystyle \text{[3]}
\displaystyle \text{ Hence factorize the resulting polynomial completely.}
\displaystyle \text{(ii) The following letters }A,D,M,N,O,S,U,Y\text{ of the English alphabet are written } \displaystyle \text{[3]}
\displaystyle \text{on separate cards and put in a box. The cards are well shuffled and one card is drawn }
\displaystyle \text{at random. What is the probability that the card drawn is a letter of the word,}
\displaystyle \text{(a) MONDAY?}
\displaystyle \text{(b) which does not appear in MONDAY?}
\displaystyle \text{(c) which appears both in SUNDAY and MONDAY?}
\displaystyle \text{(iii) Oil is stored in a spherical vessel occupying }\frac{3}{4}\text{ of its full capacity. Radius of}     \displaystyle \text{[4]}
\displaystyle \text{this spherical vessel is }28\text{ cm. This oil is then poured into a cylindrical vessel with a radius of}
\displaystyle 21\text{ cm.} \text{Find the height of the oil in the cylindrical vessel (correct to the nearest cm).}
\displaystyle \text{Take }\pi=\frac{22}{7}

\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Since }OM\perp AC,\ \text{the perpendicular from the centre to chord }AC\text{ bisects the chord}
\displaystyle \text{Hence }M\text{ is the mid-point of }AC
\displaystyle \text{In isosceles }\triangle AOC,\ OA=OC
\displaystyle \text{Given }\angle COM=65^\circ
\displaystyle \text{Since }OM\text{ bisects }\angle AOC,\ \angle AOM=\angle MOC=65^\circ
\displaystyle \therefore\ \angle AOC=130^\circ
\displaystyle \text{Also, }AB=AC\Rightarrow \text{equal chords subtend equal angles at the centre}
\displaystyle \therefore\ \angle AOB=\angle AOC=130^\circ
\displaystyle \text{Now, } \angle AOB+\angle BOC+\angle COA=360^\circ
\displaystyle 130^\circ+\angle BOC+130^\circ=360^\circ
\displaystyle \angle BOC=100^\circ
\displaystyle \textbf{(a) }\angle ABC=\frac{1}{2}\angle AOC=\frac{1}{2}\times 130^\circ=65^\circ
\displaystyle \textbf{(b) }\angle BAC=\frac{1}{2}\angle BOC=\frac{1}{2}\times 100^\circ=50^\circ
\displaystyle \textbf{(c) }\angle BCQ=\angle BAC=50^\circ\ \text{(angle between tangent and chord)}
\displaystyle \text{Answer: }\angle ABC=65^\circ,\ \angle BAC=50^\circ,\ \angle BCQ=50^\circ

\displaystyle \textbf{(ii)}
\displaystyle -3+x\leq \frac{7x}{2}+2<8+2x,\ x\in I
\displaystyle \text{First solve }-3+x\leq \frac{7x}{2}+2
\displaystyle 2x-6\leq 7x+4
\displaystyle -10\leq 5x
\displaystyle -2\leq x
\displaystyle \text{Now solve }\frac{7x}{2}+2<8+2x
\displaystyle 7x+4<16+4x
\displaystyle 3x<12
\displaystyle x<4
\displaystyle \text{Combining, }-2\leq x<4
\displaystyle \text{Since }x\in I,\ \text{solution set }=\{-2,-1,0,1,2,3\}
\displaystyle \text{On the number line, mark solid dot at }-2,\ \text{open dot at }4,\ \text{and integers in between.}

\displaystyle \textbf{(iii)}
\displaystyle \text{(a) From the graph, }A=(0,6)\text{ and }D=(-3,0)
\displaystyle \text{(b) Centroid of }\triangle ABC=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)
\displaystyle =\left(\frac{0+4+(-4)}{3},\frac{6+(-4)+(-2)}{3}\right)=\left(0,0\right)
\displaystyle \text{Hence, centroid }=(0,0)
\displaystyle \text{(c) Let }D\text{ divide }AC\text{ in the ratio }k:1
\displaystyle A=(0,6),\ C=(-4,-2),\ D=(-3,0)
\displaystyle \text{By section formula, }D=\left(\frac{k(-4)+1(0)}{k+1},\frac{k(-2)+1(6)}{k+1}\right)
\displaystyle \left(\frac{-4k}{k+1},\frac{-2k+6}{k+1}\right)=(-3,0)
\displaystyle \frac{-4k}{k+1}=-3
\displaystyle -4k=-3k-3
\displaystyle k=3
\displaystyle \text{(also, }\frac{-2k+6}{k+1}=0\Rightarrow k=3\text{)}
\displaystyle \therefore\ k=3
\displaystyle \text{(d) Equation of line }BD\text{ through }B(4,-4)\text{ and }D(-3,0)
\displaystyle \text{Slope }m=\frac{0-(-4)}{-3-4}=\frac{4}{-7}=-\frac{4}{7}
\displaystyle y+4=-\frac{4}{7}(x-4)
\displaystyle 7y+28=-4x+16
\displaystyle 4x+7y+12=0
\displaystyle \text{Hence, equation of }BD\text{ is }4x+7y+12=0
\\
\displaystyle \textbf{Question 9}
\displaystyle \text{(i) The figure shows a circle of radius }9\text{ cm with }O\text{ as the centre. The diameter}    \displaystyle \text{[3]}
\displaystyle AB \text{ produced meets the tangent }PQ\text{ at }P.\text{ If }PA=24\text{ cm, find the length of tangent }PQ.

\displaystyle \text{(ii) Mr. Gupta invested Rs }33000\text{ in buying Rs }100\text{ shares of a company at }
\displaystyle 10\%\text{ premium. The }\text{dividend declared by the company is }12\%.\text{ Find:}                            \displaystyle \text{[3]}
\displaystyle \text{(a) the number of shares purchased by him.}
\displaystyle \text{(b) his annual dividend.}
\displaystyle \text{(iii) A life insurance agent found the following data for distribution of ages of }           \displaystyle \text{[4]}
\displaystyle 100\text{ policy } \text{holders:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Age in years} & \text{Policy Holders} & \text{Cumulative frequency} \\ & \text{(frequency)} & \\\hline 20-25 & 2 & 2 \\ 25-30 & 4 & 6 \\ 30-35 & 12 & 18 \\ 35-40 & 20 & 38 \\ 40-45 & 28 & 66 \\ 45-50 & 22 & 88 \\ 50-55 & 8 & 96 \\ 55-60 & 4 & 100 \\\hline \end{array}
\displaystyle \text{On a graph sheet draw an ogive using the given data. Take }2\text{ cm }=5\text{ years along one }
\displaystyle \text{axis and }2\text{ cm }=10\text{ policy holders along the other axis. Use your graph to find:}
\displaystyle \text{(a) The median age.}
\displaystyle \text{(b) Number of policy holders whose age is above }52\text{ years.}
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Radius }=9\text{ cm}\Rightarrow AB=18\text{ cm}
\displaystyle \text{Given }PA=24\text{ cm}\Rightarrow PB=PA-AB=24-18=6\text{ cm}
\displaystyle \text{From external point, }PQ^2=PA\cdot PB
\displaystyle PQ^2=24\times 6=144
\displaystyle PQ=12\text{ cm}
\displaystyle \text{Answer: }12\text{ cm}

\displaystyle \textbf{(ii)}
\displaystyle \text{Face value }=100,\ \text{premium }10\%\Rightarrow \text{market price }=110
\displaystyle \text{Total investment }=33000
\displaystyle \text{(a) Number of shares }=\frac{33000}{110}=300
\displaystyle \text{(b) Dividend }=12\%\text{ of face value }=12\text{ per share}
\displaystyle \text{Annual dividend }=300\times 12=3600
\displaystyle \text{Answer: Shares }=300,\ \text{Dividend }=3600

\displaystyle \textbf{(iii)}
\displaystyle \text{Total }N=100,\ \frac{N}{2}=50
\displaystyle \text{Median class is }40\text{--}45\ (\text{cf just exceeds }50)
\displaystyle l=40,\ h=5,\ f=28,\ cf=38
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle =40+\frac{50-38}{28}\times 5
\displaystyle =40+\frac{12}{28}\times 5
\displaystyle =40+2.14=42.14
\displaystyle \therefore\ \text{Median age }\approx 42\text{ years}
\displaystyle \text{(b) For age above 52 years}
\displaystyle \text{From ogive, number below }52\approx 92
\displaystyle \text{So above }52=100-92=8
\displaystyle \text{Answer: Median }\approx 42\text{ years, Above 52 years }=8
\\
\displaystyle \textbf{Question 10}
\displaystyle \text{(i) Rohan bought the following eatables for his friends:}                                                    \displaystyle \text{[3]}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{\text{Soham Sweet Mart : Bill}} \\\hline \text{S. No.} & \text{Item} & \text{Price} & \text{Quantity} & \text{Rate of GST} \\\hline 1 & \text{Laddu} & \text{Rs }500\text{ per kg} & 2\text{ kg} & 5\% \\ 2 & \text{Pastries} & \text{Rs }100\text{ per piece} & 12\text{ pieces} & 18\% \\\hline \end{array}
\displaystyle \text{Calculate:}
\displaystyle \text{(a) Total GST paid.}
\displaystyle \text{(b) Total bill amount including GST.}
\displaystyle \text{(ii) (a) If the lines }kx-y+4=0\text{ and }2y=6x+7\text{ are perpendicular to each other, }       \displaystyle \text{[3]}
\displaystyle \text{find the value of }k.
\displaystyle \text{(b) Find the equation of a line parallel to }2y=6x+7\text{ and passing through }(-1,1).
\displaystyle \text{(iii) Use ruler and compass to answer this question. Construct } \angle ABC=90^{\circ},\text{ where}
\displaystyle AB=6\text{ cm, }BC=8\text{ cm.}                                                                                                       \displaystyle \text{[4]}
\displaystyle \text{(a) Construct the locus of points equidistant from }B\text{ and }C.
\displaystyle \text{(b) Construct the locus of points equidistant from }A\text{ and }B.
\displaystyle \text{(c) Mark the point which satisfies both the conditions (a) and (b) as }O.\text{ Construct the}
\displaystyle \text{locus of points keeping a fixed distance }OA\text{ from the fixed point }O.
\displaystyle \text{(d) Construct the locus of points which are equidistant from }BA\text{ and }BC.
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Laddu: }500\times 2=1000,\ \text{GST }=5\%\Rightarrow 50
\displaystyle \text{Pastries: }100\times 12=1200,\ \text{GST }=18\%\Rightarrow 216
\displaystyle \text{(a) Total GST }=50+216=266
\displaystyle \text{(b) Total bill }=1000+1200+266=2466
\displaystyle \text{Answer: GST }=266,\ \text{Total bill }=2466

\displaystyle \textbf{(ii)}
\displaystyle \text{(a) }kx-y+4=0\Rightarrow y=kx+4\ (\text{slope }=k)
\displaystyle 2y=6x+7\Rightarrow y=3x+\frac{7}{2}\ (\text{slope }=3)
\displaystyle \text{For perpendicular lines: }k\times 3=-1
\displaystyle k=-\frac{1}{3}
\displaystyle \text{(b) Required line parallel to }2y=6x+7\Rightarrow y=3x+c
\displaystyle \text{Passing through }(-1,1):\ 1=3(-1)+c\Rightarrow c=4
\displaystyle \therefore\ y=3x+4

\displaystyle \textbf{(iii)}
\displaystyle \text{(a) Locus equidistant from }B\text{ and }C:\ \text{Perpendicular bisector of }BC
\displaystyle \text{(b) Locus equidistant from }A\text{ and }B:\ \text{Perpendicular bisector of }AB
\displaystyle \text{(c) Intersection point }O:\ \text{Circumcentre of }\triangle ABC
\displaystyle \text{Locus with fixed distance }OA:\ \text{Circle with centre }O\text{ and radius }OA
\displaystyle \text{(d) Locus equidistant from }BA\text{ and }BC:\ \text{Angle bisector of }\angle ABC

 

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.