\displaystyle \textbf{Series 4JLZ3E/C} \hspace{3.0cm}  \textbf{SET } \sim 1  \hspace{4.0cm} \textbf{Code No. }30/3/1

\displaystyle \text{Candidates must write the Code on the title page of the answer-book.}

\displaystyle \textbf{NOTE}

\displaystyle \text{(i) Please check that this question paper contains 11 printed pages.}

\displaystyle \text{(ii) Code number given on the right hand side of the question paper should be written on}
\displaystyle \text{the title page of the answer-book by the candidate.}

\displaystyle \text{(iii) Please check that this question paper contains 36 questions.}

\displaystyle \text{(iv) Please write down the serial number of the question in the answer-book before}
\displaystyle \text{attempting it.}

\displaystyle \text{(v) 15 minute time has been allotted to read this question paper. The question paper}
\displaystyle \text{will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read}
\displaystyle \text{the question paper only and will not write any answer on the answer-book during this}
\displaystyle \text{period.}

 

\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \text{Time allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 80}

\displaystyle \textbf{General Instructions :}

\displaystyle \text{Read the following instructions very carefully and strictly follow them :}

\displaystyle \text{(i) This question paper contains two parts A and B.}

\displaystyle \text{(ii) Both Part A and Part B have internal choices.}

\displaystyle \textbf{Part A}

\displaystyle \text{(i) It consists of two Sections, I and II.}

\displaystyle \text{(ii) Section I has 16 questions of 1 mark each. Internal choices are provided in 5}
\displaystyle \text{questions.}

\displaystyle \text{(iii) Section II has 4 questions on case study (Q.No. 17 to 20). Each question has 5}
\displaystyle \text{sub-parts. An examinee is to attempt any 4 out of 5 sub-parts. Each sub-part is of}
\displaystyle \text{1 mark.}

\displaystyle \textbf{Part B}

\displaystyle \text{(i) It consists of three sections, III, IV and V.}

\displaystyle \text{(ii) Section III has 6 questions No. 21 to 26 of Very-short Answer Type of 2 marks}
\displaystyle \text{each.}

\displaystyle \text{(iii) Section IV has 7 questions No. 27 to 33 of Short Answer Type of 3 marks each.}

\displaystyle \text{(iv) Section V has 3 questions No. 34 to 36 of Long Answer Type of 5 marks each.}

\displaystyle \text{(v) Internal choice is provided in 2 questions in Section III, 2 questions in Section IV}
\displaystyle \text{and 1 question in Section V.}

\displaystyle \textbf{PART A}
\displaystyle \textbf{SECTION I}

\displaystyle \textbf{Question 1. }\text{Write the quadratic equation in }x\text{ whose roots are }2\text{ and }-5.
\displaystyle \text{Answer:}
\displaystyle \text{Required quadratic equation is }(x-2)(x+5)=0
\displaystyle x^{2}+3x-10=0
\displaystyle \therefore \text{Required equation is }x^{2}+3x-10=0
\\

\displaystyle \textbf{Question 2. }\text{Find the exponent of }2\text{ in the prime factorisation of }288.
\displaystyle \text{Answer:}
\displaystyle 288=2^{5}\times3^{2}
\displaystyle \therefore \text{Exponent of }2\text{ is }5.
\\

\displaystyle \textbf{Question 3(a). }\text{If }\alpha\text{ and }\beta\text{ are the zeroes of the quadratic polynomial }
\displaystyle f(x)=x^{2}-x-4, \ \text{find the value of }\frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta=1,\qquad \alpha\beta=-4
\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta=\frac{\alpha+\beta}{\alpha\beta}-\alpha\beta
\displaystyle =\frac{1}{-4}-(-4)
\displaystyle =-\frac{1}{4}+4
\displaystyle =\frac{15}{4}

\textbf{OR}

\displaystyle \textbf{Question 3(b). }\text{If one zero of the quadratic polynomial }x^{2}+3x+k\text{ is }2,\text{ then find the value of }k.
\displaystyle \text{Answer:}
\displaystyle \text{Since }2\text{ is a zero,}
\displaystyle 2^{2}+3(2)+k=0
\displaystyle 4+6+k=0
\displaystyle k=-10
\\

\displaystyle \textbf{Question 4(a). }\text{If }\frac{3}{5},a,4\text{ are three consecutive terms of an A.P., then find the value of }a.
\displaystyle \text{Answer:}
\displaystyle a=\frac{\frac{3}{5}+4}{2}
\displaystyle =\frac{\frac{3}{5}+\frac{20}{5}}{2}
\displaystyle =\frac{23}{10}

\textbf{OR}

\displaystyle \textbf{Question 4(b). }\text{In an A.P., if the common difference }d=-3\text{ and the eleventh term }
\displaystyle a_{11}=15,  \text{then find the first term.}
\displaystyle \text{Answer:}
\displaystyle a_{11}=a+10d
\displaystyle 15=a+10(-3)
\displaystyle 15=a-30
\displaystyle a=45
\\

\displaystyle \textbf{Question 5. }\text{A man goes }5\text{ metres due West and then }12\text{ metres due North. How far is he from the}
\displaystyle \text{starting point?}
\displaystyle \text{Answer:}
\displaystyle \text{Distance from starting point}=\sqrt{5^{2}+12^{2}}
\displaystyle =\sqrt{25+144}
\displaystyle =\sqrt{169}
\displaystyle =13\text{ metres}
\\

\displaystyle \textbf{Question 6. }\text{PQ is a tangent to a circle with centre O at the point P on the circle. If}
\displaystyle \triangle OPQ\text{ is an isosceles triangle, then find }\angle OQP.
\displaystyle \text{Answer:}
\displaystyle OP\perp PQ
\displaystyle \therefore \angle OPQ=90^\circ
\displaystyle \triangle OPQ\text{ is isosceles and }OP=PQ
\displaystyle \angle OQP=\angle QOP
\displaystyle \angle OQP+\angle QOP+90^\circ=180^\circ
\displaystyle 2\angle OQP=90^\circ
\displaystyle \angle OQP=45^\circ
\\

\displaystyle \textbf{Question 7. }\text{Two concentric circles have radii }10\text{ cm and }6\text{ cm. Find the }
\displaystyle \text{length of the chord of the larger circle which touches the smaller circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of larger circle}=10\text{ cm}
\displaystyle \text{Distance of chord from centre}=6\text{ cm}
\displaystyle \text{Half length of chord}=\sqrt{10^{2}-6^{2}}
\displaystyle =\sqrt{100-36}
\displaystyle =\sqrt{64}=8\text{ cm}
\displaystyle \therefore \text{Length of chord}=2\times8=16\text{ cm}
\\

\displaystyle \textbf{Question 8(a). }\text{If }3\sin A=1,\text{ then find the value of }\sec A.
\displaystyle \text{Answer:}
\displaystyle 3\sin A=1
\displaystyle \sin A=\frac{1}{3}
\displaystyle \cos A=\sqrt{1-\sin^{2}A}
\displaystyle =\sqrt{1-\frac{1}{9}}
\displaystyle =\sqrt{\frac{8}{9}}
\displaystyle =\frac{2\sqrt{2}}{3}
\displaystyle \sec A=\frac{1}{\cos A}
\displaystyle =\frac{3}{2\sqrt{2}}
\displaystyle =\frac{3\sqrt{2}}{4}

\textbf{OR}

\displaystyle \textbf{Question 8(b). }\text{Show that :}   \frac{1+\cot^{2}\theta}{1+\tan^{2}\theta}=\cot^{2}\theta
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\frac{1+\cot^{2}\theta}{1+\tan^{2}\theta}
\displaystyle =\frac{\mathrm{cosec}^{2}\theta}{\sec^{2}\theta}
\displaystyle =\frac{\frac{1}{\sin^{2}\theta}}{\frac{1}{\cos^{2}\theta}}
\displaystyle =\frac{\cos^{2}\theta}{\sin^{2}\theta}
\displaystyle =\cot^{2}\theta
\displaystyle =\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}
\\

\displaystyle \textbf{Question 9. }\text{From a point on the ground, }20\text{ m away from the foot of a vertical tower, }
\displaystyle \text{the angle of elevation of the top of the tower is }60^\circ.\text{ Find the height of the tower.}
\displaystyle \text{Answer:}
\displaystyle \tan60^\circ=\frac{\text{Height of tower}}{20}
\displaystyle \sqrt{3}=\frac{h}{20}
\displaystyle h=20\sqrt{3}\text{ m}
\displaystyle \therefore \text{Height of the tower}=20\sqrt{3}\text{ m}
\\

\displaystyle \textbf{Question 10(a). }\text{Find the area of a circle whose circumference is }66\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle 2\pi r=66
\displaystyle 2\times\frac{22}{7}\times r=66
\displaystyle r=\frac{66\times7}{44}=10.5\text{ cm}
\displaystyle \text{Area}=\pi r^{2}
\displaystyle =\frac{22}{7}\times(10.5)^{2}
\displaystyle =346.5\text{ cm}^{2}
\therefore \text{Area of the circle}=346.5\text{ cm}^{2}

\textbf{OR}

\displaystyle \textbf{Question 10(b). }\text{The perimeter of a semi-circular protractor is }108\text{ cm. Find its diameter.}
\displaystyle \text{Answer:}
\displaystyle \pi r+2r=108
\displaystyle \frac{22}{7}r+2r=108
\displaystyle \frac{36}{7}r=108
\displaystyle r=21\text{ cm}
\displaystyle \text{Diameter}=2r=42\text{ cm}
\displaystyle \therefore \text{Diameter}=42\text{ cm}
\\

\displaystyle \textbf{Question 11. }\text{Write the relationship between three measures of central tendency - }
\displaystyle \text{Mean, Median and Mode.}
\displaystyle \text{Answer:}
\displaystyle \text{Mode}=3\text{ Median}-2\text{ Mean}
\\

\displaystyle \textbf{Question 12. }\text{In a }\triangle ABC,\text{ if DE is parallel to BC, }\frac{AD}{DB}=\frac{4}{5}\text{ and }
\displaystyle AC=15\text{ cm, then find the length of } AE.
\displaystyle \text{Answer:}
\displaystyle \frac{AD}{DB}=\frac{4}{5}
\displaystyle \frac{AD}{AB}=\frac{4}{4+5}=\frac{4}{9}
\displaystyle \text{Since }DE\parallel BC,
\displaystyle \frac{AE}{AC}=\frac{AD}{AB}
\displaystyle \frac{AE}{15}=\frac{4}{9}
\displaystyle AE=\frac{15\times4}{9}=\frac{20}{3}\text{ cm}
\\

\displaystyle \textbf{Question 13. }\text{Simplify:}  \  \mathrm{cosec}^{2}60^\circ\sin^{2}30^\circ-\sec^{2}60^\circ
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}^{2}60^\circ\sin^{2}30^\circ-\sec^{2}60^\circ
\displaystyle =\left(\frac{2}{\sqrt{3}}\right)^{2}\left(\frac{1}{2}\right)^{2}-2^{2}
\displaystyle =\frac{4}{3}\times\frac{1}{4}-4
\displaystyle =\frac{1}{3}-4
\displaystyle =-\frac{11}{3}
\\

\displaystyle \textbf{Question 14. }\text{If }\tan\theta+\cot\theta=\frac{4\sqrt{3}}{3},\text{ then find the value of }\tan^{2}\theta+\cot^{2}\theta.
\displaystyle \text{Answer:}
\displaystyle \tan\theta+\cot\theta=\frac{4\sqrt{3}}{3}
\displaystyle \left(\tan\theta+\cot\theta\right)^{2}=\tan^{2}\theta+\cot^{2}\theta+2\tan\theta\cot\theta
\displaystyle \left(\frac{4\sqrt{3}}{3}\right)^{2}=\tan^{2}\theta+\cot^{2}\theta+2
\displaystyle \frac{16}{3}=\tan^{2}\theta+\cot^{2}\theta+2
\displaystyle \tan^{2}\theta+\cot^{2}\theta=\frac{10}{3}
\\

\displaystyle \textbf{Question 15. }\text{If tangents PA and PB from an external point P to a circle with }
\displaystyle \text{centre O are inclined to each other at an angle of }70^\circ,\text{ then find }\angle POA.
\displaystyle \text{Answer:}
\displaystyle \angle APB=70^\circ
\displaystyle OA\perp PA,\qquad OB\perp PB
\displaystyle \angle AOB+70^\circ=180^\circ
\displaystyle \angle AOB=110^\circ
\displaystyle OP\text{ bisects }\angle AOB
\displaystyle \angle POA=\frac{110^\circ}{2}=55^\circ
\\

\displaystyle \textbf{Question 16(a). }\text{How many outcomes are possible when three dice are thrown together ?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of outcomes}=6\times6\times6
\displaystyle =216

\textbf{OR}

\displaystyle \textbf{Question 16(b). }\text{If }P(E)=0.015,\text{ then find }P(\text{not }E).
\displaystyle \text{Answer:}
\displaystyle P(\text{not }E)=1-P(E)
\displaystyle =1-0.015
\displaystyle =0.985
\\

\displaystyle \textbf{SECTION II}
\displaystyle \text{Case study based questions (Q. No. 17 - 20) are compulsory. Attempt any 4 sub-parts}
\displaystyle \text{from each question. Each sub-part carries 1 mark.}

\displaystyle \textbf{Question 17. }\text{The residents of a housing society, on the occasion of environment }
\displaystyle \text{day, decided to build two straight paths in the central park of the society and }
\displaystyle \text{also plant trees along the boundary lines of each path.}
\displaystyle \text{Taking one corner of the park as origin and the two mutually perpendicular lines as the}
\displaystyle x\text{-axis and }y\text{-axis, the paths were represented by the two linear equations }
\displaystyle 2x-3y=5 \text{and }-6x+9y=7.
\displaystyle \text{Based on the above, answer the following questions :}

\displaystyle \text{(i) Two paths represented by the two equations here are}
\displaystyle \text{(A) intersecting}\qquad \text{(B) overlapping}\qquad \text{(C) parallel}\qquad \text{(D) mutually perpendicular}

\displaystyle \text{(ii) Which one of the following points lie on the line }2x-3y=5\text{ ?}
\displaystyle \text{(A) }(-4,1)\qquad \text{(B) }(4,-1)\qquad \text{(C) }(4,1)\qquad \text{(D) }(-4,-1)

\displaystyle \text{(iii) If the line }-6x+9y=7\text{ intersects the }y\text{-axis at a point, then its coordinates are :}
\displaystyle \text{(A) }\left(0,\frac{7}{9}\right)\qquad \text{(B) }\left(\frac{7}{9},0\right)\qquad \text{(C) }\left(-\frac{7}{6},0\right)\qquad \text{(D) }\left(0,-\frac{7}{6}\right)

\displaystyle \text{(iv) If a pair of equations }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0\text{ has a unique}
\displaystyle \text{solution, then}
\displaystyle \text{(A) }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\qquad \text{(B) }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}
\displaystyle \text{(C) }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\qquad \text{(D) }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

\displaystyle \text{(v) If }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},\text{ then the two lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0\text{ are}
\displaystyle \text{(A) parallel}\qquad \text{(B) coincident}\qquad \text{(C) intersecting}\qquad \text{(D) perpendicular to each other}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2x-3y-5=0,\qquad -6x+9y-7=0
\displaystyle \frac{a_1}{a_2}=\frac{2}{-6}=-\frac{1}{3},\qquad \frac{b_1}{b_2}=\frac{-3}{9}=-\frac{1}{3},\qquad \frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}
\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}
\displaystyle \therefore \text{The two paths are parallel.}
\displaystyle \text{Correct option : (C)}

\displaystyle \text{(ii) For }(4,1),\quad 2x-3y=2(4)-3(1)=8-3=5
\displaystyle \therefore (4,1)\text{ lies on the line }2x-3y=5.
\displaystyle \text{Correct option : (C)}

\displaystyle \text{(iii) On the }y\text{-axis, }x=0
\displaystyle -6(0)+9y=7
\displaystyle y=\frac{7}{9}
\displaystyle \therefore \text{Coordinates are }\left(0,\frac{7}{9}\right).
\displaystyle \text{Correct option : (A)}

\displaystyle \text{(iv) For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}
\displaystyle \text{Correct option : (B)}

\displaystyle \text{(v) If }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2},\text{ then the two lines are coincident.}
\displaystyle \text{Correct option : (B)}
\\

\displaystyle \textbf{Question 18. }\text{Students of a school are standing in rows and columns in }
\displaystyle \text{their school playground to celebrate their annual sports day. A, B, C and D are }
\displaystyle \text{the positions  of four students as shown in the figure.} \displaystyle \text{Based on the above, answer the following questions :}

\displaystyle \text{(i) The figure formed by the four points A, B, C and D is a}
\displaystyle \text{(A) square}\qquad \text{(B) parallelogram}\qquad \text{(C) rhombus}\qquad \text{(D) quadrilateral}

\displaystyle \text{(ii) If the sports teacher is sitting at the origin, then which of the four students is}
\displaystyle \text{closest to him ?}
\displaystyle \text{(A) A}\qquad \text{(B) B}\qquad \text{(C) C}\qquad \text{(D) D}

\displaystyle \text{(iii) The distance between A and C is}
\displaystyle \text{(A) }\sqrt{37}\text{ units}\qquad \text{(B) }\sqrt{35}\text{ units}\qquad \text{(C) 6 units}\qquad \text{(D) 5 units}

\displaystyle \text{(iv) The coordinates of the mid-point of line segment AC are}
\displaystyle \text{(A) }\left(\frac{5}{2},11\right)\qquad \text{(B) }\left(\frac{5}{2},\frac{11}{2}\right)\qquad \text{(C) }\left(5,\frac{11}{2}\right)\qquad \text{(D) }(5,11)

\displaystyle \text{(v) If a point P divides the line segment AD in the ratio }1:2,\text{ then coordinates of P}
\displaystyle \text{are}
\displaystyle \text{(A) }\left(\frac{8}{3},\frac{8}{3}\right)\qquad \text{(B) }\left(\frac{10}{3},\frac{13}{3}\right)\qquad \text{(C) }\left(\frac{13}{3},\frac{10}{3}\right)\qquad \text{(D) }\left(\frac{16}{3},\frac{11}{3}\right)
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, }A(2,5),\quad B(5,7),\quad C(8,6),\quad D(6,3)

\displaystyle \text{(i) The figure formed by joining A, B, C and D is a quadrilateral.}
\displaystyle \text{Correct option : (D)}

\displaystyle \text{(ii) Distance of A from origin}=\sqrt{2^{2}+5^{2}}=\sqrt{29}
\displaystyle \text{Distance of B from origin}=\sqrt{5^{2}+7^{2}}=\sqrt{74}
\displaystyle \text{Distance of C from origin}=\sqrt{8^{2}+6^{2}}=10
\displaystyle \text{Distance of D from origin}=\sqrt{6^{2}+3^{2}}=\sqrt{45}
\displaystyle \sqrt{29}\text{ is the least distance.}
\displaystyle \therefore \text{Student A is closest to the origin.}
\displaystyle \text{Correct option : (A)}

\displaystyle \text{(iii) } AC=\sqrt{(8-2)^{2}+(6-5)^{2}}
\displaystyle =\sqrt{36+1}
\displaystyle =\sqrt{37}\text{ units}
\displaystyle \text{Correct option : (A)}

\displaystyle \text{(iv) Mid-point of }AC=\left(\frac{2+8}{2},\frac{5+6}{2}\right)
\displaystyle =\left(5,\frac{11}{2}\right)
\displaystyle \text{Correct option : (C)}

\displaystyle \text{(v) } P\text{ divides }AD\text{ in the ratio }1:2
\displaystyle P=\left(\frac{1(6)+2(2)}{1+2},\frac{1(3)+2(5)}{1+2}\right)
\displaystyle =\left(\frac{10}{3},\frac{13}{3}\right)
\displaystyle \text{Correct option : (B)}
\\

\displaystyle \textbf{19. }\text{During the annual sports meet in a school, all the athletes were very enthusiastic.}
\displaystyle \text{They all wanted to be the winner so that their house could stand first. The instructor}
\displaystyle \text{noted down the time taken by a group of students to complete a certain race. The data}
\displaystyle \text{recorded is given below :}

\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Time (in sec.) :} & 0-20 & 20-40 & 40-60 & 60-80 & 80-100\\ \hline \text{Number of students :} & 1 & 4 & 3 & 7 & 5\\ \hline \end{array}

\displaystyle \text{Based on the above, answer the following questions :}

\displaystyle \text{(i) What is the class mark of the modal class ?}
\displaystyle \text{(A) }60 \qquad \text{(B) }70 \qquad \text{(C) }80 \qquad \text{(D) }140

\displaystyle \text{(ii) The mode of the given data is}
\displaystyle \text{(A) }70.33 \qquad \text{(B) }71.33 \qquad \text{(C) }72.33 \qquad \text{(D) }73.33

\displaystyle \text{(iii) The median class of the given data is}
\displaystyle \text{(A) }20-40 \qquad \text{(B) }40-60 \qquad \text{(C) }80-100 \qquad \text{(D) }60-80

\displaystyle \text{(iv) The sum of the lower limits of median class and modal class is}
\displaystyle \text{(A) }80 \qquad \text{(B) }140 \qquad \text{(C) }120 \qquad \text{(D) }100

\displaystyle \text{(v) The median time (in seconds) of the given data is}
\displaystyle \text{(A) }65.7 \qquad \text{(B) }85.7 \qquad \text{(C) }45.7 \qquad \text{(D) }25.7

\\
\displaystyle \text{Answer:}

\displaystyle \text{(i) Modal class}=60-80
\displaystyle \text{Class mark}=\frac{60+80}{2}=70
\displaystyle \text{Correct option : (B)}

\displaystyle \text{(ii) Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =60+\frac{7-3}{2(7)-3-5}\times20
\displaystyle =60+\frac{4}{6}\times20
\displaystyle =73.33
\displaystyle \text{Correct option : (D)}

\displaystyle \text{(iii) }N=20,\qquad \frac{N}{2}=10
\displaystyle \text{Median class}=60-80
\displaystyle \text{Correct option : (D)}

\displaystyle \text{(iv) Lower limit of median class}=60
\displaystyle \text{Lower limit of modal class}=60
\displaystyle \text{Required sum}=60+60=120
\displaystyle \text{Correct option : (C)}

\displaystyle \text{(v) Median}=l+\frac{\frac{N}{2}-c.f.}{f}\times h
\displaystyle =60+\frac{10-8}{7}\times20
\displaystyle =60+\frac{40}{7}
\displaystyle =65.7
\displaystyle \text{Correct option : (A)}
\\

\displaystyle \textbf{20. }\text{During summer break, Harish wanted to play with his friends but it was too hot}
\displaystyle \text{outside, so he decided to play some indoor game with his friends. He collects 20 identical}
\displaystyle \text{cards and writes the numbers 1 to 20 on them (one number on one card). He puts them}
\displaystyle \text{in a box. He and his friends make a bet for the chances of drawing various cards out of}
\displaystyle \text{the box. Each was given a chance to tell the probability of picking one card out of the}
\displaystyle \text{box.}
\displaystyle \text{Based on the above, answer the following questions :}

\displaystyle \text{(i) The probability that the number on the card drawn is an odd prime number, is}
\displaystyle \text{(A) }\frac{3}{5} \qquad \text{(B) }\frac{2}{5} \qquad \text{(C) }\frac{9}{20} \qquad \text{(D) }\frac{7}{20}

\displaystyle \text{(ii) The probability that the number on the card drawn is a composite number is}
\displaystyle \text{(A) }\frac{11}{20} \qquad \text{(B) }\frac{3}{5} \qquad \text{(C) }\frac{4}{5} \qquad \text{(D) }\frac{1}{2}

\displaystyle \text{(iii) The probability that the number on the card drawn is a multiple of 3, 6 and 9 is}
\displaystyle \text{(A) }\frac{1}{20} \qquad \text{(B) }\frac{1}{10} \qquad \text{(C) }\frac{3}{20} \qquad \text{(D) }0

\displaystyle \text{(iv) The probability that the number on the card drawn is a multiple of 3 and 7 is}
\displaystyle \text{(A) }\frac{3}{10} \qquad \text{(B) }\frac{1}{10} \qquad \text{(C) }0 \qquad \text{(D) }\frac{2}{5}

\displaystyle \text{(v) If all cards having odd numbers written on them are removed from the box and then}
\displaystyle \text{one card is drawn from the remaining cards, the probability of getting a card having}
\displaystyle \text{a prime number is}
\displaystyle \text{(A) }\frac{1}{20} \qquad \text{(B) }\frac{1}{10} \qquad \text{(C) }0 \qquad \text{(D) }\frac{1}{5}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=20

\displaystyle \text{(i) Odd prime numbers from }1\text{ to }20\text{ are }3,5,7,11,13,17,19
\displaystyle \text{Number of favourable outcomes}=7
\displaystyle \text{Required probability}=\frac{7}{20}
\displaystyle \text{Correct option : (D)}

\displaystyle \text{(ii) Composite numbers from }1\text{ to }20\text{ are }4,6,8,9,10,12,14,15,16,18,20
\displaystyle \text{Number of favourable outcomes}=11
\displaystyle \text{Required probability}=\frac{11}{20}
\displaystyle \text{Correct option : (A)}

\displaystyle \text{(iii) Multiple of }3,6\text{ and }9\text{ from }1\text{ to }20\text{ is }18
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle \text{Required probability}=\frac{1}{20}
\displaystyle \text{Correct option : (A)}

\displaystyle \text{(iv) Multiple of }3\text{ and }7\text{ from }1\text{ to }20\text{ is none.}
\displaystyle \text{Number of favourable outcomes}=0
\displaystyle \text{Required probability}=0
\displaystyle \text{Correct option : (C)}

\displaystyle \text{(v) After removing all odd-numbered cards, remaining cards are }2,4,6,8,10,12,14,16,18,20
\displaystyle \text{Only prime number among them is }2
\displaystyle \text{Required probability}=\frac{1}{10}
\displaystyle \text{Correct option : (B)}
\\

\displaystyle \textbf{PART B}
\displaystyle \textbf{SECTION III}
\displaystyle \text{All questions are compulsory. In case of internal choices, attempt any one.}

\displaystyle \textbf{Question 21(a). }\text{Check whether the points }P(5,-2),\ Q(6,4)\text{ and }R(7,-2)\text{ are the vertices of an}
\displaystyle \text{isosceles triangle PQR.}
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(6-5)^2+(4+2)^2}=\sqrt{1+36}=\sqrt{37}
\displaystyle QR=\sqrt{(7-6)^2+(-2-4)^2}=\sqrt{1+36}=\sqrt{37}
\displaystyle PR=\sqrt{(7-5)^2+(-2+2)^2}=\sqrt{4}=2
\displaystyle PQ=QR
\displaystyle \therefore PQR\text{ is an isosceles triangle.}

\textbf{OR}

\displaystyle \textbf{Question 21(b). }\text{Find the ratio in which }P(4,5)\text{ divides the join of }A(2,3)\text{ and }B(7,8).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(4,5)\text{ divide }AB\text{ in the ratio }m:n.
\displaystyle 4=\frac{7m+2n}{m+n}
\displaystyle 4m+4n=7m+2n
\displaystyle 2n=3m
\displaystyle \frac{m}{n}=\frac{2}{3}
\displaystyle \therefore \text{Required ratio}=2:3
\\

\displaystyle \textbf{Question 22(a). }\text{The sum of the numerator and the denominator of a fraction is }18.
\displaystyle \text{If the denominator is increased by }2,\text{ the fraction reduces to }\frac{1}{3}.\text{ Find the fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numerator be }x\text{ and denominator be }y.
\displaystyle x+y=18 \qquad (i)
\displaystyle \frac{x}{y+2}=\frac{1}{3}
\displaystyle 3x=y+2 \qquad (ii)
\displaystyle \text{From (i), }y=18-x
\displaystyle 3x=18-x+2
\displaystyle 4x=20
\displaystyle x=5
\displaystyle y=13
\displaystyle \therefore \text{Required fraction}=\frac{5}{13}

\textbf{OR}

\displaystyle \textbf{Question 22(b). }\text{Find the value of }k\text{ for which the system of equations }
\displaystyle x+2y=5\text{ and } 3x+ky+15=0\text{ has no solution.}
\displaystyle \text{Answer:}
\displaystyle x+2y-5=0,\qquad 3x+ky+15=0
\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}
\displaystyle \frac{1}{3}=\frac{2}{k}
\displaystyle k=6
\displaystyle \therefore k=6
\\

\displaystyle \textbf{Question 23. }\text{Explain why }2\times3\times5+5\text{ and }5\times7\times11+7\times5\text{ are composite numbers.}
\displaystyle \text{Answer:}
\displaystyle 2\times3\times5+5=5(2\times3+1)
\displaystyle =5(6+1)
\displaystyle =5\times7
\displaystyle =35
\displaystyle \therefore 2\times3\times5+5\text{ is a composite number.}
\displaystyle 5\times7\times11+7\times5=5\times7(11+1)
\displaystyle =5\times7\times12
\displaystyle =420
\displaystyle \therefore 5\times7\times11+7\times5\text{ is a composite number.}
\\

\displaystyle \textbf{Question 24. }\text{Find the mean of first }10\text{ composite numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{First }10\text{ composite numbers are }4,6,8,9,10,12,14,15,16,18
\displaystyle \text{Sum}=4+6+8+9+10+12+14+15+16+18
\displaystyle =112
\displaystyle \text{Mean}=\frac{112}{10}=11.2
\\

\displaystyle \textbf{Question 25. }\text{ABC is right triangle, right-angled at B, with BC}=6\text{ cm and} 
\displaystyle \text{AB}=8\text{ cm. A circle with }\text{centre O and radius }r\text{ cm has been inscribed in } \\ \triangle ABC\text{ as shown in the figure. Find the } \text{value of }r.  \displaystyle \text{Answer:}
\displaystyle AB=8\text{ cm},\qquad BC=6\text{ cm}
\displaystyle AC=\sqrt{AB^{2}+BC^{2}}
\displaystyle =\sqrt{8^{2}+6^{2}}
\displaystyle =\sqrt{64+36}
\displaystyle =10\text{ cm}
\displaystyle \text{Inradius of a right triangle}=\frac{a+b-c}{2}
\displaystyle r=\frac{8+6-10}{2}
\displaystyle =\frac{4}{2}
\displaystyle =2\text{ cm}
\\

\displaystyle \textbf{Question 26. }\text{Draw a circle of radius }5\text{ cm. From a point }8\text{ cm away from }
\displaystyle \text{its centre, construct a pair of tangents to the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw a circle with centre O and radius }5\text{ cm.}
\displaystyle \text{2. Mark a point P such that }OP=8\text{ cm.}
\displaystyle \text{3. Draw the perpendicular bisector of }OP\text{ and let it meet }OP\text{ at M.}
\displaystyle \text{4. With centre M and radius MO, draw a circle cutting the given circle at A and B.}
\displaystyle \text{5. Join PA and PB.}
\displaystyle \therefore PA\text{ and }PB\text{ are the required tangents.}
\\

\displaystyle \textbf{SECTION IV}

\displaystyle \textbf{Question 27. }\text{Divide the polynomial }f(x)=5x^{3}+10x^{2}-30x-15\text{ by the polynomial }
\displaystyle g(x)=x^{2}+1+x \text{and hence, find the quotient and the remainder.}
\displaystyle \text{Answer:}
\displaystyle g(x)=x^{2}+x+1
\displaystyle 5x^{3}+10x^{2}-30x-15=(x^{2}+x+1)(5x+5)+(-40x-20)
\displaystyle \therefore \text{Quotient}=5x+5
\displaystyle \therefore \text{Remainder}=-40x-20
\\

\displaystyle \textbf{Question 28. }\text{Prove that }3+\sqrt{2}\text{ is an irrational number, given that } \\ \sqrt{2}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }3+\sqrt{2}\text{ be a rational number.}
\displaystyle \text{Then }3+\sqrt{2}=r,\text{ where }r\text{ is rational.}
\displaystyle \sqrt{2}=r-3
\displaystyle \text{Since }r\text{ and }3\text{ are rational, }r-3\text{ is rational.}
\displaystyle \therefore \sqrt{2}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 3+\sqrt{2}\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 29. }\text{In the given figure, PT and PS are tangents to a circle with }
\displaystyle \text{centre O, from a point P, such that }PT=4\text{ cm and }\angle TPS=60^\circ.\text{ Find the }
\displaystyle \text{length of  the chord TS. Also, find the radius of circle.} \displaystyle \text{Answer:}
\displaystyle PT=PS=4\text{ cm}
\displaystyle \angle TPS=60^\circ
\displaystyle TS^{2}=PT^{2}+PS^{2}-2(PT)(PS)\cos60^\circ
\displaystyle =4^{2}+4^{2}-2(4)(4)\times\frac{1}{2}
\displaystyle =16+16-16=16
\displaystyle TS=4\text{ cm}
\displaystyle \angle TOS=180^\circ-\angle TPS=120^\circ
\displaystyle TS=2r\sin\frac{120^\circ}{2}
\displaystyle 4=2r\sin60^\circ
\displaystyle 4=2r\times\frac{\sqrt{3}}{2}
\displaystyle r=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\text{ cm}
\\

\displaystyle \textbf{Question 30. }\text{The areas of two similar triangles are }121\text{ cm}^{2}\text{ and }64\text{ cm}^{2}
\displaystyle \text{ respectively. If one median of } \text{the first triangle is }12.1\text{ cm long, then find the length }
\displaystyle \text{of the corresponding median of the other triangle.}
\displaystyle \text{Answer:}
\displaystyle \frac{\text{Area of first triangle}}{\text{Area of second triangle}}=\left(\frac{\text{Corresponding median of first triangle}}{\text{Corresponding median of second triangle}}\right)^{2}
\displaystyle \frac{121}{64}=\left(\frac{12.1}{x}\right)^{2}
\displaystyle \frac{11}{8}=\frac{12.1}{x}
\displaystyle x=\frac{12.1\times8}{11}
\displaystyle x=8.8\text{ cm}
\displaystyle \therefore \text{Corresponding median of the other triangle}=8.8\text{ cm}
\\

\displaystyle \textbf{Question 31(a). }\text{Prove :}
\displaystyle \frac{1}{(\cot A)(\sec A)-\cot A}-\mathrm{cosec}\ A=\mathrm{cosec}\ A-\frac{1}{(\cot A)(\sec A)+\cot A}
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\frac{1}{(\cot A)(\sec A)-\cot A}-\mathrm{cosec}\ A
\displaystyle =\frac{1}{\cot A(\sec A-1)}-\mathrm{cosec}\ A
\displaystyle =\frac{\tan A}{\sec A-1}-\mathrm{cosec}\ A
\displaystyle =\frac{\tan A(\sec A+1)}{(\sec A-1)(\sec A+1)}-\mathrm{cosec}\ A
\displaystyle =\frac{\tan A(\sec A+1)}{\sec^{2}A-1}-\mathrm{cosec}\ A
\displaystyle =\frac{\tan A(\sec A+1)}{\tan^{2}A}-\mathrm{cosec}\ A
\displaystyle =\frac{\sec A+1}{\tan A}-\mathrm{cosec}\ A
\displaystyle =\frac{\frac{1}{\cos A}+1}{\frac{\sin A}{\cos A}}-\frac{1}{\sin A}
\displaystyle =\frac{1+\cos A}{\sin A}-\frac{1}{\sin A}
\displaystyle =\frac{\cos A}{\sin A}
\displaystyle =\cot A
\displaystyle \text{R.H.S.}=\mathrm{cosec}\ A-\frac{1}{(\cot A)(\sec A)+\cot A}
\displaystyle =\mathrm{cosec}\ A-\frac{1}{\cot A(\sec A+1)}
\displaystyle =\mathrm{cosec}\ A-\frac{\tan A}{\sec A+1}
\displaystyle =\frac{1}{\sin A}-\frac{\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+1}
\displaystyle =\frac{1}{\sin A}-\frac{\sin A}{1+\cos A}
\displaystyle =\frac{1+\cos A-\sin^{2}A}{\sin A(1+\cos A)}
\displaystyle =\frac{1+\cos A-(1-\cos^{2}A)}{\sin A(1+\cos A)}
\displaystyle =\frac{\cos A+\cos^{2}A}{\sin A(1+\cos A)}
\displaystyle =\frac{\cos A(1+\cos A)}{\sin A(1+\cos A)}
\displaystyle =\cot A
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}

\textbf{OR}

\displaystyle \textbf{Question 31(b). }\text{Prove: }   \sin^{6}A+3\sin^{2}A\cos^{2}A=1-\cos^{6}A
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\sin^{6}A+3\sin^{2}A\cos^{2}A
\displaystyle =(1-\cos^{2}A)^{3}+3(1-\cos^{2}A)\cos^{2}A
\displaystyle =1-3\cos^{2}A+3\cos^{4}A-\cos^{6}A+3\cos^{2}A-3\cos^{4}A
\displaystyle =1-\cos^{6}A
\displaystyle =\text{R.H.S.}
\displaystyle \therefore \text{L.H.S.}=\text{R.H.S.}
\\

\displaystyle \textbf{Question 32(a). }\text{One root of the quadratic equation }2x^{2}-8x-k=0\text{ is }\frac{5}{2}.
\displaystyle \text{ Find the value of }k.\text{ Also, find the other root.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\frac{5}{2}\text{ is a root,}
\displaystyle 2\left(\frac{5}{2}\right)^{2}-8\left(\frac{5}{2}\right)-k=0
\displaystyle 2\cdot\frac{25}{4}-20-k=0
\displaystyle \frac{25}{2}-20-k=0
\displaystyle -\frac{15}{2}-k=0
\displaystyle k=-\frac{15}{2}
\displaystyle \text{Now, sum of roots}=-\frac{b}{a}=\frac{8}{2}=4
\displaystyle \frac{5}{2}+\text{other root}=4
\displaystyle \text{Other root}=4-\frac{5}{2}=\frac{3}{2}

\textbf{OR}

\displaystyle \textbf{Question 32(b). }\text{Using quadratic formula, solve the following equation for }x:
\displaystyle abx^{2}+(b^{2}-ac)x-bc=0
\displaystyle \text{Answer:}
\displaystyle x=\frac{-(b^{2}-ac)\pm\sqrt{(b^{2}-ac)^{2}-4(ab)(-bc)}}{2ab}
\displaystyle =\frac{ac-b^{2}\pm\sqrt{b^{4}+a^{2}c^{2}-2abc^{2}+4ab^{2}c}}{2ab}
\displaystyle =\frac{ac-b^{2}\pm\sqrt{(b^{2}+ac)^{2}}}{2ab}
\displaystyle =\frac{ac-b^{2}\pm(b^{2}+ac)}{2ab}
\displaystyle x=\frac{ac-b^{2}+b^{2}+ac}{2ab}\text{ or }x=\frac{ac-b^{2}-b^{2}-ac}{2ab}
\displaystyle x=\frac{2ac}{2ab}\text{ or }x=\frac{-2b^{2}}{2ab}
\displaystyle x=\frac{c}{b}\text{ or }x=-\frac{b}{a}
\\

\displaystyle \textbf{Question 33. }\text{With vertices A, B and C of a triangle ABC as centres, arcs are}
\displaystyle \text{drawn with radii }2\text{ cm} \text{ each as shown in the figure. If }AB=6\text{ cm, }BC=8\text{ cm and}
\displaystyle AC=10\text{ cm, then find the area} \text{ of the shaded region.} \displaystyle \text{Answer:}
\displaystyle AB^{2}+BC^{2}=6^{2}+8^{2}=36+64=100=AC^{2}
\displaystyle \therefore \triangle ABC\text{ is right-angled at B.}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times6\times8=24\text{ cm}^{2}
\displaystyle \text{Sum of angles of }\triangle ABC=180^\circ
\displaystyle \text{Area of three sectors}=\frac{180^\circ}{360^\circ}\times\pi r^{2}
\displaystyle =\frac{1}{2}\times\pi\times2^{2}
\displaystyle =2\pi\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=24-2\pi
\displaystyle =24-\frac{44}{7}
\displaystyle =\frac{124}{7}\text{ cm}^{2}
\\

\displaystyle \textbf{SECTION V}

\displaystyle \textbf{Question 34. }\text{Water is being pumped out through a circular pipe whose internal diameter is }8\text{ cm. If}
\displaystyle \text{the rate of flow of water is }80\text{ cm/s, then how many litres of water is being pumped out}
\displaystyle \text{through this pipe in one hour ?}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of pipe}=4\text{ cm}
\displaystyle \text{Area of cross-section}=\pi r^{2}
\displaystyle =\pi(4)^{2}=16\pi\text{ cm}^{2}
\displaystyle \text{Rate of flow}=80\text{ cm/s}
\displaystyle \text{Volume of water pumped in }1\text{ second}=16\pi\times80
\displaystyle =1280\pi\text{ cm}^{3}
\displaystyle \text{Volume of water pumped in }1\text{ hour}=1280\pi\times3600
\displaystyle =4608000\pi\text{ cm}^{3}
\displaystyle =4608000\times\frac{22}{7}\text{ cm}^{3}
\displaystyle =14482285.71\text{ cm}^{3}
\displaystyle =14482.29\text{ litres}
\\

\displaystyle \textbf{Question 35(a). }\text{A man on the top of a vertical tower observes a car moving at a uniform speed}
\displaystyle \text{coming directly towards it. If it takes }18\text{ minutes for the angle of depression to}
\displaystyle \text{change from }30^\circ\text{ to }60^\circ,\text{ how soon after this will the car reach the tower ?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of tower be }h\text{ and initial distance of car from tower be }x.
\displaystyle \tan30^\circ=\frac{h}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}
\displaystyle x=h\sqrt{3}
\displaystyle \text{Let the distance of car from tower after }18\text{ minutes be }y.
\displaystyle \tan60^\circ=\frac{h}{y}
\displaystyle \sqrt{3}=\frac{h}{y}
\displaystyle y=\frac{h}{\sqrt{3}}
\displaystyle \text{Distance covered in }18\text{ minutes}=x-y
\displaystyle =h\sqrt{3}-\frac{h}{\sqrt{3}}=\frac{2h}{\sqrt{3}}
\displaystyle \text{Remaining distance}=y=\frac{h}{\sqrt{3}}
\displaystyle \text{Time required}=\frac{\frac{h}{\sqrt{3}}}{\frac{2h}{\sqrt{3}}}\times18
\displaystyle =\frac{1}{2}\times18
\displaystyle =9\text{ minutes}

\textbf{OR}

\displaystyle \textbf{Question 35(b). }\text{A girl on a ship standing on a wooden platform, which is }50\text{ m above water level,}
\displaystyle \text{observes the angle of elevation of the top of a hill as }30^\circ\text{ and the angle of}
\displaystyle \text{depression of the base of the hill as }60^\circ.\text{ Calculate the distance of the hill from}
\displaystyle \text{the platform and the height of the hill.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the horizontal distance of the hill from the platform be }x\text{ m.}
\displaystyle \tan60^\circ=\frac{50}{x}
\displaystyle \sqrt{3}=\frac{50}{x}
\displaystyle x=\frac{50}{\sqrt{3}}\text{ m}
\displaystyle \text{Let the height of the hill be }h\text{ m.}
\displaystyle \tan30^\circ=\frac{h-50}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h-50}{\frac{50}{\sqrt{3}}}
\displaystyle h-50=\frac{50}{3}
\displaystyle h=\frac{200}{3}\text{ m}
\displaystyle \therefore \text{Distance of hill from platform}=\frac{50}{\sqrt{3}}\text{ m}=\frac{50\sqrt{3}}{3}\text{ m}
\displaystyle \therefore \text{Height of hill}=\frac{200}{3}\text{ m}
\\

\displaystyle \textbf{Question 36. }\text{If }S_n\text{ denotes the sum of first }n\text{ terms of an A.P., prove that }S_{12}=3(S_8-S_4).
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]
\displaystyle S_{12}=\frac{12}{2}[2a+11d]
\displaystyle =6(2a+11d)
\displaystyle =12a+66d
\displaystyle S_8=\frac{8}{2}[2a+7d]=4(2a+7d)=8a+28d
\displaystyle S_4=\frac{4}{2}[2a+3d]=2(2a+3d)=4a+6d
\displaystyle 3(S_8-S_4)=3[(8a+28d)-(4a+6d)]
\displaystyle =3(4a+22d)
\displaystyle =12a+66d
\displaystyle =S_{12}
\displaystyle \therefore S_{12}=3(S_8-S_4)
\\

 

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