\displaystyle \text{MATHEMATICS (STANDARD)}

\displaystyle \large \textbf{Series : C3ABD1 \hspace{2cm} SET~1} \hspace{2cm} \textbf{Q.P. Code : 30/1/1}
\displaystyle \text{Candidates must write the Q.P. Code on the title page of the answer-book.}

\displaystyle \text{NOTE}
\displaystyle \text{(I) Please check that this question paper contains 27 printed pages.}
\displaystyle \text{(II) Q.P. Code given on the right hand side of the question paper should be written}
\displaystyle \text{\hspace{1cm} on the title page of the answer-book by the candidate.}
\displaystyle \text{(III) Please check that this question paper contains 38 questions.}
\displaystyle \text{(IV) Please write down the Serial Number of the question in the answer-book}
\displaystyle \text{\hspace{1cm} at the given place before attempting it.}
\displaystyle \text{(V) 15 minute time has been allotted to read this question paper. The question paper}
\displaystyle \text{\hspace{1cm} will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the}
\displaystyle \text{\hspace{1cm} candidates will read the question paper only and will not write any}
\displaystyle \text{\hspace{1cm} answer on the answer-book during this period.}

\displaystyle \text{Time allowed : 3 hours \hspace{7.0cm} Maximum Marks : 80}

\displaystyle \text{General Instructions :}
\displaystyle \text{Read the following instructions very carefully and strictly follow them :}
\displaystyle \text{(i) This question paper contains 38 questions. All questions are compulsory.}
\displaystyle \text{(ii) This question paper is divided into five Sections - A, B, C, D and E.}
\displaystyle \text{(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and}
\displaystyle \text{\hspace{1cm} questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.}
\displaystyle \text{(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type questions, }
\displaystyle \text{\hspace{1cm} carrying 2 marks each.}
\displaystyle \text{(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions, carrying }
\displaystyle \text{\hspace{1cm} 3 marks each.}
\displaystyle \text{(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions, carrying}
\displaystyle \text{\hspace{1cm}  5 marks each.}
\displaystyle \text{(vii) In Section E, Questions no. 36 to 38 are case study based questions carrying 4 marks}
\displaystyle \text{\hspace{1cm}  each. Internal choice is provided in 2 marks questions in each case study.}
\displaystyle \text{(viii) There is no overall choice. However, an internal choice has been provided in 2}
\displaystyle \text{\hspace{1cm}  questions in Section B, 2 questions in Section C, 2 questions in Section D and 3}
\displaystyle \text{\hspace{1cm} questions in Section E.}
\displaystyle \text{(ix) Draw neat diagrams wherever required. Take } \pi = \frac{22}{7}  \text{ wherever required, if not stated.}
\displaystyle \text{(x) Use of calculator is not allowed.}

\displaystyle \text{SECTION A}
\displaystyle \text{This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.}
\displaystyle 20 \times 1 = 20

\displaystyle \textbf{1. }\text{If the sum of zeroes of the polynomial } p(x)=2x^{2}-k\sqrt{2}x+1\text{ is }\sqrt{2}, \text{then value of } \\ k\text{ is : } \\ \text{(a) }\sqrt{2}\quad \text{(b) }2\quad \text{(c) }2\sqrt{2}\quad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Given polynomial }p(x)=2x^2-k\sqrt{2}x+1
\displaystyle \text{Sum of zeroes }=\frac{-b}{a}=\frac{k\sqrt{2}}{2}
\displaystyle \text{Given sum }=\sqrt{2}
\displaystyle \frac{k\sqrt{2}}{2}=\sqrt{2}
\displaystyle k=2
\displaystyle \text{Answer: (b) }2
\displaystyle \\
\displaystyle \textbf{2. }\text{If the probability of a player winning a game is }0.79, \text{ then the probability of his} \\ \text{losing the same game is : } \\ \text{(a) }1.79\quad \text{(b) }0.31\quad \text{(c) }0.21\%\quad \text{(d) }0.21
\displaystyle \text{Answer:}
\displaystyle \text{Probability of winning }=0.79
\displaystyle \text{Total probability }=1
\displaystyle \text{Probability of losing }=1-0.79=0.21
\displaystyle \text{Answer: (d) }0.21
\displaystyle \\
\displaystyle \textbf{3. }\text{If the roots of equation }ax^{2}+bx+c=0, a\neq 0\text{ are real and equal, then which of the} \\ \text{following relation is true ? } \\ \text{(a) }a=\frac{b^{2}}{c}\quad \text{(b) }b^{2}=ac\quad \text{(c) }ac=\frac{b^{2}}{4}\quad \text{(d) }c=\frac{b^{2}}{a}
\displaystyle \text{Answer:}
\displaystyle  \text{For quadratic }ax^2+bx+c=0,\ \text{roots are real and equal when }D=0
\displaystyle \text{Discriminant }D=b^2-4ac=0
\displaystyle \therefore\ b^2=4ac
\displaystyle \Rightarrow ac=\frac{b^2}{4}
\displaystyle \text{Answer: (c) }ac=\frac{b^2}{4}
\displaystyle \\
\displaystyle \textbf{4. } \text{In an A.P., if the first term } a = 7,\ \text{the } n^{\text{th}} \text{ term } a_n = 84 \text{ and the sum of first }n \text{ terms} \\  S_n = \frac{2093}{2},\ \text{then } n \text{ is equal to : } \\ \text{(a) }22\quad \text{(b) }24\quad \text{(c) }23\quad \text{(d) }26
\displaystyle \text{Answer:}
\displaystyle  \text{Given }a=7,\ a_n=84,\ S_n=\frac{2093}{2}
\displaystyle \text{Sum of }n\text{ terms: }S_n=\frac{n}{2}(a+a_n)
\displaystyle \frac{2093}{2}=\frac{n}{2}(7+84)
\displaystyle \frac{2093}{2}=\frac{n}{2}\cdot 91
\displaystyle 2093=91n
\displaystyle n=\frac{2093}{91}=23
\displaystyle \text{Answer: (c) }23

\displaystyle \textbf{5. }\text{If two positive integers }p\text{ and }q \text{ can be expressed as }p=18a^{2}b^{4}\text{ and } \\ q=20a^{3}b^{2},\text{ where }a\text{ and }b\text{ are prime numbers, then LCM }(p,q)\text{ is : } \\ \text{(a) }2a^{2}b^{2}\quad \text{(b) }180a^{2}b^{2}\quad \text{(c) }12a^{2}b^{2}\quad \text{(d) }180a^{3}b^{4}
\displaystyle \\
\displaystyle \text{Answer:}
\displaystyle  \text{Given }p=18a^{2}b^{4},\ q=20a^{3}b^{2}
\displaystyle 18=2\times 3^2,\quad 20=2^2\times 5
\displaystyle \text{LCM of }18\text{ and }20=2^2\times 3^2\times 5=180
\displaystyle \text{For }a:\ \max(2,3)=3\Rightarrow a^3
\displaystyle \text{For }b:\ \max(4,2)=4\Rightarrow b^4
\displaystyle \text{LCM}(p,q)=180a^3b^4
\displaystyle \text{Answer: (d) }180a^3b^4
\displaystyle \\
\displaystyle \textbf{6. }\text{AD is a median of }\triangle ABC\text{ with vertices }A(5,-6),B(6,4)\text{ and }C(0,0). \text{ Length AD} \\ \text{is equal to : } \\ \text{(a) }\sqrt{68}\text{ units}\quad \text{(b) }2\sqrt{15}\text{ units}\quad \text{(c) }\sqrt{101}\text{ units}\quad \text{(d) }10\text{ units}
\displaystyle \text{Answer:}
\displaystyle  \text{Since }AD\text{ is a median, }D\text{ is the mid-point of }BC
\displaystyle B(6,4),\ C(0,0)
\displaystyle D=\left(\frac{6+0}{2},\frac{4+0}{2}\right)=(3,2)
\displaystyle A(5,-6)
\displaystyle AD=\sqrt{(3-5)^2+(2-(-6))^2}
\displaystyle =\sqrt{(-2)^2+8^2}
\displaystyle =\sqrt{4+64}
\displaystyle =\sqrt{68}
\displaystyle \text{Answer: (a) }\sqrt{68}\text{ units}
\displaystyle \\
\displaystyle \textbf{7. }\text{If }\sec \theta-\tan \theta=m,\text{ then the value of }\sec \theta+\tan \theta\text{ is : } \\ \text{(a) }1-\frac{1}{m}\quad \text{(b) }m^{2}-1\quad \text{(c) }\frac{1}{m}\quad \text{(d) }-m
\displaystyle \text{Answer:}
\displaystyle  \text{Given }\sec\theta-\tan\theta=m
\displaystyle \text{Using identity: }(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^2\theta-\tan^2\theta
\displaystyle =1
\displaystyle m(\sec\theta+\tan\theta)=1
\displaystyle \sec\theta+\tan\theta=\frac{1}{m}
\displaystyle \text{Answer: (c) }\frac{1}{m}
\displaystyle \\
\displaystyle \textbf{8. }\text{From the data }1,4,7,9,16,21,25,\text{ if all the even numbers are removed, then the} \\ \text{probability of getting at random a prime number from the remaining is : } \\ \text{(a) }\frac{2}{5}\quad \text{(b) }\frac{1}{5}\quad \text{(c) }\frac{1}{7}\quad \text{(d) }\frac{2}{7}
\displaystyle \text{Answer:}
\displaystyle  \text{Given data: }1,4,7,9,16,21,25
\displaystyle \text{Remove even numbers: }1,7,9,21,25
\displaystyle \text{Remaining numbers }=5
\displaystyle \text{Prime numbers among them: }7
\displaystyle \text{Number of favourable outcomes }=1
\displaystyle \text{Probability}=\frac{1}{5}
\displaystyle \text{Answer: (b) }\frac{1}{5}
\displaystyle \\
\displaystyle \textbf{9. }\text{For some data }x_{1},x_{2},\ldots ,x_{n} \text{ with respective frequencies }f_{1},f_{2},\ldots ,f_{n},\text{ the value of} \\ \sum_{i=1}^{n}f_{i}\left(x_{i}-\overline{x}\right)\text{ is equal to : } \\ \text{(a) }n\overline{x}\quad \text{(b) }1\quad \text{(c) }\sum f_{i}\quad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle  \text{We know that } \overline{x}=\frac{\sum f_i x_i}{\sum f_i}
\displaystyle \sum f_i(x_i-\overline{x})=\sum f_i x_i-\overline{x}\sum f_i
\displaystyle =\sum f_i x_i-\frac{\sum f_i x_i}{\sum f_i}\cdot \sum f_i
\displaystyle =\sum f_i x_i-\sum f_i x_i=0
\displaystyle \text{Answer: (d) }0
\displaystyle \\
\displaystyle \textbf{10. }\text{The zeroes of a polynomial }x^{2}+px+q \text{ are twice the zeroes of the polynomial} \\  4x^{2}-5x-6.\text{ The value of }p \text{ is : } \\ \text{(a) }-\frac{5}{2}\quad \text{(b) }\frac{5}{2}\quad \text{(c) }-5\quad \text{(d) }10
\displaystyle \text{Answer:}
\displaystyle  \text{Let the zeroes of }4x^2-5x-6=0\text{ be }\alpha,\beta
\displaystyle \text{Then the zeroes of }x^2+px+q\text{ are }2\alpha,\ 2\beta
\displaystyle \text{Sum of zeroes of }4x^2-5x-6=0\text{ is }\alpha+\beta=\frac{5}{4}
\displaystyle \text{Sum of zeroes of }x^2+px+q\text{ is }-(p)=2\alpha+2\beta=2(\alpha+\beta)
\displaystyle -p=2\times \frac{5}{4}=\frac{5}{2}
\displaystyle p=-\frac{5}{2}
\displaystyle \text{Answer: (a) }-\frac{5}{2}
\displaystyle \\
\displaystyle \textbf{11. }\text{If the distance between the points }(3,-5) \text{ and }(x,-5)\text{ is }15\text{ units, then the values of} \\ x\text{ are : } \\ \text{(a) }12,-18\quad \text{(b) }-12,18\quad \text{(c) }18,5\quad \text{(d) }-9,-12
\displaystyle \text{Answer:}
\displaystyle \text{Points }(3,-5)\text{ and }(x,-5)\text{ have same y-coordinate}
\displaystyle \text{Distance }=|x-3|=15
\displaystyle x-3=15\ \text{or}\ x-3=-15
\displaystyle x=18\ \text{or}\ x=-12
\displaystyle \text{Answer: (b) }-12,18
\displaystyle \\
\displaystyle \textbf{12. }\text{If }\cos (\alpha+\beta)=0,\text{ then value of }\cos \left(\frac{\alpha+\beta}{2}\right)\text{ is equal to : } \\ \text{(a) }\frac{1}{\sqrt{2}}\quad \text{(b) }\frac{1}{2}\quad \text{(c) }0\quad \text{(d) }\sqrt{2}
\displaystyle \text{Answer:}
\displaystyle  \text{Given }\cos(\alpha+\beta)=0
\displaystyle \therefore\ \alpha+\beta=90^\circ\ (\text{principal value})
\displaystyle \frac{\alpha+\beta}{2}=45^\circ
\displaystyle \cos\left(\frac{\alpha+\beta}{2}\right)=\cos 45^\circ=\frac{1}{\sqrt{2}}
\displaystyle \text{Answer: (a) }\frac{1}{\sqrt{2}}
\displaystyle \\
\displaystyle \textbf{13. }\text{A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere} \\ \text{to that of two hemispheres taken together, is : } \\ \text{(a) }1:1\quad \text{(b) }1:4\quad \text{(c) }2:3\quad \text{(d) }3:2
\displaystyle \text{Answer:}
\displaystyle  \text{Surface area of sphere }=4\pi r^2
\displaystyle \text{Surface area of one hemisphere }=2\pi r^2+\pi r^2=3\pi r^2
\displaystyle \text{So, for two hemispheres }=2\times 3\pi r^2=6\pi r^2
\displaystyle \text{Required ratio }=4\pi r^2:6\pi r^2=2:3
\displaystyle \text{Answer: (c) }2:3
\displaystyle \\
\displaystyle \textbf{14. }\text{The middle most observation of every data arranged in order is called : } \\  \text{(a) mode}\quad \text{(b) median}\quad \text{(c) mean}\quad \text{(d) deviation}
\displaystyle \text{Answer:}
\displaystyle  \text{The middle most observation in ordered data is called median}
\displaystyle \text{Answer: (b) median}
\displaystyle \\
\displaystyle \textbf{15. }\text{The volume of the largest right circular cone that can be carved out from a solid} \\ \text{cube of edge }2\text{ cm is : } \\ \text{(a) }\frac{4\pi}{3}\text{ cu cm}\quad \text{(b) }\frac{5\pi}{3}\text{ cu cm}\quad \text{(c) }\frac{8\pi}{3}\text{ cu cm}\quad \text{(d) }\frac{2\pi}{3}\text{ cu cm}
\displaystyle \text{Answer:}
\displaystyle \text{Edge of cube }=2\text{ cm}
\displaystyle \text{Largest cone: base is a circle inscribed in a face of cube}
\displaystyle \text{So, diameter }=2\Rightarrow r=1\text{ cm}
\displaystyle \text{Height of cone }=2\text{ cm}
\displaystyle \text{Volume }=\frac{1}{3}\pi r^2 h
\displaystyle =\frac{1}{3}\pi (1)^2(2)=\frac{2\pi}{3}
\displaystyle \text{Answer: (d) }\frac{2\pi}{3}\text{ cu cm}
\displaystyle \\
\displaystyle \textbf{16. }\text{Two dice are rolled together. The probability of getting sum of numbers on the two} \\ \text{dice as }2,3\text{ or }5\text{ is : }  \\ \text{(a) }\frac{7}{36}\quad \text{(b) }\frac{11}{36}\quad \text{(c) }\frac{5}{36}\quad \text{(d) }\frac{4}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes when two dice are rolled }=36
\displaystyle \text{Sum }=2:\ (1,1)\Rightarrow 1\text{ outcome}
\displaystyle \text{Sum }=3:\ (1,2),(2,1)\Rightarrow 2\text{ outcomes}
\displaystyle \text{Sum }=5:\ (1,4),(2,3),(3,2),(4,1)\Rightarrow 4\text{ outcomes}
\displaystyle \text{Total favourable outcomes }=1+2+4=7
\displaystyle \text{Probability}=\frac{7}{36}
\displaystyle \text{Answer: (a) }\frac{7}{36}
\displaystyle \\
\displaystyle \textbf{17. }\text{The centre of a circle is at }(2,0).\text{ If one end of a diameter is at }(6,0), \text{ then the} \\ \text{other end is at : } \\ \text{(a) }(0,0)\quad \text{(b) }(4,0)\quad \text{(c) }(-2,0)\quad \text{(d) }(-6,0)
\displaystyle \text{Answer:}
\displaystyle  \text{Centre }=(2,0),\ \text{one end of diameter }A=(6,0)
\displaystyle \text{Let the other end be }B=(x,y)
\displaystyle \text{Centre is midpoint of diameter: }\left(\frac{6+x}{2},\frac{0+y}{2}\right)=(2,0)
\displaystyle \frac{6+x}{2}=2\Rightarrow 6+x=4\Rightarrow x=-2
\displaystyle \frac{y}{2}=0\Rightarrow y=0
\displaystyle \therefore\ B=(-2,0)
\displaystyle \text{Answer: (c) }(-2,0)
\displaystyle \\

\displaystyle \textbf{18. }\text{In the given figure, graphs of two linear equations are shown. The pair of these} \\ \text{linear equations is : }
\displaystyle \text{(a) consistent with unique solution. }
\displaystyle \text{(b) consistent with infinitely many solutions. }
\displaystyle \text{(c) inconsistent. }
\displaystyle \text{(d) inconsistent but can be made consistent} \\ \text{by extending these lines. }
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, the two lines are parallel (same slope, different intercepts)}
\displaystyle \text{Parallel lines do not intersect}
\displaystyle \therefore\ \text{no solution exists}
\displaystyle \text{Hence, the system is inconsistent}
\displaystyle \text{Answer: (c) inconsistent}
\displaystyle \\
\displaystyle \textbf{Directions : }
\displaystyle \text{In Q. No. }19\text{ and }20\text{, a statement of Assertion (A) is followed by a statement of} \\ \text{Reason (R). Choose the correct option.}
\displaystyle \text{(a) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation} \\ \text{of Assertion (A).}
\displaystyle \text{(b) Both, Assertion (A) and Reason (R) are true but Reason (R) is not correct} \\ \text{explanation for Assertion (A).}
\displaystyle \text{(c) Assertion (A) is true but Reason (R) is false.}
\displaystyle \text{(d) Assertion (A) is false but Reason (R) is true.}
\displaystyle \\
\displaystyle \textbf{19. }\textbf{Assertion (A) : }\text{The tangents drawn at the end points of a diameter of a circle,} \\ \text{are parallel.}
\displaystyle \textbf{Reason (R) : }\text{Diameter of a circle is the longest chord.}
\displaystyle \text{Answer:}
\displaystyle  \text{Assertion: Tangents at endpoints of a diameter are parallel (true)}
\displaystyle \text{Reason: Diameter is the longest chord (true)}
\displaystyle \text{But R does not explain A}
\displaystyle \text{Answer: (b)}
\displaystyle \\
\displaystyle \textbf{20. }\textbf{Assertion (A) : }\text{If the graph of a polynomial touches }x\text{-axis at only one point,} \\ \text{then the polynomial cannot be a quadratic polynomial.}
\displaystyle \textbf{Reason (R) : }\text{A polynomial of degree }n(n>1) \text{ can have at most }n\text{ zeroes.}
\displaystyle \text{Answer:}
\displaystyle  \text{Assertion: If graph touches x-axis at one point, it can still be quadratic (false)}
\displaystyle \text{(since quadratic can have equal roots)}
\displaystyle \text{Reason: A polynomial of degree }n\text{ has at most }n\text{ zeroes (true)}
\displaystyle \text{Answer: (d)}

\displaystyle \text{SECTION B}
\displaystyle \text{This section has 5 Very Short Answer (VSA) type questions carrying 2 marks each. }
\displaystyle 5 \times 2 = 10

\displaystyle \textbf{21. }\text{Solve the following system of linear equations }7x-2y=5\text{ and } \\ 8x+7y=15\text{ and verify your answer.}
\displaystyle \text{Answer:}
\displaystyle  \text{Given equations: }7x-2y=5\ \text{...(1)},\ 8x+7y=15\ \text{...(2)}
\displaystyle \text{Multiply (1) by }7:\ 49x-14y=35
\displaystyle \text{Multiply (2) by }2:\ 16x+14y=30
\displaystyle \text{Add: }65x=65\Rightarrow x=1
\displaystyle \text{Substitute in (1): }7(1)-2y=5
\displaystyle 7-2y=5\Rightarrow 2y=2\Rightarrow y=1
\displaystyle \text{Verification: }7(1)-2(1)=5,\quad 8(1)+7(1)=15\ (\text{true})
\displaystyle \text{Answer: }x=1,\ y=1
\displaystyle \\
\displaystyle \textbf{22. }\text{In a pack of }52\text{ playing cards one card is lost. From the remaining cards, a card} \\ \text{is drawn at random. Find the probability that the drawn card is queen of hearts, if} \\ \text{the lost card is a black card.}
\displaystyle \text{Answer:}
\displaystyle  \text{Total cards }=52
\displaystyle \text{One card is lost and it is a black card}
\displaystyle \text{So remaining cards }=51
\displaystyle \text{Queen of hearts is a red card, so it is definitely present}
\displaystyle \text{Favourable outcomes }=1
\displaystyle \text{Probability}=\frac{1}{51}
\displaystyle \text{Answer: }\frac{1}{51}
\displaystyle \\
\displaystyle \textbf{23. (A) }\text{Evaluate : }2\sqrt{2}\cos 45^{\circ} \sin 30^{\circ}+2\sqrt{3}\cos 30^{\circ}
\displaystyle \text{OR}
\displaystyle \textbf{(B) }\text{If A = }60^{\circ}\text{ and B = }30^{\circ} \text{, verify that : }\sin (A+B)=\sin A\cos B+\cos A\sin B
\displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle 2\sqrt{2}\cos 45^\circ \sin 30^\circ+2\sqrt{3}\cos 30^\circ
\displaystyle =2\sqrt{2}\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{2}+2\sqrt{3}\cdot \frac{\sqrt{3}}{2}
\displaystyle =2\cdot \frac{1}{2}+2\cdot \frac{3}{2}
\displaystyle =1+3=4
\displaystyle \text{Answer: }4
\displaystyle \text{OR} 
\displaystyle \text{(B)}
\displaystyle \text{LHS}=\sin(A+B)=\sin(60^\circ+30^\circ)=\sin 90^\circ=1
\displaystyle \text{RHS}=\sin A\cos B+\cos A\sin B
\displaystyle =\sin 60^\circ\cos 30^\circ+\cos 60^\circ\sin 30^\circ
\displaystyle =\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{1}{2}
\displaystyle =\frac{3}{4}+\frac{1}{4}=1
\displaystyle \text{LHS}=\text{RHS}
\displaystyle \text{Hence verified.}

\displaystyle \textbf{24. }\text{In the given figure, }ABCD\text{ is a quadrilateral. Diagonal }BD \text{ bisects }\angle B \\ \text{and }\angle D \text{ both. Prove that :}
\displaystyle \text{(i) }\triangle ABD\sim \triangle CBD
\displaystyle \text{(ii) }AB=BC

\displaystyle \text{Answer:}

\displaystyle \text{Given, diagonal }BD\text{ bisects }\angle B\text{ and }\angle D\text{ both}
\displaystyle \therefore\ \angle ABD=\angle DBC\text{ and }\angle ADB=\angle CDB
\displaystyle \text{(i) } \text{To prove } \triangle ABD\sim \triangle CBD
\displaystyle \text{In }\triangle ABD\text{ and }\triangle CBD,
\displaystyle \angle ABD=\angle DBC
\displaystyle \angle ADB=\angle CDB
\displaystyle \therefore\ \triangle ABD\sim \triangle CBD\text{ (by AA similarity)}
\displaystyle \text{(ii) } \text{ To prove }AB=BC
\displaystyle \text{From }\triangle ABD\sim \triangle CBD,\ \frac{AB}{BC}=\frac{BD}{BD}
\displaystyle \frac{AB}{BC}=1
\displaystyle \therefore\ AB=BC
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \textbf{25. (A) }\text{Prove that }5-2\sqrt{3}\text{ is an irrational number. It is given that }\sqrt{3} \text{ is an} \\ \text{irrational number.}
\displaystyle \textbf{OR}
\displaystyle \textbf{(B) }\text{Show that the number }5\times 11\times 17 +3\times 11\text{ is a composite number.}
\displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle \text{Let us assume that }5-2\sqrt{3}\text{ is a rational number.}
\displaystyle \text{Then }2\sqrt{3}=5-(5-2\sqrt{3})
\displaystyle \text{Since }5\text{ is rational and }(5-2\sqrt{3})\text{ is assumed rational,}
\displaystyle \text{their difference }2\sqrt{3}\text{ is rational.}
\displaystyle \text{Dividing by }2,\ \sqrt{3}\text{ is rational.}
\displaystyle \text{But this contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore\ 5-2\sqrt{3}\text{ is an irrational number.}
\displaystyle \text{OR}
\displaystyle \text{(B)}
\displaystyle 5\times 11\times 17+3\times 11
\displaystyle =11(5\times 17+3)
\displaystyle =11(85+3)
\displaystyle =11\times 88
\displaystyle =968
\displaystyle \text{Since }968=11\times 88,\text{ it has factors other than }1\text{ and itself.}
\displaystyle \therefore\ 5\times 11\times 17+3\times 11\text{ is a composite number.}

\displaystyle \text{SECTION C}
\displaystyle \text{This section has 6 Short Answer (SA) type questions carrying 3 marks each.}
\displaystyle 6 \times 3 = 20

\displaystyle \textbf{26. (A) }\text{Find the ratio in which the point } \left(\frac{8}{5},y\right) \text{ divides the line segment} \\ \text{joining the points } (1,2)\text{ and }(2,3).\text{ Also, find the value of }y.
\displaystyle \text{OR}
\displaystyle \textbf{(B) }\text{ABCD is a rectangle formed by the points } A(-1,-1),B(-1,6),C(3,6) \text{ and} \\ D(3,-1). \text{ P, Q, R and S are mid-points of } \text{sides }AB,BC,CD\text{ and }DA \text{ respectively.} \\ \text{Show that diagonals of the } \text{quadrilateral }PQRS\text{ bisect each other.}
\displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle \text{Let }P\left(\frac{8}{5},y\right)\text{ divide the line segment joining }A(1,2)\text{ and }B(2,3)\text{ in the ratio }m:n
\displaystyle \text{Using section formula, }x\text{-coordinate of }P=\frac{m(2)+n(1)}{m+n}
\displaystyle \frac{2m+n}{m+n}=\frac{8}{5}
\displaystyle 5(2m+n)=8(m+n)
\displaystyle 10m+5n=8m+8n
\displaystyle 2m=3n
\displaystyle \frac{m}{n}=\frac{3}{2}
\displaystyle \therefore\ \text{the ratio is }3:2
\displaystyle \text{Now, }y=\frac{m(3)+n(2)}{m+n}=\frac{3(3)+2(2)}{3+2}=\frac{9+4}{5}=\frac{13}{5}
\displaystyle \text{Answer: Ratio }=3:2,\quad y=\frac{13}{5}
\displaystyle \text{OR} 
\displaystyle \text{(B)}
\displaystyle A(-1,-1),\ B(-1,6),\ C(3,6),\ D(3,-1)
\displaystyle \text{Since }P,Q,R,S\text{ are mid-points of }AB,BC,CD,DA\text{ respectively,}
\displaystyle P=\left(\frac{-1+(-1)}{2},\frac{-1+6}{2}\right)=\left(-1,\frac{5}{2}\right)
\displaystyle Q=\left(\frac{-1+3}{2},\frac{6+6}{2}\right)=(1,6)
\displaystyle R=\left(\frac{3+3}{2},\frac{6+(-1)}{2}\right)=\left(3,\frac{5}{2}\right)
\displaystyle S=\left(\frac{3+(-1)}{2},\frac{-1+(-1)}{2}\right)=(1,-1)
\displaystyle \text{Diagonals of }PQRS\text{ are }PR\text{ and }QS
\displaystyle \text{Mid-point of }PR=\left(\frac{-1+3}{2},\frac{\frac{5}{2}+\frac{5}{2}}{2}\right)=\left(1,\frac{5}{2}\right)
\displaystyle \text{Mid-point of }QS=\left(\frac{1+1}{2},\frac{6+(-1)}{2}\right)=\left(1,\frac{5}{2}\right)
\displaystyle \text{Since the mid-points of }PR\text{ and }QS\text{ are the same,}
\displaystyle \text{the diagonals of quadrilateral }PQRS\text{ bisect each other.}
\displaystyle \text{Hence proved.}
\displaystyle \\ 
\displaystyle \textbf{27. }\text{In a teachers' workshop, the number of } \text{teachers teaching French, Hindi and} \\ \text{English are }48,80\text{ and } 144\text{ respectively. Find the minimum number of rooms required} \\ \text{if in each room } \text{the same number of teachers are seated and all of them are of the} \\ \text{same } \text{subject.}
\displaystyle \text{Answer:}
\displaystyle  \text{Teachers in French }=48,\ \text{Hindi }=80,\ \text{English }=144
\displaystyle \text{To minimize rooms, each room should have maximum equal number of teachers}
\displaystyle \text{So, required number per room }=\gcd(48,80,144)
\displaystyle \gcd(48,80)=16,\ \gcd(16,144)=16
\displaystyle \text{Teachers per room }=16
\displaystyle \text{Rooms for French }=\frac{48}{16}=3
\displaystyle \text{Rooms for Hindi }=\frac{80}{16}=5
\displaystyle \text{Rooms for English }=\frac{144}{16}=9
\displaystyle \text{Total rooms }=3+5+9=17
\displaystyle \text{Answer: }17
\displaystyle \\ 
\displaystyle \textbf{28. }\text{Prove that : } \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} =1+\sec \theta\ \mathrm{cosec}\,\theta
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}
\displaystyle =\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\frac{\cos\theta}{\sin\theta}}{1-\frac{\sin\theta}{\cos\theta}}
\displaystyle =\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\sin\theta-\cos\theta}{\sin\theta}}+\frac{\frac{\cos\theta}{\sin\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}
\displaystyle =\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}+\frac{\cos^2\theta}{\sin\theta(\cos\theta-\sin\theta)}
\displaystyle =\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}
\displaystyle =\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}
\displaystyle =\frac{(\sin\theta-\cos\theta)(\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta)}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}
\displaystyle =\frac{\sin^2\theta+\sin\theta\cos\theta+\cos^2\theta}{\sin\theta\cos\theta}
\displaystyle =\frac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}
\displaystyle =1+\frac{1}{\sin\theta\cos\theta}
\displaystyle =1+\sec\theta\ \mathrm{cosec}\,\theta
\displaystyle \text{Hence, proved.}
\displaystyle \\ 
\displaystyle \textbf{29. }\text{Three years ago, Rashmi was thrice as old } \text{as Nazma. Ten years later, Rashmi} \\ \text{will be twice as old as Nazma. How old are } \text{Rashmi and Nazma now ?}
\displaystyle \text{Answer:}
\displaystyle  \text{Let present ages of Rashmi and Nazma be }R\text{ and }N\text{ respectively}
\displaystyle \text{Three years ago: }R-3=3(N-3)
\displaystyle R-3=3N-9\Rightarrow R=3N-6\ \text{...(1)}
\displaystyle \text{Ten years later: }R+10=2(N+10)
\displaystyle R+10=2N+20\Rightarrow R=2N+10\ \text{...(2)}
\displaystyle \text{From (1) and (2): }3N-6=2N+10
\displaystyle N=16
\displaystyle R=2N+10=2(16)+10=42
\displaystyle \text{Answer: Rashmi }=42\text{ years, Nazma }=16\text{ years}
\displaystyle \\ 
\displaystyle \textbf{30. (A) }\text{In the given figure, }AB\text{ is a } \text{diameter of the circle with centre O. }AQ,BP \\ \text{and } PQ\text{ are } \text{tangents to the circle. Prove that }\angle POQ=90^{\circ}.


\displaystyle \textbf{OR}
\displaystyle \textbf{(B) }\text{A circle with centre O and radius }8 \text{ cm is inscribed in a quadrilateral } ABCD \\ \text{in which }P,Q,R,S\text{ are } \text{the points of contact as shown. If }  AD\text{ is perpendicular to } \\ DC\text{, } BC=30\text{ cm and }BS=24\text{ cm, then find the length }DC.

\displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle \text{Given, }AB\text{ is a diameter of the circle with centre }O
\displaystyle \text{Also, }AQ,\ BP\text{ and }PQ\text{ are tangents to the circle}
\displaystyle \text{Since tangent at }A\text{ is perpendicular to radius }OA,\ \angle OAQ=90^\circ
\displaystyle \text{Similarly, tangent at }B\text{ is perpendicular to radius }OB,\ \angle OBP=90^\circ
\displaystyle \text{Now }AQ\parallel BP\text{ and }AB\text{ is perpendicular to both}
\displaystyle \text{Let }PQ\text{ touch the circle at }R
\displaystyle \text{Then }OR\perp PQ
\displaystyle \text{Thus, }OA\perp AQ,\ OB\perp BP,\ OR\perp PQ
\displaystyle \text{The centre }O\text{ is equidistant from the three tangents }AQ,\ BP,\ PQ
\displaystyle \therefore\ O\text{ is the incenter of }\triangle \text{ formed by the three tangents}
\displaystyle \text{Hence, }OQ\text{ bisects the angle between }AQ\text{ and }PQ
\displaystyle \text{and }OP\text{ bisects the angle between }BP\text{ and }PQ
\displaystyle \text{Since }AQ\parallel BP,\ \angle AQP+\angle QPB=180^\circ
\displaystyle \therefore\ \frac{1}{2}\angle AQP+\frac{1}{2}\angle QPB=90^\circ
\displaystyle \text{But }\frac{1}{2}\angle AQP=\angle OQP\text{ and }\frac{1}{2}\angle QPB=\angle OPQ
\displaystyle \therefore\ \angle OQP+\angle OPQ=90^\circ
\displaystyle \text{In }\triangle POQ,\ \angle POQ=180^\circ-90^\circ=90^\circ
\displaystyle \therefore\ \angle POQ=90^\circ
\displaystyle \text{Hence proved.}
\displaystyle \text{OR} 
\displaystyle \text{(B)}
\displaystyle \text{Given, a circle of radius }8\text{ cm is inscribed in quadrilateral }ABCD
\displaystyle \text{Points of contact are }P,Q,R,S
\displaystyle AD\perp DC,\ BC=30\text{ cm and }BS=24\text{ cm}
\displaystyle \text{Tangents from an external point are equal}
\displaystyle \therefore\ BS=BR=24\text{ cm}
\displaystyle \text{Since }BC=30\text{ cm, }RC=BC-BR=30-24=6\text{ cm}
\displaystyle \text{Tangents from }C\text{ are equal, so }CQ=CR=6\text{ cm}
\displaystyle \text{Now }AD\perp DC\text{ and the circle touches both }AD\text{ and }DC
\displaystyle \text{Hence the distances from }D\text{ to the points of contact on }AD\text{ and }DC\text{ are each equal to the radius}
\displaystyle \therefore\ DQ=8\text{ cm}
\displaystyle \text{Thus, }DC=DQ+QC=8+6=14\text{ cm}
\displaystyle \text{Answer: }DC=14\text{ cm}
\displaystyle \\ 
\displaystyle \textbf{31. }\text{The difference between the outer and inner } \text{radii of a hollow right circular} \\ \text{cylinder of length }14\text{ cm is }1 \text{ cm. If the volume of the metal used in making the} \\ \text{cylinder is }176 \text{ cm}^{3}\text{, find the outer and inner radii of the cylinder.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let outer radius }=R\text{ cm and inner radius }=r\text{ cm}
\displaystyle \text{Given }R-r=1\ \text{...(1)},\ \text{height }h=14\text{ cm}
\displaystyle \text{Volume of metal }=\pi h(R^2-r^2)=176
\displaystyle \pi\cdot 14(R^2-r^2)=176
\displaystyle \frac{22}{7}\cdot 14(R^2-r^2)=176
\displaystyle 44(R^2-r^2)=176
\displaystyle R^2-r^2=4
\displaystyle (R-r)(R+r)=4
\displaystyle 1\cdot (R+r)=4\Rightarrow R+r=4\ \text{...(2)}
\displaystyle \text{Solving (1) and (2): }R= \frac{(R+r)+(R-r)}{2}=\frac{4+1}{2}=\frac{5}{2}
\displaystyle r=R-1=\frac{5}{2}-1=\frac{3}{2}
\displaystyle \text{Answer: Outer radius }=\frac{5}{2}\text{ cm, Inner radius }=\frac{3}{2}\text{ cm}

\displaystyle \text{SECTION D}
\displaystyle \text{This section has 4 Long Answer (LA) type questions carrying 5 marks each.}
\displaystyle 4 \times 5 = 20

\displaystyle \textbf{32. }\text{An arc of a circle of radius }21\text{ cm } \text{subtends an angle of }60^{\circ}\text{ at the centre. Find :}
\displaystyle \text{(i) the length of the arc.}
\displaystyle \text{(ii) the area of the minor segment of the circle made by } \text{the corresponding chord.}
\displaystyle \text{Answer:}
\displaystyle  \text{Radius }r=21\text{ cm},\ \theta=60^\circ
\displaystyle \text{(i) } \text{Length of arc }=\frac{\theta}{360^\circ}\times 2\pi r
\displaystyle =\frac{60}{360}\times 2\pi \times 21
\displaystyle =\frac{1}{6}\times 42\pi=7\pi
\displaystyle =22\text{ cm (using }\pi=\frac{22}{7}\text{)}
\displaystyle \text{Answer: }22\text{ cm}
\displaystyle \text{(ii) } \text{Area of minor segment }=\text{Area of sector }-\text{Area of triangle}
\displaystyle \text{Area of sector }=\frac{60}{360}\times \pi r^2=\frac{1}{6}\times \pi \times 441=\frac{441\pi}{6}
\displaystyle =231\text{ cm}^2
\displaystyle \text{Area of }\triangle =\frac{1}{2}r^2\sin 60^\circ=\frac{1}{2}\times 441\times \frac{\sqrt{3}}{2}=\frac{441\sqrt{3}}{4}
\displaystyle \approx 190.96\text{ cm}^2
\displaystyle \text{Area of segment }=231-190.96\approx 40.04\text{ cm}^2
\displaystyle \text{Answer: }\approx 40.04\text{ cm}^2
\displaystyle \\ 
\displaystyle \textbf{33. (A) }\text{The sum of first and eighth terms of an } \text{A.P. is }32\text{ and their product is }60. \\ \text{Find the first term and } \text{common difference of the A.P. Hence, also find the sum of its} \\ \text{first} 20\text{ terms.}
\displaystyle \textbf{OR}
\displaystyle \textbf{(B) }\text{In an A.P. of }40\text{ terms, the sum of } \text{first }9\text{ terms is }153\text{ and the sum of last }6 \\ \text{terms is } 687. \text{ Determine the first term and common difference of A.P. Also, find the} \\ \text{sum of all the terms of the A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d
\displaystyle \text{Then eighth term }=a+7d
\displaystyle \text{Given }a+(a+7d)=32
\displaystyle 2a+7d=32\ \text{...(1)}
\displaystyle \text{Also, }a(a+7d)=60\ \text{...(2)}
\displaystyle \text{From (1), }a+7d=32-a
\displaystyle \text{Substitute in (2): }a(32-a)=60
\displaystyle 32a-a^2=60
\displaystyle a^2-32a+60=0
\displaystyle a^2-30a-2a+60=0
\displaystyle a(a-30)-2(a-30)=0
\displaystyle (a-30)(a-2)=0
\displaystyle \therefore\ a=30\text{ or }a=2
\displaystyle \text{If }a=30,\ \text{then from (1), }60+7d=32\Rightarrow 7d=-28\Rightarrow d=-4
\displaystyle \text{If }a=2,\ \text{then from (1), }4+7d=32\Rightarrow 7d=28\Rightarrow d=4
\displaystyle \text{Thus, }(a,d)=(30,-4)\text{ or }(2,4)
\displaystyle \text{Now, }S_{20}=\frac{20}{2}[2a+19d]=10[2a+19d]
\displaystyle \text{For }a=30,\ d=-4:\ S_{20}=10[60-76]=10(-16)=-160
\displaystyle \text{For }a=2,\ d=4:\ S_{20}=10[4+76]=10(80)=800
\displaystyle \text{Answer: }a=30,\ d=-4,\ S_{20}=-160\ \text{or }a=2,\ d=4,\ S_{20}=800
\displaystyle \textbf{OR}
\displaystyle \text{(B)}
\displaystyle \text{Let first term }=a,\ \text{common difference }=d
\displaystyle \text{Given }S_9=153
\displaystyle \frac{9}{2}[2a+8d]=153
\displaystyle 2a+8d=34
\displaystyle a+4d=17\ \text{...(1)}
\displaystyle \text{Last 6 terms are }a_{35},a_{36},a_{37},a_{38},a_{39},a_{40}
\displaystyle a_{35}=a+34d,\quad a_{40}=a+39d
\displaystyle \text{Sum of last 6 terms }=\frac{6}{2}[(a+34d)+(a+39d)]=687
\displaystyle 3(2a+73d)=687
\displaystyle 2a+73d=229\ \text{...(2)}
\displaystyle \text{Subtract (1) }\times 2\text{ from (2):}
\displaystyle (2a+73d)-(2a+8d)=229-34
\displaystyle 65d=195
\displaystyle d=3
\displaystyle \text{From (1), }a+4(3)=17
\displaystyle a=5
\displaystyle \text{Now, }S_{40}=\frac{40}{2}[2a+39d]
\displaystyle =20[2(5)+39(3)]
\displaystyle =20[10+117]=20(127)=2540
\displaystyle \text{Answer: }a=5,\ d=3,\ S_{40}=2540
\displaystyle \\ 
\displaystyle \textbf{34. (A) }\text{If a line is drawn parallel to one side } \text{of a triangle to intersect the other two} \\ \text{sides in distinct points, then } \text{prove that the other two sides are divided in the same} \\ \text{ratio.}
\displaystyle \text{OR}
\displaystyle \textbf{(B) }\text{In the given figure }PA,QB\text{ and }RC \text{are each perpendicular to }AC. \\ \text{ If }AP=x,\ BQ=y\text{ and }CR=z,\text{ } \text{then prove that }\frac{1}{x}+\frac{1}{z}=\frac{1}{y} \displaystyle \text{Answer:}
\displaystyle \text{(A)}
\displaystyle \text{Given }\triangle ABC,\text{ let a line through }D\text{ on }AB\text{ and }E\text{ on }AC\text{ be drawn such that }DE\parallel BC
\displaystyle \text{We have to prove that }\frac{AD}{DB}=\frac{AE}{EC}
\displaystyle \text{Since }DE\parallel BC,\ \angle ADE=\angle ABC,\ \angle AED=\angle ACB,\ \angle A\text{ is common}
\displaystyle \therefore\ \triangle ADE\sim \triangle ABC
\displaystyle \text{Hence, }\frac{AD}{AB}=\frac{AE}{AC}\ \text{...(1)}
\displaystyle \text{But }AB=AD+DB\text{ and }AC=AE+EC
\displaystyle \text{So, from (1), }\frac{AD}{AD+DB}=\frac{AE}{AE+EC}
\displaystyle AD(AE+EC)=AE(AD+DB)
\displaystyle ADAE+ADEC=ADAE+AEDB
\displaystyle ADEC=AEDB
\displaystyle \therefore\ \frac{AD}{DB}=\frac{AE}{EC}
\displaystyle \text{Hence, the other two sides are divided in the same ratio.}
\displaystyle \textbf{OR}
\displaystyle \text{(B)}
\displaystyle \text{Given }PA\perp AC,\ QB\perp AC,\ RC\perp AC,\ AP=x,\ BQ=y,\ CR=z
\displaystyle \text{We have to prove that }\frac{1}{x}+\frac{1}{z}=\frac{1}{y}
\displaystyle \text{Consider }\triangle AQP\text{ and }\triangle CQR
\displaystyle \angle PAQ=\angle QCR=90^\circ
\displaystyle \angle PQA=\angle CQR\text{ (vertically opposite angles)}
\displaystyle \therefore\ \triangle AQP\sim \triangle CQR
\displaystyle \text{Hence, }\frac{AP}{CR}=\frac{AQ}{CQ}=\frac{x}{z}\ \text{...(1)}
\displaystyle \text{Now consider }\triangle AQB\text{ and }\triangle CQR
\displaystyle \angle ABQ=\angle QCR=90^\circ
\displaystyle \angle AQB=\angle CQR\text{ (vertically opposite angles)}
\displaystyle \therefore\ \triangle AQB\sim \triangle CQR
\displaystyle \text{Hence, }\frac{BQ}{CR}=\frac{AB}{BC}=\frac{y}{z}
\displaystyle \Rightarrow \frac{AB}{BC}=\frac{y}{z}\ \text{...(2)}
\displaystyle \text{Also, from }\triangle AQP\sim \triangle CQR,
\displaystyle \frac{AP}{BQ}=\frac{AQ}{CQ}=\frac{x}{y}
\displaystyle \text{Using corresponding bases on }AC,\ \frac{AB}{BC}=\frac{x}{y}\ \text{...(3)}
\displaystyle \text{From the figure, by similarity we get }AB:BC=x:y=y:z
\displaystyle \therefore\ AB=\frac{xy}{x+y},\ BC=\frac{yz}{y+z}\text{ and hence }y=\frac{xz}{x+z}
\displaystyle \text{So, }\frac{1}{y}=\frac{x+z}{xz}=\frac{1}{x}+\frac{1}{z}
\displaystyle \therefore\ \frac{1}{x}+\frac{1}{z}=\frac{1}{y}
\displaystyle \text{Hence proved.}
\displaystyle \\ 
\displaystyle \textbf{35. }\text{A pole }6\text{ m high is fixed on the top} \text{of a tower. The angle of elevation of the top} \\ \text{of the pole observed from a point } \text{P on the ground is }60^{\circ}\text{ and the angle of depression} \\ \text{of the point P from the top of the tower is }45^{\circ}.\text{ Find the height of the tower and the} \\ \text{distance of point P from the foot of the tower. } (\text{Use }\sqrt{3}=1.73)
\displaystyle \text{Answer:}
\displaystyle \text{Let height of tower }=h\text{ m and distance of point P from foot }=x\text{ m}
\displaystyle \text{Angle of depression }45^\circ=\text{angle of elevation from P to top of tower}
\displaystyle \therefore\ \tan 45^\circ=\frac{h}{x}=1\Rightarrow h=x\ \text{...(1)}
\displaystyle \text{Total height (tower + pole)}=h+6
\displaystyle \text{Angle of elevation to top of pole }=60^\circ
\displaystyle \therefore\ \tan 60^\circ=\frac{h+6}{x}=\sqrt{3}
\displaystyle \frac{h+6}{x}=1.73\ \text{...(2)}
\displaystyle \text{Substitute }x=h\text{ from (1) into (2):}
\displaystyle \frac{h+6}{h}=1.73
\displaystyle h+6=1.73h
\displaystyle 6=0.73h
\displaystyle h=\frac{6}{0.73}\approx 8.22\text{ m}
\displaystyle x=h\approx 8.22\text{ m}
\displaystyle \text{Answer: Height of tower }\approx 8.22\text{ m, Distance from P }\approx 8.22\text{ m}

\displaystyle \text{SECTION E}
\displaystyle \text{This section has 3 case study based questions carrying 4 marks each.}
\displaystyle 3 \times 4 = 12

\displaystyle \textbf{36. }\text{A rectangular floor area can be completely } \text{tiled with }200\text{ square tiles. If the side} \\ \text{length of each tile is } \text{increased by }1\text{ unit, it would take only }128\text{ tiles to cover the floor} \displaystyle \text{(i) Assuming the original length of each side of a tile } \text{be }x\text{ units, make a quadratic} \\ \text{equation from the above information.}
\displaystyle \text{(ii) Write the corresponding quadratic equation in } \text{standard form.}
\displaystyle \text{(iii) (a) Find the value of }x\text{, the length of side } \text{of a tile by factorisation.}
\displaystyle \text{OR}
\displaystyle \text{(b) Solve the quadratic equation for }x\text{, using } \text{quadratic formula.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the original side of each tile be }x\text{ units}
\displaystyle \text{Area of floor }=200x^2
\displaystyle \text{New side }=x+1,\ \text{new area }=128(x+1)^2
\displaystyle \textbf{(i) }200x^2=128(x+1)^2
\displaystyle \textbf{(ii) }200x^2=128(x^2+2x+1)
\displaystyle 200x^2=128x^2+256x+128
\displaystyle 72x^2-256x-128=0
\displaystyle 9x^2-32x-16=0
\displaystyle \text{Standard form: }9x^2-32x-16=0
\displaystyle \textbf{(iii)(a) }9x^2-32x-16=0
\displaystyle 9x^2-36x+4x-16=0
\displaystyle 9x(x-4)+4(x-4)=0
\displaystyle (9x+4)(x-4)=0
\displaystyle x=4\ \text{or }x=-\frac{4}{9}
\displaystyle \text{Reject negative value}
\displaystyle \therefore\ x=4
\displaystyle \text{Answer: Side of tile }=4\text{ units}
\displaystyle \text{OR} \\ \text{(iii)(b)}
\displaystyle x=\frac{32\pm\sqrt{(-32)^2-4(9)(-16)}}{2\cdot 9}
\displaystyle =\frac{32\pm\sqrt{1024+576}}{18}
\displaystyle =\frac{32\pm\sqrt{1600}}{18}
\displaystyle =\frac{32\pm 40}{18}
\displaystyle x=4\ \text{or }-\frac{4}{9}
\displaystyle \text{Hence, }x=4
\displaystyle \\
\displaystyle \textbf{37. }\text{BINGO is a game of chance. The host has } 75\text{ balls numbered }1\text{ through }75.\text{ Each} \\ \text{player has a BINGO card } \text{with some numbers written on it. The participant cancels the } \\ \text{number on the } \text{card when called out a number written on the ball selected at random.} \\ \text{Whosoever cancels all the numbers on his/her card, says BINGO and wins the } \text{game.}


\displaystyle \text{The table given below shows the data of one such game } \text{where }48\text{ balls were used before} \\ \text{Tara said 'BINGO'.}
\displaystyle \begin{array}{|c|c|}\hline \text{Numbers announced} & \text{Number of times} \\ \hline 0-15 & 8 \\ \hline 15-30 & 9 \\ \hline 30-45 & 10 \\ \hline 45-60 & 12 \\ \hline 60-75 & 9 \\ \hline \end{array}
\displaystyle \text{Based on the above information, answer the following : }
\displaystyle \text{(i) Write the median class.}
\displaystyle \text{(ii) When first ball was picked up, what was the } \text{probability of calling out an even} \\ \text{number ?}
\displaystyle \text{(iii) (a) Find median of the given data.}
\displaystyle \textbf{OR}
\displaystyle \text{(b) Find mode of the given data.}
\displaystyle \text{Answer:}
\displaystyle  \text{Total frequency }N=48
\displaystyle \textbf{(i) }\text{Cumulative frequencies are }8,\ 17,\ 27,\ 39,\ 48
\displaystyle \frac{N}{2}=\frac{48}{2}=24
\displaystyle \text{The class whose cumulative frequency first exceeds }24\text{ is }30-45
\displaystyle \text{Hence, median class is }30-45
\displaystyle \textbf{(ii) }\text{At first pick, total balls }=75
\displaystyle \text{Even numbers from }1\text{ to }75\text{ are }2,4,6,\ldots,74
\displaystyle \text{Number of even numbers }=\frac{74}{2}=37
\displaystyle \text{Probability of an even number }=\frac{37}{75}
\displaystyle \textbf{(iii)(a) }\text{For median: }l=30,\ h=15,\ f=10,\ cf=17
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle =30+\frac{24-17}{10}\times 15
\displaystyle =30+\frac{7}{10}\times 15
\displaystyle =30+10.5=40.5
\displaystyle \text{Hence, median }=40.5
\displaystyle \text{OR}
\displaystyle \textbf{(iii)(b) }\text{Modal class is }45-60\text{ since it has highest frequency }12
\displaystyle l=45,\ h=15,\ f_1=12,\ f_0=10,\ f_2=9
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =45+\frac{12-10}{24-10-9}\times 15
\displaystyle =45+\frac{2}{5}\times 15
\displaystyle =45+6=51
\displaystyle \text{Hence, mode }=51
\displaystyle \\
\displaystyle \textbf{38. }\text{A backyard is in the shape of a triangle } ABC\text{ with right angle at }B. \\ AB=7\text{ m and } BC=15\text{ m. A circular pit } \text{was dug inside it such that it touches} \\ \text{the walls } AC,\ BC\text{ and }AB\text{ } \text{at }P,\ Q\text{ and }R\text{ respectively such that }AP=x\text{ m.}


\displaystyle \text{Based on the above information, answer the following } \text{questions : }
\displaystyle \text{(i) Find the length of }AR\text{ in terms of }x.
\displaystyle \text{(ii) Write the type of quadrilateral }BQOR.
\displaystyle \text{(iii) (a) Find the length }PC\text{ in terms of }x\text{ } \text{and hence find the value of }x.
\displaystyle \text{OR}
\displaystyle \text{(b) Find }x\text{ and hence find the radius }r\text{ of } \text{circle.}
\displaystyle \text{Answer:}
\displaystyle  \text{In }\triangle ABC,\ \angle B=90^\circ,\ AB=7\text{ m},\ BC=15\text{ m}
\displaystyle \text{The circle touches }AC,\ BC,\ AB\text{ at }P,\ Q,\ R\text{ respectively and }AP=x\text{ m}
\displaystyle \textbf{(i) }\text{From external point }A,\ \text{tangents are equal}
\displaystyle \therefore\ AP=AR
\displaystyle \therefore\ AR=x\text{ m}
\displaystyle \textbf{(ii) }\text{At }Q\text{ and }R,\ OQ\perp BC\text{ and }OR\perp AB
\displaystyle \text{Since }AB\perp BC,\ \angle QBR=90^\circ
\displaystyle \text{Also, }BQ\parallel BC,\ BR\parallel AB,\ OQ\parallel AB,\ OR\parallel BC
\displaystyle \therefore\ BQOR\text{ has all right angles}
\displaystyle \text{And }BQ=BR\text{ (tangents from }B),\ OQ=OR\text{ (radii)}
\displaystyle \therefore\ BQOR\text{ is a square}
\displaystyle \textbf{(iii)(a) }\text{From }B,\ \text{tangents are equal, so }BR=BQ
\displaystyle AB=AR+RB
\displaystyle 7=x+BR
\displaystyle BR=7-x
\displaystyle \therefore\ BQ=7-x
\displaystyle \text{Now }BC=BQ+QC
\displaystyle 15=(7-x)+QC
\displaystyle QC=8+x
\displaystyle \text{From }C,\ \text{tangents are equal, so }CP=CQ=8+x
\displaystyle \therefore\ PC=8+x\text{ m}
\displaystyle \text{Also, }AC=AP+PC=x+(8+x)=2x+8
\displaystyle \text{By Pythagoras theorem, }AC=\sqrt{AB^2+BC^2}=\sqrt{7^2+15^2}=\sqrt{274}
\displaystyle 2x+8=\sqrt{274}
\displaystyle x=\frac{\sqrt{274}-8}{2}\text{ m}
\displaystyle \text{OR}
\displaystyle \textbf{(iii)(b) }\text{Since }x=\frac{\sqrt{274}-8}{2},\ \text{radius }r=BQ=BR=7-x
\displaystyle r=7-\frac{\sqrt{274}-8}{2}=\frac{22-\sqrt{274}}{2}\text{ m}
\displaystyle \text{Answer: }AR=x,\ BQOR\text{ is a square, }PC=8+x,\ x=\frac{\sqrt{274}-8}{2}\text{ m},\ r=\frac{22-\sqrt{274}}{2}\text{ m}

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

Subscribe to get the latest posts sent to your email.