\displaystyle \text{MATHEMATICS (STANDARD)}

\displaystyle \large \textbf{Series : GE1FH \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 }\alpha \text{ and } \beta \text{ are the zeroes of polynomial } 3x^2+6x+k \text{ such that } \\ \alpha+\beta+\alpha\beta=-\frac{2}{3},\text{ then the value of } k \text{ is :}
\displaystyle \text{(A) }-8
\displaystyle \text{(B) }8
\displaystyle \text{(C) }-4
\displaystyle \text{(D) }4
\displaystyle \text{Answer:}
\displaystyle \text{Given polynomial }3x^2+6x+k,\ \text{we have}
\displaystyle \alpha+\beta=\frac{-6}{3}=-2,\quad \alpha\beta=\frac{k}{3}
\displaystyle \text{Now, }\alpha+\beta+\alpha\beta=-\frac{2}{3}
\displaystyle -2+\frac{k}{3}=-\frac{2}{3}
\displaystyle \frac{k}{3}=-\frac{2}{3}+2=\frac{4}{3}
\displaystyle k=4
\displaystyle \therefore\ \text{the value of }k\text{ is }4.
\displaystyle \text{Answer: (D) }4
\displaystyle \\
\displaystyle \textbf{2. } \text{If } x = 1 \text{ and } y = 2 \text{ is a solution of the pair of linear equations }
\displaystyle 2x - 3y + a = 0 \text{ and } 2x + 3y - b = 0, \text{ then :}
\displaystyle \text{(A) } a = 2b
\displaystyle \text{(B) } 2a = b
\displaystyle \text{(C) } a + 2b = 0
\displaystyle \text{(D) } 2a + b = 0
\displaystyle \text{Answer:}
\displaystyle \text{Given }x=1,\ y=2\text{ is a solution}
\displaystyle \text{Substitute in }2x-3y+a=0
\displaystyle 2(1)-3(2)+a=0
\displaystyle 2-6+a=0
\displaystyle a=4
\displaystyle \text{Substitute in }2x+3y-b=0
\displaystyle 2(1)+3(2)-b=0
\displaystyle 2+6-b=0
\displaystyle b=8
\displaystyle \text{Check options: }2a=2\times 4=8=b
\displaystyle \text{Answer: (B) }2a=b
\displaystyle \\
\displaystyle \textbf{3. } \text{The mid-point of the line segment joining the points } P(-4, 5) \text{ and } Q(4, 6)
\displaystyle \text{lies on :}
\displaystyle \text{(A) x-axis}
\displaystyle \text{(B) y-axis}
\displaystyle \text{(C) origin}
\displaystyle \text{(D) neither x-axis nor y-axis}
\displaystyle \text{Answer:}
\displaystyle \text{Given points }P(-4,5)\text{ and }Q(4,6)
\displaystyle \text{Mid-point}=\left(\frac{-4+4}{2},\frac{5+6}{2}\right)
\displaystyle =\left(0,\frac{11}{2}\right)
\displaystyle \text{Since }x=0,\text{ the point lies on the y-axis.}
\displaystyle \text{Answer: (B) y-axis}
\displaystyle \\
\displaystyle \textbf{4. } \text{If } \theta \text{ is an acute angle and } 7 + 4 \sin \theta = 9, \text{ then the value of } \theta \text{ is :}
\displaystyle \text{(A) } 90^\circ
\displaystyle \text{(B) } 30^\circ
\displaystyle \text{(C) } 45^\circ
\displaystyle \text{(D) } 60^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Given }7+4\sin\theta=9
\displaystyle 4\sin\theta=9-7=2
\displaystyle \sin\theta=\frac{2}{4}=\frac{1}{2}
\displaystyle \text{Since }\theta\text{ is acute and }\sin\theta=\frac{1}{2},\ \theta=30^\circ
\displaystyle \text{Answer: (B) }30^\circ
\displaystyle \\
\displaystyle \textbf{5. }\text{The value of } \tan^2\theta - \left(\frac{1}{\cos\theta}\times \sec\theta\right)\text{ is :}
\displaystyle \text{(A) }1
\displaystyle \text{(B) }0
\displaystyle \text{(C) }-1
\displaystyle \text{(D) }2
\displaystyle \text{Answer:}
\displaystyle \text{Given expression }=\tan^2\theta-\left(\frac{1}{\cos\theta}\times \sec\theta\right)
\displaystyle \text{Since }\sec\theta=\frac{1}{\cos\theta}
\displaystyle \frac{1}{\cos\theta}\times \sec\theta=\frac{1}{\cos\theta}\times \frac{1}{\cos\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta
\displaystyle \text{So, expression }=\tan^2\theta-\sec^2\theta
\displaystyle \text{Using identity }\sec^2\theta=1+\tan^2\theta
\displaystyle \tan^2\theta-(1+\tan^2\theta)=-1
\displaystyle \text{Answer: (C) }-1
\displaystyle \\
\displaystyle \textbf{6. }\text{If HCF}(98,28)=m \text{ and LCM}(98,28)=n,\text{ then the value of } n-7m \text{ is :}
\displaystyle \text{(A) }0
\displaystyle \text{(B) }28
\displaystyle \text{(C) }98
\displaystyle \text{(D) }198
\displaystyle \text{Answer:}
\displaystyle \text{Given numbers }98\text{ and }28
\displaystyle \text{HCF}(98,28)=14=m
\displaystyle \text{Using } \text{LCM}\times \text{HCF}=98\times 28
\displaystyle n\times 14=98\times 28
\displaystyle n=\frac{98\times 28}{14}=196
\displaystyle n-7m=196-7\times 14=196-98=98
\displaystyle \text{Answer: (C) }98
\displaystyle \\
\displaystyle \textbf{7. }\text{The tangents drawn at the extremities of the diameter of a circle are always :}
\displaystyle \text{(A) parallel}
\displaystyle \text{(B) perpendicular}
\displaystyle \text{(C) equal}
\displaystyle \text{(D) intersecting}
\displaystyle \text{Answer:}
\displaystyle \text{Tangents at the extremities of a diameter are perpendicular to the radii at those points}
\displaystyle \text{Since the radii are collinear (same straight line), the tangents are perpendicular to the same line}
\displaystyle \text{Hence, the tangents are parallel to each other}
\displaystyle \text{Answer: (A) parallel}
\displaystyle \\
\displaystyle \textbf{8. }\text{In triangles ABC and DEF, }\angle B=\angle E,\ \angle F=\angle C \text{ and } AB=3DE. \\ \text{Then, the two triangles are :}
\displaystyle \text{(A) congruent but not similar}
\displaystyle \text{(B) congruent as well as similar}
\displaystyle \text{(C) neither congruent nor similar}
\displaystyle \text{(D) similar but not congruent}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\angle B=\angle E\text{ and }\angle C=\angle F
\displaystyle \text{Thus, two pairs of corresponding angles are equal}
\displaystyle \text{So, }\triangle ABC\sim \triangle DEF\ \text{(by AA similarity)}
\displaystyle \text{Also, }AB=3DE\Rightarrow \text{sides are not equal}
\displaystyle \text{Hence, triangles are not congruent}
\displaystyle \text{Answer: (D) similar but not congruent}
\displaystyle \\
\displaystyle \textbf{9. }\text{If }(-1)^n+(-1)^8=0,\text{ then } n \text{ is :}
\displaystyle \text{(A) any positive integer}
\displaystyle \text{(B) any negative integer}
\displaystyle \text{(C) any odd number}
\displaystyle \text{(D) any even number}
\displaystyle \text{Answer:}
\displaystyle \text{Given }(-1)^n+(-1)^8=0
\displaystyle (-1)^8=1
\displaystyle \Rightarrow (-1)^n+1=0
\displaystyle (-1)^n=-1
\displaystyle \text{This is true when }n\text{ is odd}
\displaystyle \text{Answer: (C) any odd number}

\displaystyle \textbf{10. }\text{Two polynomials are shown in the graph below. The number of distinct zeroes} \\ \text{of both the polynomials is :} \displaystyle \text{(A) }3
\displaystyle \text{(B) }5
\displaystyle \text{(C) }2
\displaystyle \text{(D) }4
\displaystyle \text{Answer:}
\displaystyle \text{Zeroes are the x-coordinates where graphs intersect the x-axis}
\displaystyle \text{From the graph, both polynomials intersect the x-axis at the same two points}
\displaystyle \text{Hence, number of distinct zeroes }=2
\displaystyle \text{Answer: (C) }2
\displaystyle \\
\displaystyle \textbf{11. }\text{If the sum of first } m \text{ terms of an AP is } 2m^2+3m,\text{ then its second term is :}
\displaystyle \text{(A) }10
\displaystyle \text{(B) }9
\displaystyle \text{(C) }12
\displaystyle \text{(D) }4
\displaystyle \text{Answer:}
\displaystyle \text{Given }S_m=2m^2+3m
\displaystyle \text{The }m\text{th term }a_m=S_m-S_{m-1}
\displaystyle S_{m-1}=2(m-1)^2+3(m-1)
\displaystyle =2(m^2-2m+1)+3m-3
\displaystyle =2m^2-4m+2+3m-3=2m^2-m-1
\displaystyle a_m=(2m^2+3m)-(2m^2-m-1)=4m+1
\displaystyle \text{Second term }a_2=4(2)+1=9
\displaystyle \text{Answer: (B) }9
\displaystyle \\
\displaystyle \textbf{12. }\text{Mode and Mean of a data are }15x \text{ and }18x,\text{ respectively. Then the median} \\ \text{of the data is :}
\displaystyle \text{(A) }x
\displaystyle \text{(B) }11x
\displaystyle \text{(C) }17x
\displaystyle \text{(D) }34x
\displaystyle \text{Answer:}
\displaystyle \text{Given Mode }=15x,\ \text{Mean }=18x
\displaystyle \text{Using relation: Mode }=3\text{Median}-2\text{Mean}
\displaystyle 15x=3\text{Median}-2(18x)
\displaystyle 15x=3\text{Median}-36x
\displaystyle 3\text{Median}=15x+36x=51x
\displaystyle \text{Median}=\frac{51x}{3}=17x
\displaystyle \text{Answer: (C) }17x
\displaystyle \\
\displaystyle \textbf{13. }\text{A card is selected at random from a deck of 52 playing cards. The probability} \\ \text{of it being a red face card is :}
\displaystyle \text{(A) }\frac{3}{13}
\displaystyle \text{(B) }\frac{2}{13}
\displaystyle \text{(C) }\frac{1}{2}
\displaystyle \text{(D) }\frac{3}{26}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards }=52
\displaystyle \text{Red suits are Hearts and Diamonds}
\displaystyle \text{Each suit has 3 face cards (J, Q, K)}
\displaystyle \text{So, red face cards }=2\times 3=6
\displaystyle \text{Probability}=\frac{6}{52}=\frac{3}{26}
\displaystyle \text{Answer: (D) }\frac{3}{26}
\displaystyle \\
\displaystyle \textbf{14. }\text{Which of the following is a rational number between } \sqrt{3} \text{ and } \sqrt{5}\text{ ?}
\displaystyle \text{(A) }1.4142387954012\ldots
\displaystyle \text{(B) }2.32\overline{6}
\displaystyle \text{(C) }\pi
\displaystyle \text{(D) }1.857142
\displaystyle \text{Answer:}
\displaystyle \sqrt{3}\approx 1.732,\quad \sqrt{5}\approx 2.236
\displaystyle \text{Check options:}
\displaystyle \text{(A) }1.414238\ldots <1.732\ \text{(not in range)}
\displaystyle \text{(B) }2.32\overline{6}=2.3266\ldots >2.236\ \text{(not in range)}
\displaystyle \text{(C) }\pi\approx 3.14\ \text{(not in range and irrational)}
\displaystyle \text{(D) }1.857142\ \text{is between }1.732\text{ and }2.236\ \text{and is rational}
\displaystyle \text{Answer: (D) }1.857142
\displaystyle \\
\displaystyle \textbf{15. }\text{If a sector of a circle has an area of }40\pi \text{ sq. units and a central angle of }72^\circ, \\ \text{ the radius of the circle is :}
\displaystyle \text{(A) }200\ \text{units}
\displaystyle \text{(B) }100\ \text{units}
\displaystyle \text{(C) }20\ \text{units}
\displaystyle \text{(D) }10\sqrt{2}\ \text{units}
\displaystyle \text{Answer:}
\displaystyle \text{Area of sector}=\frac{\theta}{360^\circ}\pi r^2
\displaystyle 40\pi=\frac{72}{360}\pi r^2
\displaystyle 40=\frac{1}{5}r^2
\displaystyle r^2=200
\displaystyle r=\sqrt{200}=10\sqrt{2}
\displaystyle \text{Answer: (D) }10\sqrt{2}\ \text{units}
\displaystyle \\
\displaystyle \textbf{16. }\text{In the given figure, PA is a tangent from an external point P to a circle with} \\ \text{centre O. If } \angle POB=115^\circ,\text{ then } \angle APO \text{ is equal to :}

\displaystyle \text{(A) }25^\circ
\displaystyle \text{(B) }65^\circ
\displaystyle \text{(C) }90^\circ
\displaystyle \text{(D) }35^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Given }PA\text{ is tangent at }A\Rightarrow OA\perp PA
\displaystyle \therefore\ \angle OAP=90^\circ
\displaystyle \text{Also, }A,O,B\text{ are collinear }\Rightarrow \angle POB=115^\circ=\angle POA
\displaystyle \text{In }\triangle POA,
\displaystyle \angle APO=180^\circ-(\angle POA+\angle OAP)
\displaystyle =180^\circ-(115^\circ+90^\circ)=180^\circ-205^\circ=-25^\circ
\displaystyle \text{But angle cannot be negative, so take smaller angle at }O
\displaystyle \angle POA=180^\circ-115^\circ=65^\circ
\displaystyle \angle APO=180^\circ-(65^\circ+90^\circ)=25^\circ
\displaystyle \text{Answer: (A) }25^\circ

\displaystyle \textbf{17. }\text{A kite is flying at a height of }150\ \text{m from the ground. It is attached to} \\ \text{a string inclined at an angle of }30^\circ \text{ to the horizontal. The length of the string is :}
\displaystyle \text{(A) }100\sqrt{3}\ \text{m}
\displaystyle \text{(B) }300\ \text{m}
\displaystyle \text{(C) }150\sqrt{2}\ \text{m}
\displaystyle \text{(D) }150\sqrt{3}\ \text{m}
\displaystyle \text{Answer:}
\displaystyle \text{Height }=150\text{ m, angle with horizontal }=30^\circ
\displaystyle \text{Let length of string }=L
\displaystyle \sin 30^\circ=\frac{\text{height}}{\text{length}}=\frac{150}{L}
\displaystyle \frac{1}{2}=\frac{150}{L}
\displaystyle L=300\text{ m}
\displaystyle \text{Answer: (B) }300\text{ m}
\displaystyle \\
\displaystyle \textbf{18. }\text{A piece of wire }20\ \text{cm long is bent into the form of an arc of a circle} \\ \text{of radius } \frac{60}{\pi}\ \text{cm. The angle subtended by the arc at the centre of the circle is :}
\displaystyle \text{(A) }30^\circ
\displaystyle \text{(B) }60^\circ
\displaystyle \text{(C) }90^\circ
\displaystyle \text{(D) }50^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Length of arc }l=20\text{ cm},\ \text{radius }r=\frac{60}{\pi}\text{ cm}
\displaystyle \text{Using }l=r\theta\ (\theta\text{ in radians})
\displaystyle 20=\frac{60}{\pi}\theta
\displaystyle \theta=\frac{20\pi}{60}=\frac{\pi}{3}
\displaystyle \text{Convert to degrees: }\frac{\pi}{3}=60^\circ
\displaystyle \text{Answer: (B) }60^\circ
\displaystyle \\
\displaystyle \text{Questions number 19 and 20 are Assertion and Reason based questions. Two statements} \\ \text{are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the} \\ \text{correct answer to these questions from the codes (A), (B), (C) and (D) as given below.}
\displaystyle \text{(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct} \\ \text{explanation of the Assertion (A).}
\displaystyle \text{(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct} \\ \text{explanation of the 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. }\text{Assertion (A) : The probability of selecting a number at random from the} \\ \text{numbers 1 to 20 is 1.}
\displaystyle \text{Reason (R) : For any event } E,\text{ if } P(E)=1,\text{ then } E \text{ is called a sure event.}
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): Probability of selecting a number from 1 to 20 is }1\ (\text{true})
\displaystyle \text{Reason (R): If }P(E)=1,\text{ then }E\text{ is a sure event (true)}
\displaystyle \text{But R does not explain why probability here is }1
\displaystyle \text{Answer: (B)}
\displaystyle \\
\displaystyle \textbf{20. }\text{Assertion (A) : If we join two hemispheres of same radius along their bases,} \\ \text{then we get a sphere.}
\displaystyle \text{Reason (R) : Total Surface Area of a sphere of radius } r \text{ is } 3\pi r^2.
\displaystyle \text{Answer:}
\displaystyle \text{Assertion (A): Two hemispheres joined form a sphere (true)}
\displaystyle \text{Reason (R): Total Surface Area of sphere is }4\pi r^2\ (\text{given }3\pi r^2\text{ is false})
\displaystyle \text{So, Assertion is true but Reason is false}
\displaystyle \text{Answer: (C)}

\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. (a) }\text{If } x\cos 60^\circ + y\cos 0^\circ + \sin 30^\circ - \cot 45^\circ = 5,\text{ then find the value of } \\ x+2y.
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Evaluate : } \frac{\tan^2 60^\circ}{\sin^2 60^\circ + \cos^2 30^\circ}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle x\cos 60^\circ+y\cos 0^\circ+\sin 30^\circ-\cot 45^\circ=5
\displaystyle x\left(\frac{1}{2}\right)+y(1)+\frac{1}{2}-1=5
\displaystyle \frac{x}{2}+y-\frac{1}{2}=5
\displaystyle \frac{x}{2}+y=\frac{11}{2}
\displaystyle x+2y=11
\displaystyle \text{Answer: }11
\displaystyle \text{(b)}
\displaystyle \frac{\tan^2 60^\circ}{\sin^2 60^\circ+\cos^2 30^\circ}
\displaystyle =\frac{(\sqrt{3})^2}{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}
\displaystyle =\frac{3}{\frac{3}{4}+\frac{3}{4}}
\displaystyle =\frac{3}{\frac{6}{4}}=\frac{3}{\frac{3}{2}}=2
\displaystyle \text{Answer: }2
\displaystyle \\
\displaystyle \textbf{22. }\text{Find the zeroes of the polynomial } p(x)=x^2+\frac{4}{3}x-\frac{4}{3}.
\displaystyle \text{Answer:}
\displaystyle  \text{Given }p(x)=x^2+\frac{4}{3}x-\frac{4}{3}
\displaystyle \text{Multiply by }3:\ 3x^2+4x-4=0
\displaystyle \text{Factorise: }3x^2+6x-2x-4=0
\displaystyle 3x(x+2)-2(x+2)=0
\displaystyle (3x-2)(x+2)=0
\displaystyle x=\frac{2}{3}\ \text{or}\ x=-2
\displaystyle \text{Zeroes are }\frac{2}{3}\text{ and }-2
\displaystyle \\
\displaystyle \textbf{23. }\text{The coordinates of the centre of a circle are } (2a, a-7).\text{ Find the value(s) of } \\   a \text{ if the circle passes through the point } (11,-9)\text{ and has diameter } 10\sqrt{2}\ \text{units.}
\displaystyle \text{Answer:}
\displaystyle \text{Centre of the circle }=(2a,\ a-7)
\displaystyle \text{Diameter }=10\sqrt{2}\ \text{units}
\displaystyle \text{So, radius }r=\frac{10\sqrt{2}}{2}=5\sqrt{2}
\displaystyle \text{Since the circle passes through }(11,-9),\text{ distance from centre to this point }=r
\displaystyle \sqrt{(11-2a)^2+[-9-(a-7)]^2}=5\sqrt{2}
\displaystyle \sqrt{(11-2a)^2+(-a-2)^2}=5\sqrt{2}
\displaystyle (11-2a)^2+(-a-2)^2=50
\displaystyle (121-44a+4a^2)+(a^2+4a+4)=50
\displaystyle 5a^2-40a+125=50
\displaystyle 5a^2-40a+75=0
\displaystyle a^2-8a+15=0
\displaystyle (a-3)(a-5)=0
\displaystyle a=3\ \text{or}\ a=5
\displaystyle \text{Answer: }a=3\text{ or }a=5
\displaystyle \\
\displaystyle \textbf{24. (a) }\text{If } \triangle ABC \sim \triangle PQR \text{ in which } AB=6\ \text{cm},\ BC=4\ \text{cm}, \\ AC=8\ \text{cm and } PR=6\ \text{cm},\text{ then find the length of } (PQ+QR).
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{In the given figure, } \frac{QR}{QS}=\frac{QT}{PR} \text{ and } \angle 1=\angle 2,\text{ show that } \triangle PQS \sim \triangle TQR.

\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle \triangle ABC\sim \triangle PQR
\displaystyle \text{So, }\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}
\displaystyle \text{Given }AB=6\text{ cm},\ BC=4\text{ cm},\ AC=8\text{ cm},\ PR=6\text{ cm}
\displaystyle \frac{AC}{PR}=\frac{8}{6}=\frac{4}{3}
\displaystyle \therefore\ \frac{AB}{PQ}=\frac{4}{3}\Rightarrow \frac{6}{PQ}=\frac{4}{3}\Rightarrow PQ=\frac{18}{4}=\frac{9}{2}\text{ cm}
\displaystyle \text{Also, }\frac{BC}{QR}=\frac{4}{3}\Rightarrow \frac{4}{QR}=\frac{4}{3}\Rightarrow QR=3\text{ cm}
\displaystyle PQ+QR=\frac{9}{2}+3=\frac{9}{2}+\frac{6}{2}=\frac{15}{2}\text{ cm}
\displaystyle \text{Answer: }PQ+QR=\frac{15}{2}\text{ cm}
\displaystyle \\
\displaystyle \text{OR}
\displaystyle \\
\displaystyle \text{(b)}
\displaystyle \text{Given }\frac{QR}{QS}=\frac{QT}{PR}\text{ and }\angle 1=\angle 2
\displaystyle \text{Now, }\angle 1=\angle PQS\text{ and }\angle 2=\angle TRQ
\displaystyle \therefore\ \angle PQS=\angle TRQ
\displaystyle \text{Also, }\frac{QR}{QS}=\frac{QT}{PR}\Rightarrow \frac{QS}{QR}=\frac{PR}{QT}
\displaystyle \text{In }\triangle PQS\text{ and }\triangle TQR,\text{ one angle is equal and the including sides are proportional}
\displaystyle \therefore\ \triangle PQS\sim \triangle TQR\text{ (by SAS similarity)}
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \textbf{25. }\text{A person is standing at P outside a circular ground at a distance of 26 m from} \\ \text{the centre of the ground. He found that his distances from the points A and B on} \\ \text{the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the} \\ \text{circular ground.}

\displaystyle \text{Answer:}
\displaystyle  \text{Given }PO=26\text{ m and }PA=PB=10\text{ m}
\displaystyle \text{Since }PA\text{ is tangent to the circle at }A,\ OA\perp PA
\displaystyle \text{So, in right }\triangle OAP
\displaystyle OP^2=OA^2+AP^2
\displaystyle 26^2=r^2+10^2
\displaystyle 676=r^2+100
\displaystyle r^2=576
\displaystyle r=24
\displaystyle \therefore\ \text{radius of the circular ground }=24\text{ m}

\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{In the given figure, } O \text{ is the centre of the circle and } BCD \text{ is tangent to it at } \\ C.\text{ Prove that } \angle BAC + \angle ACD = 90^\circ.

\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Prove that opposite sides of a quadrilateral circumscribing a circle subtend} \\ \text{supplementary angles at the centre of the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle \text{Given, }O\text{ is the centre of the circle and }BCD\text{ is tangent at }C
\displaystyle \text{We have to prove that }\angle BAC+\angle ACD=90^\circ
\displaystyle \text{Since }BC\text{ is tangent at }C,\ \ OC\perp BC
\displaystyle \therefore\ \angle OCB=90^\circ
\displaystyle \text{Also, }A,\ O,\ B\text{ are collinear, so }AB\text{ is along }AO
\displaystyle \text{Hence, }\angle BAC=\angle OAC
\displaystyle \text{Now in }\triangle AOC,\ OA=OC\text{ (radii of the same circle)}
\displaystyle \therefore\ \angle OAC=\angle ACO
\displaystyle \text{So, }\angle BAC=\angle ACO
\displaystyle \text{At point }C,\ \angle ACD=\angle \angle ACB
\displaystyle \text{and }\angle ACO+\angle ACB=\angle OCB=90^\circ
\displaystyle \therefore\ \angle ACO+\angle ACD=90^\circ
\displaystyle \text{But }\angle ACO=\angle BAC
\displaystyle \therefore\ \angle BAC+\angle ACD=90^\circ
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{OR}
\displaystyle \\
\displaystyle \text{(b)}
\displaystyle \text{Let }ABCD\text{ be a quadrilateral circumscribing a circle with centre }O.
\displaystyle \text{Suppose the circle touches }AB,BC,CD,DA\text{ at }P,Q,R,S\text{ respectively.}
\displaystyle \text{We have to prove that opposite sides subtend supplementary angles at the centre.}
\displaystyle \text{That is, }\angle AOB+\angle COD=180^\circ\text{ and }\angle BOC+\angle DOA=180^\circ
\displaystyle \text{Since radius is perpendicular to tangent at the point of contact,}
\displaystyle OP\perp AB,\ OQ\perp BC,\ OR\perp CD,\ OS\perp DA
\displaystyle \text{Now in quadrilateral }APOQ,\ \angle APO=90^\circ,\ \angle AQO=90^\circ
\displaystyle \therefore\ \angle PAQ+\angle POQ=180^\circ
\displaystyle \text{But }\angle PAQ=\angle A,\ \angle POQ=\angle AOB
\displaystyle \therefore\ \angle A+\angle AOB=180^\circ\ \text{...(1)}
\displaystyle \text{Similarly, in quadrilateral }CROS,\ \angle CRS=90^\circ,\ \angle CSO=90^\circ
\displaystyle \therefore\ \angle RCS+\angle ROS=180^\circ
\displaystyle \text{But }\angle RCS=\angle C,\ \angle ROS=\angle COD
\displaystyle \therefore\ \angle C+\angle COD=180^\circ\ \text{...(2)}
\displaystyle \text{Adding (1) and (2),}
\displaystyle \angle A+\angle C+\angle AOB+\angle COD=360^\circ
\displaystyle \text{But in any quadrilateral, }\angle A+\angle B+\angle C+\angle D=360^\circ
\displaystyle \text{Hence, from the figure, }\angle A+\angle C=180^\circ-\text{(sum of the other suitable pair),}
\displaystyle \text{and directly from (1) and (2), the angles at the centre corresponding to opposite sides are supplementary.}
\displaystyle \therefore\ \angle AOB+\angle COD=180^\circ
\displaystyle \text{Similarly, }\angle BOC+\angle DOA=180^\circ
\displaystyle \text{Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.}
\displaystyle \\
\displaystyle \textbf{27. (a) }\text{Prove that : } \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta} = 1+\sec\theta\ \mathrm{cosec}\ \theta
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{Prove that : } \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2\sin^2 A - 1}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\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 \text{OR}
\displaystyle \text{(b)}
\displaystyle \text{LHS}=\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A}
\displaystyle =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)}
\displaystyle =\frac{\sin^2 A+\cos^2 A+2\sin A\cos A+\sin^2 A+\cos^2 A-2\sin A\cos A}{\sin^2 A-\cos^2 A}
\displaystyle =\frac{2(\sin^2 A+\cos^2 A)}{\sin^2 A-\cos^2 A}
\displaystyle =\frac{2}{\sin^2 A-\cos^2 A}
\displaystyle =\frac{2}{\sin^2 A-(1-\sin^2 A)}
\displaystyle =\frac{2}{2\sin^2 A-1}
\displaystyle \text{Hence, proved.}
\displaystyle \\
\displaystyle \textbf{28. }\text{Find the ratio in which the y-axis divides the line segment joining the points } \\ (5,-6) \text{ and } (-1,-4).\text{ Also find the point of intersection.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }A(5,-6)\text{ and }B(-1,-4)
\displaystyle \text{Let the y-axis divide }AB\text{ in the ratio }m:n\ (\text{internally})
\displaystyle \text{Coordinates of point }P=\left(\frac{m(-1)+n(5)}{m+n},\frac{m(-4)+n(-6)}{m+n}\right)
\displaystyle \text{Since }P\text{ lies on y-axis, its x-coordinate is }0
\displaystyle \frac{-m+5n}{m+n}=0
\displaystyle -m+5n=0\Rightarrow m=5n
\displaystyle \text{Hence, ratio }m:n=5:1
\displaystyle \text{Now, }P=\left(0,\frac{5(-4)+1(-6)}{5+1}\right)
\displaystyle =\left(0,\frac{-20-6}{6}\right)=\left(0,-\frac{26}{6}\right)=\left(0,-\frac{13}{3}\right)
\displaystyle \text{Answer: Ratio }=5:1,\ \text{Point of intersection }=\left(0,-\frac{13}{3}\right)
\displaystyle \\
\displaystyle \textbf{29. }\text{Prove that } \frac{1}{\sqrt{5}} \text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle  \text{Let us assume that }\frac{1}{\sqrt{5}}\text{ is rational.}
\displaystyle \text{Then }\frac{1}{\sqrt{5}}=\frac{p}{q},\text{ where }p,q\text{ are integers and }q\neq 0.
\displaystyle \text{Taking reciprocal on both sides, we get }\sqrt{5}=\frac{q}{p}.
\displaystyle \text{Thus }\sqrt{5}\text{ is rational, which is not possible since }\sqrt{5}\text{ is irrational.}
\displaystyle \text{Hence, our assumption is wrong.}
\displaystyle \therefore\ \frac{1}{\sqrt{5}}\text{ is an irrational number.}
\displaystyle \\
\displaystyle \textbf{30. }\text{A room is in the form of a cylinder surmounted by a hemispherical dome. } \\ \text{The base radius of the hemisphere is half of the height of the cylindrical part. } \\ \text{If the room contains } \frac{1408}{21}\ \text{m}^3 \text{ of air, find the height of the cylindrical part. }\\ \text{(Use } \pi=\frac{22}{7}\text{).}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the height of the cylindrical part be }h\text{ m.}
\displaystyle \text{Then, radius of the hemisphere }r=\frac{h}{2}\text{ m}
\displaystyle \text{Volume of room}=\text{Volume of cylinder}+\text{Volume of hemisphere}
\displaystyle \pi r^2 h+\frac{2}{3}\pi r^3=\frac{1408}{21}
\displaystyle \pi\left(\frac{h}{2}\right)^2h+\frac{2}{3}\pi\left(\frac{h}{2}\right)^3=\frac{1408}{21}
\displaystyle \pi\cdot \frac{h^3}{4}+\frac{2}{3}\pi\cdot \frac{h^3}{8}=\frac{1408}{21}
\displaystyle \pi\cdot \frac{h^3}{4}+\pi\cdot \frac{h^3}{12}=\frac{1408}{21}
\displaystyle \pi\left(\frac{3h^3+h^3}{12}\right)=\frac{1408}{21}
\displaystyle \pi\cdot \frac{4h^3}{12}=\frac{1408}{21}
\displaystyle \pi\cdot \frac{h^3}{3}=\frac{1408}{21}
\displaystyle \frac{22}{7}\cdot \frac{h^3}{3}=\frac{1408}{21}
\displaystyle \frac{22h^3}{21}=\frac{1408}{21}
\displaystyle 22h^3=1408
\displaystyle h^3=64
\displaystyle h=4
\displaystyle \therefore\ \text{height of the cylindrical part }=4\text{ m}.
\displaystyle \\
\displaystyle \textbf{31. }\text{Two dice are thrown at the same time. Determine the probability that the} \\ \text{difference of the numbers on the two dice is }2.
\displaystyle \text{Answer:}
\displaystyle  \text{Total outcomes when two dice are thrown }=6\times 6=36
\displaystyle \text{Favourable outcomes where difference is }2:
\displaystyle (1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4)
\displaystyle \text{Number of favourable outcomes }=8
\displaystyle \text{Probability}=\frac{8}{36}=\frac{2}{9}
\displaystyle \text{Answer: }\frac{2}{9}

\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{Vijay invested certain amounts of money in two schemes A and B, } \\ \text{which offer interest at the rate of }8\%\ \text{per annum and }9\%\ \text{per annum, respectively.  } \\ \text{He received Rs 1860 as the total annual interest. However, had he interchanged } \\ \text{the amounts of investments in the two schemes, he would have received } \\ \text{Rs 20 more as annual interest. How much money did he invest in each scheme ?}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the amounts invested in schemes A and B be }x\text{ and }y\text{ respectively.}
\displaystyle \text{Interest rates: }8\%\text{ and }9\%
\displaystyle \text{Total interest: }0.08x+0.09y=1860\ \text{...(1)}
\displaystyle \text{If interchanged: }0.09x+0.08y=1880\ \text{...(2)}
\displaystyle \text{Subtract (1) from (2):}
\displaystyle (0.09x+0.08y)-(0.08x+0.09y)=1880-1860
\displaystyle 0.01x-0.01y=20
\displaystyle x-y=2000\ \text{...(3)}
\displaystyle \text{From (3): }x=y+2000
\displaystyle \text{Substitute in (1): }0.08(y+2000)+0.09y=1860
\displaystyle 0.08y+160+0.09y=1860
\displaystyle 0.17y=1700
\displaystyle y=10000
\displaystyle x=y+2000=12000
\displaystyle \text{Answer: }x=12000,\ y=10000
\displaystyle \\
\displaystyle \textbf{33. (a) }\text{The diagonal BD of a parallelogram ABCD intersects the line } \\ \text{segment AE at the point F, where E is any point on the side BC. Prove that } \\ DF\times EF = FB\times FA.
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{In } \triangle ABC,\text{ if } AD \perp BC \text{ and } AD^2 = BD\times DC,\text{ then prove that } \\ \angle BAC = 90^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle \text{In parallelogram }ABCD,\ AB\parallel CD\text{ and }BC\parallel AD
\displaystyle \text{Since }E\text{ lies on }BC,\ \text{we have }BE\parallel CD
\displaystyle \text{Diagonal }BD\text{ intersects }AE\text{ at }F
\displaystyle \text{Consider }\triangle AFB\text{ and }\triangle EFD
\displaystyle \angle AFB=\angle EFD\text{ since they are vertically opposite angles}
\displaystyle \angle FAB=\angle FED\text{ since }AB\parallel DE\text{ and }AE\text{ is a transversal}
\displaystyle \angle FBA=\angle FDE\text{ since }AB\parallel DE\text{ and }BD\text{ is a transversal}
\displaystyle \therefore\ \triangle AFB\sim \triangle EFD
\displaystyle \text{Hence, }\frac{AF}{FE}=\frac{FB}{FD}
\displaystyle \text{On cross-multiplication, }AF\times FD=FB\times FE
\displaystyle \therefore\ DF\times EF=FB\times FA
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \text{OR}
\displaystyle \\
\displaystyle \text{(b)}
\displaystyle \text{Given, in }\triangle ABC,\ AD\perp BC\text{ and }AD^2=BD\times DC
\displaystyle \text{We have to prove that }\angle BAC=90^\circ
\displaystyle \text{In right }\triangle ADB,\ AB^2=AD^2+BD^2\ \text{(by Pythagoras theorem)}
\displaystyle \text{In right }\triangle ADC,\ AC^2=AD^2+DC^2\ \text{(by Pythagoras theorem)}
\displaystyle \text{Adding, }AB^2+AC^2=2AD^2+BD^2+DC^2
\displaystyle \text{Since }AD^2=BD\times DC
\displaystyle AB^2+AC^2=2BD\times DC+BD^2+DC^2
\displaystyle AB^2+AC^2=BD^2+2BD\times DC+DC^2
\displaystyle AB^2+AC^2=(BD+DC)^2
\displaystyle \text{But }BD+DC=BC
\displaystyle \therefore\ AB^2+AC^2=BC^2
\displaystyle \text{Hence, by converse of Pythagoras theorem, }\angle BAC=90^\circ
\displaystyle \text{Hence proved.}
\displaystyle \\
\displaystyle \textbf{34. (a) }\text{The perimeter of a right triangle is }60\ \text{cm and its hypotenuse is 25 cm. } \\ \text{Find the lengths of other two sides of the triangle.}
\displaystyle \text{OR}
\displaystyle \textbf{(b) }\text{A train travels a distance of }480\ \text{km at a uniform speed. If the speed had been} \\ 8\ \text{km/h less, then it would have taken }3\ \text{hours more to cover the same distance. Find } \\ \text{the speed of the train.}
\displaystyle \text{Answer:}
\displaystyle \text{(a)}
\displaystyle \text{Let the two perpendicular sides be }a\text{ and }b
\displaystyle \text{Hypotenuse }=25\text{ cm}
\displaystyle \text{Perimeter }=a+b+25=60
\displaystyle a+b=35\ \text{...(1)}
\displaystyle \text{By Pythagoras theorem: }a^2+b^2=25^2=625\ \text{...(2)}
\displaystyle (a+b)^2=a^2+b^2+2ab
\displaystyle 35^2=625+2ab
\displaystyle 1225=625+2ab
\displaystyle 2ab=600\Rightarrow ab=300
\displaystyle \text{Thus, }a\text{ and }b\text{ are roots of }x^2-35x+300=0
\displaystyle (x-15)(x-20)=0
\displaystyle a=15,\ b=20
\displaystyle \text{Answer: }15\text{ cm and }20\text{ cm}
\displaystyle \\
\displaystyle \text{OR}
\displaystyle \\
\displaystyle \text{(b)}
\displaystyle \text{Let speed of train }=x\text{ km/h}
\displaystyle \text{Time taken }=\frac{480}{x}
\displaystyle \text{If speed is }(x-8),\ \text{time }=\frac{480}{x-8}
\displaystyle \text{Given: }\frac{480}{x-8}=\frac{480}{x}+3
\displaystyle \frac{480}{x-8}-\frac{480}{x}=3
\displaystyle \frac{480x-480(x-8)}{x(x-8)}=3
\displaystyle \frac{480x-480x+3840}{x(x-8)}=3
\displaystyle \frac{3840}{x(x-8)}=3
\displaystyle 3840=3x(x-8)
\displaystyle 1280=x^2-8x
\displaystyle x^2-8x-1280=0
\displaystyle (x-40)(x+32)=0
\displaystyle x=40\ (\text{reject }-32)
\displaystyle \text{Answer: }40\text{ km/h}
\displaystyle \\
\displaystyle \textbf{35. }\text{Find the missing frequency } f \text{ in the following table, if the mean of the given data} \\ \text{is }18.\text{ Hence find the mode.}
\displaystyle \begin{array}{|c|c|}\hline \text{Daily Allowance} & \text{Number of Children} \\ \hline 11-13 & 7 \\ \hline 13-15 & 6 \\ \hline 15-17 & 9 \\ \hline 17-19 & 13 \\ \hline 19-21 & f \\ \hline 21-23 & 5 \\ \hline 23-25 & 4 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \\
\displaystyle  \text{Class intervals and mid-points:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class} & f_i & x_i & f_ix_i \\ \hline 11-13 & 7 & 12 & 84 \\ \hline 13-15 & 6 & 14 & 84 \\ \hline 15-17 & 9 & 16 & 144 \\ \hline 17-19 & 13 & 18 & 234 \\ \hline 19-21 & f & 20 & 20f \\ \hline 21-23 & 5 & 22 & 110 \\ \hline 23-25 & 4 & 24 & 96 \\ \hline \end{array}
\displaystyle \sum f_i = 44+f,\quad \sum f_ix_i = 752+20f
\displaystyle \text{Mean }=18=\frac{752+20f}{44+f}
\displaystyle 18(44+f)=752+20f
\displaystyle 792+18f=752+20f
\displaystyle 40=2f\Rightarrow f=20
\displaystyle \text{Modal class is }19-21\ (\text{highest frequency }=20)
\displaystyle l=19,\ h=2,\ f_1=20,\ f_0=13,\ f_2=5
\displaystyle \text{Mode }=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =19+\frac{20-13}{40-13-5}\times 2
\displaystyle =19+\frac{7}{22}\times 2=19+\frac{14}{22}=19+\frac{7}{11}
\displaystyle =19.64\ (\text{approx})
\displaystyle \text{Answer: }f=20,\ \text{Mode }\approx 19.64

\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{Case Study - 1}
\displaystyle \text{Answer:}
\displaystyle \\
\displaystyle \textbf{36. }\text{A school is organizing a charity run to raise funds for a local hospital. The run} \\ \text{is planned as a series of rounds around a track, with each round being 300 meters} \\ \text{To make the event more challenging and engaging, the organizers decide to increase} \\ \text{the distance of each subsequent round by 50 metres. For example, the second round} \\ \text{will be 350 metres, the third round will be 400 metres and so on. The total number of } \\ \text{rounds planned is }10.

\displaystyle \text{Based on the information given above, answer the following questions :}
\displaystyle \text{(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.}
\displaystyle \text{(ii) Determine the distance of the 8th round.}
\displaystyle \text{(iii) (a) Find the total distance run after completing all 10 rounds.}
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) If a runner completes only the first 6 rounds, what is the total distance} \\ \text{run by the runner ?}
\displaystyle \text{Answer:}
\displaystyle  \text{First round distance }a=300\text{ m, common difference }d=50\text{ m}
\displaystyle \text{So, the distances form an A.P. }300,\ 350,\ 400,\ 450,\ldots
\displaystyle \textbf{(i) }\text{Fourth term }=a+3d=300+3(50)=450\text{ m}
\displaystyle \text{Fifth term }=a+4d=300+4(50)=500\text{ m}
\displaystyle \text{Sixth term }=a+5d=300+5(50)=550\text{ m}
\displaystyle \text{Answer: Fourth term }=450\text{ m, Fifth term }=500\text{ m, Sixth term }=550\text{ m}
\displaystyle \textbf{(ii) }\text{Eighth term }=a+7d=300+7(50)=650\text{ m}
\displaystyle \text{Answer: Distance of 8th round }=650\text{ m}
\displaystyle \textbf{(iii)(a) }\text{Total distance after 10 rounds }=S_{10}=\frac{10}{2}[2(300)+(10-1)(50)]
\displaystyle =5[600+450]=5(1050)=5250\text{ m}
\displaystyle \text{Answer: Total distance after 10 rounds }=5250\text{ m}
\displaystyle \text{OR}
\displaystyle \textbf{(iii)(b) }\text{Total distance after first 6 rounds }=S_6=\frac{6}{2}[2(300)+(6-1)(50)]
\displaystyle =3[600+250]=3(850)=2550\text{ m}
\displaystyle \text{Answer: Total distance after first 6 rounds }=2550\text{ m}
\displaystyle \\
\displaystyle \textbf{Case Study - 2}
\displaystyle \textbf{37. }\text{A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses} \\ \text{to add elegance. Made from precious metals and decorated with gemstones, brooches} \\ \text{come in many shapes and designs.}

\displaystyle \text{One such brooch is made with silver wire in the form of a circle with diameter 35 mm.} \\ \text{The wire is also used in making }5\ \text{diameters which divide the circle into 10 equal } \\ \text{sectors as shown in the figure.} \displaystyle \text{Based on the above given information, answer the following questions :}
\displaystyle \text{(i) Find the central angle of each sector.}
\displaystyle \text{(ii) Find the length of the arc ACB.}
\displaystyle \text{(iii) (a) Find the area of each sector of the brooch.}
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) Find the total length of the silver wire used.}
\displaystyle \text{Answer:}
\displaystyle  \text{Diameter of the brooch }=35\text{ mm}
\displaystyle \text{Radius }r=\frac{35}{2}=17.5\text{ mm}
\displaystyle \text{The circle is divided into }10\text{ equal sectors}
\displaystyle \textbf{(i) }\text{Central angle of each sector }=\frac{360^\circ}{10}=36^\circ
\displaystyle \text{Answer: }36^\circ
\displaystyle \textbf{(ii) }\text{Arc ACB covers one sector, so its angle is }36^\circ
\displaystyle \text{Length of arc ACB }=\frac{36^\circ}{360^\circ}\times 2\pi r
\displaystyle =\frac{1}{10}\times 2\pi \times 17.5
\displaystyle =\frac{35\pi}{10}=3.5\pi\text{ mm}
\displaystyle =11\text{ mm}\ \left(\text{using }\pi=\frac{22}{7}\right)
\displaystyle \text{Answer: Length of arc ACB }=11\text{ mm}
\displaystyle \textbf{(iii)(a) }\text{Area of each sector }=\frac{36^\circ}{360^\circ}\times \pi r^2
\displaystyle =\frac{1}{10}\times \pi \times (17.5)^2
\displaystyle =\frac{1}{10}\times \pi \times 306.25
\displaystyle =30.625\pi\text{ mm}^2
\displaystyle =96.25\text{ mm}^2\ \left(\text{using }\pi=\frac{22}{7}\right)
\displaystyle \text{Answer: Area of each sector }=96.25\text{ mm}^2
\displaystyle \text{OR}
\displaystyle \textbf{(iii)(b) }\text{Total silver wire used }=\text{circumference of circle }+ \text{length of 5 diameters}
\displaystyle =\pi d+5d
\displaystyle =35\pi+5(35)
\displaystyle =35\pi+175
\displaystyle =110+175\ \left(\text{using }\pi=\frac{22}{7}\right)
\displaystyle =285\text{ mm}
\displaystyle \text{Answer: Total length of silver wire used }=285\text{ mm}

\displaystyle \textbf{Case Study - 3}
\displaystyle \textbf{38. }\text{Amrita stood near the base of a lighthouse, gazing up at its towering height.} \\ \text{She measured the angle of elevation to the top and found it to be }60^\circ.\text{ Then, she} \\ \text{climbed a nearby observation deck, 40 metres higher than her original position and} \\ \text{noticed the angle of elevation to the top of the lighthouse to be }45^\circ.


\displaystyle \text{Based on the above given information, answer the following questions :}
\displaystyle \text{(i) If CD is } h \text{ metres, find the distance BD in terms of } h.
\displaystyle \text{(ii) Find distance BC in terms of } h.
\displaystyle \text{(iii) (a) Find the height CE of the lighthouse [Use } \sqrt{3}=1.73\text{].}
\displaystyle \text{OR}
\displaystyle \text{(iii) (b) Find distance AE, if AC = 100\ \text{m}.}
\displaystyle \text{Answer:}
\displaystyle  \text{Here }AB=40\text{ m},\ \angle CAE=60^\circ,\ \angle CBD=45^\circ,\ CD=h\text{ m}
\displaystyle \text{Also, }AE=BD\text{ and }DE=AB=40\text{ m}
\displaystyle \textbf{(i) }\text{In }\triangle BCD,\ \tan 45^\circ=\frac{CD}{BD}
\displaystyle 1=\frac{h}{BD}
\displaystyle \therefore\ BD=h
\displaystyle \text{Answer: }BD=h\text{ m}
\displaystyle \textbf{(ii) }\text{In }\triangle BCD,\ \cos 45^\circ=\frac{BD}{BC}
\displaystyle \frac{1}{\sqrt{2}}=\frac{h}{BC}
\displaystyle \therefore\ BC=h\sqrt{2}
\displaystyle \text{Answer: }BC=h\sqrt{2}\text{ m}
\displaystyle \textbf{(iii)(a) }\text{Height of lighthouse }CE=CD+DE=h+40
\displaystyle \text{Now in }\triangle ACE,\ \tan 60^\circ=\frac{CE}{AE}=\frac{h+40}{h}
\displaystyle \sqrt{3}=\frac{h+40}{h}
\displaystyle h\sqrt{3}=h+40
\displaystyle h(\sqrt{3}-1)=40
\displaystyle h=\frac{40}{\sqrt{3}-1}=\frac{40(\sqrt{3}+1)}{2}=20(\sqrt{3}+1)
\displaystyle \text{Using }\sqrt{3}=1.73,\ h=20(1.73+1)=20(2.73)=54.6
\displaystyle \therefore\ CE=h+40=54.6+40=94.6
\displaystyle \text{Answer: Height of lighthouse }CE=94.6\text{ m}
\displaystyle \text{OR}
\displaystyle \textbf{(iii)(b) }\text{If }AC=100\text{ m, then in }\triangle ACE
\displaystyle \cos 60^\circ=\frac{AE}{AC}
\displaystyle \frac{1}{2}=\frac{AE}{100}
\displaystyle \therefore\ AE=50
\displaystyle \text{Answer: }AE=50\text{ m}

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