\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series PPQQC/2}  \hspace{3.0cm} \textbf{SET~1 } \hspace{4.0cm} \textbf{Q.P. Code }30/2/1
\displaystyle \text{Roll No.}

\displaystyle \textbf{NOTE}

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

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

\displaystyle \text{(III) Please check that this question paper contains 14 questions.}

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

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

\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \text{Time allowed : 2 hours}                                                                 \displaystyle \text{Maximum marks : 40}

\displaystyle \textbf{General Instructions :}

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

\displaystyle \text{(i) This question paper contains 14 questions. All questions are compulsory.}

\displaystyle \text{(ii) This question paper is divided into three sections - Sections A, B and C.}

\displaystyle \text{(iii) Section A comprises of 6 questions (Q.no. 1 to 6) of 2 marks each. Internal}
\displaystyle \text{choice has been provided in two questions.}

\displaystyle \text{(iv) Section B comprises of 4 questions (Q.no. 7 to 10) of 3 marks each. Internal}
\displaystyle \text{choice has been provided in one question.}

\displaystyle \text{(v) Section C comprises of 4 questions (Q.no. 11 to 14) of 4 marks each. Internal}
\displaystyle \text{choice has been provided in one question. It also contains two case study based}
\displaystyle \text{questions.}

\displaystyle \text{(vi) Use of calculator is not permitted.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Question numbers 1 to 6 carry 2 marks each.}

\displaystyle \textbf{Question 1. }\text{Solve the quadratic equation: }x^2+2\sqrt{2}x-6=0\text{ for }x.
\displaystyle \text{Answer:}
\displaystyle x^2+2\sqrt{2}x-6=0
\displaystyle x=\frac{-2\sqrt{2}\pm\sqrt{(2\sqrt{2})^2-4(1)(-6)}}{2}
\displaystyle =\frac{-2\sqrt{2}\pm\sqrt{8+24}}{2}
\displaystyle =\frac{-2\sqrt{2}\pm\sqrt{32}}{2}
\displaystyle =\frac{-2\sqrt{2}\pm4\sqrt{2}}{2}
\displaystyle x=\sqrt{2}\quad \text{or}\quad x=-3\sqrt{2}
\displaystyle \therefore x=\sqrt{2},-3\sqrt{2}
\\

\displaystyle \textbf{Question 2. }\text{(a) Which term of the A.P. }-\frac{11}{2},-3,-\frac{1}{2},\ldots\text{ is }\frac{49}{2}\text{?}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Find }a\text{ and }b\text{ so that the numbers }a,7,b,23\text{ are in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }a=-\frac{11}{2},\quad d=-3-\left(-\frac{11}{2}\right)
\displaystyle d=-3+\frac{11}{2}=\frac{5}{2}
\displaystyle a_n=a+(n-1)d
\displaystyle \frac{49}{2}=-\frac{11}{2}+(n-1)\frac{5}{2}
\displaystyle 49=-11+5(n-1)
\displaystyle 60=5(n-1)
\displaystyle n-1=12
\displaystyle n=13
\displaystyle \therefore \frac{49}{2}\text{ is the }13^{\text{th}}\text{ term.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Since }a,7,b,23\text{ are in A.P.,}
\displaystyle 7-a=b-7=23-b
\displaystyle b-7=23-b
\displaystyle 2b=30
\displaystyle b=15
\displaystyle 7-a=b-7
\displaystyle 7-a=15-7=8
\displaystyle a=-1
\displaystyle \therefore a=-1,\quad b=15
\\

\displaystyle \textbf{Question 3. }\text{A solid piece of metal in the form of a cuboid of dimensions }
\displaystyle 11\text{ cm}\times7\text{ cm}\times 7\text{ cm is melted to form }n\text{ number of solid spheres of radii }\frac{7}{2}\text{ cm each.}
\displaystyle \text{ Find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{Volume of cuboid}=11\times7\times7
\displaystyle =539\text{ cm}^3
\displaystyle \text{Radius of each sphere}=\frac{7}{2}\text{ cm}
\displaystyle \text{Volume of one sphere}=\frac{4}{3}\pi r^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times\left(\frac{7}{2}\right)^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times\frac{343}{8}
\displaystyle =\frac{539}{3}\text{ cm}^3
\displaystyle n=\frac{\text{Volume of cuboid}}{\text{Volume of one sphere}}
\displaystyle =\frac{539}{\frac{539}{3}}
\displaystyle =3
\displaystyle \therefore n=3
\\

\displaystyle \textbf{Question 4. }\text{(a) In Fig. }1,\text{ }AB\text{ is diameter of a circle centered at }O.
\displaystyle BC\text{ is tangent to the circle at }B.\text{ If }OP\text{ bisects the chord }AD\text{ and }
\displaystyle \angle AOP=60^\circ,  \text{then find }m\angle C.

\displaystyle \textbf{OR}

\displaystyle \text{(b) In Fig. }2,\text{ }XAY\text{ is a tangent to the circle centered at }O.\text{ If }\angle ABO=40^\circ,
\displaystyle \text{then find }m\angle BAY\text{ and }m\angle AOB. \displaystyle \text{Answer:}
\displaystyle \text{(a) Since perpendicular from centre to a chord bisects the chord,}
\displaystyle OP\perp AD
\displaystyle \therefore \angle APO=90^\circ
\displaystyle \text{In }\triangle AOP,\quad \angle AOP=60^\circ
\displaystyle \therefore \angle OAP=180^\circ-90^\circ-60^\circ
\displaystyle =30^\circ
\displaystyle \text{Since }BC\text{ is tangent at }B,\quad AB\perp BC
\displaystyle \therefore \angle C=90^\circ-30^\circ
\displaystyle =60^\circ
\displaystyle \therefore m\angle C=60^\circ

\displaystyle \textbf{OR}

\displaystyle \text{(b) }OA=OB\text{ radii of the same circle}
\displaystyle \therefore \angle OAB=\angle ABO=40^\circ
\displaystyle \angle AOB=180^\circ-40^\circ-40^\circ
\displaystyle =100^\circ
\displaystyle \text{Since }XAY\text{ is tangent at }A,\quad OA\perp AY
\displaystyle \therefore \angle OAY=90^\circ
\displaystyle \angle BAY=90^\circ-\angle OAB
\displaystyle =90^\circ-40^\circ=50^\circ
\displaystyle \therefore m\angle BAY=50^\circ,\quad m\angle AOB=100^\circ
\\

\displaystyle \textbf{Question 5. }\text{If mode of the following frequency distribution is }55,\text{ then find the value}
\displaystyle \text{of }x.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class} & 0-15 & 15-30 & 30-45 & 45-60 & 60-75 & 75-90\\ \hline \text{Frequency} & 10 & 7 & x & 15 & 10 & 12\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Mode}=55
\displaystyle \therefore \text{Modal class}=45-60
\displaystyle l=45,\quad h=15,\quad f_1=15,\quad f_0=x,\quad f_2=10
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle 55=45+\frac{15-x}{2(15)-x-10}\times15
\displaystyle 10=\frac{15(15-x)}{20-x}
\displaystyle 10(20-x)=15(15-x)
\displaystyle 200-10x=225-15x
\displaystyle 5x=25
\displaystyle x=5
\displaystyle \therefore \text{Value of }x=5
\\

\displaystyle \textbf{Question 6. }\text{Find the sum of first }20\text{ terms of an A.P. whose }n^{\text{th}}\text{ term is given as}
\displaystyle a_n=5-2n.
\displaystyle \text{Answer:}
\displaystyle a_n=5-2n
\displaystyle a_1=5-2(1)=3
\displaystyle a_{20}=5-2(20)=-35
\displaystyle S_{20}=\frac{20}{2}(a_1+a_{20})
\displaystyle =10(3-35)
\displaystyle =10(-32)
\displaystyle =-320
\displaystyle \therefore \text{Sum of first }20\text{ terms}=-320
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question numbers 7 to 10 carry 3 marks each.}

\displaystyle \textbf{Question 7. }\text{Draw two concentric circles of radii }2\text{ cm and }5\text{ cm. From a point on the}
\displaystyle \text{outer circle, construct a pair of tangents to the inner circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw two concentric circles with centre }O\text{ and radii }2\text{ cm and }5\text{ cm.}
\displaystyle \text{2. Mark any point }P\text{ on the outer circle.}
\displaystyle \text{3. Join }OP.
\displaystyle \text{4. Draw the perpendicular bisector of }OP\text{ to get its mid-point }M.
\displaystyle \text{5. With centre }M\text{ and radius }MO,\text{ draw a circle.}
\displaystyle \text{6. This circle cuts the inner circle at }A\text{ and }B.
\displaystyle \text{7. Join }PA\text{ and }PB.
\displaystyle \therefore PA\text{ and }PB\text{ are the required tangents to the inner circle.}
\\

\displaystyle \textbf{Question 8. }\text{In Fig. }3,\text{ }AB\text{ is tower of height }50\text{ m. A man standing on its top, observes}
\displaystyle \text{two cars on the opposite sides of the tower with angles of depression }30^\circ\text{ and }45^\circ
\displaystyle \text{respectively. Find the distance between the two cars.} \displaystyle \text{Answer:}
\displaystyle AB=50\text{ m}
\displaystyle \text{Angles of depression are }30^\circ\text{ and }45^\circ.
\displaystyle \therefore \angle ACB=30^\circ\text{ and }\angle ADB=45^\circ
\displaystyle \text{In }\triangle ABC,
\displaystyle \tan 30^\circ=\frac{AB}{BC}
\displaystyle \frac{1}{\sqrt{3}}=\frac{50}{BC}
\displaystyle BC=50\sqrt{3}\text{ m}
\displaystyle \text{In }\triangle ABD,
\displaystyle \tan 45^\circ=\frac{AB}{BD}
\displaystyle 1=\frac{50}{BD}
\displaystyle BD=50\text{ m}
\displaystyle \text{Distance between the two cars}=BC+BD
\displaystyle =50\sqrt{3}+50
\displaystyle =50(\sqrt{3}+1)\text{ m}
\displaystyle \therefore \text{Distance between the two cars}=50(\sqrt{3}+1)\text{ m}
\\

\displaystyle \textbf{Question 9. }\text{(a) The mean of the following frequency distribution is }25.\text{ Find the}
\displaystyle \text{value of }f.
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50\\ \hline \text{Frequency} & 5 & 18 & 15 & f & 6\\ \hline \end{array}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Find the mean of the following data using assumed mean method:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25\\ \hline \text{Frequency} & 8 & 7 & 10 & 13 & 12\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }\begin{array}{|c|c|c|c|}\hline \text{Class} & f_i & x_i & f_ix_i\\ \hline 0-10 & 5 & 5 & 25\\ \hline 10-20 & 18 & 15 & 270\\ \hline 20-30 & 15 & 25 & 375\\ \hline 30-40 & f & 35 & 35f\\ \hline 40-50 & 6 & 45 & 270\\ \hline \end{array}
\displaystyle \sum f_i=44+f,\qquad \sum f_ix_i=940+35f
\displaystyle \overline{x}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle 25=\frac{940+35f}{44+f}
\displaystyle 25(44+f)=940+35f
\displaystyle 1100+25f=940+35f
\displaystyle 10f=160
\displaystyle f=16
\displaystyle \therefore \text{Required value of }f=16

\displaystyle \textbf{OR}

\displaystyle \text{(b) }\begin{array}{|c|c|c|c|}\hline \text{Class} & f_i & x_i & f_ix_i\\ \hline 0-5 & 8 & 2.5 & 20\\ \hline 5-10 & 7 & 7.5 & 52.5\\ \hline 10-15 & 10 & 12.5 & 125\\ \hline 15-20 & 13 & 17.5 & 227.5\\ \hline 20-25 & 12 & 22.5 & 270\\ \hline \end{array}
\displaystyle \sum f_i=50,\qquad \sum f_ix_i=695
\displaystyle \overline{x}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle =\frac{695}{50}
\displaystyle =13.9
\displaystyle \therefore \text{Mean}=13.9
\\

\displaystyle \textbf{Question 10. }\text{Heights of }50\text{ students of class X of a school are recorded and following}
\displaystyle \text{data is obtained. Find the median height of the students.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Height (in cm)} & 130-135 & 135-140 & 140-145 & 145-150 & 150-155 & 155-160\\ \hline \text{Number of Students} & 4 & 11 & 12 & 7 & 10 & 6\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class} & f & c.f.\\ \hline 130-135 & 4 & 4\\ \hline 135-140 & 11 & 15\\ \hline 140-145 & 12 & 27\\ \hline 145-150 & 7 & 34\\ \hline 150-155 & 10 & 44\\ \hline 155-160 & 6 & 50\\ \hline \end{array}
\displaystyle N=50,\qquad \frac{N}{2}=25
\displaystyle \text{Median class}=140-145
\displaystyle l=140,\quad h=5,\quad f=12,\quad c.f.=15
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-c.f.}{f}\times h
\displaystyle =140+\frac{25-15}{12}\times5
\displaystyle =140+\frac{10}{12}\times5
\displaystyle =140+\frac{50}{12}
\displaystyle =140+4.17
\displaystyle =144.17
\displaystyle \therefore \text{Median height}=144.17\text{ cm}
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question numbers 11 to 14 carry 4 marks each.}

\displaystyle \textbf{Question 11. }\text{In Fig. }4,\text{ }PQ\text{ is a chord of length }8\text{ cm of a circle of radius }5\text{ cm. The}
\displaystyle \text{tangents at }P\text{ and }Q\text{ meet at a point }T.\text{ Find the length of }TP. \displaystyle \text{Answer:}
\displaystyle PQ=8\text{ cm},\quad OP=OQ=5\text{ cm}
\displaystyle \text{Let }R\text{ be the midpoint of chord }PQ.
\displaystyle PR=\frac{PQ}{2}=4\text{ cm}
\displaystyle OR\perp PQ
\displaystyle \text{In }\triangle OPR,
\displaystyle OP^2=OR^2+PR^2
\displaystyle 5^2=OR^2+4^2
\displaystyle OR^2=25-16=9
\displaystyle OR=3\text{ cm}
\displaystyle \text{Since tangents from }T\text{ are equal, }TP=TQ.
\displaystyle \therefore T\text{ lies on the perpendicular bisector of }PQ.
\displaystyle \therefore O,R,T\text{ are collinear.}
\displaystyle \text{Let }TR=x\text{ cm. Then }TP^2=TR^2+PR^2=x^2+16
\displaystyle OT=OR+TR=3+x
\displaystyle \text{Since }OP\perp TP,\text{ in }\triangle OPT,
\displaystyle OT^2=OP^2+TP^2
\displaystyle (x+3)^2=5^2+(x^2+16)
\displaystyle x^2+6x+9=x^2+41
\displaystyle 6x=32
\displaystyle x=\frac{16}{3}
\displaystyle TP^2=x^2+16
\displaystyle =\left(\frac{16}{3}\right)^2+16
\displaystyle =\frac{256}{9}+\frac{144}{9}=\frac{400}{9}
\displaystyle TP=\frac{20}{3}\text{ cm}
\displaystyle \therefore \text{Length of }TP=\frac{20}{3}\text{ cm}
\\

\displaystyle \textbf{Question 12. }\text{(a) A }2\text{-digit number is such that the product of its digits is }24.\text{ If }18\text{ is}
\displaystyle \text{subtracted from the number, the digits interchange their places. Find the number.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) The difference of the squares of two numbers is }180.\text{ The square of the}
\displaystyle \text{smaller number is }8\text{ times the greater number. Find the two numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the ten's digit be }x\text{ and unit's digit be }y.
\displaystyle \text{Number}=10x+y
\displaystyle xy=24 \qquad (i)
\displaystyle 10x+y-18=10y+x
\displaystyle 9x-9y=18
\displaystyle x-y=2
\displaystyle x=y+2 \qquad (ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle y(y+2)=24
\displaystyle y^2+2y-24=0
\displaystyle (y+6)(y-4)=0
\displaystyle y=4\quad \text{or}\quad y=-6
\displaystyle \text{Since digit cannot be negative, }y=4
\displaystyle x=y+2=6
\displaystyle \therefore \text{Required number}=64

\displaystyle \textbf{OR}

\displaystyle \text{(b) Let the greater number be }x\text{ and the smaller number be }y.
\displaystyle x^2-y^2=180 \qquad (i)
\displaystyle y^2=8x \qquad (ii)
\displaystyle \text{From }(i),\quad x^2-8x=180
\displaystyle x^2-8x-180=0
\displaystyle (x-18)(x+10)=0
\displaystyle x=18\quad \text{or}\quad x=-10
\displaystyle \text{Taking positive value, }x=18
\displaystyle y^2=8(18)=144
\displaystyle y=12
\displaystyle \therefore \text{The two numbers are }18\text{ and }12
\\

\displaystyle \textbf{Question 13. }\textbf{Case Study - 1 : Kite Festival}
\displaystyle \text{Kite festival is celebrated in many countries at different times of the year.}
\displaystyle \text{In India, every year }14^{\text{th}}\text{ January is celebrated as International Kite Day.}
\displaystyle \text{On this day many people visit India and participate in the festival by flying various}
\displaystyle \text{kinds of kites.}
\displaystyle \text{The picture given below, shows three kites flying together.} \displaystyle \text{In Fig. }5,\text{ the angles of elevation of two kites (Points A and B) from the hands}
\displaystyle \text{of a man (Point C) are found to be }30^\circ\text{ and }60^\circ\text{ respectively.}
\displaystyle \text{Taking }AD=50\text{ m and }BE=60\text{ m, find:}
\displaystyle \text{(i) the lengths of strings used (take them straight) for kites A and B.}
\displaystyle \text{(ii) the distance }d\text{ between the two kites.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\triangle ACD,
\displaystyle \sin 30^\circ=\frac{AD}{AC}
\displaystyle \frac{1}{2}=\frac{50}{AC}
\displaystyle AC=100\text{ m}
\displaystyle \text{Hence, length of string for kite A}=100\text{ m}
\displaystyle \text{In }\triangle BCE,
\displaystyle \sin 60^\circ=\frac{BE}{BC}
\displaystyle \frac{\sqrt{3}}{2}=\frac{60}{BC}
\displaystyle BC=\frac{120}{\sqrt{3}}=40\sqrt{3}\text{ m}
\displaystyle \text{Hence, length of string for kite B}=40\sqrt{3}\text{ m}
\displaystyle \text{(ii) In }\triangle ACD,
\displaystyle \tan 30^\circ=\frac{AD}{DC}
\displaystyle \frac{1}{\sqrt{3}}=\frac{50}{DC}
\displaystyle DC=50\sqrt{3}\text{ m}
\displaystyle \text{In }\triangle BCE,
\displaystyle \tan 60^\circ=\frac{BE}{CE}
\displaystyle \sqrt{3}=\frac{60}{CE}
\displaystyle CE=\frac{60}{\sqrt{3}}=20\sqrt{3}\text{ m}
\displaystyle DE=DC+CE
\displaystyle =50\sqrt{3}+20\sqrt{3}=70\sqrt{3}\text{ m}
\displaystyle AB^2=(DE)^2+(BE-AD)^2
\displaystyle =(70\sqrt{3})^2+(60-50)^2
\displaystyle =14700+100
\displaystyle =14800
\displaystyle AB=\sqrt{14800}=20\sqrt{37}\text{ m}
\displaystyle \therefore d=AB=20\sqrt{37}\text{ m}\approx121.66\text{ m}
\\

\displaystyle \textbf{Question 14. }\textbf{Case Study - 2}
\displaystyle \text{A 'circus' is a company of performers who put on shows of acrobats, clowns etc. to}
\displaystyle \text{entertain people started around 250 years back, in open fields, now generally}
\displaystyle \text{performed in tents.}
\displaystyle \text{One such 'Circus Tent' is shown below.} \displaystyle \text{The tent is in the shape of a cylinder surmounted by a conical top. If the height and}
\displaystyle \text{diameter of cylindrical part are }9\text{ m and }30\text{ m respectively and height of conical}
\displaystyle \text{part is }8\text{ m with same diameter as that of the cylindrical part, then find:}
\displaystyle \text{(i) the area of the canvas used in making the tent.}
\displaystyle \text{(ii) the cost of the canvas bought for the tent at the rate Rs. }200\text{ per sq m,}
\displaystyle \text{if }30\text{ sq m canvas was wasted during stitching.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter}=30\text{ m}
\displaystyle \therefore r=15\text{ m}
\displaystyle \text{Height of cylindrical part}=9\text{ m}
\displaystyle \text{Height of conical part}=8\text{ m}
\displaystyle \text{Slant height of cone}=l=\sqrt{15^2+8^2}
\displaystyle =\sqrt{225+64}=\sqrt{289}=17\text{ m}
\displaystyle \text{Area of canvas used}=\text{CSA of cylinder}+\text{CSA of cone}
\displaystyle =2\pi rh+\pi rl
\displaystyle =2\pi(15)(9)+\pi(15)(17)
\displaystyle =270\pi+255\pi
\displaystyle =525\pi
\displaystyle =525\times\frac{22}{7}
\displaystyle =1650\text{ sq m}
\displaystyle \therefore \text{Area of canvas used}=1650\text{ sq m}
\displaystyle \text{Canvas wasted}=30\text{ sq m}
\displaystyle \text{Total canvas bought}=1650+30=1680\text{ sq m}
\displaystyle \text{Cost}=1680\times200
\displaystyle =336000
\displaystyle \therefore \text{Cost of canvas bought = Rs. }336000
\\

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