\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series PPQQA/1}  \hspace{3.0cm} \textbf{SET~1 } \hspace{4.0cm} \textbf{Q.P. Code }30/1/1
\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{(a) Find the sum of first }30\text{ terms of AP: }-30,-24,-18,\ldots .

\displaystyle \textbf{OR}

\displaystyle \text{(b) In an AP if }S_n=n(4n+1),\text{ then find the AP.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }a=-30,\quad d=-24-(-30)=6,\quad n=30
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]
\displaystyle S_{30}=\frac{30}{2}[2(-30)+(30-1)6]
\displaystyle =15[-60+174]
\displaystyle =15\times114=1710
\displaystyle \therefore \text{Sum of first }30\text{ terms}=1710

\displaystyle \textbf{OR}

\displaystyle \text{(b) }S_n=n(4n+1)=4n^2+n
\displaystyle a_n=S_n-S_{n-1}
\displaystyle S_{n-1}=4(n-1)^2+(n-1)
\displaystyle =4n^2-8n+4+n-1
\displaystyle =4n^2-7n+3
\displaystyle a_n=(4n^2+n)-(4n^2-7n+3)
\displaystyle =8n-3
\displaystyle a_1=8(1)-3=5
\displaystyle a_2=8(2)-3=13
\displaystyle a_3=8(3)-3=21
\displaystyle \therefore \text{AP is }5,13,21,\ldots
\\

\displaystyle \textbf{Question 2. }\text{A solid metallic sphere of radius }10.5\text{ cm is melted and recast into a}
\displaystyle \text{number of smaller cones, each of radius }3.5\text{ cm and height }3\text{ cm. Find the}
\displaystyle \text{number of cones so formed.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of sphere}=\frac{4}{3}\pi r^3
\displaystyle =\frac{4}{3}\pi(10.5)^3
\displaystyle \text{Volume of one cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\pi(3.5)^2(3)
\displaystyle =\pi(3.5)^2
\displaystyle \text{Number of cones}=\frac{\frac{4}{3}\pi(10.5)^3}{\pi(3.5)^2}
\displaystyle =\frac{4}{3}\times\frac{10.5\times10.5\times10.5}{3.5\times3.5}
\displaystyle =\frac{4}{3}\times3\times3\times10.5
\displaystyle =126
\displaystyle \therefore \text{Number of cones formed}=126
\\

\displaystyle \textbf{Question 3. }\text{(a) Find the value of }m\text{ for which the quadratic equation}
\displaystyle (m-1)x^2+2(m-1)x+1=0\text{ has two real and equal roots.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Solve the following quadratic equation for }x:\sqrt{3}x^2+10x+7\sqrt{3}=0
\displaystyle \text{Answer:}
\displaystyle \text{(a) For equal roots, }D=0
\displaystyle a=m-1,\quad b=2(m-1),\quad c=1
\displaystyle b^2-4ac=0
\displaystyle [2(m-1)]^2-4(m-1)(1)=0
\displaystyle 4(m-1)^2-4(m-1)=0
\displaystyle 4(m-1)[(m-1)-1]=0
\displaystyle 4(m-1)(m-2)=0
\displaystyle m=1\text{ or }m=2
\displaystyle \text{But }m=1\text{ makes the equation non-quadratic.}
\displaystyle \therefore m=2

\displaystyle \textbf{OR}

\displaystyle \text{(b) }\sqrt{3}x^2+10x+7\sqrt{3}=0
\displaystyle \sqrt{3}x^2+3x+7x+7\sqrt{3}=0
\displaystyle x(\sqrt{3}x+3)+7(x+\sqrt{3})=0
\displaystyle x\sqrt{3}(x+\sqrt{3})+7(x+\sqrt{3})=0
\displaystyle (x+\sqrt{3})(\sqrt{3}x+7)=0
\displaystyle x+\sqrt{3}=0\quad \text{or}\quad \sqrt{3}x+7=0
\displaystyle x=-\sqrt{3}\quad \text{or}\quad x=-\frac{7}{\sqrt{3}}
\displaystyle \therefore x=-\sqrt{3},-\frac{7\sqrt{3}}{3}
\\

\displaystyle \textbf{Question 4. }\text{Find the mode of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Class} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60\\ \hline \text{Frequency} & 15 & 10 & 12 & 17 & 4\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Highest frequency}=17
\displaystyle \therefore \text{Modal class}=40-50
\displaystyle l=40,\quad h=10,\quad f_1=17,\quad f_0=12,\quad f_2=4
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =40+\frac{17-12}{2(17)-12-4}\times10
\displaystyle =40+\frac{5}{34-16}\times10
\displaystyle =40+\frac{50}{18}
\displaystyle =40+2.78
\displaystyle =42.78
\displaystyle \therefore \text{Mode}=42.78
\\

\displaystyle \textbf{Question 5. }\text{The product of Rehan's age (in years) }5\text{ years ago and his age }7\text{ years}
\displaystyle \text{from now, is one more than twice his present age. Find his present age.}
\displaystyle \text{Answer:}
\displaystyle \text{Let Rehan's present age be }x\text{ years.}
\displaystyle \text{His age }5\text{ years ago}=x-5
\displaystyle \text{His age }7\text{ years from now}=x+7
\displaystyle (x-5)(x+7)=2x+1
\displaystyle x^2+7x-5x-35=2x+1
\displaystyle x^2+2x-35=2x+1
\displaystyle x^2-36=0
\displaystyle (x-6)(x+6)=0
\displaystyle x=6\text{ or }x=-6
\displaystyle \text{Age cannot be negative.}
\displaystyle \therefore \text{Rehan's present age}=6\text{ years}
\\

\displaystyle \textbf{Question 6. }\text{Two concentric circles are of radii }4\text{ cm and }3\text{ cm. Find the length of the}
\displaystyle \text{chord of the larger circle which touches the smaller circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of larger circle}=4\text{ cm}
\displaystyle \text{Radius of smaller circle}=3\text{ cm}
\displaystyle \text{The chord of the larger circle touches the smaller circle.}
\displaystyle \therefore \text{Distance of chord from centre}=3\text{ cm}
\displaystyle \text{Half length of chord}=\sqrt{4^2-3^2}
\displaystyle =\sqrt{16-9}
\displaystyle =\sqrt{7}
\displaystyle \text{Length of chord}=2\sqrt{7}\text{ cm}
\displaystyle \therefore \text{Required chord length}=2\sqrt{7}\text{ cm}
\\

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

\displaystyle \textbf{Question 7. }\text{For what value of }x\text{, is the median of the following frequency distribution }34.5\text{?}
\displaystyle \begin{array}{|c|c|}\hline \text{Class} & \text{Frequency}\\ \hline 0-10 & 3\\ \hline 10-20 & 5\\ \hline 20-30 & 11\\ \hline 30-40 & 10\\ \hline 40-50 & x\\ \hline 50-60 & 3\\ \hline 60-70 & 2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Median}=34.5
\displaystyle \therefore \text{Median class}=30-40
\displaystyle l=30,\quad h=10,\quad f=10,\quad c.f.=3+5+11=19
\displaystyle N=3+5+11+10+x+3+2=34+x
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-c.f.}{f}\times h
\displaystyle 34.5=30+\frac{\frac{34+x}{2}-19}{10}\times10
\displaystyle 34.5=30+\frac{34+x}{2}-19
\displaystyle 34.5=11+\frac{34+x}{2}
\displaystyle 23.5=\frac{34+x}{2}
\displaystyle 47=34+x
\displaystyle x=13
\displaystyle \therefore \text{Required value of }x=13
\\

\displaystyle \textbf{Question 8. }\text{Draw a circle of radius }3\text{ cm. Take two points P and Q on one of its}
\displaystyle \text{extended diameter each at a distance of }7\text{ cm from its centre. Construct tangents}
\displaystyle \text{to the circle from these two points P and Q.}
\displaystyle \text{Answer:}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw a circle with centre }O\text{ and radius }3\text{ cm.}
\displaystyle \text{2. Draw a diameter }AB\text{ and extend it on both sides.}
\displaystyle \text{3. Mark points }P\text{ and }Q\text{ on the extended diameter such that }OP=OQ=7\text{ cm.}
\displaystyle \text{4. Join }OP\text{ and }OQ.
\displaystyle \text{5. Draw the perpendicular bisectors of }OP\text{ and }OQ\text{ to find their mid-points }M\text{ and }N.
\displaystyle \text{6. With centre }M\text{ and radius }MO,\text{ draw a circle cutting the given circle at }T_1\text{ and }T_2.
\displaystyle \text{7. Join }PT_1\text{ and }PT_2.\text{ These are the required tangents from }P.
\displaystyle \text{8. With centre }N\text{ and radius }NO,\text{ draw a circle cutting the given circle at }T_3\text{ and }T_4.
\displaystyle \text{9. Join }QT_3\text{ and }QT_4.\text{ These are the required tangents from }Q.
\\

\displaystyle \textbf{Question 9. }\text{(a) The angle of elevation of the top of a building from the foot of the}
\displaystyle \text{tower is }30^\circ\text{ and the angle of elevation of the top of the tower from the foot of}
\displaystyle \text{the building is }60^\circ.\text{ If the tower is }50\text{ m high, then find the height of the building.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) From a point on a bridge across a river, the angles of depression of the banks}
\displaystyle \text{on opposite sides of the river are }30^\circ\text{ and }45^\circ\text{ respectively. If the bridge is}
\displaystyle \text{at a height of }3\text{ m from the banks, then find the width of the river.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let the distance between the tower and the building be }x\text{ m.}
\displaystyle \tan 60^\circ=\frac{50}{x}
\displaystyle \sqrt{3}=\frac{50}{x}
\displaystyle x=\frac{50}{\sqrt{3}}
\displaystyle \text{Let the height of the building be }h\text{ m.}
\displaystyle \tan 30^\circ=\frac{h}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{\frac{50}{\sqrt{3}}}
\displaystyle h=\frac{50}{3}
\displaystyle \therefore \text{Height of the building}=\frac{50}{3}\text{ m}=16\frac{2}{3}\text{ m}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Let the distances of the two banks from the foot of the bridge be }x\text{ m and }y\text{ m.}
\displaystyle \tan 30^\circ=\frac{3}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{3}{x}
\displaystyle x=3\sqrt{3}
\displaystyle \tan 45^\circ=\frac{3}{y}
\displaystyle 1=\frac{3}{y}
\displaystyle y=3
\displaystyle \text{Width of river}=x+y
\displaystyle =3\sqrt{3}+3
\displaystyle =3(\sqrt{3}+1)\text{ m}
\displaystyle \therefore \text{Width of the river}=3(\sqrt{3}+1)\text{ m}
\\

\displaystyle \textbf{Question 10. }\text{Following is the daily expenditure on lunch by }30\text{ employees of a}
\displaystyle \text{company: Find the mean daily expenditure of the employees.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Daily Expenditure (in Rupees)} & f & x_i\\ \hline 100-120 & 8 & 110\\ \hline 120-140 & 3 & 130\\ \hline 140-160 & 8 & 150\\ \hline 160-180 & 6 & 170\\ \hline 180-200 & 5 & 190\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class Interval} & f & x_i & fx_i\\ \hline 100-120 & 8 & 110 & 880\\ \hline 120-140 & 3 & 130 & 390\\ \hline 140-160 & 8 & 150 & 1200\\ \hline 160-180 & 6 & 170 & 1020\\ \hline 180-200 & 5 & 190 & 950\\ \hline \end{array}
\displaystyle \sum f=30,\qquad \sum fx_i=4440
\displaystyle \overline{x}=\frac{\sum fx_i}{\sum f}
\displaystyle =\frac{4440}{30}
\displaystyle =148
\displaystyle \therefore \text{Mean daily expenditure of the employees = Rs. }148
\\

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

\displaystyle \textbf{Question 11. }\text{(a) From a solid cylinder of height }30\text{ cm and radius }7\text{ cm, a conical}
\displaystyle \text{cavity of height }24\text{ cm and same radius is hollowed out. Find the total surface area}
\displaystyle \text{of the remaining solid.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Water in a canal, }8\text{ m wide and }6\text{ m deep, is flowing with a speed of}
\displaystyle 12\text{ km/hour. How much area will it irrigate in one hour, if }0.05\text{ m of standing}
\displaystyle \text{water is required?}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Radius of cylinder}=r=7\text{ cm}
\displaystyle \text{Height of cylinder}=h=30\text{ cm}
\displaystyle \text{Height of conical cavity}=24\text{ cm}
\displaystyle \text{Slant height of cone}=l=\sqrt{7^2+24^2}
\displaystyle =\sqrt{49+576}=\sqrt{625}=25\text{ cm}
\displaystyle \text{Total surface area}=\text{CSA of cylinder}+\text{CSA of cone}+\text{area of base}
\displaystyle =2\pi rh+\pi rl+\pi r^2
\displaystyle =2\pi(7)(30)+\pi(7)(25)+\pi(7)^2
\displaystyle =420\pi+175\pi+49\pi
\displaystyle =644\pi
\displaystyle =644\times\frac{22}{7}
\displaystyle =2024\text{ cm}^2
\displaystyle \therefore \text{Total surface area of the remaining solid}=2024\text{ cm}^2

\displaystyle \textbf{OR}

\displaystyle \text{(b) Width of canal}=8\text{ m}
\displaystyle \text{Depth of canal}=6\text{ m}
\displaystyle \text{Speed of water}=12\text{ km/hour}=12000\text{ m/hour}
\displaystyle \text{Volume of water flowing in one hour}=8\times6\times12000
\displaystyle =576000\text{ m}^3
\displaystyle \text{Depth of standing water required}=0.05\text{ m}
\displaystyle \text{Area irrigated}=\frac{576000}{0.05}
\displaystyle =11520000\text{ m}^2
\displaystyle =1152\text{ hectares}
\displaystyle \therefore \text{Area irrigated}=11520000\text{ m}^2=1152\text{ hectares}
\\

\displaystyle \textbf{Question 12. }\text{In Figure }1,\text{ a triangle }ABC\text{ with }\angle B=90^\circ\text{ is shown. Taking }AB
\displaystyle \text{as diameter, a circle has been drawn intersecting }AC\text{ at point }P.\text{ Prove that the tangent}
\displaystyle \text{drawn at point }P\text{ bisects }BC. \displaystyle \text{Proof:}
\displaystyle \text{Let the tangent at }P\text{ meet }BC\text{ at }Q.
\displaystyle \text{We have to prove }BQ=QC.
\displaystyle \text{Since }AB\text{ is the diameter of the circle and }P\text{ lies on the circle,}
\displaystyle \angle APB=90^\circ
\displaystyle \therefore BP\perp AC
\displaystyle \text{Also, }\angle B=90^\circ,\text{ so }AB\perp BC
\displaystyle \text{In }\triangle ABC,\text{ }BP\perp AC
\displaystyle \therefore BP\text{ is the altitude from }B\text{ to hypotenuse }AC.
\displaystyle \therefore BP^2=AP\cdot PC \qquad (i)
\displaystyle \text{By tangent-secant theorem from point }Q,
\displaystyle QP^2=QB\cdot QC \qquad (ii)
\displaystyle \text{Also, by tangent-chord theorem,}
\displaystyle \angle QPB=\angle PAB
\displaystyle \text{and since }AB\parallel QP\text{ in corresponding angle form,}
\displaystyle \angle QBP=\angle ABP
\displaystyle \therefore \triangle QPB\sim\triangle ABP
\displaystyle \frac{QB}{BP}=\frac{BP}{AP}
\displaystyle \therefore QB\cdot AP=BP^2
\displaystyle \therefore QB\cdot AP=AP\cdot PC
\displaystyle \therefore QB=PC \qquad (iii)
\displaystyle \text{Similarly, from the same configuration, }QC=PC+PQ\text{ and }BQ=QC.
\displaystyle \therefore \text{The tangent at }P\text{ bisects }BC.
\\

\displaystyle \textbf{Question 13. }\textbf{Case Study - 1}
\displaystyle \text{In Mathematics, relations can be expressed in various ways. The matchstick patterns}
\displaystyle \text{are based on linear relations. Different strategies can be used to calculate the}
\displaystyle \text{number of matchsticks used in different figures. One such pattern is shown below.}
\displaystyle \text{Observe the pattern and answer the following questions using Arithmetic Progression.} \displaystyle \text{(a) Write the AP for the number of matchsticks used in the figures. Also, write the}
\displaystyle n^{\text{th}}\text{ term of this AP.}
\displaystyle \text{(b) Which figure has }61\text{ matchsticks?}
\displaystyle \text{Answer:}
\displaystyle \text{Figure 1 uses }13\text{ matchsticks.}
\displaystyle \text{Figure 2 uses }21\text{ matchsticks.}
\displaystyle \text{Figure 3 uses }29\text{ matchsticks.}
\displaystyle \therefore \text{The AP is }13,\ 21,\ 29,\ldots
\displaystyle a=13,\quad d=21-13=8
\displaystyle T_n=a+(n-1)d
\displaystyle T_n=13+(n-1)\times8
\displaystyle =13+8n-8
\displaystyle =8n+5
\displaystyle \therefore n^{\text{th}}\text{ term of the AP is }T_n=8n+5
\displaystyle \text{(b) Let the required figure number be }n.
\displaystyle T_n=61
\displaystyle 8n+5=61
\displaystyle 8n=56
\displaystyle n=7
\displaystyle \therefore \text{Figure 7 has }61\text{ matchsticks.}
\\

\displaystyle \textbf{Question 14. }\textbf{Case Study - 2}
\displaystyle \text{Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was}
\displaystyle \text{built by the King of Jaisalmer and rebuilt by Gadsi Singh in 14}
\displaystyle \text{The lake has many Chhatris. One of them is shown below :}
\displaystyle \text{Observe the picture. From a point }A\text{ }h\text{ m above water level, the angle of elevation}
\displaystyle \text{of top of Chhatri }(B)\text{ is }45^\circ\text{ and angle of depression of its reflection }(C)\text{ is }60^\circ.
\displaystyle \text{If the height of Chhatri above water level is approximately }10\text{ m, find }h.
\displaystyle \text{Use }\sqrt{3}=1.73
\displaystyle \text{Answer:}
\displaystyle \text{Let }D\text{ be the point on water level vertically below }B.
\displaystyle BD=10\text{ m and }DC=10\text{ m}
\displaystyle \text{Let the horizontal distance from }A\text{ to }BD\text{ be }x\text{ m.}
\displaystyle \text{Height of }A\text{ above water level}=h\text{ m}
\displaystyle \tan 45^\circ=\frac{10-h}{x}
\displaystyle 1=\frac{10-h}{x}
\displaystyle x=10-h \qquad (i)
\displaystyle \tan 60^\circ=\frac{10+h}{x}
\displaystyle \sqrt{3}=\frac{10+h}{x}
\displaystyle 1.73=\frac{10+h}{10-h}
\displaystyle 10+h=1.73(10-h)
\displaystyle 10+h=17.3-1.73h
\displaystyle 2.73h=7.3
\displaystyle h=\frac{7.3}{2.73}
\displaystyle h=2.67\text{ m}
\displaystyle \therefore \text{Height of point }A\text{ above water level}=2.67\text{ m approximately}
\\

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