\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series RHB/1}                                                                 \displaystyle \textbf{Code No. }30/1/1
\displaystyle \text{Roll No.}

  • \displaystyle \text{Candidates must write the Code on the title page of the answer-book.}
  • \displaystyle \text{Please check that this question paper contains 16 printed pages.}
  • \displaystyle \text{Code number 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{Please check that this question paper contains 34 questions.}
  • \displaystyle \text{Please write down the serial number of the question before attempting it.}
  • \displaystyle \text{15 minutes time has been allotted to read this question paper. The question paper will be}
  • \displaystyle \text{distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question}
  • \displaystyle \text{paper only and will not write any answer on the answer-script during this period.}

\displaystyle \textbf{SUMMATIVE ASSESSMENT-II}
\displaystyle \textbf{MATHEMATICS}

\displaystyle \text{Time allowed : 3 hours}                                                                 \displaystyle \text{Maximum marks : 90}

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of 34 questions divided into four sections A,}
\displaystyle \text{B, C and D.}
\displaystyle \text{(iii) Section A contains 8 questions of one mark each, which are multiple}
\displaystyle \text{choice type questions, Section B contains 6 questions of two marks each,}
\displaystyle \text{Section C contains 10 questions of three marks each, and Section D}
\displaystyle \text{contains 10 questions of four marks each.}
\displaystyle \text{(iv) Use of calculators is not permitted.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Question Numbers 1 to 10 carry one mark each. In each of these questions, four}
\displaystyle \text{alternative choices have been provided of which only one is correct. Select the}
\displaystyle \text{correct choice.}
\\

\displaystyle \textbf{Question 1. }\text{The roots of the equation }x^2-3x-m(m+3)=0,\text{ where }m\text{ is a constant, are}
\displaystyle \text{(A) }m,\ m+3
\displaystyle \text{(B) }-m,\ m+3
\displaystyle \text{(C) }m,\ -(m+3)
\displaystyle \text{(D) }-m,\ -(m+3)
\displaystyle \text{Answer:}
\displaystyle x^{2}-3x-m(m+3)=0
\displaystyle x^{2}-3x-m^{2}-3m=0
\displaystyle x^{2}-(m+3-m)x-m(m+3)=0
\displaystyle x^{2}-(m+3)x+mx-m(m+3)=0
\displaystyle x\{x-(m+3)\}+m\{x-(m+3)\}=0
\displaystyle (x+m)\{x-(m+3)\}=0
\displaystyle x=-m\text{ or }x=m+3
\displaystyle \therefore \text{Roots are }-m,\ m+3
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 2. }\text{If the common difference of an A.P. is }3,\text{ then }a_{20}-a_{15}\text{ is}
\displaystyle \text{(A) }5
\displaystyle \text{(B) }3
\displaystyle \text{(C) }15
\displaystyle \text{(D) }20
\displaystyle \text{Answer:}
\displaystyle a_{n}=a+(n-1)d
\displaystyle a_{20}=a+19d
\displaystyle a_{15}=a+14d
\displaystyle a_{20}-a_{15}=(a+19d)-(a+14d)
\displaystyle =5d
\displaystyle =5\times3=15
\displaystyle \therefore a_{20}-a_{15}=15
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 3. }\text{In Figure }1,\ O\text{ is the centre of a circle, }PQ\text{ is a chord and }PT\text{ is the}
\displaystyle \text{tangent at }P.\text{ If }\angle POQ=70^\circ,\text{ then }\angle TPQ\text{ is equal to} \displaystyle \text{(A) }55^\circ
\displaystyle \text{(B) }70^\circ
\displaystyle \text{(C) }45^\circ
\displaystyle \text{(D) }35^\circ
\displaystyle \text{Answer:}
\displaystyle OP=OQ\text{ }(\text{radii of the same circle})
\displaystyle \therefore \angle OPQ=\angle OQP
\displaystyle \angle POQ=70^{\circ}
\displaystyle \angle OPQ+\angle OQP+\angle POQ=180^{\circ}
\displaystyle 2\angle OPQ+70^{\circ}=180^{\circ}
\displaystyle 2\angle OPQ=110^{\circ}
\displaystyle \angle OPQ=55^{\circ}
\displaystyle OP\perp PT\text{ }(\text{radius is perpendicular to tangent at the point of contact})
\displaystyle \therefore \angle OPT=90^{\circ}
\displaystyle \angle TPQ=90^{\circ}-\angle OPQ
\displaystyle =90^{\circ}-55^{\circ}=35^{\circ}
\displaystyle \therefore \angle TPQ=35^{\circ}
\displaystyle \text{Hence, the correct option is (D).}
\\

\displaystyle \textbf{Question 4. }\text{In Figure }2,\ AB\text{ and }AC\text{ are tangents to the circle with centre }O\text{ such}
\displaystyle \text{that }\angle BAC=40^\circ.\text{ Then }\angle BOC\text{ is equal to} \displaystyle \text{(A) }40^\circ
\displaystyle \text{(B) }50^\circ
\displaystyle \text{(C) }140^\circ
\displaystyle \text{(D) }150^\circ
\displaystyle \text{Answer:}
\displaystyle OB\perp AB\text{ and }OC\perp AC
\displaystyle \therefore \angle ABO=90^{\circ}\text{ and }\angle ACO=90^{\circ}
\displaystyle \text{In quadrilateral }ABOC,
\displaystyle \angle ABO+\angle BOC+\angle ACO+\angle BAC=360^{\circ}
\displaystyle 90^{\circ}+\angle BOC+90^{\circ}+40^{\circ}=360^{\circ}
\displaystyle \angle BOC+220^{\circ}=360^{\circ}
\displaystyle \angle BOC=140^{\circ}
\displaystyle \therefore \angle BOC=140^{\circ}
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 5. }\text{The perimeter (in cm) of a square circumscribing a circle of radius }a\text{ cm, is}
\displaystyle \text{(A) }8a
\displaystyle \text{(B) }4a
\displaystyle \text{(C) }2a
\displaystyle \text{(D) }16a
\displaystyle \text{Answer:}
\displaystyle \text{Since the square circumscribes the circle, the circle is inscribed in the square.}
\displaystyle \text{Therefore, side of the square}=\text{diameter of the circle}=2a
\displaystyle \text{Perimeter of the square}=4\times\text{side}
\displaystyle =4\times2a
\displaystyle =8a
\displaystyle \therefore \text{The perimeter of the square is }8a\text{ cm}
\displaystyle \text{Hence, the correct option is (A).}
\\

\displaystyle \textbf{Question 6. }\text{The radius (in cm) of the largest right circular cone that can be cut out}
\displaystyle \text{from a cube of edge }4.2\text{ cm is}
\displaystyle \text{(A) }4.2
\displaystyle \text{(B) }2.1
\displaystyle \text{(C) }8.4
\displaystyle \text{(D) }1.05
\displaystyle \text{Answer:}
\displaystyle \text{The largest right circular cone that can be cut from a cube has its base}
\displaystyle \text{inscribed in one face of the cube.}
\displaystyle \text{Diameter of the base circle}=\text{edge of the cube}=4.2\text{ cm}
\displaystyle \text{Radius of the base circle}=\frac{4.2}{2}=2.1\text{ cm}
\displaystyle \therefore \text{Radius of the largest cone}=2.1\text{ cm}
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 7. }\text{A tower stands vertically on the ground. From a point on the ground which is }25\text{ m away}
\displaystyle \text{from the foot of the tower, the angle of elevation of the top of the tower is found to be }45^\circ.
\displaystyle \text{Then the height (in meters) of the tower is}
\displaystyle \text{(A) }25\sqrt{2}
\displaystyle \text{(B) }25\sqrt{3}
\displaystyle \text{(C) }25
\displaystyle \text{(D) }12.5
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m.}
\displaystyle \tan45^{\circ}=\frac{\text{height of tower}}{\text{distance from foot of tower}}
\displaystyle \tan45^{\circ}=\frac{h}{25}
\displaystyle 1=\frac{h}{25}
\displaystyle h=25
\displaystyle \therefore \text{Height of the tower}=25\text{ m}
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 8. }\text{If }P\left(\frac{a}{2},4\right)\text{ is the mid-point of the line-segment joining the points}
\displaystyle A(-6,5)\text{ and }B(-2,3),\text{ then the value of }a\text{ is}
\displaystyle \text{(A) }-8
\displaystyle \text{(B) }3
\displaystyle \text{(C) }-4
\displaystyle \text{(D) }4
\displaystyle \text{Answer:}
\displaystyle \text{Mid-point of }A(-6,\ 5)\text{ and }B(-2,\ 3)\text{ is}
\displaystyle \left(\frac{-6+(-2)}{2},\ \frac{5+3}{2}\right)
\displaystyle =\left(\frac{-8}{2},\ \frac{8}{2}\right)
\displaystyle =(-4,\ 4)
\displaystyle P\left(\frac{a}{2},\ 4\right)=(-4,\ 4)
\displaystyle \therefore \frac{a}{2}=-4
\displaystyle a=-8
\displaystyle \therefore \text{The value of }a=-8
\displaystyle \text{Hence, the correct option is (A).}
\\

\displaystyle \textbf{Question 9. }\text{If }A\text{ and }B\text{ are the points }(-6,7)\text{ and }(-1,-5)\text{ respectively, then the}
\displaystyle \text{distance }2AB\text{ is equal to}
\displaystyle \text{(A) }13
\displaystyle \text{(B) }26
\displaystyle \text{(C) }169
\displaystyle \text{(D) }238
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(-1+6)^2+(-5-7)^2}
\displaystyle =\sqrt{5^2+(-12)^2}
\displaystyle =\sqrt{25+144}
\displaystyle =\sqrt{169}
\displaystyle =13
\displaystyle \therefore 2AB=2\times13=26
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 10. }\text{A card is drawn from a well-shuffled deck of }52\text{ playing cards. The}
\displaystyle \text{probability that the card will not be an ace is}
\displaystyle \text{(A) }\frac{1}{13}
\displaystyle \text{(B) }\frac{1}{4}
\displaystyle \text{(C) }\frac{12}{13}
\displaystyle \text{(D) }\frac{3}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=52
\displaystyle \text{Number of aces}=4
\displaystyle \text{Number of cards which are not aces}=52-4=48
\displaystyle \text{Probability of not drawing an ace}=\frac{48}{52}
\displaystyle =\frac{12}{13}
\displaystyle \therefore \text{The required probability is }\frac{12}{13}
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question Numbers 11 to 18 carry two marks each.}

\displaystyle \textbf{Question 11. }\text{Find the value of }m\text{ so that the quadratic equation }mx(x-7)+49=0
\displaystyle \text{has two equal roots.}
\displaystyle \text{Answer:}
\displaystyle mx^{2}-7mx+49=0
\displaystyle a=m,\quad b=-7m,\quad c=49
\displaystyle \text{For equal roots, }D=b^{2}-4ac=0
\displaystyle (-7m)^{2}-4(m)(49)=0
\displaystyle 49m^{2}-196m=0
\displaystyle 49m(m-4)=0
\displaystyle m=0\text{ or }m=4
\displaystyle m\neq0\text{ since the equation must remain quadratic}
\displaystyle \therefore m=4
\\

\displaystyle \textbf{Question 12. }\text{Find how many two-digit numbers are divisible by }6.
\displaystyle \text{Answer:}
\displaystyle \text{Smallest two-digit multiple of }6=12
\displaystyle \text{Largest two-digit multiple of }6=96
\displaystyle \text{The numbers form an A.P. }12,\ 18,\ 24,\ldots,\ 96
\displaystyle a=12,\quad l=96,\quad d=6
\displaystyle l=a+(n-1)d
\displaystyle 96=12+(n-1)6
\displaystyle 84=6(n-1)
\displaystyle n-1=14
\displaystyle n=15
\displaystyle \therefore \text{There are }15\text{ two-digit numbers divisible by }6.
\\

\displaystyle \textbf{Question 13. }\text{In Figure 3, a circle touches all the four sides of a quadrilateral }ABCD
\displaystyle \text{whose sides are }AB=6\text{ cm, }BC=9\text{ cm and }CD=8\text{ cm. Find the length}
\displaystyle \text{of side }AD. \displaystyle \text{Answer:}
\displaystyle \text{Since a circle touches all the four sides of the quadrilateral,}
\displaystyle \text{it is a tangential quadrilateral.}
\displaystyle AB+CD=BC+AD
\displaystyle 6+8=9+AD
\displaystyle 14=9+AD
\displaystyle AD=5
\displaystyle \therefore \text{Length of side }AD=5\text{ cm}
\\

\displaystyle \textbf{Question 14. }\text{Draw a line segment }AB\text{ of length }7\text{ cm. Using ruler and compasses,}
\displaystyle \text{find a point }P\text{ on }AB\text{ such that }\frac{AP}{AB}=\frac{3}{5}.
\displaystyle \text{Answer:}
\displaystyle \text{Construction Steps:}
\displaystyle \text{1. Draw a line segment }AB=7\text{ cm.}
\displaystyle \text{2. From }A,\text{ draw a ray }AX\text{ making an acute angle with }AB.
\displaystyle \text{3. Mark five equal points }A_{1},A_{2},A_{3},A_{4},A_{5}\text{ on }AX.
\displaystyle \text{4. Join }A_{5}\text{ to }B.
\displaystyle \text{5. Through }A_{3},\text{ draw a line parallel to }A_{5}B\text{ meeting }AB\text{ at }P.
\displaystyle \text{6. Then }P\text{ divides }AB\text{ in the ratio }3:5.
\displaystyle \therefore \frac{AP}{AB}=\frac{3}{5}
\\

\displaystyle \textbf{Question 15. }\text{Find the perimeter of the shaded region in Figure 4, if }ABCD\text{ is a}
\displaystyle \text{square of side }14\text{ cm and }APB\text{ and }CPD\text{ are semicircles. }\left[\text{Use }\pi=\frac{22}{7}\right] \displaystyle \text{Answer:}
\displaystyle \text{Side of square}=14\text{ cm}
\displaystyle \text{Radius of each semicircle}=\frac{14}{2}=7\text{ cm}
\displaystyle \text{Perimeter of shaded region}=\text{AD}+\text{BC}+\text{arc APB}+\text{arc CPD}
\displaystyle =14+14+\pi r+\pi r
\displaystyle =28+2\pi(7)
\displaystyle =28+14\pi
\displaystyle =28+14\times\frac{22}{7}
\displaystyle =28+44
\displaystyle =72\text{ cm}
\displaystyle \therefore \text{Perimeter of the shaded region}=72\text{ cm}
\\

\displaystyle \textbf{Question 16. }\text{Two cubes each of volume }27\text{ cm}^3\text{ are joined end to end to form a solid.}
\displaystyle \text{Find the surface area of the resulting cuboid.}

\displaystyle \textbf{OR}

\displaystyle \text{A cone of height }20\text{ cm and radius of base }5\text{ cm is made up of modelling clay.}
\displaystyle \text{A child reshapes it in the form of a sphere. Find the diameter of the sphere.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of each cube}=27\text{ cm}^{3}
\displaystyle a^{3}=27
\displaystyle a=3\text{ cm}
\displaystyle \text{Dimensions of resulting cuboid}=6\text{ cm}\times3\text{ cm}\times3\text{ cm}
\displaystyle \text{Surface area of cuboid}=2(lb+bh+hl)
\displaystyle =2(6\times3+3\times3+3\times6)
\displaystyle =2(18+9+18)
\displaystyle =90\text{ cm}^{2}
\displaystyle \therefore \text{Surface area of the resulting cuboid}=90\text{ cm}^{2}

\displaystyle \textbf{OR}

\displaystyle \text{Volume of cone}=\frac{1}{3}\pi r^{2}h
\displaystyle =\frac{1}{3}\pi(5)^{2}(20)
\displaystyle =\frac{500\pi}{3}\text{ cm}^{3}
\displaystyle \text{Let the radius of the sphere be }R\text{ cm.}
\displaystyle \frac{4}{3}\pi R^{3}=\frac{500\pi}{3}
\displaystyle 4R^{3}=500
\displaystyle R^{3}=125
\displaystyle R=5\text{ cm}
\displaystyle \text{Diameter}=2R=10\text{ cm}
\displaystyle \therefore \text{Diameter of the sphere}=10\text{ cm}
\\

\displaystyle \textbf{Question 17. }\text{Find the value of }y\text{ for which the distance between the points }A(3,-1)
\displaystyle \text{and }B(11,y)\text{ is }10\text{ units.}
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(11-3)^{2}+\{y-(-1)\}^{2}}
\displaystyle 10=\sqrt{8^{2}+(y+1)^{2}}
\displaystyle 100=64+(y+1)^{2}
\displaystyle (y+1)^{2}=36
\displaystyle y+1=\pm6
\displaystyle y=5\text{ or }y=-7
\displaystyle \therefore \text{The value of }y\text{ is }5\text{ or }-7.
\\

\displaystyle \textbf{Question 18. }\text{A ticket is drawn at random from a bag containing tickets numbered}
\displaystyle \text{from }1\text{ to }40.\text{ Find the probability that the selected ticket has a}
\displaystyle \text{number which is a multiple of }5.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of tickets}=40
\displaystyle \text{Multiples of }5\text{ between }1\text{ and }40\text{ are}
\displaystyle 5,\ 10,\ 15,\ 20,\ 25,\ 30,\ 35,\ 40
\displaystyle \text{Number of favourable outcomes}=8
\displaystyle \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}
\displaystyle =\frac{8}{40}
\displaystyle =\frac{1}{5}
\displaystyle \therefore \text{The required probability is }\frac{1}{5}
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question Numbers 19 to 28 carry three marks each.}

\displaystyle \textbf{Question 19. }\text{Find the roots of the following quadratic equation:}
\displaystyle x^2-3\sqrt{5}x+10=0
\displaystyle \text{Answer:}
\displaystyle x^{2}-3\sqrt{5}\,x+10=0
\displaystyle a=1,\quad b=-3\sqrt{5},\quad c=10
\displaystyle D=b^{2}-4ac
\displaystyle =(-3\sqrt{5})^{2}-4(1)(10)
\displaystyle =45-40=5
\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}
\displaystyle =\frac{3\sqrt{5}\pm\sqrt{5}}{2}
\displaystyle =\frac{\sqrt{5}(3\pm1)}{2}
\displaystyle x=\frac{4\sqrt{5}}{2}=2\sqrt{5}\quad \text{or}\quad x=\frac{2\sqrt{5}}{2}=\sqrt{5}
\displaystyle \therefore \text{The roots are }\sqrt{5}\text{ and }2\sqrt{5}.
\\

\displaystyle \textbf{Question 20. }\text{Find an A.P. whose fourth term is }9\text{ and the sum of its sixth term and}
\displaystyle \text{thirteenth term is }40.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle T_{4}=a+3d=9 \qquad (1)
\displaystyle T_{6}+T_{13}=40
\displaystyle (a+5d)+(a+12d)=40
\displaystyle 2a+17d=40 \qquad (2)
\displaystyle \text{From (1), }a=9-3d
\displaystyle \text{Substituting in (2),}
\displaystyle 2(9-3d)+17d=40
\displaystyle 18-6d+17d=40
\displaystyle 11d=22
\displaystyle d=2
\displaystyle a=9-3(2)=3
\displaystyle \therefore \text{The required A.P. is }3,\ 5,\ 7,\ 9,\ 11,\ 13,\ldots
\\

\displaystyle \textbf{Question 21. }\text{In Figure }5,\text{ a triangle }PQR\text{ is drawn to circumscribe a circle of radius}
\displaystyle 6\text{ cm such that the segments }QT\text{ and }TR\text{ into which }QR\text{ is divided by}
\displaystyle \text{the point of contact }T,\text{ are of lengths }12\text{ cm and }9\text{ cm respectively. If}
\displaystyle \text{the area of }\triangle PQR=189\text{ cm}^2,\text{ then find the lengths of sides }PQ\text{ and }PR. \displaystyle \text{Answer:}
\displaystyle QR=QT+TR=12+9=21\text{ cm}
\displaystyle \text{Area of }\triangle PQR=rs
\displaystyle 189=6s
\displaystyle s=\frac{189}{6}=31.5\text{ cm}
\displaystyle \therefore PQ+QR+PR=2s=63\text{ cm}
\displaystyle PQ+PR+21=63
\displaystyle PQ+PR=42 \qquad (1)
\displaystyle \text{Tangents drawn from an external point are equal.}
\displaystyle \text{Let the tangent lengths from }P\text{ be }x\text{ cm each.}
\displaystyle PQ=x+12
\displaystyle PR=x+9
\displaystyle (x+12)+(x+9)=42
\displaystyle 2x+21=42
\displaystyle 2x=21
\displaystyle x=10.5
\displaystyle PQ=10.5+12=22.5\text{ cm}
\displaystyle PR=10.5+9=19.5\text{ cm}
\displaystyle \therefore PQ=22.5\text{ cm and }PR=19.5\text{ cm}
\\

\displaystyle \textbf{Question 22. }\text{Draw a pair of tangents to a circle of radius }3\text{ cm, which are inclined}
\displaystyle \text{to each other at an angle of }60^\circ.

\displaystyle \textbf{OR}

\displaystyle \text{Draw a right triangle in which the sides (other than hypotenuse) are of lengths }4\text{ cm and }3\text{ cm.}
\displaystyle \text{Then construct another triangle whose sides are }\frac{3}{5}\text{ times the corresponding sides of the given triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Construction Steps:}
\displaystyle \text{1. Draw a circle with centre }O\text{ and radius }3\text{ cm.}
\displaystyle \text{2. Draw two radii }OA\text{ and }OB\text{ such that }\angle AOB=120^{\circ}.
\displaystyle \text{3. At }A,\text{ draw a line perpendicular to }OA.
\displaystyle \text{4. At }B,\text{ draw a line perpendicular to }OB.
\displaystyle \text{5. Let these two perpendicular lines meet at }P.
\displaystyle \text{6. Then }PA\text{ and }PB\text{ are the required tangents inclined at }60^{\circ}.

\displaystyle \textbf{OR}

\displaystyle \text{Construction Steps:}
\displaystyle \text{1. Draw }AB=4\text{ cm.}
\displaystyle \text{2. At }A,\text{ draw }AC\perp AB\text{ and cut }AC=3\text{ cm.}
\displaystyle \text{3. Join }BC.\text{ Then }\triangle ABC\text{ is the required right triangle.}
\displaystyle \text{4. From }A,\text{ draw a ray }AX\text{ making an acute angle with }AB.
\displaystyle \text{5. Mark five equal points }A_{1},A_{2},A_{3},A_{4},A_{5}\text{ on }AX.
\displaystyle \text{6. Join }A_{5}\text{ to }B.
\displaystyle \text{7. Through }A_{3},\text{ draw a line parallel to }A_{5}B\text{ meeting }AB\text{ at }B'.
\displaystyle \text{8. Through }B',\text{ draw a line parallel to }BC\text{ meeting }AC\text{ at }C'.
\displaystyle \text{9. Then }\triangle AB'C'\text{ is the required triangle whose sides are }\frac{3}{5}\text{ times the}
\displaystyle \text{corresponding sides of }\triangle ABC.
\\

\displaystyle \textbf{Question 23. }\text{A chord of a circle of radius }14\text{ cm subtends an angle of }120^\circ\text{ at the}
\displaystyle \text{centre. Find the area of the corresponding minor segment of the circle.}
\displaystyle \left[\text{Use }\pi=\frac{22}{7}\text{ and }\sqrt{3}=1.73\right]
\displaystyle \text{Answer:}
\displaystyle \text{Area of sector }=\frac{120}{360}\times\frac{22}{7}\times14\times14
\displaystyle =\frac{616}{3}\text{ cm}^{2}
\displaystyle \text{Area of }\triangle=\frac{1}{2}r^{2}\sin120^{\circ}
\displaystyle =\frac{1}{2}\times14\times14\times\frac{\sqrt{3}}{2}
\displaystyle =49\sqrt{3}
\displaystyle =49\times1.73=84.77\text{ cm}^{2}
\displaystyle \text{Area of minor segment}=\frac{616}{3}-84.77
\displaystyle =120.56\text{ cm}^{2}\ (\text{approx.})
\displaystyle \therefore \text{Area of the minor segment}=120.56\text{ cm}^{2}
\\

\displaystyle \textbf{Question 24. }\text{An open metal bucket is in the shape of a frustum of a cone of height}
\displaystyle 21\text{ cm with radii of its lower and upper ends as }10\text{ cm and }20\text{ cm}
\displaystyle \text{respectively. Find the cost of milk which can completely fill the bucket}
\displaystyle \text{at Rs. }30\text{ per litre. }\left[\text{Use }\pi=\frac{22}{7}\right]
\displaystyle \text{Answer:}
\displaystyle R=20\text{ cm},\quad r=10\text{ cm},\quad h=21\text{ cm}
\displaystyle \text{Volume of frustum}=\frac{1}{3}\pi h(R^{2}+r^{2}+Rr)
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times21\times(400+100+200)
\displaystyle =22\times700
\displaystyle =15400\text{ cm}^{3}
\displaystyle =15.4\text{ litres}
\displaystyle \text{Cost}=15.4\times30
\displaystyle =\text{Rs. }462
\displaystyle \therefore \text{Cost of milk}= \text{Rs. }462
\\

\displaystyle \textbf{Question 25. }\text{Point }P(x,4)\text{ lies on the line segment joining the points }A(-5,8)\text{ and}
\displaystyle B(4,-10).\text{ Find the ratio in which point }P\text{ divides the line segment }AB.
\displaystyle \text{Also find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ divide }AB\text{ in the ratio }m:n.
\displaystyle \text{Using section formula for y-coordinate,}
\displaystyle 4=\frac{m(-10)+n(8)}{m+n}
\displaystyle 4m+4n=-10m+8n
\displaystyle 14m=4n
\displaystyle 7m=2n
\displaystyle m:n=2:7
\displaystyle \text{Using section formula for x-coordinate,}
\displaystyle x=\frac{2(4)+7(-5)}{2+7}
\displaystyle =\frac{8-35}{9}
\displaystyle =-3
\displaystyle \therefore P\text{ divides }AB\text{ in the ratio }2:7\text{ and }x=-3
\\

\displaystyle \textbf{Question 26. }\text{Find the area of the quadrilateral }ABCD,\text{ whose vertices are }A(-3,-1),
\displaystyle B(-2,-4),\ C(4,-1)\text{ and }D(3,4).

\displaystyle \textbf{OR}

\displaystyle \text{Find the area of the triangle formed by joining the mid-points of the sides of the triangle}
\displaystyle \text{whose vertices are }A(2,1),\ B(4,3)\text{ and }C(2,5).
\displaystyle \text{Answer:}
\displaystyle \text{Area of quadrilateral }ABCD=\frac{1}{2}\left|x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{4}+x_{4}y_{1}\right.
\displaystyle \left.-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{4}+y_{4}x_{1})\right|
\displaystyle =\frac{1}{2}\left|(-3)(-4)+(-2)(-1)+(4)(4)+(3)(-1)\right.
\displaystyle \left.-\{(-1)(-2)+(-4)(4)+(-1)(3)+(4)(-3)\}\right|
\displaystyle =\frac{1}{2}\left|12+2+16-3-\{2-16-3-12\}\right|
\displaystyle =\frac{1}{2}\left|27-(-29)\right|
\displaystyle =\frac{56}{2}=28
\displaystyle \therefore \text{Area of quadrilateral }ABCD=28\text{ square units}

\displaystyle \textbf{OR}

\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|2(3-5)+4(5-1)+2(1-3)\right|
\displaystyle =\frac{1}{2}\left|-4+16-4\right|
\displaystyle =4\text{ square units}
\displaystyle \text{The triangle formed by joining the mid-points has area }=\frac{1}{4}\times\text{area of }\triangle ABC
\displaystyle =\frac{1}{4}\times4=1\text{ square unit}
\displaystyle \therefore \text{Required area}=1\text{ square unit}
\\

\displaystyle \textbf{Question 27. }\text{From the top of a vertical tower, the angles of depression of two cars,}
\displaystyle \text{in the same straight line with the base of the tower, at an instant are found}
\displaystyle \text{to be }45^\circ\text{ and }60^\circ.\text{ If the cars are }100\text{ m apart and are on the same side of}
\displaystyle \text{the tower, find the height of the tower. }\left[\text{Use }\sqrt{3}=1.73\right]
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m.}
\displaystyle \text{The car nearer to the tower has angle of depression }60^{\circ}.
\displaystyle \text{The car farther from the tower has angle of depression }45^{\circ}.
\displaystyle \text{Distance of nearer car from the tower}=\frac{h}{\tan60^{\circ}}=\frac{h}{\sqrt{3}}
\displaystyle \text{Distance of farther car from the tower}=\frac{h}{\tan45^{\circ}}=h
\displaystyle h-\frac{h}{\sqrt{3}}=100
\displaystyle h\left(1-\frac{1}{\sqrt{3}}\right)=100
\displaystyle h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)=100
\displaystyle h=\frac{100\sqrt{3}}{\sqrt{3}-1}
\displaystyle =\frac{100\times1.73}{1.73-1}
\displaystyle =\frac{173}{0.73}=236.99
\displaystyle \therefore \text{Height of the tower}=236.99\text{ m}\text{ approximately}
\\

\displaystyle \textbf{Question 28. }\text{Two dice are rolled once. Find the probability of getting such numbers}
\displaystyle \text{on the two dice, whose product is }12.

\displaystyle \textbf{OR}

\displaystyle \text{A box contains }80\text{ discs which are numbered from }1\text{ to }80.\text{ If one disc}
\displaystyle \text{is drawn at random from the box, find the probability that it bears a perfect square number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes when two dice are rolled}=36
\displaystyle \text{Favourable outcomes for product }12\text{ are }(2,6),(3,4),(4,3),(6,2)
\displaystyle \text{Number of favourable outcomes}=4
\displaystyle \text{Probability}=\frac{4}{36}=\frac{1}{9}
\displaystyle \therefore \text{Required probability}=\frac{1}{9}

\displaystyle \textbf{OR}

\displaystyle \text{Total number of discs}=80
\displaystyle \text{Perfect squares from }1\text{ to }80\text{ are }1,4,9,16,25,36,49,64
\displaystyle \text{Number of favourable outcomes}=8
\displaystyle \text{Probability}=\frac{8}{80}=\frac{1}{10}
\displaystyle \therefore \text{Required probability}=\frac{1}{10}
\\

\displaystyle \textbf{SECTION D}
\displaystyle \text{Question Numbers 29 to 34 carry four marks each.}

\displaystyle \textbf{Question 29. }\text{Prove that the tangent at any point of a circle is perpendicular to the}
\displaystyle \text{radius through the point of contact.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }O\text{ be the centre of the circle and }PT\text{ be the tangent at point }P.
\displaystyle \text{Let }Q\text{ be any point on }PT\text{ other than }P.
\displaystyle \text{Join }OQ.
\displaystyle \text{Since }Q\text{ lies outside the circle, }OQ>OP
\displaystyle \therefore OP\text{ is the shortest distance from }O\text{ to the tangent }PT.
\displaystyle \text{The shortest distance from a point to a line is the perpendicular distance.}
\displaystyle \therefore OP\perp PT
\displaystyle \text{Hence, the tangent at any point of a circle is perpendicular to the radius}
\displaystyle \text{through the point of contact.}
\\

\displaystyle \textbf{Question 30. }\text{The first and the last terms of an A.P. are }8\text{ and }350\text{ respectively. If its}
\displaystyle \text{common difference is }9,\text{ how many terms are there and what is their sum?}

\displaystyle \textbf{OR}

\displaystyle \text{How many multiples of }4\text{ lie between }10\text{ and }250?\text{ Also find their sum.}
\displaystyle \text{Answer:}
\displaystyle a=8,\quad l=350,\quad d=9
\displaystyle l=a+(n-1)d
\displaystyle 350=8+(n-1)9
\displaystyle 342=9(n-1)
\displaystyle n-1=38
\displaystyle n=39
\displaystyle S_{n}=\frac{n}{2}(a+l)
\displaystyle S_{39}=\frac{39}{2}(8+350)
\displaystyle =\frac{39}{2}\times358
\displaystyle =39\times179=6981
\displaystyle \therefore \text{Number of terms}=39\text{ and their sum}=6981

\displaystyle \textbf{OR}

\displaystyle \text{Multiples of }4\text{ between }10\text{ and }250\text{ are }12,16,20,\ldots,248
\displaystyle a=12,\quad l=248,\quad d=4
\displaystyle 248=12+(n-1)4
\displaystyle 236=4(n-1)
\displaystyle n-1=59
\displaystyle n=60
\displaystyle S_{60}=\frac{60}{2}(12+248)
\displaystyle =30\times260=7800
\displaystyle \therefore \text{There are }60\text{ multiples of }4\text{ and their sum is }7800
\\

\displaystyle \textbf{Question 31. }\text{A train travels }180\text{ km at a uniform speed. If the speed had been}
\displaystyle 9\text{ km/hour more, it would have taken }1\text{ hour less for the same journey.}
\displaystyle \text{Find the speed of the train.}

\displaystyle \textbf{OR}

\displaystyle \text{Find the roots of the equation }\frac{1}{2x-3}+\frac{1}{x-5}=1,\quad x\neq\frac{3}{2},5.
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the train be }x\text{ km/hour.}
\displaystyle \text{Time taken}=\frac{180}{x}\text{ hours}
\displaystyle \text{New speed}=(x+9)\text{ km/hour}
\displaystyle \text{New time taken}=\frac{180}{x+9}\text{ hours}
\displaystyle \frac{180}{x}-\frac{180}{x+9}=1
\displaystyle \frac{180(x+9)-180x}{x(x+9)}=1
\displaystyle \frac{1620}{x(x+9)}=1
\displaystyle x(x+9)=1620
\displaystyle x^{2}+9x-1620=0
\displaystyle (x+45)(x-36)=0
\displaystyle x=-45\text{ or }x=36
\displaystyle \text{Speed cannot be negative.}
\displaystyle \therefore \text{Speed of the train}=36\text{ km/hour}

\displaystyle \textbf{OR}

\displaystyle \frac{1}{2x-3}+\frac{1}{x-5}=1
\displaystyle \frac{x-5+2x-3}{(2x-3)(x-5)}=1
\displaystyle \frac{3x-8}{(2x-3)(x-5)}=1
\displaystyle 3x-8=(2x-3)(x-5)
\displaystyle 3x-8=2x^{2}-10x-3x+15
\displaystyle 3x-8=2x^{2}-13x+15
\displaystyle 2x^{2}-16x+23=0
\displaystyle x=\frac{16\pm\sqrt{(-16)^{2}-4(2)(23)}}{2(2)}
\displaystyle =\frac{16\pm\sqrt{256-184}}{4}
\displaystyle =\frac{16\pm\sqrt{72}}{4}
\displaystyle =\frac{16\pm6\sqrt{2}}{4}
\displaystyle =\frac{8\pm3\sqrt{2}}{2}
\displaystyle \therefore \text{Roots are }\frac{8+3\sqrt{2}}{2}\text{ and }\frac{8-3\sqrt{2}}{2}
\\

\displaystyle \textbf{Question 32. }\text{In Figure }6,\text{ three circles each of radius }3.5\text{ cm are drawn in such a}
\displaystyle \text{way that each of them touches the other two. Find the area enclosed}
\displaystyle \text{between these three circles (shaded region). }\left[\text{Use }\pi=\frac{22}{7}\right] \displaystyle \text{Answer:}
\displaystyle \text{Radius of each circle}=3.5\text{ cm}
\displaystyle \text{Distance between centres of any two circles}=7\text{ cm}
\displaystyle \text{The centres form an equilateral triangle of side }7\text{ cm.}
\displaystyle \text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}(7)^2
\displaystyle =\frac{1.732}{4}\times49=21.217\text{ cm}^{2}
\displaystyle \text{Area of three sectors}=\frac{60}{360}\times\frac{22}{7}\times(3.5)^2\times3
\displaystyle =19.25\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=21.217-19.25
\displaystyle =1.967\text{ cm}^{2}
\displaystyle \therefore \text{Area of shaded region}=1.97\text{ cm}^{2}\text{ approximately}
\\

\displaystyle \textbf{Question 33. }\text{Water is flowing at the rate of }15\text{ km/hour through a pipe of diameter}
\displaystyle 14\text{ cm into a cuboidal pond which is }50\text{ m long and }44\text{ m wide. In what}
\displaystyle \text{time will the level of water in the pond rise by }21\text{ cm?}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of pipe}=7\text{ cm}
\displaystyle 15\text{ km/hour}=1500000\text{ cm/hour}
\displaystyle \text{Volume of water flowing in }1\text{ hour}=\pi r^{2}h
\displaystyle =\frac{22}{7}\times7\times7\times1500000
\displaystyle =231000000\text{ cm}^{3}
\displaystyle \text{Volume of water required in pond}=5000\times4400\times21
\displaystyle =462000000\text{ cm}^{3}
\displaystyle \text{Time}=\frac{462000000}{231000000}=2\text{ hours}
\displaystyle \therefore \text{Required time}=2\text{ hours}
\\

\displaystyle \textbf{Question 34. }\text{The angle of elevation of the top of a vertical tower from a point on the}
\displaystyle \text{ground is }60^\circ.\text{ From another point }10\text{ m vertically above the first, its}
\displaystyle \text{angle of elevation is }30^\circ.\text{ Find the height of the tower.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m and horizontal distance be }x\text{ m.}
\displaystyle \tan60^{\circ}=\frac{h}{x}
\displaystyle \sqrt{3}=\frac{h}{x}
\displaystyle x=\frac{h}{\sqrt{3}} \qquad (1)
\displaystyle \tan30^{\circ}=\frac{h-10}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h-10}{x}
\displaystyle x=\sqrt{3}(h-10) \qquad (2)
\displaystyle \frac{h}{\sqrt{3}}=\sqrt{3}(h-10)
\displaystyle h=3h-30
\displaystyle 2h=30
\displaystyle h=15
\displaystyle \therefore \text{Height of the tower}=15\text{ m}
\\

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