\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series RSH/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 8 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 common difference of the AP }\frac{1}{p},\frac{1-p}{p},\frac{1-2p}{p},\ldots\text{ is :}
\displaystyle \text{(A) }p
\displaystyle \text{(B) }-p
\displaystyle \text{(C) }-1
\displaystyle \text{(D) }1
\displaystyle \text{Answer:}
\displaystyle a_1=\frac{1}{p},\quad a_2=\frac{1-p}{p},\quad a_3=\frac{1-2p}{p}
\displaystyle d=a_2-a_1=\frac{1-p}{p}-\frac{1}{p}
\displaystyle =\frac{1-p-1}{p}
\displaystyle =\frac{-p}{p}=-1
\displaystyle \text{Also, }a_3-a_2=\frac{1-2p}{p}-\frac{1-p}{p}
\displaystyle =\frac{-p}{p}=-1
\displaystyle \therefore \text{Common difference}=-1
\displaystyle \therefore \text{Correct option is (C).}
\\

\displaystyle \textbf{Question 2. }\text{In Fig. 1, PA and PB are two tangents drawn from an external point P to a}
\displaystyle \text{circle with centre C and radius 4 cm. If PA }\perp\text{ PB, then the length of each}
\displaystyle \text{tangent is :} \displaystyle \text{(A) }3\ \text{cm}
\displaystyle \text{(B) }4\ \text{cm}
\displaystyle \text{(C) }5\ \text{cm}
\displaystyle \text{(D) }6\ \text{cm}
\displaystyle \text{Answer:}
\displaystyle PA=PB\text{ since tangents drawn from an external point are equal.}
\displaystyle \text{Also, }CA\perp PA\text{ and }CB\perp PB.
\displaystyle \text{Since }PA\perp PB,\text{ quadrilateral }CAPB\text{ is a square.}
\displaystyle \therefore PA=CA=4\text{ cm}
\displaystyle \therefore \text{Length of each tangent}=4\text{ cm}
\displaystyle \therefore \text{Correct option is (B).}
\\

\displaystyle \textbf{Question 3. }\text{In Fig. 2, a circle with centre O is inscribed in a quadrilateral ABCD such}
\displaystyle \text{that, it touches the sides BC, AB, AD and CD at points P, Q, R and S}
\displaystyle \text{respectively. If AB=29 cm, AD=23 cm, }\angle B=90^\circ\text{ and DS}=5\text{ cm, then the}
\displaystyle \text{radius of the circle (in cm.) is :} \displaystyle \text{(A) }11
\displaystyle \text{(B) }18
\displaystyle \text{(C) }6
\displaystyle \text{(D) }15
\displaystyle \text{Answer:}
\displaystyle DS=DR=5\text{ cm}
\displaystyle AD=AR+DR
\displaystyle \therefore 23=AR+5
\displaystyle \therefore AR=18\text{ cm}
\displaystyle \text{Also, }AQ=AR=18\text{ cm}
\displaystyle AB=AQ+BQ
\displaystyle \therefore 29=18+BQ
\displaystyle \therefore BQ=11\text{ cm}
\displaystyle \text{Since }\angle B=90^\circ,\text{ the radius }r=BQ=11\text{ cm}
\displaystyle \therefore \text{Correct option is (A).}
\\

\displaystyle \textbf{Question 4. }\text{The angle of depression of a car, standing on the ground, from the top of a}
\displaystyle 75\text{ m high tower, is }30^\circ\text{. The distance of the car from the base of the tower}
\displaystyle \text{(in m.) is :}
\displaystyle \text{(A) }25\sqrt{3}
\displaystyle \text{(B) }50\sqrt{3}
\displaystyle \text{(C) }75\sqrt{3}
\displaystyle \text{(D) }150
\displaystyle \text{Answer:}
\displaystyle \text{Angle of depression }=30^\circ
\displaystyle \therefore \text{Angle of elevation of the top of the tower from the car }=30^\circ
\displaystyle \text{Let the distance of the car from the base of the tower be }x\text{ m.}
\displaystyle \tan30^\circ=\frac{\text{Height of tower}}{\text{Distance from base}}=\frac{75}{x}
\displaystyle \frac{1}{\sqrt3}=\frac{75}{x}
\displaystyle \therefore x=75\sqrt3
\displaystyle \therefore \text{Distance of the car from the base of the tower }=75\sqrt3\text{ m}
\displaystyle \therefore \text{Correct option is (C).}
\\

\displaystyle \textbf{Question 5. }\text{The probability of getting an even number, when a die is thrown once, is :}
\displaystyle \text{(A) }\frac{1}{2}
\displaystyle \text{(B) }\frac{1}{3}
\displaystyle \text{(C) }\frac{1}{6}
\displaystyle \text{(D) }\frac{5}{6}
\displaystyle \text{Answer:}
\displaystyle \text{Sample space }S=\{1,2,3,4,5,6\}
\displaystyle \text{Even numbers are }\{2,4,6\}
\displaystyle \therefore \text{Number of favourable outcomes}=3
\displaystyle \text{Total number of outcomes}=6
\displaystyle \therefore P(\text{getting an even number})=\frac{3}{6}=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (A).}
\\

\displaystyle \textbf{Question 6. }\text{A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at}
\displaystyle \text{random from the box, the probability that it bears a prime-number less than}
\displaystyle 23\text{, is :}
\displaystyle \text{(A) }\frac{7}{90}
\displaystyle \text{(B) }\frac{10}{90}
\displaystyle \text{(C) }\frac{4}{45}
\displaystyle \text{(D) }\frac{9}{89}
\displaystyle \text{Answer:}
\displaystyle \text{Prime numbers less than }23\text{ are }2,3,5,7,11,13,17,19,23?
\displaystyle \text{Since the prime number must be less than }23,\ 23\text{ is not included.}
\displaystyle \therefore \text{Favourable outcomes }=2,3,5,7,11,13,17,19=\ 8
\displaystyle \text{Total outcomes }=90
\displaystyle \therefore P(\text{prime number less than }23)=\frac{8}{90}=\frac{4}{45}
\displaystyle \therefore \text{Correct option is (C).}
\\

\displaystyle \textbf{Question 7. }\text{In Fig. 3, the area of triangle ABC (in sq. units) is :} \displaystyle \text{(A) }15
\displaystyle \text{(B) }10
\displaystyle \text{(C) }7.5
\displaystyle \text{(D) }2.5
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, }A(1,3),\ B(-1,0)\text{ and }C(4,0).
\displaystyle BC=4-(-1)=5\text{ units}
\displaystyle \text{Height of the triangle from }A\text{ to }BC=3\text{ units}
\displaystyle \therefore \text{Area of }\triangle ABC=\frac{1}{2}\times\text{base}\times\text{height}
\displaystyle =\frac{1}{2}\times5\times3
\displaystyle =7.5\text{ sq. units}
\displaystyle \therefore \text{Correct option is (C).}
\\

\displaystyle \textbf{Question 8. }\text{If the difference between the circumference and the radius of a circle is}
\displaystyle 37\text{ cm, then using }\pi=\frac{22}{7}\text{, the circumference (in cm) of the circle is :}
\displaystyle \text{(A) }154
\displaystyle \text{(B) }44
\displaystyle \text{(C) }14
\displaystyle \text{(D) }7
\displaystyle \text{Answer:}
\displaystyle \text{Let the radius of the circle be }r\text{ cm.}
\displaystyle \text{Circumference }=2\pi r=\frac{44}{7}r
\displaystyle \text{Given, }\frac{44}{7}r-r=37
\displaystyle \therefore \frac{37}{7}r=37
\displaystyle \therefore r=7\text{ cm}
\displaystyle \therefore \text{Circumference }=\frac{44}{7}\times7=44\text{ cm}
\displaystyle \therefore \text{Correct option is (B).}
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question Numbers 9 to 14 carry two marks each.}

\displaystyle \textbf{Question 9. }\text{Solve the following quadratic equation for }x\text{ :}
\displaystyle 4\sqrt{3}x^{2}+5x-2\sqrt{3}=0
\displaystyle \text{Answer:}
\displaystyle a=4\sqrt3,\quad b=5,\quad c=-2\sqrt3
\displaystyle \text{Using the quadratic formula,}
\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\displaystyle =\frac{-5\pm\sqrt{25-4(4\sqrt3)(-2\sqrt3)}}{8\sqrt3}
\displaystyle =\frac{-5\pm\sqrt{25+96}}{8\sqrt3}
\displaystyle =\frac{-5\pm11}{8\sqrt3}
\displaystyle \therefore x=\frac{6}{8\sqrt3}=\frac{\sqrt3}{4}
\displaystyle \text{or}\quad x=\frac{-16}{8\sqrt3}=-\frac{2}{\sqrt3}=-\frac{2\sqrt3}{3}
\displaystyle \therefore x=\frac{\sqrt3}{4}\text{ or }x=-\frac{2\sqrt3}{3}
\\

\displaystyle \textbf{Question 10. }\text{How many three-digit natural numbers are divisible by 7?}
\displaystyle \text{Answer:}
\displaystyle \text{The smallest three-digit number divisible by }7\text{ is }105=7\times15.
\displaystyle \text{The largest three-digit number divisible by }7\text{ is }994=7\times142.
\displaystyle \text{Thus, the required numbers form an A.P.}
\displaystyle 105,\ 112,\ 119,\ \ldots,\ 994
\displaystyle \text{with first term }a=105,\text{ common difference }d=7\text{ and last term }l=994.
\displaystyle l=a+(n-1)d
\displaystyle 994=105+(n-1)7
\displaystyle 889=7(n-1)
\displaystyle n-1=127
\displaystyle n=128
\displaystyle \therefore \text{There are }128\text{ three-digit natural numbers divisible by }7.
\\

\displaystyle \textbf{Question 11. }\text{In Fig. 4, a circle inscribed in triangle ABC touches}
\displaystyle \text{its sides AB, BC and AC at points D, E}
\displaystyle \text{and F respectively. If AB = 12 cm, BC = 8 cm and}
\displaystyle \text{AC = 10 cm, then find the lengths of AD, BE and CF.} \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle \text{Let }AD=AF=x,\quad BD=BE=y,\quad CE=CF=z
\displaystyle AB=AD+BD
\displaystyle \therefore x+y=12 \quad ...(1)
\displaystyle BC=BE+CE
\displaystyle \therefore y+z=8 \quad ...(2)
\displaystyle AC=AF+CF
\displaystyle \therefore x+z=10 \quad ...(3)
\displaystyle \text{Adding (1), (2) and (3),}
\displaystyle 2(x+y+z)=30
\displaystyle \therefore x+y+z=15
\displaystyle \therefore z=15-(x+y)=15-12=3
\displaystyle y=8-z=8-3=5
\displaystyle x=10-z=10-3=7
\displaystyle \therefore AD=7\text{ cm},\quad BE=5\text{ cm},\quad CF=3\text{ cm}
\\

\displaystyle \textbf{Question 12. }\text{Prove that the parallelogram circumscribing a circle is a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be a parallelogram circumscribing a circle.}
\displaystyle \text{We know that tangents drawn from an external point to a circle are equal.}
\displaystyle \therefore AB+CD=AD+BC
\displaystyle \text{But in a parallelogram, opposite sides are equal.}
\displaystyle \therefore AB=CD\text{ and }AD=BC
\displaystyle \therefore AB+AB=AD+AD
\displaystyle \therefore 2AB=2AD
\displaystyle \therefore AB=AD
\displaystyle \text{Thus, adjacent sides of the parallelogram are equal.}
\displaystyle \therefore AB=BC=CD=DA
\displaystyle \therefore \text{The parallelogram is a rhombus.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 13. }\text{A card is drawn at random from a well shuffled pack of 52 playing cards.}
\displaystyle \text{Find the probability that the drawn card is neither a king nor a queen.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards in a pack}=52
\displaystyle \text{Number of kings}=4
\displaystyle \text{Number of queens}=4
\displaystyle \text{Number of cards that are neither king nor queen}=52-(4+4)=44
\displaystyle \therefore P(\text{neither a king nor a queen})=\frac{44}{52}
\displaystyle =\frac{11}{13}
\displaystyle \therefore \text{Required probability}=\frac{11}{13}
\\

\displaystyle \textbf{Question 14. }\text{Two circular pieces of equal radii and maximum area, touching each other}
\displaystyle \text{are cut out from a rectangular card board of dimensions }14\text{ cm}\times7\text{ cm. Find}
\displaystyle \text{the area of the remaining card board. [Use }\pi=\frac{22}{7}\text{]}
\displaystyle \text{Answer:}
\displaystyle \text{Since two equal circles of maximum area fit in a }14\text{ cm}\times7\text{ cm rectangle,}
\displaystyle \text{diameter of each circle}=7\text{ cm}
\displaystyle \therefore r=\frac{7}{2}\text{ cm}
\displaystyle \text{Area of rectangle}=14\times7=98\text{ cm}^2
\displaystyle \text{Area of two circles}=2\times\pi r^2
\displaystyle =2\times\frac{22}{7}\times\left(\frac{7}{2}\right)^2
\displaystyle =2\times\frac{22}{7}\times\frac{49}{4}=77\text{ cm}^2
\displaystyle \therefore \text{Area of remaining card board}=98-77=21\text{ cm}^2
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question Numbers 15 to 24 carry three marks each.}

\displaystyle \textbf{Question 15. }\text{For what value of }k\text{, are the roots of the quadratic equation }
\displaystyle kx(x-2)+6=0  \text{equal ?}
\displaystyle \text{Answer:}
\displaystyle kx(x-2)+6=0
\displaystyle \therefore kx^2-2kx+6=0
\displaystyle \text{For equal roots, discriminant }D=0.
\displaystyle b^2-4ac=0
\displaystyle (-2k)^2-4(k)(6)=0
\displaystyle 4k^2-24k=0
\displaystyle 4k(k-6)=0
\displaystyle \therefore k=0\text{ or }k=6
\displaystyle \text{Since the equation is quadratic, }k\ne0.
\displaystyle \therefore k=6
\\

\displaystyle \textbf{Question 16. }\text{Find the number of terms of the AP }18,\ 15\frac{1}{2},\ 13,\ \ldots,\ -49\frac{1}{2}\text{ and find the}
\displaystyle \text{sum of all its terms.}
\displaystyle \text{Answer:}
\displaystyle a=18,\quad d=15\frac{1}{2}-18=\frac{31}{2}-18=-\frac{5}{2},\quad l=-49\frac{1}{2}=-\frac{99}{2}
\displaystyle l=a+(n-1)d
\displaystyle -\frac{99}{2}=18+(n-1)\left(-\frac{5}{2}\right)
\displaystyle -99=36-5(n-1)
\displaystyle -135=-5(n-1)
\displaystyle n-1=27
\displaystyle n=28
\displaystyle S_n=\frac{n}{2}(a+l)
\displaystyle S_{28}=\frac{28}{2}\left(18-\frac{99}{2}\right)
\displaystyle =14\left(\frac{36-99}{2}\right)
\displaystyle =14\left(-\frac{63}{2}\right)=-441
\displaystyle \therefore \text{Number of terms}=28\text{ and sum of all terms}=-441.
\\

\displaystyle \textbf{Question 17. }\text{Construct a triangle with sides }5\ \text{cm, }4\ \text{cm and }6\ \text{cm. Then construct}
\displaystyle \text{another triangle whose sides are }\frac{2}{3}\text{ times the corresponding sides of first triangle.}
\displaystyle \text{Answer: Construction Steps}
\displaystyle \textbf{Step 1. }\text{Draw a line segment }BC=6\text{ cm.}
\displaystyle \textbf{Step 2. }\text{With centre }B\text{ and radius }5\text{ cm, draw an arc.}
\displaystyle \textbf{Step 3. }\text{With centre }C\text{ and radius }4\text{ cm, draw another arc intersecting the first arc at }A.
\displaystyle \textbf{Step 4. }\text{Join }AB\text{ and }AC.\text{ Thus, }\triangle ABC\text{ is the required triangle.}
\displaystyle \textbf{Step 5. }\text{From }B,\text{ draw a ray }BX\text{ making an acute angle with }BC.
\displaystyle \textbf{Step 6. }\text{Mark three equal points }B_1,\ B_2,\ B_3\text{ on ray }BX.
\displaystyle \textbf{Step 7. }\text{Join }B_3\text{ to }C.
\displaystyle \textbf{Step 8. }\text{Through }B_2,\text{ draw a line parallel to }B_3C\text{ meeting }BC\text{ at }C'.
\displaystyle \textbf{Step 9. }\text{Through }C',\text{ draw a line parallel to }CA\text{ meeting }BA\text{ at }A'.
\displaystyle \textbf{Step 10. }\text{Join }A'C'.
\displaystyle \textbf{Result: }\triangle A'BC'\sim\triangle ABC
\displaystyle \frac{BC'}{BC}=\frac{BB_2}{BB_3}=\frac{2}{3}
\displaystyle \therefore \frac{A'B}{AB}=\frac{BC'}{BC}=\frac{A'C'}{AC}=\frac{2}{3}
\displaystyle \therefore \triangle A'BC'\text{ is the required triangle whose sides are }\frac{2}{3}\text{ times the}
\displaystyle \text{corresponding sides of }\triangle ABC.
\\

\displaystyle \textbf{Question 18. }\text{The horizontal distance between two poles is }15\ \text{m. The angle of depression}
\displaystyle \text{of the top of first pole as seen from the top of second pole is }30^\circ\text{. If the}
\displaystyle \text{height of the second pole is }24\ \text{m, find the height of the first}
\displaystyle \text{pole. [Use }\sqrt{3}=1.732\text{]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the first pole be }h\text{ m.}
\displaystyle \text{Difference between the heights of the poles}=24-h
\displaystyle \tan30^\circ=\frac{24-h}{15}
\displaystyle \frac{1}{\sqrt3}=\frac{24-h}{15}
\displaystyle 24-h=\frac{15}{1.732}
\displaystyle 24-h=8.66
\displaystyle h=24-8.66=15.34
\displaystyle \therefore \text{Height of the first pole}=15.34\text{ m}
\\

\displaystyle \textbf{Question 19. }\text{Prove that the points }(7,10),\ (-2,5)\text{ and }(3,-4)\text{ are the vertices of an}
\displaystyle \text{isosceles right triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(7,10),\ B(-2,5)\text{ and }C(3,-4).
\displaystyle AB^2=(-2-7)^2+(5-10)^2
\displaystyle =(-9)^2+(-5)^2=81+25=106
\displaystyle BC^2=(3+2)^2+(-4-5)^2
\displaystyle =5^2+(-9)^2=25+81=106
\displaystyle AC^2=(3-7)^2+(-4-10)^2
\displaystyle =(-4)^2+(-14)^2=16+196=212
\displaystyle \therefore AB^2=BC^2
\displaystyle \therefore AB=BC
\displaystyle \text{Also, }AB^2+BC^2=106+106=212=AC^2
\displaystyle \therefore \angle B=90^\circ
\displaystyle \therefore \triangle ABC\text{ is an isosceles right triangle.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 20. }\text{Find the ratio in which the }y\text{-axis divides the line segment joining the points}
\displaystyle (-4,-6)\text{ and }(10,12)\text{. Also find the coordinates of the point of division.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the }y\text{-axis divide the line segment in the ratio }m:n.
\displaystyle \text{Since the point lies on the }y\text{-axis, its }x\text{-coordinate is }0.
\displaystyle \therefore \frac{10m+(-4)n}{m+n}=0
\displaystyle 10m-4n=0
\displaystyle \therefore 10m=4n
\displaystyle \therefore \frac{m}{n}=\frac{2}{5}
\displaystyle \therefore \text{Required ratio}=2:5
\displaystyle y=\frac{12m+(-6)n}{m+n}
\displaystyle =\frac{12(2)-6(5)}{2+5}
\displaystyle =\frac{24-30}{7}=-\frac{6}{7}
\displaystyle \therefore \text{Point of division}=\left(0,-\frac{6}{7}\right)
\\

\displaystyle \textbf{Question 21. }\text{In Fig.5, AB and CD are two diameters of a circle with centre O, which are}
\displaystyle \text{perpendicular to each other. OB is the diameter of the smaller circle. If}
\displaystyle OA=7\text{ cm, find the area of the shaded region. [Use }\pi=\frac{22}{7}\text{]} \displaystyle \text{Answer:}
\displaystyle OA=OB=OC=OD=7\text{ cm}
\displaystyle \text{Area of right semicircle of bigger circle}=\frac{1}{2}\pi r^2
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times7^2=77\text{ cm}^2
\displaystyle \text{Area of }\triangle ACD=\frac{1}{2}\times CD\times OA
\displaystyle =\frac{1}{2}\times14\times7=49\text{ cm}^2
\displaystyle \text{Area of shaded part on right side}=77-49=28\text{ cm}^2
\displaystyle \text{Since }OB=7\text{ cm is the diameter of the smaller circle,}
\displaystyle \text{radius of smaller circle}=\frac{7}{2}\text{ cm}
\displaystyle \text{Area of smaller circle}=\pi r^2
\displaystyle =\frac{22}{7}\times\left(\frac{7}{2}\right)^2=\frac{77}{2}=38.5\text{ cm}^2
\displaystyle \therefore \text{Total shaded area}=28+38.5=66.5\text{ cm}^2
\\

\displaystyle \textbf{Question 22. }\text{A vessel is in the form of a hemispherical bowl surmounted by a hollow}
\displaystyle \text{cylinder of same diameter. The diameter of the hemispherical bowl is}
\displaystyle 14\text{ cm and the total height of the vessel is }13\text{ cm. Find the total surface area}
\displaystyle \text{of the vessel. [Use }\pi=\frac{22}{7}\text{]}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of hemispherical bowl}=r=7\text{ cm}
\displaystyle \text{Height of cylinder}=13-7=6\text{ cm}
\displaystyle \text{Total surface area of vessel}=\text{CSA of hemisphere}+\text{CSA of cylinder}
\displaystyle =2\pi r^2+2\pi rh
\displaystyle =2\pi r(r+h)
\displaystyle =2\times\frac{22}{7}\times7(7+6)
\displaystyle =44\times13=572\text{ cm}^2
\displaystyle \therefore \text{Total surface area of the vessel}=572\text{ cm}^2
\\

\displaystyle \textbf{Question 23. }\text{A wooden toy was made by scooping out a hemisphere of same radius from}
\displaystyle \text{each end of a solid cylinder. If the height of the cylinder is }10\text{ cm, and its}
\displaystyle \text{base is of radius }3.5\text{ cm, find the volume of wood in the toy. [Use }\pi=\frac{22}{7}\text{]}
\displaystyle \text{Answer:}
\displaystyle r=3.5\text{ cm},\quad h=10\text{ cm}
\displaystyle \text{Volume of wood}=\text{Volume of cylinder}-\text{Volume of two hemispheres}
\displaystyle =\pi r^2h-\frac{4}{3}\pi r^3
\displaystyle =\frac{22}{7}\times(3.5)^2\times10-\frac{4}{3}\times\frac{22}{7}\times(3.5)^3
\displaystyle =385-\frac{539}{3}
\displaystyle =\frac{1155-539}{3}=\frac{616}{3}
\displaystyle =205\frac{1}{3}\text{ cm}^3
\displaystyle \therefore \text{Volume of wood in the toy}=205\frac{1}{3}\text{ cm}^3
\\

\displaystyle \textbf{Question 24. }\text{In a circle of radius }21\text{ cm, an arc subtends an angle of }60^\circ\text{ at the centre.}
\displaystyle \text{Find : (i) the length of the arc (ii) area of the sector formed by the arc.}
\displaystyle \text{[Use }\pi=\frac{22}{7}\text{]}
\displaystyle \text{Answer:}
\displaystyle r=21\text{ cm},\quad \theta=60^\circ
\displaystyle \text{(i) Length of arc}=\frac{\theta}{360^\circ}\times2\pi r
\displaystyle =\frac{60}{360}\times2\times\frac{22}{7}\times21
\displaystyle =22\text{ cm}
\displaystyle \text{(ii) Area of sector}=\frac{\theta}{360^\circ}\times\pi r^2
\displaystyle =\frac{60}{360}\times\frac{22}{7}\times21^2
\displaystyle =231\text{ cm}^2
\displaystyle \therefore \text{Length of arc}=22\text{ cm and area of sector}=231\text{ cm}^2
\\

\displaystyle \textbf{SECTION D}
\displaystyle \text{Question Numbers 25 to 34 carry four marks each.}

\displaystyle \textbf{Question 25. }\text{Solve the following for }x\text{ :}
\displaystyle \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}
\displaystyle \text{Answer:}
\displaystyle \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}
\displaystyle =\frac{bx+2ax+ab}{2abx}
\displaystyle \therefore 2abx=(2a+b+2x)(bx+2ax+ab)
\displaystyle \therefore 2abx=(2a+b+2x)\{x(2a+b)+ab\}
\displaystyle \therefore 2abx=(2a+b)x(2a+b)+(2a+b)ab+2x^2(2a+b)+2abx
\displaystyle \therefore 0=(2a+b)\{x(2a+b)+ab+2x^2\}
\displaystyle \therefore 2x^2+(2a+b)x+ab=0
\displaystyle \therefore 2x^2+2ax+bx+ab=0
\displaystyle \therefore 2x(x+a)+b(x+a)=0
\displaystyle \therefore (x+a)(2x+b)=0
\displaystyle \therefore x=-a\text{ or }x=-\frac{b}{2}
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\displaystyle \textbf{Question 26. }\text{Sum of the areas of two squares is }400\text{ cm}^{2}\text{. If the difference of their}
\displaystyle \text{perimeters is }16\text{ cm, find the sides of the two squares.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the sides of the two squares be }x\text{ cm and }y\text{ cm, where }x>y.
\displaystyle x^2+y^2=400 \quad ...(1)
\displaystyle 4x-4y=16
\displaystyle \therefore x-y=4 \quad ...(2)
\displaystyle \therefore x=y+4
\displaystyle \text{Substituting in (1),}
\displaystyle (y+4)^2+y^2=400
\displaystyle y^2+8y+16+y^2=400
\displaystyle 2y^2+8y-384=0
\displaystyle y^2+4y-192=0
\displaystyle (y+16)(y-12)=0
\displaystyle \therefore y=12\quad(\text{since side cannot be negative})
\displaystyle \therefore x=12+4=16
\displaystyle \therefore \text{Sides of the two squares are }16\text{ cm and }12\text{ cm.}
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\displaystyle \textbf{Question 27. }\text{If the sum of first }7\text{ terms of an AP is }49\text{ and that of first }17\text{ terms is }289\text{,}
\displaystyle \text{find the sum of its first }n\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]
\displaystyle S_7=49
\displaystyle \therefore \frac{7}{2}[2a+6d]=49
\displaystyle \therefore 7(a+3d)=49
\displaystyle \therefore a+3d=7 \quad ...(1)
\displaystyle S_{17}=289
\displaystyle \therefore \frac{17}{2}[2a+16d]=289
\displaystyle \therefore 17(a+8d)=289
\displaystyle \therefore a+8d=17 \quad ...(2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle 5d=10
\displaystyle \therefore d=2
\displaystyle \text{From (1), }a+3(2)=7
\displaystyle \therefore a=1
\displaystyle \therefore S_n=\frac{n}{2}[2(1)+(n-1)2]
\displaystyle =\frac{n}{2}[2+2n-2]
\displaystyle =n^2
\displaystyle \therefore \text{Sum of first }n\text{ terms is }n^2.
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\displaystyle \textbf{Question 28. }\text{Prove that the tangent at any point of a circle is perpendicular to the radius}
\displaystyle \text{through the point of contact.}
\displaystyle \text{Answer:}
\displaystyle \text{Let a tangent }l\text{ touch the circle with centre }O\text{ at point }P.
\displaystyle \text{We have to prove that }OP\perp l.
\displaystyle \text{Take any point }Q\text{ on the tangent }l\text{ other than }P.
\displaystyle \text{Since }Q\text{ lies outside the circle, }OQ>OP.
\displaystyle \therefore OP\text{ is the shortest distance from }O\text{ to the line }l.
\displaystyle \text{The shortest distance from a point to a line is the perpendicular distance.}
\displaystyle \therefore OP\perp l
\displaystyle \therefore \text{The tangent at any point of a circle is perpendicular to the radius through the point of contact.}
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 29. }\text{In Fig. 6, }l\text{ and }m\text{ are two parallel tangents to a circle with centre O,}
\displaystyle \text{touching the circle at A and B respectively. Another tangent at C intersects}
\displaystyle \text{the line }l\text{ at D and }m\text{ at E. Prove that }\angle DOE=90^\circ\text{.} \displaystyle \text{Answer:}
\displaystyle \text{From point }D,\text{ the two tangents drawn to the circle are }DA\text{ and }DC.
\displaystyle \therefore DA=DC
\displaystyle OA=OC\text{ are radii of the circle.}
\displaystyle OD=OD\text{ is common.}
\displaystyle \therefore \triangle OAD\cong\triangle OCD
\displaystyle \therefore \angle ADO=\angle ODC
\displaystyle \text{Thus, }OD\text{ bisects }\angle ADC. \quad ...(1)
\displaystyle \text{Similarly, from point }E,\text{ the two tangents drawn to the circle are }EB\text{ and }EC.
\displaystyle \therefore EB=EC
\displaystyle OB=OC\text{ are radii of the circle.}
\displaystyle OE=OE\text{ is common.}
\displaystyle \therefore \triangle OBE\cong\triangle OCE
\displaystyle \therefore \angle BEO=\angle OEC
\displaystyle \text{Thus, }OE\text{ bisects }\angle BEC. \quad ...(2)
\displaystyle \text{Since }l\parallel m,\text{ the angles }\angle ADC\text{ and }\angle BEC\text{ are supplementary.}
\displaystyle \therefore \angle ADC+\angle BEC=180^\circ
\displaystyle \therefore \frac{1}{2}\angle ADC+\frac{1}{2}\angle BEC=90^\circ
\displaystyle \therefore \angle CDO+\angle CEO=90^\circ
\displaystyle \text{In }\triangle DOE,\ \angle CDO+\angle CEO+\angle DOE=180^\circ
\displaystyle \therefore 90^\circ+\angle DOE=180^\circ
\displaystyle \therefore \angle DOE=90^\circ
\displaystyle \text{Hence proved.}
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\displaystyle \textbf{Question 30. }\text{The angle of elevation of the top of a building from the foot of the tower is}
\displaystyle 30^\circ\text{ and the angle of elevation of the top of the tower from the foot of the}
\displaystyle \text{building is }60^\circ\text{. If the tower is }60\text{ m high, find the height of the building.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance between the tower and the building be }x\text{ m.}
\displaystyle \text{Let the height of the building be }h\text{ m.}
\displaystyle \tan60^\circ=\frac{60}{x}
\displaystyle \therefore \sqrt3=\frac{60}{x}
\displaystyle \therefore x=\frac{60}{\sqrt3}=20\sqrt3
\displaystyle \tan30^\circ=\frac{h}{x}
\displaystyle \therefore \frac{1}{\sqrt3}=\frac{h}{20\sqrt3}
\displaystyle \therefore h=20
\displaystyle \therefore \text{Height of the building}=20\text{ m}
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\displaystyle \textbf{Question 31. }\text{A group consists of }12\text{ persons, of which }3\text{ are extremely patient, other }6
\displaystyle \text{are extremely honest and rest are extremely kind. A person from the group}
\displaystyle \text{is selected at random. Assuming that each person is equally likely to be}
\displaystyle \text{selected, find the probability of selecting a person who is (i) extremely}
\displaystyle \text{patient (ii) extremely kind or honest. Which of the above values you prefer}
\displaystyle \text{more.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of persons}=12
\displaystyle \text{Number of extremely patient persons}=3
\displaystyle \text{Number of extremely honest persons}=6
\displaystyle \text{Number of extremely kind persons}=12-3-6=3
\displaystyle \text{(i) }P(\text{extremely patient})=\frac{3}{12}=\frac{1}{4}
\displaystyle \text{(ii) }P(\text{extremely kind or honest})=\frac{3+6}{12}=\frac{9}{12}=\frac{3}{4}
\displaystyle \text{All the above values are good, but honesty should be preferred more.}
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\displaystyle \textbf{Question 32. }\text{The three vertices of a parallelogram ABCD are A}(3,-4)\text{, B}(-1,-3)\text{ and}
\displaystyle \text{C}(-6,2)\text{. Find the coordinates of vertex D and find the area of ABCD.}
\displaystyle \text{Answer:}
\displaystyle \text{In parallelogram }ABCD,\text{ diagonals bisect each other.}
\displaystyle \text{Let }D=(x,y).
\displaystyle \text{Midpoint of }AC=\text{Midpoint of }BD
\displaystyle \therefore \left(\frac{3-6}{2},\frac{-4+2}{2}\right)=\left(\frac{-1+x}{2},\frac{-3+y}{2}\right)
\displaystyle \therefore \left(-\frac{3}{2},-1\right)=\left(\frac{x-1}{2},\frac{y-3}{2}\right)
\displaystyle \therefore \frac{x-1}{2}=-\frac{3}{2},\quad \frac{y-3}{2}=-1
\displaystyle \therefore x=-2,\quad y=1
\displaystyle \therefore D=(-2,1)
\displaystyle \text{Area of parallelogram }ABCD=2\times\text{Area of }\triangle ABC
\displaystyle =2\times\frac{1}{2}\left|3(-3-2)+(-1)(2+4)+(-6)(-4+3)\right|
\displaystyle =\left|-15-6+6\right|
\displaystyle =15\text{ sq. units}
\displaystyle \therefore \text{Area of }ABCD=15\text{ sq. units}
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\displaystyle \textbf{Question 33. }\text{Water is flowing through a cylindrical pipe, of internal diameter }2\text{ cm, into}
\displaystyle \text{a cylindrical tank of base radius }40\text{ cm, at the rate of }0.4\text{ m/s. Determine the}
\displaystyle \text{rise in level of water in the tank in half an hour.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of pipe}=1\text{ cm}
\displaystyle 0.4\text{ m/s}=40\text{ cm/s}
\displaystyle \text{Time}=\frac{1}{2}\text{ hour}=30\times60=1800\text{ s}
\displaystyle \text{Length of water column}=40\times1800=72000\text{ cm}
\displaystyle \text{Volume of water flowing}=\pi r^2h=\pi(1)^2(72000)=72000\pi\text{ cm}^3
\displaystyle \text{Let the rise in water level be }H\text{ cm.}
\displaystyle \pi(40)^2H=72000\pi
\displaystyle 1600H=72000
\displaystyle H=45
\displaystyle \therefore \text{Rise in water level}=45\text{ cm}
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\displaystyle \textbf{Question 34. }\text{A bucket open at the top, and made up of a metal sheet is in the form of a}
\displaystyle \text{frustum of a cone. The depth of the bucket is }24\text{ cm and the diameters of its}
\displaystyle \text{upper and lower circular ends are }30\text{ cm and }10\text{ cm respectively. Find the}
\displaystyle \text{cost of metal sheet used in it at the rate of Rs }10\text{ per }100\text{ cm}^{2}\text{. [Use }\pi=3.14\text{]}
\displaystyle \text{Answer:}
\displaystyle R=15\text{ cm},\quad r=5\text{ cm},\quad h=24\text{ cm}
\displaystyle l=\sqrt{h^2+(R-r)^2}=\sqrt{24^2+10^2}=\sqrt{676}=26\text{ cm}
\displaystyle \text{Area of metal sheet}=\text{CSA of frustum}+\text{area of lower base}
\displaystyle =\pi(R+r)l+\pi r^2
\displaystyle =3.14(15+5)(26)+3.14(5)^2
\displaystyle =1632.8+78.5=1711.3\text{ cm}^2
\displaystyle \text{Cost}=\frac{1711.3}{100}\times10
\displaystyle =\text{Rs }171.13
\displaystyle \therefore \text{Cost of metal sheet used}=\text{Rs }171.13
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