\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series BRH}                                                                 \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 quadratic equation }2x^{2}-x-6=0\text{ are}
\displaystyle \text{(A) }-2,\ \frac{3}{2}
\displaystyle \text{(B) }2,\ -\frac{3}{2}
\displaystyle \text{(C) }-2,\ -\frac{3}{2}
\displaystyle \text{(D) }2,\ \frac{3}{2}
\displaystyle \text{Answer:}
\displaystyle 2x^{2}-x-6=0
\displaystyle 2x^{2}+3x-4x-6=0
\displaystyle x(2x+3)-2(2x+3)=0
\displaystyle (2x+3)(x-2)=0
\displaystyle 2x+3=0\quad \text{or}\quad x-2=0
\displaystyle x=-\frac{3}{2}\quad \text{or}\quad x=2
\displaystyle \therefore \text{The roots are }2\text{ and }-\frac{3}{2}
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 2. }\text{If the }n^{\text{th}}\text{ term of an A.P. is }(2n+1),\text{ then the sum of its first three}
\displaystyle \text{terms is}
\displaystyle \text{(A) }6n+3
\displaystyle \text{(B) }15
\displaystyle \text{(C) }12
\displaystyle \text{(D) }21
\displaystyle \text{Answer:}
\displaystyle T_{n}=2n+1
\displaystyle T_{1}=2(1)+1=3
\displaystyle T_{2}=2(2)+1=5
\displaystyle T_{3}=2(3)+1=7
\displaystyle \text{Sum of first three terms}=3+5+7
\displaystyle =15
\displaystyle \therefore \text{The sum of the first three terms is }15
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 3. }\text{From a point }Q,\ 13\text{ cm away from the centre of a circle, the length}
\displaystyle \text{of tangent }PQ\text{ to the circle is }12\text{ cm. The radius of the circle (in cm) is}
\displaystyle \text{(A) }25
\displaystyle \text{(B) }\sqrt{313}
\displaystyle \text{(C) }5
\displaystyle \text{(D) }1
\displaystyle \text{Answer:}
\displaystyle \text{Let }O\text{ be the centre of the circle.}
\displaystyle OQ=13\text{ cm},\quad PQ=12\text{ cm}
\displaystyle OP\perp PQ
\displaystyle \text{In right-angled }\triangle OPQ,
\displaystyle OQ^{2}=OP^{2}+PQ^{2}
\displaystyle 13^{2}=OP^{2}+12^{2}
\displaystyle 169=OP^{2}+144
\displaystyle OP^{2}=25
\displaystyle OP=5\text{ cm}
\displaystyle \therefore \text{Radius of the circle}=5\text{ cm}
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 4. }\text{In Figure 1, }AP,\ AQ\text{ and }BC\text{ are tangents to the circle. If }
\displaystyle AB=5\text{ cm,} \ AC=6\text{ cm and }BC=4\text{ cm, then the length of }AP\text{ (in cm) is} \displaystyle \text{(A) }7.5
\displaystyle \text{(B) }15
\displaystyle \text{(C) }10
\displaystyle \text{(D) }9
\displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point are equal.}
\displaystyle BP=BD\text{ and }CQ=CD
\displaystyle \text{Let }BP=BD=x\text{ cm and }CQ=CD=y\text{ cm.}
\displaystyle BC=BD+CD
\displaystyle x+y=4 \qquad (1)
\displaystyle AP=AB+BP=5+x
\displaystyle AQ=AC+CQ=6+y
\displaystyle AP=AQ
\displaystyle 5+x=6+y
\displaystyle x-y=1 \qquad (2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle 2x=5
\displaystyle x=2.5
\displaystyle AP=5+2.5=7.5\text{ cm}
\displaystyle \therefore \text{Length of }AP=7.5\text{ cm}
\displaystyle \text{Hence, the correct option is (A).}
\\

\displaystyle \textbf{Question 5. }\text{The circumference of a circle is }22\text{ cm. The area of its quadrant (in cm}^{2}\text{)}
\displaystyle \text{is}
\displaystyle \text{(A) }\frac{77}{2}
\displaystyle \text{(B) }\frac{77}{4}
\displaystyle \text{(C) }\frac{77}{8}
\displaystyle \text{(D) }\frac{77}{16}
\displaystyle \text{Answer:}
\displaystyle 2\pi r=22
\displaystyle 2\times\frac{22}{7}\times r=22
\displaystyle r=\frac{7}{2}\text{ cm}
\displaystyle \text{Area of circle}=\pi r^{2}
\displaystyle =\frac{22}{7}\times\left(\frac{7}{2}\right)^{2}
\displaystyle =\frac{77}{2}\text{ cm}^{2}
\displaystyle \text{Area of quadrant}=\frac{1}{4}\times\frac{77}{2}
\displaystyle =\frac{77}{8}\text{ cm}^{2}
\displaystyle \therefore \text{Area of the quadrant}=\frac{77}{8}\text{ cm}^{2}
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 6. }\text{A solid right circular cone is cut into two parts at the middle of its}
\displaystyle \text{height by a plane parallel to its base. The ratio of the volume of the smaller cone to }
\displaystyle \text{the whole cone is}
\displaystyle \text{(A) }1:2
\displaystyle \text{(B) }1:4
\displaystyle \text{(C) }1:6
\displaystyle \text{(D) }1:8
\displaystyle \text{Answer:}
\displaystyle \text{The plane cuts the cone at half of its height.}
\displaystyle \text{Therefore, the smaller cone is similar to the original cone.}
\displaystyle \frac{\text{Height of smaller cone}}{\text{Height of whole cone}}=\frac{1}{2}
\displaystyle \text{For similar solids, volume ratio is the cube of the ratio of}
\displaystyle \text{their corresponding linear dimensions.}
\displaystyle \frac{V_{\text{small}}}{V_{\text{whole}}}=\left(\frac{1}{2}\right)^{3}
\displaystyle =\frac{1}{8}
\displaystyle \therefore \text{Required ratio}=1:8
\displaystyle \text{Hence, the correct option is (D).}
\\

\displaystyle \textbf{Question 7. }\text{A kite is flying at a height of }30\text{ m from the ground. The length of string}
\displaystyle \text{from the kite to the ground is }60\text{ m. Assuming that there is no slack in the string, the }
\displaystyle \text{angle of elevation of the kite at the ground is}
\displaystyle \text{(A) }45^{\circ}
\displaystyle \text{(B) }30^{\circ}
\displaystyle \text{(C) }60^{\circ}
\displaystyle \text{(D) }90^{\circ}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta\text{ be the angle of elevation of the kite.}
\displaystyle \sin\theta=\frac{\text{Height of kite}}{\text{Length of string}}
\displaystyle =\frac{30}{60}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \theta=30^{\circ}
\displaystyle \text{Hence, the angle of elevation of the kite is }30^{\circ}
\displaystyle \text{Hence, the correct option is (B).}
\\

\displaystyle \textbf{Question 8. }\text{The distance of the point }(-3,\ 4)\text{ from the x-axis is}
\displaystyle \text{(A) }3
\displaystyle \text{(B) }-3
\displaystyle \text{(C) }4
\displaystyle \text{(D) }5
\displaystyle \text{Answer:}
\displaystyle \text{Distance of a point from the x-axis is the absolute value of its y-coordinate.}
\displaystyle \text{For the point }(-3,\ 4),\ y=4
\displaystyle \therefore \text{Distance from the x-axis}=|4|=4
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 9. }\text{In Figure 2, }P(5,-3)\text{ and }Q(3,y)\text{ are the points of trisection of the}
\displaystyle \text{line segment joining }A(7,-2)\text{ and }B(1,-5).\text{ Then }y\text{ equals} \displaystyle \text{(A) }2
\displaystyle \text{(B) }4
\displaystyle \text{(C) }-4
\displaystyle \text{(D) }-\frac{5}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ and }Q\text{ are points of trisection, }AQ:QB=2:1
\displaystyle \text{Using section formula for }Q,
\displaystyle y=\frac{2(-5)+1(-2)}{2+1}
\displaystyle =\frac{-10-2}{3}
\displaystyle =-4
\displaystyle \therefore y=-4
\displaystyle \text{Hence, the correct option is (C).}
\\

\displaystyle \textbf{Question 10. }\text{Cards bearing numbers }2,\ 3,\ 4,\ \ldots,\ 11\text{ are kept in a bag. A card is}
\displaystyle \text{drawn at random from the bag. The probability of getting a card with a prime number is}
\displaystyle \text{(A) }\frac{1}{2}
\displaystyle \text{(B) }\frac{2}{5}
\displaystyle \text{(C) }\frac{3}{10}
\displaystyle \text{(D) }\frac{5}{9}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards}=10
\displaystyle \text{Prime numbers among }2,\ 3,\ 4,\ \ldots,\ 11\text{ are }2,\ 3,\ 5,\ 7,\ 11
\displaystyle \text{Number of favourable outcomes}=5
\displaystyle \text{Probability}=\frac{5}{10}
\displaystyle =\frac{1}{2}
\displaystyle \therefore \text{The required probability is }\frac{1}{2}
\displaystyle \text{Hence, the correct option is (A).}
\\

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

\displaystyle \textbf{Question 11. }\text{Find the value of }p\text{ for which the roots of the equation}
\displaystyle px(x-2)+6=0,\text{ are equal.} \displaystyle \text{Answer:}
\displaystyle px(x-2)+6=0
\displaystyle px^{2}-2px+6=0
\displaystyle a=p,\quad b=-2p,\quad c=6
\displaystyle \text{For equal roots, }D=b^{2}-4ac=0
\displaystyle (-2p)^{2}-4(p)(6)=0
\displaystyle 4p^{2}-24p=0
\displaystyle 4p(p-6)=0
\displaystyle p=0\text{ or }p=6
\displaystyle p\neq0\text{ since the equation must remain quadratic}
\displaystyle \therefore p=6
\\

\displaystyle \textbf{Question 12. }\text{How many two-digit numbers are divisible by }3\text{ ?}
\displaystyle \text{Answer:}
\displaystyle \text{Smallest two-digit number divisible by }3=12
\displaystyle \text{Largest two-digit number divisible by }3=99
\displaystyle \text{The numbers form an A.P. }12,\ 15,\ 18,\ldots,\ 99
\displaystyle a=12,\quad l=99,\quad d=3
\displaystyle l=a+(n-1)d
\displaystyle 99=12+(n-1)3
\displaystyle 87=3(n-1)
\displaystyle n-1=29
\displaystyle n=30
\displaystyle \therefore \text{There are }30\text{ two-digit numbers divisible by }3.
\\

\displaystyle \textbf{Question 13. }\text{In Figure 3, a right triangle }ABC,\text{ circumscribes a circle of radius }r.
\displaystyle \text{If }AB\text{ and }BC\text{ are of lengths }8\text{ cm and }6\text{ cm respectively, find the }   \text{value of }r.
\displaystyle \text{Answer:}
\displaystyle AB=8\text{ cm},\quad BC=6\text{ cm}
\displaystyle \text{Since }\triangle ABC\text{ is right-angled at }B,
\displaystyle AC=\sqrt{AB^{2}+BC^{2}}
\displaystyle =\sqrt{8^{2}+6^{2}}
\displaystyle =\sqrt{100}=10\text{ cm}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times8\times6=24\text{ cm}^{2}
\displaystyle \text{Semi-perimeter }s=\frac{8+6+10}{2}=12\text{ cm}
\displaystyle \text{Area}=rs
\displaystyle 24=r\times12
\displaystyle r=2\text{ cm}
\displaystyle \therefore \text{The value of }r=2\text{ cm}
\\

\displaystyle \textbf{Question 14. }\text{Prove that the tangents drawn at the ends of a diameter of a circle}
\displaystyle \text{are parallel.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AB\text{ be a diameter of a circle with centre }O.
\displaystyle \text{Let tangents }l\text{ and }m\text{ be drawn at }A\text{ and }B\text{ respectively.}
\displaystyle OA\perp l
\displaystyle OB\perp m
\displaystyle \text{Since }A,\ O,\ B\text{ are collinear, }OA\text{ and }OB\text{ lie on the same straight line.}
\displaystyle \therefore l\perp AB\text{ and }m\perp AB
\displaystyle \text{Two lines perpendicular to the same line are parallel.}
\displaystyle \therefore l\parallel m
\displaystyle \text{Hence, the tangents drawn at the ends of a diameter of a circle are parallel.}
\\

\displaystyle \textbf{Question 15. }\text{In Figure 4, }ABCD\text{ is a square of side }4\text{ cm. A quadrant of a circle of}
\displaystyle \text{radius }1\text{ cm is drawn at each vertex of the square and a circle of diameter}
\displaystyle 2\text{ cm is also drawn. Find the area of the shaded region. }(\text{Use }\pi=3.14)

\displaystyle \textbf{OR}

\displaystyle \text{From a rectangular sheet of paper }ABCD\text{ with }AB=40\text{ cm and}
\displaystyle AD=28\text{ cm, a semi-circular portion with }BC\text{ as diameter is cut off. Find}
\displaystyle \text{the area of the remaining paper. }(\text{Use }\pi=\frac{22}{7})
\displaystyle \text{Answer:}
\displaystyle \text{Area of square}=4\times4=16\text{ cm}^{2}
\displaystyle \text{Area of four quadrants of radius }1\text{ cm}=\pi r^{2}=3.14\times1^{2}=3.14\text{ cm}^{2}
\displaystyle \text{Radius of inner circle}=\frac{2}{2}=1\text{ cm}
\displaystyle \text{Area of inner circle}=3.14\times1^{2}=3.14\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=16-3.14-3.14
\displaystyle =9.72\text{ cm}^{2}
\displaystyle \therefore \text{Area of shaded region}=9.72\text{ cm}^{2}

\displaystyle \textbf{OR}

\displaystyle \text{Area of rectangle}=40\times28=1120\text{ cm}^{2}
\displaystyle \text{Diameter of semicircle}=BC=AD=28\text{ cm}
\displaystyle \text{Radius of semicircle}=14\text{ cm}
\displaystyle \text{Area of semicircle}=\frac{1}{2}\pi r^{2}
\displaystyle =\frac{1}{2}\times\frac{22}{7}\times14\times14
\displaystyle =308\text{ cm}^{2}
\displaystyle \text{Area of remaining paper}=1120-308=812\text{ cm}^{2}
\displaystyle \therefore \text{Area of remaining paper}=812\text{ cm}^{2}
\\

\displaystyle \textbf{Question 16. }\text{A solid sphere of radius }10.5\text{ cm is melted and recast into smaller solid}
\displaystyle \text{cones, each of radius }3.5\text{ cm and height }3\text{ cm. Find the number of cones }\text{so formed. }
\displaystyle (\text{Use }\pi=\frac{22}{7})
\displaystyle \text{Answer:}
\displaystyle \text{Volume of sphere}=\frac{4}{3}\pi R^{3}
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times(10.5)^{3}
\displaystyle \text{Volume of one cone}=\frac{1}{3}\pi r^{2}h
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(3.5)^{2}\times3
\displaystyle \text{Number of cones}=\frac{\text{Volume of sphere}}{\text{Volume of one cone}}
\displaystyle =\frac{\frac{4}{3}\pi(10.5)^{3}}{\frac{1}{3}\pi(3.5)^{2}\times3}
\displaystyle =\frac{4(10.5)^{3}}{3(3.5)^{2}\times3}
\displaystyle =126
\displaystyle \therefore \text{Number of cones formed}=126
\\

\displaystyle \textbf{Question 17. }\text{Find the value of }k,\text{ if the point }P(2,4)\text{ is equidistant from the points}
\displaystyle A(5,k)\text{ and }B(k,7).
\displaystyle \text{Answer:}
\displaystyle PA=PB
\displaystyle PA^{2}=PB^{2}
\displaystyle (5-2)^{2}+(k-4)^{2}=(k-2)^{2}+(7-4)^{2}
\displaystyle 9+(k-4)^{2}=(k-2)^{2}+9
\displaystyle (k-4)^{2}=(k-2)^{2}
\displaystyle k^{2}-8k+16=k^{2}-4k+4
\displaystyle -4k=-12
\displaystyle k=3
\displaystyle \therefore \text{The value of }k=3
\\

\displaystyle \textbf{Question 18. }\text{A card is drawn at random from a well-shuffled pack of }52\text{ cards. }
\displaystyle \text{Find the probability of getting}
\displaystyle \text{(i) a red king.}
\displaystyle \text{(ii) a queen or a jack.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=52
\displaystyle \text{(i) Number of red kings}=2
\displaystyle P(\text{a red king})=\frac{2}{52}=\frac{1}{26}
\displaystyle \text{(ii) Number of queens}=4
\displaystyle \text{Number of jacks}=4
\displaystyle \text{Number of favourable cards}=4+4=8
\displaystyle P(\text{a queen or a jack})=\frac{8}{52}=\frac{2}{13}
\displaystyle \therefore \text{Required probabilities are }\frac{1}{26}\text{ and }\frac{2}{13}
\\

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

\displaystyle \textbf{Question 19. }\text{Solve the following quadratic equation for }x:
\displaystyle x^{2}-4ax-b^{2}+4a^{2}=0

\displaystyle \textbf{OR}

\displaystyle \text{If the sum of two natural numbers is }8\text{ and their product is }15,\text{ find }   \text{the numbers.}
\displaystyle \text{Answer:}
\displaystyle x^{2}-4ax-b^{2}+4a^{2}=0
\displaystyle x^{2}-4ax+4a^{2}-b^{2}=0
\displaystyle (x-2a)^{2}-b^{2}=0
\displaystyle (x-2a-b)(x-2a+b)=0
\displaystyle x-2a-b=0\quad \text{or}\quad x-2a+b=0
\displaystyle x=2a+b\quad \text{or}\quad x=2a-b
\displaystyle \therefore \text{The roots are }2a+b\text{ and }2a-b.

\displaystyle \textbf{OR}

\displaystyle \text{Let the two natural numbers be }x\text{ and }y.
\displaystyle x+y=8,\quad xy=15
\displaystyle \text{The numbers are }3\text{ and }5\text{ since }3+5=8\text{ and }3\times5=15.
\displaystyle \therefore \text{The required numbers are }3\text{ and }5.
\\

\displaystyle \textbf{Question 20. }\text{Find the sum of all multiples of }7\text{ lying between }500\text{ and }900.
\displaystyle \text{Answer:}
\displaystyle \text{First multiple of }7\text{ greater than }500=504
\displaystyle \text{Last multiple of }7\text{ less than }900=896
\displaystyle \text{The multiples form an A.P. }504,\ 511,\ 518,\ldots,\ 896
\displaystyle a=504,\quad l=896,\quad d=7
\displaystyle l=a+(n-1)d
\displaystyle 896=504+(n-1)7
\displaystyle 392=7(n-1)
\displaystyle n-1=56
\displaystyle n=57
\displaystyle S_{n}=\frac{n}{2}(a+l)
\displaystyle S_{57}=\frac{57}{2}(504+896)
\displaystyle =\frac{57}{2}\times1400
\displaystyle =57\times700=39900
\displaystyle \therefore \text{Required sum}=39900
\\

\displaystyle \textbf{Question 21. }\text{Draw a triangle }ABC\text{ with }BC=7\text{ cm, }\angle B=45^{\circ}\text{ and }
\displaystyle \angle C=60^{\circ}. \text{ Then construct another triangle, whose sides are }\frac{3}{5}\text{ times the}
\displaystyle \text{corresponding sides of }\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Construction Steps:}
\displaystyle \text{1. Draw }BC=7\text{ cm.}
\displaystyle \text{2. At }B,\text{ draw a ray making an angle of }45^{\circ}\text{ with }BC.
\displaystyle \text{3. At }C,\text{ draw a ray making an angle of }60^{\circ}\text{ with }CB.
\displaystyle \text{4. Let the two rays meet at }A.\text{ Thus, }\triangle ABC\text{ is formed.}
\displaystyle \text{5. From }B,\text{ draw a ray }BX\text{ making an acute angle with }BC.
\displaystyle \text{6. Mark five equal points }B_{1},B_{2},B_{3},B_{4},B_{5}\text{ on }BX.
\displaystyle \text{7. Join }B_{5}\text{ to }C.
\displaystyle \text{8. Through }B_{3},\text{ draw a line parallel to }B_{5}C\text{ meeting }BC\text{ at }C'.
\displaystyle \text{9. Through }C',\text{ draw a line parallel to }CA\text{ meeting }BA\text{ at }A'.
\displaystyle \text{10. Then }\triangle A'BC'\text{ is the required triangle whose sides are }\frac{3}{5}\text{ times}
\displaystyle \text{the corresponding sides of }\triangle ABC.
\\

\displaystyle \textbf{Question 22. }\text{In Figure 5, a circle is inscribed in a triangle }PQR\text{ with }
\displaystyle PQ=10\text{ cm,} \ QR=8\text{ cm and }PR=12\text{ cm. Find the lengths }QM,\ RN\text{ and }PL. \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point are equal.}
\displaystyle QL=QM,\quad RM=RN,\quad PL=PN
\displaystyle \text{Let }QM=QL=x,\quad RM=RN=y,\quad PL=PN=z.
\displaystyle PQ=PL+QL=z+x=10 \qquad (1)
\displaystyle QR=QM+RM=x+y=8 \qquad (2)
\displaystyle PR=PN+RN=z+y=12 \qquad (3)
\displaystyle \text{Adding (1), (2) and (3),}
\displaystyle 2(x+y+z)=30
\displaystyle x+y+z=15
\displaystyle x=15-12=3
\displaystyle y=15-10=5
\displaystyle z=15-8=7
\displaystyle \therefore QM=3\text{ cm},\ RN=5\text{ cm and }PL=7\text{ cm}
\\

\displaystyle \textbf{Question 23. }\text{In Figure 6, }O\text{ is the centre of the circle with }AC=24\text{ cm, }
\displaystyle AB=7\text{ cm} \ \text{and }\angle BOD=90^{\circ}.\text{ Find the area of the shaded region. } \\ ( \text{Use }\pi=3.14 )

\displaystyle \textbf{OR}

\displaystyle \text{In Figure 7, find the area of the shaded region, if }ABCD\text{ is a square}
\displaystyle \text{of side }14\text{ cm and }APD\text{ and }BPC\text{ are semicircles.} \displaystyle \text{Answer:}
\displaystyle AB=7\text{ cm},\quad AC=24\text{ cm}
\displaystyle BC=\sqrt{AB^{2}+AC^{2}}
\displaystyle =\sqrt{7^{2}+24^{2}}
\displaystyle =\sqrt{625}=25\text{ cm}
\displaystyle \text{Radius of circle}=\frac{25}{2}=12.5\text{ cm}
\displaystyle \text{Area of circle}=3.14\times(12.5)^{2}=490.625\text{ cm}^{2}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times7\times24=84\text{ cm}^{2}
\displaystyle \text{Area of sector }BOD=\frac{90}{360}\times3.14\times(12.5)^{2}
\displaystyle =122.65625\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=490.625-84-122.65625
\displaystyle =283.96875\text{ cm}^{2}
\displaystyle \therefore \text{Area of shaded region}=283.97\text{ cm}^{2}\text{ approximately}

\displaystyle \textbf{OR}

\displaystyle \text{Area of square}=14\times14=196\text{ cm}^{2}
\displaystyle \text{Radius of each semicircle}=7\text{ cm}
\displaystyle \text{Area of two semicircles}=\pi r^{2}
\displaystyle =\frac{22}{7}\times7\times7=154\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=196-154=42\text{ cm}^{2}
\displaystyle \therefore \text{Area of shaded region}=42\text{ cm}^{2}
\\

\displaystyle \textbf{Question 24. }\text{A hemispherical bowl of internal radius }9\text{ cm is full of water. Its}
\displaystyle \text{contents are emptied in a cylindrical vessel of internal radius }6\text{ cm. Find the height }
\displaystyle \text{of water in the cylindrical vessel.}
\displaystyle \text{Answer:}
\displaystyle \text{Volume of hemispherical bowl}=\frac{2}{3}\pi r^{3}
\displaystyle =\frac{2}{3}\pi(9)^{3}
\displaystyle =486\pi\text{ cm}^{3}
\displaystyle \text{Let the height of water in the cylindrical vessel be }h\text{ cm.}
\displaystyle \text{Volume of cylinder}=\pi R^{2}h
\displaystyle =\pi(6)^{2}h=36\pi h
\displaystyle 36\pi h=486\pi
\displaystyle h=\frac{486}{36}=13.5
\displaystyle \therefore \text{Height of water in the cylindrical vessel}=13.5\text{ cm}
\\

\displaystyle \textbf{Question 25. }\text{The angles of depression of the top and bottom of a tower as seen}
\displaystyle \text{from the top of a }60\sqrt{3}\text{ m high cliff are }45^{\circ}\text{ and }60^{\circ}\text{ respectively. Find the height }
\displaystyle \text{of the tower.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m and the horizontal distance be }x\text{ m.}
\displaystyle \tan60^{\circ}=\frac{60\sqrt{3}}{x}
\displaystyle \sqrt{3}=\frac{60\sqrt{3}}{x}
\displaystyle x=60
\displaystyle \tan45^{\circ}=\frac{60\sqrt{3}-h}{x}
\displaystyle 1=\frac{60\sqrt{3}-h}{60}
\displaystyle 60=60\sqrt{3}-h
\displaystyle h=60\sqrt{3}-60
\displaystyle =60(\sqrt{3}-1)
\displaystyle \therefore \text{Height of the tower}=60(\sqrt{3}-1)\text{ m}
\\

\displaystyle \textbf{Question 26. }\text{Find the coordinates of a point }P,\text{ which lies on the line segment}
\displaystyle \text{joining the points }A(-2,-2)\text{ and }B(2,-4)\text{ such that }AP=\frac{3}{7}AB.

\displaystyle \textbf{OR}

\displaystyle \text{Find the area of the quadrilateral }ABCD\text{ whose vertices are}
\displaystyle A(-3,-1),\ B(-2,-4),\ C(4,-1)\text{ and }D(3,4).
\displaystyle \text{Answer:}
\displaystyle AP=\frac{3}{7}AB
\displaystyle \therefore AP:PB=3:4
\displaystyle \text{Using section formula,}
\displaystyle P=\left(\frac{3(2)+4(-2)}{3+4},\frac{3(-4)+4(-2)}{3+4}\right)
\displaystyle =\left(\frac{6-8}{7},\frac{-12-8}{7}\right)
\displaystyle =\left(-\frac{2}{7},-\frac{20}{7}\right)
\displaystyle \therefore P\left(-\frac{2}{7},-\frac{20}{7}\right)

\displaystyle \textbf{OR}

\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|27-(-29)\right|
\displaystyle =28
\displaystyle \therefore \text{Area of quadrilateral }ABCD=28\text{ square units}
\\

\displaystyle \textbf{Question 27. }\text{If the points }A(x,y),\ B(3,6)\text{ and }C(-3,4)\text{ are collinear, show that}
\displaystyle x-3y+15=0.
\displaystyle \text{Answer:}
\displaystyle \text{Since }A(x,y),\ B(3,6)\text{ and }C(-3,4)\text{ are collinear,}
\displaystyle \text{Area of }\triangle ABC=0
\displaystyle \frac{1}{2}\left|x(6-4)+3(4-y)+(-3)(y-6)\right|=0
\displaystyle 2x+12-3y-3y+18=0
\displaystyle 2x-6y+30=0
\displaystyle x-3y+15=0
\displaystyle \therefore x-3y+15=0
\\

\displaystyle \textbf{Question 28. }\text{All kings, queens and aces are removed from a pack of }52\text{ cards. The}
\displaystyle \text{remaining cards are well shuffled and then a card is drawn from it. Find the probability }
\displaystyle \text{that the drawn card is}
\displaystyle \text{(i) a black face card.}
\displaystyle \text{(ii) a red card.}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards in a pack}=52
\displaystyle \text{Kings, queens and aces removed}=4+4+4=12
\displaystyle \text{Remaining cards}=52-12=40
\displaystyle \text{(i) Face cards are kings, queens and jacks.}
\displaystyle \text{Since kings and queens are removed, only jacks remain.}
\displaystyle \text{Black face cards}=2\text{ (Jack of spades and Jack of clubs)}
\displaystyle P(\text{a black face card})=\frac{2}{40}=\frac{1}{20}
\displaystyle \text{(ii) Total red cards}=26
\displaystyle \text{Red kings, red queens and red aces removed}=2+2+2=6
\displaystyle \text{Remaining red cards}=26-6=20
\displaystyle P(\text{a red card})=\frac{20}{40}=\frac{1}{2}
\displaystyle \therefore \text{Required probabilities are }\frac{1}{20}\text{ and }\frac{1}{2}
\\

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

\displaystyle \textbf{Question 29. }\text{The numerator of a fraction is }3\text{ less than its denominator. If }1\text{ is}
\displaystyle \text{added to the denominator, the fraction is decreased by }\frac{1}{15}.\text{ Find the fraction.}

\displaystyle \textbf{OR}

\displaystyle \text{In a flight of }2800\text{ km, an aircraft was slowed down due to bad weather. Its average }
\displaystyle \text{speed is reduced by }100\text{ km/h and time } \text{increased by }30\text{ minutes. Find the original }
\displaystyle \text{duration of the flight.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the denominator be }x.
\displaystyle \text{Then numerator}=x-3
\displaystyle \text{Fraction}=\frac{x-3}{x}
\displaystyle \frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15}
\displaystyle \frac{x-3}{x(x+1)}=\frac{1}{15}
\displaystyle 15x-45=x^{2}+x
\displaystyle x^{2}-14x+45=0
\displaystyle (x-5)(x-9)=0
\displaystyle x=5\text{ or }x=9
\displaystyle \therefore \text{Fraction}=\frac{2}{5}\text{ or }\frac{6}{9}=\frac{2}{3}

\displaystyle \textbf{OR}

\displaystyle \text{Let the original speed be }x\text{ km/h.}
\displaystyle \text{Original time}=\frac{2800}{x}\text{ hours}
\displaystyle \text{Reduced speed}=(x-100)\text{ km/h}
\displaystyle \frac{2800}{x-100}-\frac{2800}{x}=\frac{1}{2}
\displaystyle \frac{280000}{x(x-100)}=\frac{1}{2}
\displaystyle x(x-100)=560000
\displaystyle x^{2}-100x-560000=0
\displaystyle (x-800)(x+700)=0
\displaystyle x=800\text{ km/h}
\displaystyle \text{Original duration}=\frac{2800}{800}=3.5\text{ hours}
\displaystyle \therefore \text{Original duration of the flight}=3\text{ hours }30\text{ minutes}
\\

\displaystyle \textbf{Question 30. }\text{Find the common difference of an A.P. whose first term is }5\text{ and the}
\displaystyle \text{sum of its first four terms is half the sum of the next four terms.}
\displaystyle \text{Answer:}
\displaystyle a=5
\displaystyle S_{4}=\frac{4}{2}[2a+3d]
\displaystyle =2[10+3d]=20+6d
\displaystyle \text{Sum of next four terms}=T_{5}+T_{6}+T_{7}+T_{8}
\displaystyle =(a+4d)+(a+5d)+(a+6d)+(a+7d)
\displaystyle =4a+22d=20+22d
\displaystyle 20+6d=\frac{1}{2}(20+22d)
\displaystyle 40+12d=20+22d
\displaystyle 20=10d
\displaystyle d=2
\displaystyle \therefore \text{Common difference}=2
\\

\displaystyle \textbf{Question 31. }\text{Prove that the length of tangents drawn from an external point to a}
\displaystyle \text{circle are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ be an external point and }PA,\ PB\text{ be two tangents to a circle}
\displaystyle \text{with centre }O,\text{ touching the circle at }A\text{ and }B\text{ respectively.}
\displaystyle OA\perp PA\text{ and }OB\perp PB
\displaystyle OA=OB\text{ }(\text{radii of the same circle})
\displaystyle OP=OP\text{ }(\text{common})
\displaystyle \text{In right triangles }OPA\text{ and }OPB,
\displaystyle OP^{2}=OA^{2}+PA^{2}
\displaystyle OP^{2}=OB^{2}+PB^{2}
\displaystyle \therefore OA^{2}+PA^{2}=OB^{2}+PB^{2}
\displaystyle \therefore PA^{2}=PB^{2}
\displaystyle \therefore PA=PB
\displaystyle \text{Hence, the lengths of tangents drawn from an external point to a circle are equal.}
\\

\displaystyle \textbf{Question 32. }\text{A hemispherical tank, full of water, is emptied by a pipe at the rate}
\displaystyle \text{of }\frac{25}{7}\text{ litres per sec. How much time will it take to empty half the tank if the }
\displaystyle \text{diameter of the base of the tank is }3\text{ m ?}

\displaystyle \textbf{OR}

\displaystyle \text{A drinking glass is in the shape of the frustum of a cone of height } 14\text{ cm. The diameters }
\displaystyle \text{of its two circular ends are }4\text{ cm and }2\text{ cm. }  \text{Find the capacity of the glass. } \\ ( \text{Use } \pi=\frac{22}{7} )
\displaystyle \text{Answer:}
\displaystyle \text{Radius of hemispherical tank}=\frac{3}{2}\text{ m}
\displaystyle \text{Volume of half tank}=\frac{1}{2}\times\frac{2}{3}\pi r^{3}=\frac{1}{3}\pi r^{3}
\displaystyle =\frac{1}{3}\times\frac{22}{7}\times\left(\frac{3}{2}\right)^{3}
\displaystyle =\frac{99}{28}\text{ m}^{3}=\frac{99000}{28}\text{ litres}
\displaystyle \text{Time}=\frac{\frac{99000}{28}}{\frac{25}{7}}
\displaystyle =990\text{ seconds}=16\text{ minutes }30\text{ seconds}
\displaystyle \therefore \text{Required time}=16\text{ minutes }30\text{ seconds}

\displaystyle \textbf{OR}

\displaystyle R=2\text{ cm},\quad r=1\text{ cm},\quad h=14\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}\times14(4+1+2)
\displaystyle =\frac{308}{3}\text{ cm}^{3}
\displaystyle \therefore \text{Capacity of the glass}=\frac{308}{3}\text{ cm}^{3}
\\

\displaystyle \textbf{Question 33. }\text{A military tent of height }8.25\text{ m is in the form of a right circular}
\displaystyle \text{cylinder of base diameter }30\text{ m and height }5.5\text{ m surmounted by a right circular }
\displaystyle \text{cone of same base radius. Find the length of the canvas used in making the tent, }
\displaystyle \text{if the  breadth of the canvas is }1.5\text{ m.}
\displaystyle \text{Answer:}
\displaystyle r=15\text{ m}
\displaystyle \text{Height of cylindrical part}=5.5\text{ m}
\displaystyle \text{Height of conical part}=8.25-5.5=2.75\text{ m}
\displaystyle l=\sqrt{15^{2}+(2.75)^{2}}
\displaystyle =\sqrt{232.5625}=15.25\text{ m}
\displaystyle \text{Canvas area}=\text{CSA of cylinder}+\text{CSA of cone}
\displaystyle =2\pi rh+\pi rl
\displaystyle =2\pi(15)(5.5)+\pi(15)(15.25)
\displaystyle =165\pi+228.75\pi=393.75\pi\text{ m}^{2}
\displaystyle \text{Length of canvas}=\frac{393.75\pi}{1.5}
\displaystyle =262.5\times\frac{22}{7}=825\text{ m}
\displaystyle \therefore \text{Length of canvas}=825\text{ m}
\\

\displaystyle \textbf{Question 34. }\text{The angles of elevation and depression of the top and bottom of a}
\displaystyle \text{light-house from the top of a }60\text{ m high building are }30^{\circ}\text{ and }60^{\circ}   \text{ respectively. Find}
\displaystyle \text{(i) the difference between the heights of the light-house and the building.}
\displaystyle \text{(ii) the distance between the light-house and the building.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance between the light-house and the building be }x\text{ m.}
\displaystyle \tan60^{\circ}=\frac{60}{x}
\displaystyle \sqrt{3}=\frac{60}{x}
\displaystyle x=\frac{60}{\sqrt{3}}=20\sqrt{3}\text{ m}
\displaystyle \text{Let the difference between the heights be }h\text{ m.}
\displaystyle \tan30^{\circ}=\frac{h}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}}
\displaystyle h=20\text{ m}
\displaystyle \therefore \text{Difference between the heights}=20\text{ m}
\displaystyle \text{Distance between the light-house and the building}=20\sqrt{3}\text{ m}
\\

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