\displaystyle \text{SHORT ANSWER TYPE-1 (3 Mark Each)}

\displaystyle \textbf{Question 50: }\text{A shopkeeper marked a pressure cooker at Rs. }1800.\text{ The rate of}
\displaystyle \text{GST on the pressure cooker is }12\%.\text{ The customer has only Rs. }1792\text{ with him and}
\displaystyle \text{requests the shopkeeper to reduce the price so that he can buy the cooker for Rs. }1792.
\displaystyle \text{What percent discount must the shopkeeper give?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the price after discount, excluding GST, be Rs. }x.
\displaystyle \text{GST}=12\%\text{ of Rs. }x=\frac{12x}{100}.
\displaystyle \therefore x+\frac{12x}{100}=1792
\displaystyle \frac{112x}{100}=1792
\displaystyle x=\frac{1792\times100}{112}=1600
\displaystyle \therefore \text{Discount}=1800-1600=\text{Rs. }200
\displaystyle \text{Discount percent}=\frac{200}{1800}\times100
\displaystyle =\frac{100}{9}\%=11\frac{1}{9}\%
\displaystyle \therefore\text{The shopkeeper must give a discount of }11\frac{1}{9}\%.
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{A man opened a recurring deposit account in a branch of PNB. The man}
\displaystyle \text{deposits a certain amount of money per month such that after }2\text{ years, the interest}
\displaystyle \text{accumulated is equal to his monthly deposit. Find the rate of interest per annum that}
\displaystyle \text{the bank was paying for the recurring deposit account.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the monthly deposit be Rs. }P\text{ and the rate of interest be }r\%\text{ p.a.}
\displaystyle \text{Number of months}=2\times12=24
\displaystyle I=P\times\frac{24(24+1)}{2}\times\frac{r}{12\times100}
\displaystyle I=P\times300\times\frac{r}{1200}=\frac{Pr}{4}
\displaystyle \text{Since the interest accumulated is equal to the monthly deposit,}
\displaystyle \frac{Pr}{4}=P
\displaystyle \frac{r}{4}=1
\displaystyle r=4
\displaystyle \therefore\text{The rate of interest paid by the bank was }4\%\text{ per annum.}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Akshay buys }350\text{ shares of Rs. }50\text{ par value of a company. The}
\displaystyle \text{dividend declared by the company is }14\%.\text{ If his return percent from the shares is}
\displaystyle 10\%,\text{ find the market value of each share.}
\displaystyle \text{Answer:}
\displaystyle \text{Dividend per share}=14\%\text{ of Rs. }50
\displaystyle =\frac{14}{100}\times50=\text{Rs. }7
\displaystyle \text{Let the market value of each share be Rs. }x.
\displaystyle \text{Return percent}=\frac{\text{Dividend}}{\text{Market value}}\times100
\displaystyle 10=\frac{7}{x}\times100
\displaystyle 10x=700
\displaystyle x=70
\displaystyle \therefore\text{The market value of each share is Rs. }70.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Solve the following inequation:}
\displaystyle \frac{1}{2}(2x-1)\leq2x+\frac{1}{2}\leq5\frac{1}{2}+x
\displaystyle \text{(a) Write the maximum and minimum values of }x\text{ for }x\in R.
\displaystyle \text{(b) What will be the change in maximum and minimum values of }x\text{ if }x\in W?
\displaystyle \text{Answer:}
\displaystyle \frac{1}{2}(2x-1)\leq2x+\frac{1}{2}
\displaystyle x-\frac{1}{2}\leq2x+\frac{1}{2}
\displaystyle -1\leq x
\displaystyle \text{Also,}\qquad2x+\frac{1}{2}\leq5\frac{1}{2}+x
\displaystyle 2x+\frac{1}{2}\leq\frac{11}{2}+x
\displaystyle x\leq5
\displaystyle \therefore -1\leq x\leq5
\displaystyle \text{(a) For }x\in R,\text{ minimum value}=-1,\quad\text{maximum value}=5.
\displaystyle \text{(b) For }x\in W,\quad x\in\{0,1,2,3,4,5\}.
\displaystyle \therefore\text{Minimum value}=0,\qquad\text{maximum value}=5.
\displaystyle \text{Thus, the minimum changes from }-1\text{ to }0,\text{ while the maximum remains }5.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Solve for }x,\text{ if }\frac{5}{x}+4\sqrt3=\frac{2\sqrt3}{x^2},\ x\ne0.
\displaystyle \text{Answer:}
\displaystyle \frac{5}{x}+4\sqrt3=\frac{2\sqrt3}{x^2}
\displaystyle \text{Multiplying throughout by }x^2,
\displaystyle 5x+4\sqrt3x^2=2\sqrt3
\displaystyle 4\sqrt3x^2+5x-2\sqrt3=0
\displaystyle x=\frac{-5\pm\sqrt{25-4(4\sqrt3)(-2\sqrt3)}}{8\sqrt3}
\displaystyle =\frac{-5\pm\sqrt{121}}{8\sqrt3}
\displaystyle =\frac{-5\pm11}{8\sqrt3}
\displaystyle x=\frac{6}{8\sqrt3}=\frac{\sqrt3}{4}
\displaystyle \text{or}\qquad x=\frac{-16}{8\sqrt3}=-\frac{2\sqrt3}{3}
\displaystyle \therefore x=\frac{\sqrt3}{4}\text{ or }-\frac{2\sqrt3}{3}.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{The marked price of a toy is same as the percentage of GST that is}
\displaystyle \text{charged. The price of the toy is Rs. }24\text{ including GST. Taking the marked price as }x,
\displaystyle \text{form an equation and solve it to find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Marked price}=\text{Rs. }x,\qquad\text{Rate of GST}=x\%
\displaystyle \text{GST}=\frac{x}{100}\times x=\frac{x^2}{100}
\displaystyle \text{Since the price including GST is Rs. }24,
\displaystyle x+\frac{x^2}{100}=24
\displaystyle x^2+100x-2400=0
\displaystyle x^2+120x-20x-2400=0
\displaystyle x(x+120)-20(x+120)=0
\displaystyle (x-20)(x+120)=0
\displaystyle x=20\text{ or }x=-120
\displaystyle \text{Since the marked price cannot be negative, }x=20.
\displaystyle \therefore\text{The marked price of the toy is Rs. }20.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{The mean proportion between two numbers is }6\text{ and their third}
\displaystyle \text{proportional is }48.\text{ Find the two numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Since }6\text{ is the mean proportional,}
\displaystyle a:6=6:b
\displaystyle ab=36\qquad\qquad ...(1)
\displaystyle \text{Since }48\text{ is the third proportional,}
\displaystyle a:b=b:48
\displaystyle b^2=48a\qquad\qquad ...(2)
\displaystyle \text{From (1),}\qquad b=\frac{36}{a}
\displaystyle \text{Substituting in (2),}
\displaystyle \left(\frac{36}{a}\right)^2=48a
\displaystyle 1296=48a^3
\displaystyle a^3=27
\displaystyle a=3
\displaystyle b=\frac{36}{3}=12
\displaystyle \therefore\text{The two numbers are }3\text{ and }12.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Pamela factorized the following polynomial:}
\displaystyle 2x^3+3x^2-3x-2
\displaystyle \text{She found the result as }(x+2)(x-1)(x-2).\text{ Using remainder and factor theorem,}
\displaystyle \text{verify whether her result is correct. If incorrect, give the correct result.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=2x^3+3x^2-3x-2.
\displaystyle f(-2)=2(-2)^3+3(-2)^2-3(-2)-2
\displaystyle =-16+12+6-2=0
\displaystyle \therefore (x+2)\text{ is a factor of }f(x).
\displaystyle f(1)=2+3-3-2=0
\displaystyle \therefore (x-1)\text{ is a factor of }f(x).
\displaystyle f(2)=2(2)^3+3(2)^2-3(2)-2
\displaystyle =16+12-6-2=20\ne0
\displaystyle \therefore (x-2)\text{ is not a factor of }f(x).
\displaystyle \text{Hence, Pamela's factorization is incorrect.}
\displaystyle (x+2)(x-1)=x^2+x-2
\displaystyle 2x^3+3x^2-3x-2=(x^2+x-2)(2x+1)
\displaystyle \therefore 2x^3+3x^2-3x-2=(x+2)(x-1)(2x+1).
\displaystyle \\

\displaystyle \textbf{Question 58: }\ A=\begin{bmatrix}-6&0\\4&2\end{bmatrix}\text{ and }B=\begin{bmatrix}1&0\\1&3\end{bmatrix}.
\displaystyle \text{Find matrix }M,\text{ if }M=\frac{1}{2}A-2B+5I,\text{ where }I\text{ is the identity matrix.}
\displaystyle \text{Answer:}
\displaystyle M=\frac{1}{2}\begin{bmatrix}-6&0\\4&2\end{bmatrix} -2\begin{bmatrix}1&0\\1&3\end{bmatrix} +5\begin{bmatrix}1&0\\0&1\end{bmatrix}
\displaystyle =\begin{bmatrix}-3&0\\2&1\end{bmatrix} +\begin{bmatrix}-2&0\\-2&-6\end{bmatrix} +\begin{bmatrix}5&0\\0&5\end{bmatrix}
\displaystyle =\begin{bmatrix}-3-2+5&0\\2-2&1-6+5\end{bmatrix}
\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}
\displaystyle \therefore M=\begin{bmatrix}0&0\\0&0\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{(a) Write the }n\text{th term }(T_n)\text{ of an Arithmetic Progression (A.P.)}
\displaystyle \text{consisting of all whole numbers which are divisible by }3\text{ and }7.
\displaystyle \text{(b) How many of these are two-digit numbers? Write them.}
\displaystyle \text{(c) Find the sum of first }10\text{ terms of this A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{LCM}(3,7)=21
\displaystyle \therefore\text{The A.P. is }21,\ 42,\ 63,\ 84,\ 105,\ldots
\displaystyle \text{Here, }a=21\text{ and }d=21.
\displaystyle \text{(a) }T_n=a+(n-1)d
\displaystyle =21+(n-1)(21)
\displaystyle =21n
\displaystyle \therefore T_n=21n.
\displaystyle \text{(b) The two-digit terms are }21,\ 42,\ 63,\ 84.
\displaystyle \therefore\text{There are }4\text{ two-digit terms.}
\displaystyle \text{(c) }S_{10}=\frac{10}{2}[2(21)+(10-1)(21)]
\displaystyle =5[42+189]
\displaystyle =5(231)=1155
\displaystyle \therefore\text{The sum of the first }10\text{ terms is }1155.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Write the first five terms of the sequence given by }(\sqrt3)^n,\ n\in N.
\displaystyle \text{(a) Is the sequence an A.P. or G.P.?}
\displaystyle \text{(b) If the sum of its first ten terms is }p(3+\sqrt3),\text{ find the value of }p.
\displaystyle \text{Answer:}
\displaystyle \text{For }n=1,2,3,4,5,\text{ the first five terms are}
\displaystyle \sqrt3,\ 3,\ 3\sqrt3,\ 9,\ 9\sqrt3.
\displaystyle \text{(a) The ratio of two consecutive terms is }\sqrt3.
\displaystyle \therefore\text{The sequence is a G.P. with }a=\sqrt3\text{ and }r=\sqrt3.
\displaystyle \text{(b) }S_{10}=\frac{a(r^{10}-1)}{r-1}
\displaystyle =\frac{\sqrt3[(\sqrt3)^{10}-1]}{\sqrt3-1}
\displaystyle =\frac{\sqrt3(243-1)}{\sqrt3-1}
\displaystyle =\frac{242\sqrt3}{\sqrt3-1}
\displaystyle =\frac{242\sqrt3(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}
\displaystyle =121\sqrt3(\sqrt3+1)
\displaystyle =121(3+\sqrt3)
\displaystyle \text{Comparing with }p(3+\sqrt3),
\displaystyle \therefore p=121.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{ABC is a triangle as shown in the figure below.}
\displaystyle \text{(a) Write down the coordinates of }A,\ B\text{ and }C\text{ on reflecting through the origin.}
\displaystyle \text{(b) Write down the coordinates of the point/s which remain invariant on reflecting}
\displaystyle \text{the triangle }ABC\text{ on the }x\text{-axis and }y\text{-axis respectively.} \displaystyle \text{Answer:}
\displaystyle \text{From the graph, }A(4,5),\quad B(0,3),\quad C(3,0).
\displaystyle \text{(a) Under reflection through the origin, }(x,y)\rightarrow(-x,-y).
\displaystyle A(4,5)\rightarrow A'(-4,-5)
\displaystyle B(0,3)\rightarrow B'(0,-3)
\displaystyle C(3,0)\rightarrow C'(-3,0)
\displaystyle \therefore A'(-4,-5),\quad B'(0,-3),\quad C'(-3,0).
\displaystyle \text{(b) A point on the }x\text{-axis remains invariant under reflection in the }x\text{-axis.}
\displaystyle \therefore C(3,0)\text{ remains invariant on reflection in the }x\text{-axis.}
\displaystyle \text{A point on the }y\text{-axis remains invariant under reflection in the }y\text{-axis.}
\displaystyle \therefore B(0,3)\text{ remains invariant on reflection in the }y\text{-axis.}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Determine the ratio in which the line }y=2+3x\text{ divides the line}
\displaystyle \text{segment }AB\text{ joining the points }A(-3,9)\text{ and }B(4,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let the line divide }AB\text{ at }P\text{ in the ratio }m:n.
\displaystyle \therefore P\left(\frac{4m-3n}{m+n},\frac{2m+9n}{m+n}\right)
\displaystyle \text{Since }P\text{ lies on }y=2+3x,
\displaystyle \frac{2m+9n}{m+n}=2+3\left(\frac{4m-3n}{m+n}\right)
\displaystyle 2m+9n=2m+2n+12m-9n
\displaystyle 16n=12m
\displaystyle \frac{m}{n}=\frac{4}{3}
\displaystyle \therefore\text{The required ratio is }4:3.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Square }ABCD\text{ lies in the third quadrant of an }XY\text{ plane such}
\displaystyle \text{that its vertex }A\text{ is at }(-3,-1)\text{ and the diagonal }DB\text{ produced is equally}
\displaystyle \text{inclined to both the axes. The diagonals }AC\text{ and }BD\text{ meet at }P(-2,-2).
\displaystyle \text{Find the: (a) slope of }BD\qquad\text{(b) equation of }AC.
\displaystyle \text{Answer:}
\displaystyle \text{Since }A(-3,-1)\text{ and }P(-2,-2)\text{ lie on diagonal }AC,
\displaystyle \text{Slope of }AC=\frac{-2-(-1)}{-2-(-3)}=\frac{-1}{1}=-1
\displaystyle \text{The diagonals of a square are perpendicular to each other.}
\displaystyle \therefore (\text{slope of }AC)(\text{slope of }BD)=-1
\displaystyle (-1)(\text{slope of }BD)=-1
\displaystyle \therefore\text{(a) Slope of }BD=1.
\displaystyle \text{(b) Using point }A(-3,-1)\text{ and slope of }AC=-1,
\displaystyle y+1=-1(x+3)
\displaystyle y+1=-x-3
\displaystyle x+y+4=0
\displaystyle \therefore\text{The equation of }AC\text{ is }x+y+4=0.
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{ABCD is a rectangle where side }BC\text{ is twice side }AB.
\displaystyle \text{If }\triangle ACQ\sim\triangle BAP,\text{ find area of }\triangle BAP:\text{ area of }\triangle ACQ. \displaystyle \text{Answer:}
\displaystyle \text{Let }AB=x.\text{ Since }BC=2AB,\quad BC=2x.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle AC=\sqrt{AB^2+BC^2}
\displaystyle =\sqrt{x^2+(2x)^2}=\sqrt5\,x
\displaystyle \text{Given }\triangle ACQ\sim\triangle BAP.
\displaystyle \therefore \frac{AC}{BA}=\frac{\sqrt5\,x}{x}=\sqrt5
\displaystyle \text{For similar triangles, the ratio of their areas is the square of the ratio}
\displaystyle \text{of their corresponding sides.}
\displaystyle \frac{\text{Area of }\triangle ACQ}{\text{Area of }\triangle BAP} =\left(\frac{AC}{BA}\right)^2
\displaystyle =(\sqrt5)^2=5
\displaystyle \therefore\text{Area of }\triangle BAP:\text{Area of }\triangle ACQ=1:5.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Given a triangle }ABC,\text{ and }D\text{ is a point on }BC\text{ such that}
\displaystyle BD=4\text{ cm and }DC=x\text{ cm. If }\angle BAD=\angle C,\text{ and }AB=8\text{ cm, then:} \displaystyle \text{(a) Prove that }\triangle ABD\text{ is similar to }\triangle CBA.
\displaystyle \text{(b) Find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{(a) In }\triangle ABD\text{ and }\triangle CBA,
\displaystyle \angle BAD=\angle BCA\qquad\text{(Given)}
\displaystyle \angle ABD=\angle CBA\qquad\text{(Since }B,D,C\text{ are collinear)}
\displaystyle \therefore \triangle ABD\sim\triangle CBA\qquad\text{(AA similarity)}
\displaystyle \text{(b) From the similarity of the triangles,}
\displaystyle \frac{AB}{CB}=\frac{BD}{BA}
\displaystyle \frac{8}{CB}=\frac{4}{8}
\displaystyle CB=\frac{8\times8}{4}=16\text{ cm}
\displaystyle \text{Now, }DC=BC-BD
\displaystyle x=16-4=12\text{ cm}
\displaystyle \therefore x=12\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{In the extract of Survey of India map G43S7, prepared on a scale}
\displaystyle \text{of }2\text{ cm to }1\text{ km, a child finds the length of the cart track between two settlements}
\displaystyle \text{is }7.6\text{ cm. Find:}
\displaystyle \text{(a) the actual length of the cart track on the ground.}
\displaystyle \text{(b) actual area of a grid square, if each has an area of }4\text{ cm}^2.
\displaystyle \text{Answer:}
\displaystyle \text{Given scale: }2\text{ cm}=1\text{ km}
\displaystyle \text{(a) }7.6\text{ cm}=\frac{7.6}{2}\text{ km}=3.8\text{ km}
\displaystyle \therefore\text{The actual length of the cart track is }3.8\text{ km}.
\displaystyle \text{(b) Since }2\text{ cm represents }1\text{ km,}
\displaystyle 4\text{ cm}^2=(2\text{ cm})^2\text{ represents }(1\text{ km})^2=1\text{ km}^2.
\displaystyle \therefore\text{The actual area of the grid square is }1\text{ km}^2.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Construct a triangle }ABC\text{ such that }AB=7\text{ cm},\ BC=6\text{ cm}
\displaystyle \text{and }CA=5\text{ cm. (Use ruler and compass to do so).}
\displaystyle \text{(a) Draw the locus of the points such that:}
\displaystyle \text{(i) it is equidistant from }BC\text{ and }BA.
\displaystyle \text{(ii) it is equidistant from points }A\text{ and }B.
\displaystyle \text{(b) Mark }P\text{ where the loci (i) and (ii) meet, measure and write length of }PA.
\displaystyle \text{Answer:}
\displaystyle \text{Construct }AB=7\text{ cm}.
\displaystyle \text{With }A\text{ as centre and radius }5\text{ cm, draw an arc.}
\displaystyle \text{With }B\text{ as centre and radius }6\text{ cm, draw another arc cutting it at }C.
\displaystyle \text{Join }AC\text{ and }BC\text{ to obtain }\triangle ABC.
\displaystyle \text{(a)(i) The locus of points equidistant from }BC\text{ and }BA\text{ is the angle bisector}
\displaystyle \text{of }\angle ABC.\text{ Construct the internal angle bisector of }\angle ABC.
\displaystyle \text{(ii) The locus of points equidistant from }A\text{ and }B\text{ is the perpendicular}
\displaystyle \text{bisector of }AB.\text{ Construct the perpendicular bisector of }AB.
\displaystyle \text{(b) Let the two loci intersect at }P.
\displaystyle \text{On measuring, }PA\approx3.8\text{ cm}.
\displaystyle \therefore PA\approx3.8\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{In the given figure }O\text{ is the centre of the circle. }ABCD\text{ is a}
\displaystyle \text{quadrilateral where sides }AB,\ BC,\ CD\text{ and }DA\text{ touch the circle at }E,\ F,\ G\text{ and }H
\displaystyle \text{respectively. If }AB=15\text{ cm},\ BC=18\text{ cm and }AD=24\text{ cm, find the length of }CD. \displaystyle \text{Answer:}
\displaystyle \text{Tangents drawn from an external point to a circle are equal.}
\displaystyle AE=AH,\quad BE=BF,\quad CF=CG,\quad DG=DH
\displaystyle AB+CD=(AE+EB)+(CG+GD)
\displaystyle =(AH+BF)+(CF+DH)
\displaystyle =(BF+CF)+(AH+DH)=BC+AD
\displaystyle \therefore AB+CD=BC+AD
\displaystyle 15+CD=18+24
\displaystyle CD=42-15=27\text{ cm}
\displaystyle \therefore\text{The length of }CD\text{ is }27\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{In the given diagram }ABCDEF\text{ is a regular hexagon inscribed}
\displaystyle \text{in a circle with centre }O.\ PQ\text{ is a tangent to the circle at }D.\text{ Find the value of:}
\displaystyle \text{(a) }\angle FAG\qquad\text{(b) }\angle BCD\qquad\text{(c) }\angle PDE \displaystyle \text{Answer:}
\displaystyle \text{Since }ABCDEF\text{ is a regular hexagon, each interior angle is }120^\circ.
\displaystyle \text{(a) Since }AB\text{ is produced to }G,
\displaystyle \angle FAG+\angle FAB=180^\circ
\displaystyle \angle FAG=180^\circ-120^\circ=60^\circ
\displaystyle \therefore\angle FAG=60^\circ.
\displaystyle \text{(b) }\angle BCD\text{ is an interior angle of the regular hexagon.}
\displaystyle \therefore\angle BCD=120^\circ.
\displaystyle \text{(c) Since the hexagon is regular, }\angle DOE=60^\circ.
\displaystyle \text{The angle subtended by chord }DE\text{ at the circumference is}
\displaystyle \frac{1}{2}\angle DOE=\frac{1}{2}\times60^\circ=30^\circ.
\displaystyle \text{By the tangent-chord theorem, }\angle PDE=30^\circ.
\displaystyle \therefore\angle FAG=60^\circ,\quad\angle BCD=120^\circ,\quad\angle PDE=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{AB and CD intersect at the centre }O\text{ of the circle as shown}
\displaystyle \text{in the diagram. If }\angle EBA=33^\circ\text{ and }\angle EAC=82^\circ,\text{ find:}
\displaystyle \text{(a) }\angle BAE\qquad\text{(b) }\angle BOC\qquad\text{(c) }\angle ODB \displaystyle \text{Answer:}
\displaystyle \text{(a) Since }AB\text{ is a diameter, }\angle AEB=90^\circ.
\displaystyle \text{In }\triangle AEB,
\displaystyle \angle BAE=180^\circ-(90^\circ+33^\circ)=57^\circ
\displaystyle \therefore\angle BAE=57^\circ.
\displaystyle \text{(b) }\angle BAE=57^\circ\text{ subtends arc }BE.
\displaystyle \therefore\text{arc }BE=2\times57^\circ=114^\circ.
\displaystyle \angle EAC=82^\circ\text{ subtends arc }EC.
\displaystyle \therefore\text{arc }EC=2\times82^\circ=164^\circ.
\displaystyle \therefore\text{arc }BC=164^\circ-114^\circ=50^\circ.
\displaystyle \therefore\angle BOC=50^\circ\qquad\text{(angle at the centre)}
\displaystyle \text{(c) Since }C,\ O,\ D\text{ are collinear,}
\displaystyle \angle BOD=180^\circ-\angle BOC=180^\circ-50^\circ=130^\circ.
\displaystyle \text{Also, since }D,\ E,\ B\text{ are collinear, }\angle OBD=\angle EBA=33^\circ.
\displaystyle \text{In }\triangle OBD,
\displaystyle \angle ODB=180^\circ-(130^\circ+33^\circ)=17^\circ
\displaystyle \therefore\angle BAE=57^\circ,\quad\angle BOC=50^\circ,\quad\angle ODB=17^\circ.
\displaystyle \\

\displaystyle \textbf{Question 71: }\text{A famous sweet shop ``Madanlal Sweets'' sells tinned rasgullas.}
\displaystyle \text{The tin container is cylindrical in shape with diameter }14\text{ cm, height }16\text{ cm, and it}
\displaystyle \text{can hold }20\text{ spherical rasgullas of diameter }6\text{ cm and sweetened liquid such that}
\displaystyle \text{the can is filled and then sealed. Find how much sweetened liquid the can contains.}
\displaystyle \text{Take }\pi=3.14.
\displaystyle \text{Answer:}
\displaystyle \text{Radius of cylindrical tin}=\frac{14}{2}=7\text{ cm}
\displaystyle \text{Height of cylindrical tin}=16\text{ cm}
\displaystyle \text{Volume of cylindrical tin}=\pi r^2h
\displaystyle =\pi(7)^2(16)=784\pi\text{ cm}^3
\displaystyle \text{Radius of each rasgulla}=\frac{6}{2}=3\text{ cm}
\displaystyle \text{Volume of }20\text{ rasgullas}=20\times\frac{4}{3}\pi(3)^3
\displaystyle =20\times36\pi=720\pi\text{ cm}^3
\displaystyle \text{Volume of sweetened liquid}=784\pi-720\pi
\displaystyle =64\pi=64\times3.14=200.96\text{ cm}^3
\displaystyle \therefore\text{The can contains }200.96\text{ cm}^3\text{ of sweetened liquid.}
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{The ratio of the radius and the height of a solid metallic right}
\displaystyle \text{circular cylinder is }7:27.\text{ This is melted and made into a cone of diameter }14\text{ cm}
\displaystyle \text{and slant height }25\text{ cm. Find the height of the:}
\displaystyle \text{(a) cone}\qquad\text{(b) cylinder}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Radius of cone}=\frac{14}{2}=7\text{ cm},\qquad l=25\text{ cm}
\displaystyle l^2=r^2+h^2
\displaystyle 25^2=7^2+h^2
\displaystyle h^2=625-49=576
\displaystyle h=24\text{ cm}
\displaystyle \therefore\text{The height of the cone is }24\text{ cm}.
\displaystyle \text{(b) Let the radius and height of the cylinder be }7x\text{ and }27x\text{ respectively.}
\displaystyle \text{Since the cylinder is melted to form the cone, their volumes are equal.}
\displaystyle \pi(7x)^2(27x)=\frac{1}{3}\pi(7)^2(24)
\displaystyle 49\times27x^3=49\times8
\displaystyle 27x^3=8
\displaystyle x^3=\frac{8}{27}
\displaystyle x=\frac{2}{3}
\displaystyle \text{Height of cylinder}=27x=27\times\frac{2}{3}=18\text{ cm}
\displaystyle \therefore\text{The height of the cylinder is }18\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{An inclined plane }AC\text{ is prepared with its base }AB\text{ which is }\sqrt3
\displaystyle \text{times its vertical height }BC.\text{ The length of the inclined plane is }15\text{ m. Find:}
\displaystyle \text{(a) value of }\theta\qquad\text{(b) length of its base }AB,\text{ in nearest metre.} \displaystyle \text{Answer:}
\displaystyle \text{Given }AB=\sqrt3\,BC.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan\theta=\frac{BC}{AB}
\displaystyle =\frac{BC}{\sqrt3\,BC}=\frac{1}{\sqrt3}
\displaystyle \therefore \theta=30^\circ.
\displaystyle \text{(b) Since }AC=15\text{ m},
\displaystyle \cos30^\circ=\frac{AB}{AC}
\displaystyle \frac{\sqrt3}{2}=\frac{AB}{15}
\displaystyle AB=\frac{15\sqrt3}{2}\approx12.99\text{ m}
\displaystyle \therefore AB\approx13\text{ m, to the nearest metre.}
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{Prove that}
\displaystyle \tan^2\theta+\cos^2\theta-1=\tan^2\theta\sin^2\theta
\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\tan^2\theta+\cos^2\theta-1
\displaystyle =\tan^2\theta-(1-\cos^2\theta)
\displaystyle =\tan^2\theta-\sin^2\theta
\displaystyle =\frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta
\displaystyle =\sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right)
\displaystyle =\sin^2\theta(\sec^2\theta-1)
\displaystyle =\sin^2\theta\tan^2\theta
\displaystyle =\tan^2\theta\sin^2\theta=\text{R.H.S.}
\displaystyle \therefore\text{L.H.S.}=\text{R.H.S. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{The class mark and frequency of a data is given in the graph.}
\displaystyle \text{From the graph, find:}
\displaystyle \text{(a) the table showing the class interval and frequency.}
\displaystyle \text{(b) Calculate the mean.}
\displaystyle \text{Answer:}
\displaystyle \text{From the graph, the class marks are }13,\ 15,\ 17,\ 19,\ 21,\ 23.
\displaystyle \text{The difference between two consecutive class marks is }2.
\displaystyle \therefore\text{Class size}=2.
\displaystyle \text{(a) The required frequency distribution is:}
\displaystyle \begin{array}{c|c|c} \text{Class Interval}&\text{Class Mark }(x)&\text{Frequency }(f)\\ \hline 12-14&13&8\\ 14-16&15&2\\ 16-18&17&3\\ 18-20&19&4\\ 20-22&21&5\\ 22-24&23&6 \end{array}
\displaystyle \text{(b) For calculating the mean:}
\displaystyle \begin{array}{c|c|c} x&f&fx\\ \hline 13&8&104\\ 15&2&30\\ 17&3&51\\ 19&4&76\\ 21&5&105\\ 23&6&138\\ \hline &\sum f=28&\sum fx=504 \end{array}
\displaystyle \bar{x}=\frac{\sum fx}{\sum f}
\displaystyle =\frac{504}{28}=18
\displaystyle \therefore\text{The mean of the given distribution is }18.
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{The mean of }5,\ 7,\ 8,\ 4\text{ and }m\text{ is }n\text{ and the mean of }5,\ 7,\ 8,\ 4,
\displaystyle m\text{ and }n\text{ is }m.\text{ Find the values of }m\text{ and }n.
\displaystyle \text{Answer:}
\displaystyle \text{Mean of }5,\ 7,\ 8,\ 4\text{ and }m\text{ is }n.
\displaystyle \therefore \frac{5+7+8+4+m}{5}=n
\displaystyle 24+m=5n\qquad\qquad ...(1)
\displaystyle \text{Mean of }5,\ 7,\ 8,\ 4,\ m\text{ and }n\text{ is }m.
\displaystyle \therefore \frac{5+7+8+4+m+n}{6}=m
\displaystyle 24+m+n=6m
\displaystyle n=5m-24\qquad\qquad ...(2)
\displaystyle \text{Substituting (2) in (1),}
\displaystyle 24+m=5(5m-24)
\displaystyle 24+m=25m-120
\displaystyle 24m=144
\displaystyle m=6
\displaystyle \text{From (2),}\qquad n=5(6)-24=6
\displaystyle \therefore m=6,\qquad n=6.
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{The probability of selecting a blue marble and a red marble from a}
\displaystyle \text{bag containing red, blue and green marbles is }\frac{1}{3}\text{ and }\frac{1}{5}\text{ respectively. If the bag}
\displaystyle \text{contains }14\text{ green marbles, find:}
\displaystyle \text{(a) number of red marbles.}\qquad\text{(b) total number of marbles in the bag.}
\displaystyle \text{Answer:}
\displaystyle P(\text{blue})=\frac{1}{3},\qquad P(\text{red})=\frac{1}{5}
\displaystyle P(\text{green})=1-\left(\frac{1}{3}+\frac{1}{5}\right)
\displaystyle =1-\frac{8}{15}=\frac{7}{15}
\displaystyle \text{Let the total number of marbles be }N.
\displaystyle \therefore \frac{14}{N}=\frac{7}{15}
\displaystyle 7N=14\times15
\displaystyle N=30
\displaystyle \text{(a) Number of red marbles}=\frac{1}{5}\times30=6
\displaystyle \therefore\text{The number of red marbles is }6.
\displaystyle \text{(b) The total number of marbles in the bag is }30.
\displaystyle \\

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