\displaystyle \text{LONG ANSWER TYPE-1 (GRAPH BASED) (5 Mark Each)}

\displaystyle \textbf{Question 106: }\text{(For this question, use a graph paper. Scale: }2\text{ cm}=1\text{ unit}
\displaystyle \text{along both }x\text{ and }y\text{-axis.) Plot the points }A(2,2)\text{ and }B(6,-2)\text{ and answer:}
\displaystyle \text{(a) Reflect point }A\text{ in origin to point }D\text{ and write the coordinates of }D.
\displaystyle \text{(b) Reflect point }A\text{ in line }y=-2\text{ to point }C\text{ and write the coordinates of }C.
\displaystyle \text{(c) Find a point }P\text{ on }CD\text{ which is invariant under reflection in }x=0.
\displaystyle \text{Write its coordinates.}
\displaystyle \text{(d) Write the geometrical name of the closed figure }ABCD.
\displaystyle \text{(e) Write the coordinates of the point of intersection of the diagonals of }ABCD.
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,2)\text{ and }B(6,-2).
\displaystyle \text{(a) Under reflection in the origin, }(x,y)\rightarrow(-x,-y).
\displaystyle \therefore A(2,2)\rightarrow D(-2,-2).
\displaystyle \therefore D=(-2,-2).
\displaystyle \text{(b) The line of reflection is }y=-2.
\displaystyle \text{The perpendicular distance of }A(2,2)\text{ from }y=-2\text{ is }4\text{ units.}
\displaystyle \therefore\text{Its image lies }4\text{ units below the line }y=-2.
\displaystyle C=(2,-6).
\displaystyle \text{(c) A point invariant under reflection in }x=0\text{ must lie on the }y\text{-axis.}
\displaystyle \text{The line through }C(2,-6)\text{ and }D(-2,-2)\text{ has slope}
\displaystyle m=\frac{-2-(-6)}{-2-2}=-1.
\displaystyle \text{Hence, the equation of }CD\text{ is }y=-x-4.
\displaystyle \text{At }x=0,\qquad y=-4.
\displaystyle \therefore P=(0,-4).
\displaystyle \text{(d) }AB=BC=CD=DA=4\sqrt2\text{ units.}
\displaystyle \text{Also, }AB\perp BC.
\displaystyle \therefore ABCD\text{ is a square.}
\displaystyle \text{(e) The diagonals of a square bisect each other.}
\displaystyle \text{Midpoint of }AC=\left(\frac{2+2}{2},\frac{2+(-6)}{2}\right)
\displaystyle =(2,-2)
\displaystyle \therefore\text{The diagonals of }ABCD\text{ intersect at }(2,-2).
\displaystyle \\

\displaystyle \textbf{Question 107: }\text{(For this question, use a graph paper. Scale: }1\text{ cm}=1\text{ unit}
\displaystyle \text{along both }x\text{ and }y\text{-axis.) Plot points }A(0,3),\ B(4,0),\ C(6,2)\text{ and }D(5,0).
\displaystyle \text{Reflect the points as given below and write their coordinates:}
\displaystyle \text{(a) Reflect }A\text{ on }x\text{-axis to }A'.\qquad\text{(b) Reflect }B\text{ on }y\text{-axis to }B'.
\displaystyle \text{(c) Reflect }C\text{ on }x\text{-axis to }C'.
\displaystyle \text{(d) }D\text{ remains invariant when reflected on the line whose equation is }\underline{\hspace{1.5cm}}.
\displaystyle \text{(e) Join the points }A,\ B,\ C,\ D,\ C',\ B',\ A',\ B'\text{ and }A\text{ to form a closed figure.}
\displaystyle \text{Name the closed figure }BCDC'.
\displaystyle \text{Answer:}
\displaystyle \text{(a) Reflection in the }x\text{-axis maps }(x,y)\rightarrow(x,-y).
\displaystyle \therefore A(0,3)\rightarrow A'(0,-3).
\displaystyle \text{(b) Reflection in the }y\text{-axis maps }(x,y)\rightarrow(-x,y).
\displaystyle \therefore B(4,0)\rightarrow B'(-4,0).
\displaystyle \text{(c) Reflection in the }x\text{-axis maps }(x,y)\rightarrow(x,-y).
\displaystyle \therefore C(6,2)\rightarrow C'(6,-2).
\displaystyle \text{(d) Since }D(5,0)\text{ lies on the }x\text{-axis, it remains invariant under reflection}
\displaystyle \text{in the }x\text{-axis.}
\displaystyle \therefore\text{The required line is }y=0.
\displaystyle \text{(e) }BC=\sqrt{(6-4)^2+(2-0)^2}=2\sqrt2
\displaystyle BC'=\sqrt{(6-4)^2+(-2-0)^2}=2\sqrt2
\displaystyle CD=\sqrt{(6-5)^2+(2-0)^2}=\sqrt5
\displaystyle C'D=\sqrt{(6-5)^2+(-2-0)^2}=\sqrt5
\displaystyle \therefore BC=BC'\text{ and }CD=C'D.
\displaystyle \text{Thus, }BCDC'\text{ has two pairs of adjacent equal sides.}
\displaystyle \therefore BCDC'\text{ is a kite.}
\displaystyle \\

\displaystyle \textbf{Question 108: }\text{The following data represents the daily wages in rupees of a certain}
\displaystyle \text{number of employees of a company:}
\displaystyle \begin{array}{c|cccccccc} \text{Daily wages (Rs.)}&30-40&40-50&50-60&60-70&70-80&80-90&90-100&100-110\\ \hline \text{No. of employees}&8&14&12&17&20&26&13&10 \end{array}
\displaystyle \text{Use a graph to answer the following questions:}
\displaystyle \text{(a) Represent the above distribution by an ogive.}
\displaystyle \text{(b) Find on the graph drawn: (i) Median wage.}
\displaystyle \text{(ii) Percentage of employees who earn more than Rs. }84\text{ per day.}
\displaystyle \text{(iii) Number of employees who earn Rs. }56\text{ and below.}
\displaystyle \text{Answer:}
\displaystyle \text{For a less-than ogive, the cumulative frequency table is:}
\displaystyle \begin{array}{c|c} \text{Wages below (Rs.)}&\text{Cumulative frequency}\\ \hline 40&8\\ 50&22\\ 60&34\\ 70&51\\ 80&71\\ 90&97\\ 100&110\\ 110&120 \end{array}
\displaystyle \text{(a) Plot the points }(30,0),(40,8),(50,22),(60,34),(70,51),(80,71),
\displaystyle (90,97),(100,110),(110,120)\text{ and join them by a smooth curve.}
\displaystyle \text{This gives the required less-than ogive.}
\displaystyle \text{Total number of employees }N=120.
\displaystyle \text{(b)(i) }\frac{N}{2}=\frac{120}{2}=60
\displaystyle \text{From cumulative frequency }60,\text{ draw a horizontal line to meet the ogive and then}
\displaystyle \text{draw a vertical line to the }x\text{-axis. The median wage is approximately Rs. }74.5.
\displaystyle \therefore\text{Median wage}\approx\text{Rs. }74.5.
\displaystyle \text{(ii) From Rs. }84\text{ on the }x\text{-axis, the cumulative frequency is approximately }81.
\displaystyle \text{Number earning more than Rs. }84\approx120-81=39.
\displaystyle \text{Required percentage}\approx\frac{39}{120}\times100=32.5\%
\displaystyle \therefore\text{Percentage earning more than Rs. }84\approx32\%.
\displaystyle \text{(iii) From Rs. }56\text{ on the }x\text{-axis, read the corresponding cumulative frequency.}
\displaystyle \text{The cumulative frequency is approximately }29.
\displaystyle \therefore\text{Number of employees earning Rs. }56\text{ and below}\approx29.
\displaystyle \\

\displaystyle \textbf{Question 109: }\text{Study the graph and answer the questions that follow:}
\displaystyle \text{(a) Make a frequency table for the information provided in the graph.}
\displaystyle \text{(b) Find the number of students whose height is less than }150\text{ cm}.
\displaystyle \text{(c) Find the total number of students.}\qquad\text{(d) Find the modal height.}
\displaystyle \text{(e) Find the difference in the modal height and the mean height, if the average}
\displaystyle \text{height of the students is }145.5\text{ cm}. \displaystyle \text{Answer:}
\displaystyle \text{(a) From the histogram, the frequency distribution is:}
\displaystyle \begin{array}{c|c} \text{Height (in cm)}&\text{Number of students}\\ \hline 120-130&6\\ 130-140&29\\ 140-150&34\\ 150-160&22\\ 160-170&12 \end{array}
\displaystyle \text{(b) Number of students whose height is less than }150\text{ cm}
\displaystyle =6+29+34=69
\displaystyle \therefore\text{The number of students whose height is less than }150\text{ cm is }69.
\displaystyle \text{(c) Total number of students}=6+29+34+22+12
\displaystyle =103
\displaystyle \therefore\text{The total number of students is }103.
\displaystyle \text{(d) The highest rectangle corresponds to the class }140-150.
\displaystyle \therefore\text{The modal class is }140-150.
\displaystyle \text{From the graphical construction, the modal height is approximately }143\text{ cm}.
\displaystyle \therefore\text{Modal height}\approx143\text{ cm}.
\displaystyle \text{(e) Mean height}=145.5\text{ cm}
\displaystyle \text{Difference}=145.5-143=2.5\text{ cm}
\displaystyle \therefore\text{The difference between the mean and modal heights is approximately }2.5\text{ cm}.
\displaystyle \\

 

\displaystyle \text{LONG ANSWER TYPE-2  (5 Mark Each)}

\displaystyle \textbf{Question 110: }\text{On seeing the display board outside Pearl Stationary Shop, Chetan}
\displaystyle \text{enters the shop to buy the following items:} \displaystyle \begin{array}{c|c|c|c} &\text{Pen}&\text{Pencil}&\text{Rainbow Cover Notebook}\\ \hline \text{Price}&\text{Rs. }5\text{ each}&\text{Rs. }7\text{ each}&\text{Rs. }200\text{ each}\\ \text{Discount}&5\%\text{ on a dozen}&10\%\text{ on }20&--\\ \text{Premium}&-&-&\text{Rs. }50\text{ each}\\ \text{Items purchased}&12&20&5\\ \text{GST}&18\%&12\%&12\% \end{array}
\displaystyle \text{The shopkeeper gives a further discount of }2\%\text{ on the total bill but charges }18\%
\displaystyle \text{GST uniformly on all the items. Calculate the required amounts and determine whether}
\displaystyle \text{Chetan has been overcharged.}
\displaystyle \text{Answer:}
\displaystyle \text{(a)(i) Cost of }12\text{ pens}=12\times5=\text{Rs. }60
\displaystyle \text{Discount on pens}=5\%\text{ of }60=\text{Rs. }3
\displaystyle \text{Selling price of pens}=60-3=\text{Rs. }57
\displaystyle \text{Cost of }20\text{ pencils}=20\times7=\text{Rs. }140
\displaystyle \text{Discount on pencils}=10\%\text{ of }140=\text{Rs. }14
\displaystyle \text{Selling price of pencils}=140-14=\text{Rs. }126
\displaystyle \text{Price of each notebook after premium}=200+50=\text{Rs. }250
\displaystyle \text{Selling price of }5\text{ notebooks}=5\times250=\text{Rs. }1250
\displaystyle \text{Total selling price}=57+126+1250=\text{Rs. }1433
\displaystyle \therefore\text{Total selling price of all the items}=\text{Rs. }1433.
\displaystyle \text{(a)(ii) GST on pens}=18\%\text{ of }57=\text{Rs. }10.26
\displaystyle \text{Amount for pens including GST}=57+10.26=\text{Rs. }67.26
\displaystyle \text{GST on pencils}=12\%\text{ of }126=\text{Rs. }15.12
\displaystyle \text{Amount for pencils including GST}=126+15.12=\text{Rs. }141.12
\displaystyle \text{GST on notebooks}=12\%\text{ of }1250=\text{Rs. }150
\displaystyle \text{Amount for notebooks including GST}=1250+150=\text{Rs. }1400
\displaystyle \text{Total amount with correct GST}=67.26+141.12+1400
\displaystyle =\text{Rs. }1608.38
\displaystyle \therefore\text{Amount payable with correct GST rates}=\text{Rs. }1608.38.
\displaystyle \text{(a)(iii) Further discount}=2\%\text{ of }1433
\displaystyle =\frac{2}{100}\times1433=\text{Rs. }28.66
\displaystyle \text{Amount after further discount}=1433-28.66=\text{Rs. }1404.34
\displaystyle \text{The shopkeeper wrongly charged }18\%\text{ GST on this amount.}
\displaystyle \text{GST charged}=18\%\text{ of }1404.34
\displaystyle =\text{Rs. }252.7812
\displaystyle \text{Actual amount charged}=1404.34+252.7812
\displaystyle =\text{Rs. }1657.1212\approx\text{Rs. }1657.12
\displaystyle \therefore\text{Actual amount charged by the shopkeeper}=\text{Rs. }1657.12.
\displaystyle \text{(b) Overcharged amount}=1657.12-1608.38
\displaystyle =\text{Rs. }48.74
\displaystyle \therefore\text{Yes, the shopkeeper overcharged Chetan by Rs. }48.74.
\displaystyle \\

\displaystyle \textbf{Question 111: }\text{Using remainder and factor theorem, show that }(2x+3)\text{ is a}
\displaystyle \text{factor of the polynomial }2x^2+11x+12.\text{ Hence, factorise it completely. What must}
\displaystyle \text{be multiplied to the given polynomial so that }x^2+3x-4\text{ is a factor of the resulting}
\displaystyle \text{polynomial? Also, write the resulting polynomial.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=2x^2+11x+12.
\displaystyle \text{For }2x+3=0,\qquad x=-\frac{3}{2}.
\displaystyle f\left(-\frac{3}{2}\right)=2\left(-\frac{3}{2}\right)^2+11\left(-\frac{3}{2}\right)+12
\displaystyle =\frac{9}{2}-\frac{33}{2}+12
\displaystyle =-12+12=0
\displaystyle \therefore (2x+3)\text{ is a factor of }2x^2+11x+12.
\displaystyle 2x^2+11x+12=2x^2+3x+8x+12
\displaystyle =x(2x+3)+4(2x+3)
\displaystyle =(2x+3)(x+4)
\displaystyle \therefore\text{The complete factorisation is }(2x+3)(x+4).
\displaystyle \text{Now, }x^2+3x-4=(x+4)(x-1).
\displaystyle \text{The given polynomial already contains the factor }(x+4).
\displaystyle \therefore\text{It must be multiplied by }(x-1).
\displaystyle (x-1)(2x^2+11x+12)
\displaystyle =2x^3+9x^2+x-12
\displaystyle \therefore\text{The required multiplier is }x-1.
\displaystyle \therefore\text{The resulting polynomial is }2x^3+9x^2+x-12.
\displaystyle \\

\displaystyle \textbf{Question 112: }\text{The sequence }2,\ 9,\ 16,\ldots\text{ is given.}
\displaystyle \text{(a) Identify if the given sequence is an A.P. or a G.P. Give reasons to support your answer.}
\displaystyle \text{(b) Find the }20\text{th term of the sequence.}
\displaystyle \text{(c) Find the difference between the sum of its first }22\text{ and }25\text{ terms.}
\displaystyle \text{(d) Does the term }102\text{ belong to this sequence?}
\displaystyle \text{(e) If }k\text{ is added to each of the above terms, will the new sequence be in A.P. or G.P.?}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }9-2=7\qquad\text{and}\qquad16-9=7
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \text{Here, }a=2\text{ and }d=7.
\displaystyle \text{(b) }T_n=a+(n-1)d
\displaystyle T_{20}=2+(20-1)(7)
\displaystyle =2+133=135
\displaystyle \therefore T_{20}=135.
\displaystyle \text{(c) }S_{25}-S_{22}=T_{23}+T_{24}+T_{25}
\displaystyle T_{23}=2+22(7)=156
\displaystyle T_{24}=2+23(7)=163
\displaystyle T_{25}=2+24(7)=170
\displaystyle S_{25}-S_{22}=156+163+170=489
\displaystyle \therefore\text{The required difference is }489.
\displaystyle \text{(d) If }102\text{ is a term, then}
\displaystyle 102=2+(n-1)7
\displaystyle 100=7(n-1)
\displaystyle n-1=\frac{100}{7}
\displaystyle n=\frac{107}{7}
\displaystyle \text{Since }n\text{ is not a natural number, }102\text{ is not a term of the sequence.}
\displaystyle \text{(e) The new sequence is }2+k,\ 9+k,\ 16+k,\ldots
\displaystyle (9+k)-(2+k)=7
\displaystyle (16+k)-(9+k)=7
\displaystyle \text{The common difference remains }7.
\displaystyle \therefore\text{The new sequence is also an A.P.}
\displaystyle \\

\displaystyle \textbf{Question 113: }\text{Given the equations of two straight lines, }L_1\text{ and }L_2\text{ are}
\displaystyle x-y=1\text{ and }x+y=5\text{ respectively. If }L_1\text{ and }L_2\text{ intersect at }Q(3,2),\text{ find:}
\displaystyle \text{(a) the equation of line }L_3\text{ which is parallel to }L_1\text{ and has }y\text{-intercept }3.
\displaystyle \text{(b) the value of }k,\text{ if }L_3\text{ meets }L_2\text{ at a point }P(k,4).
\displaystyle \text{(c) the coordinate of }R\text{ and the ratio }PQ:QR,\text{ if }L_2\text{ meets the }x\text{-axis at }R.
\displaystyle \text{Answer:}
\displaystyle \text{(a) }L_1:x-y=1
\displaystyle y=x-1
\displaystyle \therefore\text{Slope of }L_1=1.
\displaystyle \text{Since }L_3\parallel L_1,\text{ slope of }L_3=1.
\displaystyle \text{Also, the }y\text{-intercept of }L_3\text{ is }3.
\displaystyle \therefore L_3:y=x+3.
\displaystyle \text{Hence, the equation of }L_3\text{ is }y=x+3.
\displaystyle \text{(b) Since }P(k,4)\text{ lies on }L_2:x+y=5,
\displaystyle k+4=5
\displaystyle k=1
\displaystyle \therefore k=1.
\displaystyle \text{(c) At }R,\text{ the line }L_2\text{ meets the }x\text{-axis, so }y=0.
\displaystyle x+0=5
\displaystyle x=5
\displaystyle \therefore R=(5,0).
\displaystyle \text{Also, }P=(1,4),\quad Q=(3,2),\quad R=(5,0).
\displaystyle PQ=\sqrt{(3-1)^2+(2-4)^2}
\displaystyle =\sqrt{4+4}=2\sqrt2
\displaystyle QR=\sqrt{(5-3)^2+(0-2)^2}
\displaystyle =\sqrt{4+4}=2\sqrt2
\displaystyle \therefore PQ:QR=1:1.
\displaystyle \therefore R=(5,0)\text{ and }PQ:QR=1:1.
\displaystyle \\

\displaystyle \textbf{Question 114: }\text{In the figure given above (not drawn to scale), }AD\parallel GE\parallel BC,
\displaystyle DE=18\text{ cm},\ EC=3\text{ cm and }AD=35\text{ cm. Find:}
\displaystyle \text{(a) }AF:FC\qquad\text{(b) length of }EF
\displaystyle \text{(c) area of trapezium }ADEF:\text{ area of }\triangle EFC\qquad\text{(d) }BC:GF
\displaystyle \text{Answer:}
\displaystyle DC=DE+EC=18+3=21\text{ cm}
\displaystyle \text{Since }EF\parallel AD,\quad\triangle CEF\sim\triangle CDA.
\displaystyle \therefore\frac{CE}{CD}=\frac{CF}{CA}=\frac{EF}{AD}
\displaystyle =\frac{3}{21}=\frac{1}{7}
\displaystyle \text{(a) }\frac{CF}{CA}=\frac{1}{7}
\displaystyle \therefore CF:CA=1:7
\displaystyle AF=AC-CF
\displaystyle \therefore AF:FC=6:1.
\displaystyle \text{(b) }\frac{EF}{AD}=\frac{1}{7}
\displaystyle \frac{EF}{35}=\frac{1}{7}
\displaystyle EF=5\text{ cm}
\displaystyle \therefore\text{The length of }EF\text{ is }5\text{ cm}.
\displaystyle \text{(c) Since }\triangle CEF\sim\triangle CDA,
\displaystyle \frac{{ar}(\triangle CEF)}{{ar}(\triangle CDA)}=\left(\frac{CE}{CD}\right)^2
\displaystyle =\left(\frac{1}{7}\right)^2=\frac{1}{49}
\displaystyle \therefore\text{ar}(\triangle CEF):\text{ar}(\triangle CDA)=1:49.
\displaystyle \text{ar}(\text{trapezium }ADEF)=49-1=48\text{ parts}
\displaystyle \therefore\text{ar}(ADEF):\text{ar}(\triangle EFC)=48:1.
\displaystyle \text{(d) Since }GF\parallel BC,\quad\triangle AFG\sim\triangle ABC.
\displaystyle \therefore\frac{GF}{BC}=\frac{AF}{AC}
\displaystyle =\frac{6}{7}
\displaystyle \therefore BC:GF=7:6.
\displaystyle \\

\displaystyle \textbf{Question 115: }\text{(Use a ruler and a compass for this question.)}
\displaystyle \text{(a) Construct the locus of a moving point which moves such that it keeps a fixed}
\displaystyle \text{distance of }4.5\text{ cm from a fixed point }O.
\displaystyle \text{(b) Draw line segment }AB\text{ of }6\text{ cm where }A\text{ and }B\text{ are two points on locus (a).}
\displaystyle \text{(c) Construct the locus of all points equidistant from }A\text{ and }B.\text{ Name the points}
\displaystyle \text{of intersection of loci (a) and (c) as }P\text{ and }Q\text{ respectively.}
\displaystyle \text{(d) Join }PA.\text{ Find the locus of all points equidistant from }AP\text{ and }AB.
\displaystyle \text{Mark the point of intersection of loci (a) and (d) as }R.\text{ Measure }AR.
\displaystyle \text{Answer:}
\displaystyle \text{(a) With }O\text{ as centre and radius }4.5\text{ cm, draw a circle.}
\displaystyle \text{This circle is the locus of all points at a distance }4.5\text{ cm from }O.
\displaystyle \text{(b) Mark two points }A\text{ and }B\text{ on the circle such that }AB=6\text{ cm}.
\displaystyle \text{(c) Construct the perpendicular bisector of }AB.
\displaystyle \text{The perpendicular bisector of }AB\text{ is the locus of points equidistant from }A\text{ and }B.
\displaystyle \text{Let it intersect the circle at }P\text{ and }Q.
\displaystyle \text{(d) Join }PA.
\displaystyle \text{The locus of points equidistant from the lines }AP\text{ and }AB\text{ is the angle bisector}
\displaystyle \text{of }\angle PAB.
\displaystyle \text{Construct the internal angle bisector of }\angle PAB\text{ and let it meet the circle at }R.
\displaystyle \text{On measuring, }AR\approx4.7\text{ cm}.
\displaystyle \therefore\text{The required length of }AR\text{ is approximately }4.7\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 116: }\text{(Use a ruler and a compass for this question.)}
\displaystyle \text{Construct a regular hexagon }ABCDEF\text{ of side }4.3\text{ cm and construct its}
\displaystyle \text{circumscribed circle. Also, construct tangents to the circle at }B\text{ and }C\text{ which}
\displaystyle \text{meet each other at }P.\text{ Measure and record }\angle BPC.
\displaystyle \text{Answer:}
\displaystyle \text{Draw a circle of radius }4.3\text{ cm with centre }O.
\displaystyle \text{Mark six equal chords of length }4.3\text{ cm successively on the circle.}
\displaystyle \text{Name the vertices }A,\ B,\ C,\ D,\ E,\ F\text{ and join consecutive vertices.}
\displaystyle \text{Thus, }ABCDEF\text{ is the required regular hexagon.}
\displaystyle \text{Join }OB\text{ and }OC.
\displaystyle \text{At }B,\text{ draw a line perpendicular to }OB,\text{ and at }C,\text{ draw a line perpendicular}
\displaystyle \text{to }OC.\text{ Let these two tangents meet at }P.
\displaystyle \text{For a regular hexagon, }\angle BOC=60^\circ.
\displaystyle \text{Also, }OB\perp PB\text{ and }OC\perp PC.
\displaystyle \therefore\angle OBP=\angle OCP=90^\circ.
\displaystyle \text{In quadrilateral }OBPC,
\displaystyle \angle BPC=360^\circ-(90^\circ+90^\circ+60^\circ)
\displaystyle =120^\circ
\displaystyle \therefore\text{On measuring, }\angle BPC=120^\circ.
\displaystyle \\

\displaystyle \textbf{Question 117: }\text{A mathematics teacher uses a certain amount of terracotta clay to form}
\displaystyle \text{different shaped solids. First, he turned it into a sphere of radius }7\text{ cm and then made a}
\displaystyle \text{right circular cone with base radius }14\text{ cm. Find the height of the cone so formed.}
\displaystyle \text{If the same clay is turned to make a right circular cylinder of height }\frac{7}{3}\text{ cm, find}
\displaystyle \text{the radius of the cylinder. Also, compare the total surface areas of sphere and cylinder.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the same amount of clay is used, the volumes of the solids are equal.}
\displaystyle \text{Volume of sphere}=\frac{4}{3}\pi(7)^3
\displaystyle =\frac{1372}{3}\pi\text{ cm}^3
\displaystyle \text{Let the height of the cone be }h\text{ cm}.
\displaystyle \text{Volume of cone}=\frac{1}{3}\pi(14)^2h
\displaystyle =\frac{196}{3}\pi h
\displaystyle \frac{196}{3}\pi h=\frac{1372}{3}\pi
\displaystyle h=\frac{1372}{196}=7\text{ cm}
\displaystyle \therefore\text{The height of the cone is }7\text{ cm}.
\displaystyle \text{Let the radius of the cylinder be }R\text{ cm}.
\displaystyle \text{Volume of cylinder}=\pi R^2\times\frac{7}{3}
\displaystyle \pi R^2\times\frac{7}{3}=\frac{1372}{3}\pi
\displaystyle 7R^2=1372
\displaystyle R^2=196
\displaystyle R=14\text{ cm}
\displaystyle \therefore\text{The radius of the cylinder is }14\text{ cm}.
\displaystyle \text{Total surface area of sphere}=4\pi r^2
\displaystyle =4\pi(7)^2=196\pi\text{ cm}^2
\displaystyle \text{Total surface area of cylinder}=2\pi R(R+h)
\displaystyle =2\pi(14)\left(14+\frac{7}{3}\right)
\displaystyle =28\pi\left(\frac{49}{3}\right)=\frac{1372}{3}\pi\text{ cm}^2
\displaystyle \text{TSA of sphere : TSA of cylinder}=196\pi:\frac{1372}{3}\pi
\displaystyle =588:1372=3:7
\displaystyle \therefore\text{TSA of sphere : TSA of cylinder}=3:7.
\displaystyle \\

\displaystyle \textbf{Question 118: }\text{A tree }(TS)\text{ of height }30\text{ m stands in front of a tall building}
\displaystyle (AB).\text{ Two friends Rohit and Neha are standing at }R\text{ and }N\text{ respectively, along the}
\displaystyle \text{same straight line joining the tree and the building. Rohit, standing at a distance of}
\displaystyle 150\text{ m from the foot of the building, observes the angle of elevation of the top of the}
\displaystyle \text{building as }30^\circ.\text{ Neha observes that the top of the building and the tree have the}
\displaystyle \text{same elevation of }60^\circ.\text{ Find:}

\displaystyle \text{(a) height of the building.}
\displaystyle \text{(b) distance between: (i) Neha and the foot of the building.}
\displaystyle \text{(ii) Rohit and Neha.}\quad\text{(iii) Neha and the tree.}\quad\text{(iv) building and the tree.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }RA=150\text{ m and }\angle BRA=30^\circ.
\displaystyle \text{(a) In right-angled }\triangle RAB,
\displaystyle \tan30^\circ=\frac{AB}{RA}
\displaystyle \frac{1}{\sqrt3}=\frac{AB}{150}
\displaystyle AB=\frac{150}{\sqrt3}=50\sqrt3\text{ m}
\displaystyle \therefore\text{Height of the building}=50\sqrt3\text{ m}\approx86.6\text{ m}.
\displaystyle \text{(b)(i) Let }NA=x\text{ m}.
\displaystyle \text{In right-angled }\triangle NAB,
\displaystyle \tan60^\circ=\frac{AB}{NA}
\displaystyle \sqrt3=\frac{50\sqrt3}{x}
\displaystyle x=50
\displaystyle \therefore NA=50\text{ m}.
\displaystyle \text{(ii) }RN=RA-NA
\displaystyle =150-50=100\text{ m}
\displaystyle \therefore RN=100\text{ m}.
\displaystyle \text{(iii) Let }NT=y\text{ m}.
\displaystyle \text{In right-angled }\triangle NTS,
\displaystyle \tan60^\circ=\frac{TS}{NT}
\displaystyle \sqrt3=\frac{30}{y}
\displaystyle y=\frac{30}{\sqrt3}=10\sqrt3\text{ m}
\displaystyle \therefore NT=10\sqrt3\text{ m}\approx17.3\text{ m}.
\displaystyle \text{(iv) }TA=NA-NT
\displaystyle =50-10\sqrt3\text{ m}
\displaystyle \approx50-17.32=32.68\text{ m}
\displaystyle \therefore TA=50-10\sqrt3\text{ m}\approx32.7\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 119: }\text{A life insurance agent found the following data of age distribution of}
\displaystyle 100\text{ policyholders where }f\text{ is an unknown frequency.}
\displaystyle \begin{array}{c|c} \text{Age in years}&\text{No. of Policyholders}\\ \hline 15-20&7\\ 20-25&12\\ 25-30&15\\ 30-35&22\\ 35-40&f\\ 40-45&14\\ 45-50&8\\ 50-55&4 \end{array}
\displaystyle \text{(a) If the mean age of the policyholders is }35.65\text{ years, find the unknown frequency }f.
\displaystyle \text{(b) Find the median class of the distribution.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Total number of policyholders}=100.
\displaystyle 7+12+15+22+f+14+8+4=100
\displaystyle 82+f=100
\displaystyle f=18
\displaystyle \therefore\text{The unknown frequency is }18.
\displaystyle \text{Verification using the given mean:}
\displaystyle \begin{array}{c|c|c|c} \text{Age}&f_i&x_i&f_ix_i\\ \hline 15-20&7&17.5&122.5\\ 20-25&12&22.5&270\\ 25-30&15&27.5&412.5\\ 30-35&22&32.5&715\\ 35-40&18&37.5&675\\ 40-45&14&42.5&595\\ 45-50&8&47.5&380\\ 50-55&4&52.5&210 \end{array}
\displaystyle \sum f_i=100,\qquad\sum f_ix_i=3380
\displaystyle \overline{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{3380}{100}=33.8\text{ years}
\displaystyle \text{This does not agree with the stated mean of }35.65\text{ years.}
\displaystyle \therefore\text{The given mean }35.65\text{ years appears to be an error in the question.}
\displaystyle \text{However, since the total number of policyholders is }100,\text{ the value }f=18\text{ is fixed.}
\displaystyle \text{(b) }N=100,\qquad\frac{N}{2}=50
\displaystyle \text{The cumulative frequencies are }7,\ 19,\ 34,\ 56,\ 74,\ 88,\ 96,\ 100.
\displaystyle \text{The first cumulative frequency greater than }50\text{ is }56,
\displaystyle \text{which corresponds to the class interval }30-35.
\displaystyle \therefore\text{The median class is }30-35\text{ years.}
\displaystyle \\

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