\displaystyle \text{SHORT ANSWER TYPE-2 (4 Mark Each)}

\displaystyle \textbf{Question 78: }\text{The following bill shows the GST rate and the marked price of items:}
\displaystyle \begin{array}{c|l|c|c|c} \text{S. No.}&\text{Item}&\text{Marked Price (Rs.)}&\text{Quantity}&\text{Rate of GST}\\ \hline 1&\text{Wheat Flour (unpacked)}&35.00&5\text{ kg}&x\%\\ 2&\text{Basmati Rice (Branded \& Packed)}&180.00&5\text{ kg}&5\%\\ 3&\text{Surf Excel Quick Wash Detergent}&220.00&y\text{ kg}&18\% \end{array}
\displaystyle \text{Find:}
\displaystyle \text{(a) the value of }x\text{ if the total GST on wheat flour and basmati rice is Rs. }45.
\displaystyle \text{(b) the value of }y,\text{ if CGST paid for detergent powder is Rs. }39.60.
\displaystyle \text{(c) total amount to be paid (including GST) for the above bill.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Cost of wheat flour}=35\times5=\text{Rs. }175
\displaystyle \text{Cost of basmati rice}=180\times5=\text{Rs. }900
\displaystyle \text{GST on basmati rice}=5\%\text{ of }900
\displaystyle =\frac{5}{100}\times900=\text{Rs. }45
\displaystyle \text{Total GST on wheat flour and basmati rice}=\text{Rs. }45.
\displaystyle \therefore\text{GST on wheat flour}=45-45=\text{Rs. }0
\displaystyle \therefore x=0\%.
\displaystyle \text{(b) GST on detergent}=18\%
\displaystyle \therefore\text{CGST}=9\%
\displaystyle \text{Given CGST paid}=\text{Rs. }39.60.
\displaystyle \therefore 9\%\text{ of }220y=39.60
\displaystyle \frac{9}{100}\times220y=39.60
\displaystyle 19.8y=39.60
\displaystyle y=2
\displaystyle \therefore y=2\text{ kg}.
\displaystyle \text{(c) Cost of detergent}=220\times2=\text{Rs. }440
\displaystyle \text{GST on detergent}=18\%\text{ of }440=\text{Rs. }79.20
\displaystyle \text{Total marked price}=175+900+440=\text{Rs. }1515
\displaystyle \text{Total GST}=0+45+79.20=\text{Rs. }124.20
\displaystyle \text{Total amount payable}=1515+124.20=\text{Rs. }1639.20
\displaystyle \therefore\text{The total amount to be paid is Rs. }1639.20.
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{Amit deposited Rs. }600\text{ per month in a recurring deposit account.}
\displaystyle \text{The bank pays a simple interest of }12\%\text{ p.a. Calculate the:}
\displaystyle \text{(a) number of monthly instalments Amit deposits to get a maturity amount of Rs. }11826.
\displaystyle \text{(b) total interest paid by the bank.}
\displaystyle \text{(c) total amount deposited by him.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of monthly instalments be }n.
\displaystyle \text{Monthly deposit}=P=\text{Rs. }600,\qquad r=12\%\text{ p.a.}
\displaystyle I=P\times\frac{n(n+1)}{2}\times\frac{r}{12\times100}
\displaystyle =600\times\frac{n(n+1)}{2}\times\frac{12}{12\times100}
\displaystyle =3n(n+1)
\displaystyle \text{Maturity amount}=\text{Total deposit}+\text{Interest}
\displaystyle 11826=600n+3n(n+1)
\displaystyle 11826=600n+3n^2+3n
\displaystyle 3n^2+603n-11826=0
\displaystyle n^2+201n-3942=0
\displaystyle (n-18)(n+219)=0
\displaystyle n=18\text{ or }n=-219
\displaystyle \text{Since the number of instalments cannot be negative, }n=18.
\displaystyle \text{(a) Number of monthly instalments}=18.
\displaystyle \text{(b) Interest}=3n(n+1)
\displaystyle =3(18)(19)=1026
\displaystyle \therefore\text{Total interest paid by the bank is Rs. }1026.
\displaystyle \text{(c) Total amount deposited}=600\times18
\displaystyle =\text{Rs. }10800
\displaystyle \therefore\text{The total amount deposited by Amit is Rs. }10800.
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Aman has }500,\text{ Rs. }100\text{ shares of a company quoted at Rs. }120,
\displaystyle \text{paying a }10\%\text{ dividend. When the share price rises to Rs. }200\text{ each, he sells all his}
\displaystyle \text{shares. He invests half of the sale proceeds in Rs. }10,\ 12\%\text{ shares at Rs. }25,\text{ and}
\displaystyle \text{the remaining sale proceeds in Rs. }400,\ 9\%\text{ shares at Rs. }500.\text{ Find his:}
\displaystyle \text{(a) sales proceeds}\qquad\text{(b) investment in Rs. }10,\ 12\%\text{ shares at Rs. }25.
\displaystyle \text{(c) original income}\qquad\text{(d) change in income.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Sales proceeds}=500\times200=\text{Rs. }100000.
\displaystyle \therefore\text{Sales proceeds}=\text{Rs. }100000.
\displaystyle \text{(b) Amount invested in Rs. }10,\ 12\%\text{ shares}=\frac{100000}{2}
\displaystyle =\text{Rs. }50000
\displaystyle \text{Number of shares purchased}=\frac{50000}{25}=2000
\displaystyle \therefore\text{He invests Rs. }50000\text{ and purchases }2000\text{ shares.}
\displaystyle \text{(c) Dividend per original share}=10\%\text{ of Rs. }100=\text{Rs. }10
\displaystyle \text{Original income}=500\times10=\text{Rs. }5000.
\displaystyle \text{(d) Income from the first new investment}
\displaystyle =2000\times12\%\text{ of Rs. }10
\displaystyle =2000\times1.20=\text{Rs. }2400
\displaystyle \text{Remaining investment}=\text{Rs. }50000
\displaystyle \text{Number of Rs. }400,\ 9\%\text{ shares}=\frac{50000}{500}=100
\displaystyle \text{Dividend per share}=9\%\text{ of Rs. }400=\text{Rs. }36
\displaystyle \text{Income from the second investment}=100\times36=\text{Rs. }3600
\displaystyle \text{New income}=2400+3600=\text{Rs. }6000
\displaystyle \text{Change in income}=6000-5000=\text{Rs. }1000
\displaystyle \therefore\text{His income increases by Rs. }1000.
\displaystyle \\

\displaystyle \textbf{Question 81: }\text{Solve the following inequation:}
\displaystyle \frac{11+3x}{5}\geq3-x>-\frac{3}{2},\quad x\in R
\displaystyle \text{(a) Write the solution set.}
\displaystyle \text{(b) Represent the solution on the number line.}
\displaystyle \text{Answer:}
\displaystyle \frac{11+3x}{5}\geq3-x
\displaystyle 11+3x\geq15-5x
\displaystyle 8x\geq4
\displaystyle x\geq\frac{1}{2}\qquad ...(1)
\displaystyle \text{Also,}\qquad3-x>-\frac{3}{2}
\displaystyle -x>-\frac{3}{2}-3
\displaystyle -x>-\frac{9}{2}
\displaystyle x<\frac{9}{2}\qquad ...(2)
\displaystyle \text{From (1) and (2),}\qquad\frac{1}{2}\leq x<\frac{9}{2}
\displaystyle \text{(a) Solution set}=\left\{x:x\in R,\ \frac{1}{2}\leq x<\frac{9}{2}\right\}
\displaystyle =\left[\frac{1}{2},\frac{9}{2}\right)
\displaystyle \text{(b) On the number line, shade the interval from }\frac{1}{2}\text{ to }\frac{9}{2}.
\displaystyle \text{Use a closed circle at }\frac{1}{2}\text{ and an open circle at }\frac{9}{2}.
\displaystyle \\

\displaystyle \textbf{Question 82: }\text{Determine whether the following quadratic equation has real roots.}
\displaystyle 5x^2-9x+4=0
\displaystyle \text{(a) Give reasons for your answer.}
\displaystyle \text{(b) If the equation has real roots, identify them.}
\displaystyle \text{Answer:}
\displaystyle \text{For }5x^2-9x+4=0,\quad a=5,\quad b=-9,\quad c=4.
\displaystyle \text{(a) Discriminant }D=b^2-4ac
\displaystyle =(-9)^2-4(5)(4)
\displaystyle =81-80=1
\displaystyle \text{Since }D>0,\text{ the equation has two distinct real roots.}
\displaystyle \text{(b) }x=\frac{-b\pm\sqrt D}{2a}
\displaystyle =\frac{9\pm1}{10}
\displaystyle x=\frac{10}{10}=1\qquad\text{or}\qquad x=\frac{8}{10}=\frac{4}{5}
\displaystyle \therefore\text{The real roots are }1\text{ and }\frac{4}{5}.
\displaystyle \\

\displaystyle \textbf{Question 83: }\text{The profit in rupees in a local restaurant and the number of customers}
\displaystyle \text{who visited the restaurant are tabulated below for each week for one month.}
\displaystyle \begin{array}{c|c|c|c|c} \text{Week number}&\text{Week 1}&\text{Week 2}&\text{Week 3}&\text{Week 4}\\ \hline \text{Number of customers}&1400&5600&x&3212\\ \text{Profit in Rs.}&28000&112000&32140&y \end{array}
\displaystyle \text{Find:}
\displaystyle \text{(a) if the number of customers and profit per week are in continued proportion or not.}
\displaystyle \text{Justify your answer.}\qquad\text{(b) the value of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{(a) For Week 1,}\qquad\frac{\text{Profit}}{\text{Number of customers}}
\displaystyle =\frac{28000}{1400}=20
\displaystyle \text{For Week 2,}\qquad\frac{112000}{5600}=20
\displaystyle \therefore\text{The number of customers and profit are in the same proportion.}
\displaystyle 1400:28000=5600:112000=x:32140=3212:y
\displaystyle \text{(b) }\frac{x}{32140}=\frac{1400}{28000}=\frac{1}{20}
\displaystyle x=\frac{32140}{20}=1607
\displaystyle \therefore x=1607.
\displaystyle \text{Also,}\qquad\frac{3212}{y}=\frac{1}{20}
\displaystyle y=3212\times20=64240
\displaystyle \therefore y=64240.
\displaystyle \\

\displaystyle \textbf{Question 84: }\text{Given, }9x^2-4\text{ is a factor of }9x^3-mx^2-nx+8.
\displaystyle \text{(a) Find the value of }m\text{ and }n\text{ using the remainder and factor theorem.}
\displaystyle \text{(b) Factorise the given polynomial completely.}
\displaystyle \text{Answer:}
\displaystyle 9x^2-4=(3x-2)(3x+2)
\displaystyle \text{Therefore, }x=\frac{2}{3}\text{ and }x=-\frac{2}{3}\text{ are zeros of the polynomial.}
\displaystyle \text{Let }f(x)=9x^3-mx^2-nx+8.
\displaystyle \text{By the factor theorem, }f\left(\frac{2}{3}\right)=0.
\displaystyle 9\left(\frac{2}{3}\right)^3-m\left(\frac{2}{3}\right)^2-n\left(\frac{2}{3}\right)+8=0
\displaystyle \frac{8}{3}-\frac{4m}{9}-\frac{2n}{3}+8=0
\displaystyle 2m+3n=48\qquad ...(1)
\displaystyle \text{Also, }f\left(-\frac{2}{3}\right)=0.
\displaystyle 9\left(-\frac{2}{3}\right)^3-m\left(-\frac{2}{3}\right)^2-n\left(-\frac{2}{3}\right)+8=0
\displaystyle -\frac{8}{3}-\frac{4m}{9}+\frac{2n}{3}+8=0
\displaystyle -2m+3n=-24\qquad ...(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle 6n=24
\displaystyle n=4
\displaystyle \text{Substituting }n=4\text{ in (1),}
\displaystyle 2m+12=48
\displaystyle m=18
\displaystyle \therefore m=18,\qquad n=4.
\displaystyle \text{(b) The polynomial becomes}
\displaystyle 9x^3-18x^2-4x+8
\displaystyle =(9x^2-4)(x-2)
\displaystyle =(3x-2)(3x+2)(x-2)
\displaystyle \therefore\text{The complete factorisation is }(3x-2)(3x+2)(x-2).
\displaystyle \\

\displaystyle \textbf{Question 85: }\text{The marks scored by }100\text{ students are given below:}
\displaystyle \begin{array}{c|c} \text{Marks scored}&\text{No. of students}\\ \hline 0-10&4\\ 10-20&5\\ 20-30&9\\ 30-40&7\\ 40-50&13\\ 50-60&12\\ 60-70&15\\ 70-80&11\\ 80-90&14\\ 90-100&10 \end{array}
\displaystyle \text{A student in the class is selected at random. Find the probability that the student has}
\displaystyle \text{scored: (a) less than }20.\quad\text{(b) below }60\text{ but }30\text{ or more.}
\displaystyle \text{(c) more than or equal to }70.\qquad\text{(d) above }89.
\displaystyle \text{Answer:}
\displaystyle \text{Total number of students}=100.
\displaystyle \text{(a) Number of students scoring less than }20=4+5=9
\displaystyle P(\text{less than }20)=\frac{9}{100}
\displaystyle \therefore P(\text{less than }20)=\frac{9}{100}.
\displaystyle \text{(b) Number of students scoring }30\text{ or more but below }60
\displaystyle =7+13+12=32
\displaystyle P(30\leq\text{score}<60)=\frac{32}{100}=\frac{8}{25}
\displaystyle \therefore P(30\leq\text{score}<60)=\frac{8}{25}.
\displaystyle \text{(c) Number of students scoring }70\text{ or more}=11+14+10=35
\displaystyle P(\text{score}\geq70)=\frac{35}{100}=\frac{7}{20}
\displaystyle \therefore P(\text{score}\geq70)=\frac{7}{20}.
\displaystyle \text{(d) Number of students scoring above }89=10
\displaystyle P(\text{above }89)=\frac{10}{100}=\frac{1}{10}
\displaystyle \therefore P(\text{above }89)=\frac{1}{10}.
\displaystyle \\

\displaystyle \textbf{Question 86: }\text{Given, matrix }A=\begin{bmatrix}x&1\\y&2\end{bmatrix}\text{ and }B=\begin{bmatrix}x\\x-2\end{bmatrix}
\displaystyle \text{such that }AB\text{ is a null matrix. Find:}
\displaystyle \text{(a) order of the null matrix.}\qquad\text{(b) possible values of }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle \text{(a) }A\text{ is of order }2\times2\text{ and }B\text{ is of order }2\times1.
\displaystyle \therefore AB\text{ is of order }2\times1.
\displaystyle \therefore\text{The order of the null matrix is }2\times1.
\displaystyle \text{(b) }AB=\begin{bmatrix}x&1\\y&2\end{bmatrix}\begin{bmatrix}x\\x-2\end{bmatrix}
\displaystyle =\begin{bmatrix}x^2+x-2\\xy+2x-4\end{bmatrix}
\displaystyle \text{Since }AB\text{ is a null matrix,}
\displaystyle x^2+x-2=0
\displaystyle (x+2)(x-1)=0
\displaystyle x=1\text{ or }x=-2.
\displaystyle \text{Also,}\qquad xy+2x-4=0.
\displaystyle \text{When }x=1,\qquad y+2-4=0
\displaystyle y=2.
\displaystyle \text{When }x=-2,\qquad -2y-4-4=0
\displaystyle y=-4.
\displaystyle \therefore (x,y)=(1,2)\text{ or }(-2,-4).
\displaystyle \\

\displaystyle \textbf{Question 87: }\text{The sum of a certain number of terms of the Arithmetic Progression}
\displaystyle \text{(A.P.) }20,\ 17,\ 14,\ldots\text{ is }65.\text{ Find the:}
\displaystyle \text{(a) number of terms.}\qquad\text{(b) last term.}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=20,\qquad d=17-20=-3,\qquad S_n=65.
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]
\displaystyle 65=\frac{n}{2}[40-3(n-1)]
\displaystyle 130=n(43-3n)
\displaystyle 3n^2-43n+130=0
\displaystyle (3n-13)(n-10)=0
\displaystyle n=\frac{13}{3}\text{ or }n=10
\displaystyle \text{Since the number of terms must be a whole number, }n=10.
\displaystyle \therefore\text{(a) Number of terms}=10.
\displaystyle \text{(b) }T_n=a+(n-1)d
\displaystyle T_{10}=20+(10-1)(-3)
\displaystyle =20-27=-7
\displaystyle \therefore\text{The last term is }-7.
\displaystyle \\

\displaystyle \textbf{Question 88: }\text{(a) Point }P(2,-3)\text{ on reflection becomes }P'(2,3).\text{ Name the}
\displaystyle \text{line of reflection (say }L_1\text{).}
\displaystyle \text{(b) Point }P'\text{ is reflected to }P''\text{ along the line }(L_2),\text{ which is perpendicular to}
\displaystyle \text{the line }L_1\text{ and passes through the point, which is invariant along both axes. Write}
\displaystyle \text{the coordinates of }P''.
\displaystyle \text{(c) Name and write the coordinates of the point of intersection of the lines }L_1\text{ and }L_2.
\displaystyle \text{(d) Point }P\text{ is reflected to }P'''\text{ on reflection through the point named in the answer}
\displaystyle \text{of above part of this question. Write the coordinates of }P'''.\text{ Comment on the location}
\displaystyle \text{of the points }P''\text{ and }P'''.
\displaystyle \text{Answer:}
\displaystyle \text{(a) }P(2,-3)\rightarrow P'(2,3).
\displaystyle \text{Only the sign of the }y\text{-coordinate changes.}
\displaystyle \therefore L_1\text{ is the }x\text{-axis, i.e. }y=0.
\displaystyle \text{(b) The point invariant along both axes is the origin }O(0,0).
\displaystyle L_2\text{ is perpendicular to the }x\text{-axis and passes through }O.
\displaystyle \therefore L_2\text{ is the }y\text{-axis, i.e. }x=0.
\displaystyle \text{Reflection in the }y\text{-axis maps }(x,y)\rightarrow(-x,y).
\displaystyle \therefore P'(2,3)\rightarrow P''(-2,3).
\displaystyle \text{(c) }L_1\text{ and }L_2\text{ intersect at the origin }O(0,0).
\displaystyle \text{(d) Reflection through the origin maps }(x,y)\rightarrow(-x,-y).
\displaystyle \therefore P(2,-3)\rightarrow P'''(-2,3).
\displaystyle \text{Thus, }P''(-2,3)\text{ and }P'''(-2,3)\text{ are the same point.}
\displaystyle \therefore P''\text{ and }P'''\text{ coincide.}
\displaystyle \\

\displaystyle \textbf{Question 89: }\text{In the given figure, if the line segment }AB\text{ is intercepted by the}
\displaystyle y\text{-axis and }x\text{-axis at }C\text{ and }D\text{ respectively, such that }AC:AD=1:4\text{ and }D
\displaystyle \text{is the midpoint of }CB.\text{ Find the coordinates of }D,\ C\text{ and }B.
\displaystyle \text{Answer:}
\displaystyle \text{From the figure, }A=(-2,6).
\displaystyle \text{Given }AC:AD=1:4.
\displaystyle \text{Let }AC=k.\text{ Then }AD=4k.
\displaystyle \therefore CD=AD-AC=3k.
\displaystyle \text{Since }D\text{ is the midpoint of }CB,\quad CD=DB=3k.
\displaystyle \therefore AC:CD:DB=1:3:3.
\displaystyle \text{Hence, }AC:CB=1:6.
\displaystyle \text{Let }C=(0,c)\text{ and }B=(p,q).
\displaystyle \text{Since }C\text{ divides }AB\text{ internally in the ratio }1:6,
\displaystyle C=\left(\frac{1(p)+6(-2)}{7},\frac{1(q)+6(6)}{7}\right)
\displaystyle \therefore 0=\frac{p-12}{7}
\displaystyle p=12
\displaystyle \text{Since }D\text{ is the midpoint of }CB,
\displaystyle D=\left(\frac{0+12}{2},\frac{c+q}{2}\right)
\displaystyle \text{As }D\text{ lies on the }x\text{-axis, its }y\text{-coordinate is }0.
\displaystyle \therefore D=(6,0)\text{ and }c+q=0\qquad ...(1)
\displaystyle \text{Also, from the }y\text{-coordinate of }C,
\displaystyle c=\frac{q+36}{7}\qquad ...(2)
\displaystyle \text{Using }q=-c\text{ from (1) in (2),}
\displaystyle c=\frac{-c+36}{7}
\displaystyle 8c=36
\displaystyle c=\frac{9}{2}
\displaystyle q=-\frac{9}{2}
\displaystyle \therefore C\left(0,\frac{9}{2}\right),\quad D(6,0),\quad B\left(12,-\frac{9}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 90: }\text{Find the equation of the straight line perpendicular to the line}
\displaystyle x+2y=4,\text{ which cuts an intercept of }2\text{ units from the positive }y\text{-axis. Hence, find}
\displaystyle \text{the intersection point of the two lines.}
\displaystyle \text{Answer:}
\displaystyle x+2y=4
\displaystyle 2y=-x+4
\displaystyle y=-\frac{1}{2}x+2
\displaystyle \therefore\text{Slope of the given line}=-\frac{1}{2}.
\displaystyle \text{Slope of a line perpendicular to it}=2.
\displaystyle \text{The required line cuts a positive }y\text{-intercept of }2.
\displaystyle \therefore\text{It passes through }(0,2).
\displaystyle \text{Using }y=mx+c,
\displaystyle y=2x+2
\displaystyle \therefore\text{The required equation is }y=2x+2.
\displaystyle \text{For the point of intersection, substitute }y=2x+2\text{ in }x+2y=4.
\displaystyle x+2(2x+2)=4
\displaystyle 5x+4=4
\displaystyle x=0
\displaystyle y=2(0)+2=2
\displaystyle \therefore\text{The point of intersection of the two lines is }(0,2).
\displaystyle \\

\displaystyle \textbf{Question 91: }\text{While preparing a PowerPoint presentation, }\triangle ABC\text{ is enlarged along}
\displaystyle \text{the side }BC\text{ to }\triangle AB'C',\text{ as shown in the diagram, such that }BC:B'C'=3:5.
\displaystyle \text{Find: (a) }AB:BB'\qquad\text{(b) length }AB,\text{ if }BB'=4\text{ cm}.
\displaystyle \text{(c) Is }\triangle ABC\sim\triangle AB'C'?\text{ Justify your answer.}
\displaystyle \text{(d) ar }(\triangle ABC):\text{ ar }(\text{quad. }BB'C'C).
\displaystyle \text{Answer:}
\displaystyle \text{Since }BC\parallel B'C',\quad\triangle ABC\sim\triangle AB'C'.
\displaystyle \therefore\frac{AB}{AB'}=\frac{BC}{B'C'}=\frac{3}{5}
\displaystyle \text{Let }AB=3k\text{ and }AB'=5k.
\displaystyle BB'=AB'-AB=5k-3k=2k
\displaystyle \therefore\text{(a) }AB:BB'=3k:2k=3:2.
\displaystyle \text{(b) Given }BB'=4\text{ cm}.
\displaystyle \frac{AB}{BB'}=\frac{3}{2}
\displaystyle AB=\frac{3}{2}\times4=6\text{ cm}
\displaystyle \therefore AB=6\text{ cm}.
\displaystyle \text{(c) Since }BC\parallel B'C',
\displaystyle \angle ABC=\angle AB'C'\quad\text{and}\quad\angle ACB=\angle AC'B'.
\displaystyle \therefore\triangle ABC\sim\triangle AB'C'\qquad\text{(AA similarity)}
\displaystyle \text{(d) For similar triangles,}
\displaystyle \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle AB'C')} =\left(\frac{BC}{B'C'}\right)^2
\displaystyle =\left(\frac{3}{5}\right)^2=\frac{9}{25}
\displaystyle \therefore \text{ar}(\triangle ABC): \text{ar}(\triangle AB'C')=9:25.
\displaystyle \text{Area of quad. }BB'C'C=25-9=16\text{ parts}.
\displaystyle \therefore \text{ar}(\triangle ABC): \text{ar}(\text{quad. }BB'C'C)=9:16.
\displaystyle \\

\displaystyle \textbf{Question 92: }\text{The approximate volume of a human eye is }6.5\text{ cm}^3.\text{ The volume}
\displaystyle \text{of a laboratory model (excluding base and stand) of the human eye is }1404\text{ cm}^3.
\displaystyle \text{(a) State whether the scale factor }k\text{ is less than, equals to or greater than }1.
\displaystyle \text{(b) Calculate the: (i) value of }k.
\displaystyle \text{(ii) diameter of the human eye if the radius of the model is }7.2\text{ cm}.
\displaystyle \text{(iii) external surface area of the human eye if the surface area of the model is}
\displaystyle 651.6\text{ cm}^2.
\displaystyle \text{Answer:}
\displaystyle \text{(a) Since the model is larger than the human eye, the scale factor }k>1.
\displaystyle \text{(b)(i) For similar solids, the ratio of volumes is the cube of the scale factor.}
\displaystyle k^3=\frac{\text{Volume of model}}{\text{Volume of human eye}}
\displaystyle =\frac{1404}{6.5}=216
\displaystyle k=\sqrt[3]{216}=6
\displaystyle \therefore k=6.
\displaystyle \text{(ii) Radius of model}=7.2\text{ cm}.
\displaystyle \text{Radius of human eye}=\frac{7.2}{6}=1.2\text{ cm}
\displaystyle \text{Diameter of human eye}=2(1.2)=2.4\text{ cm}
\displaystyle \therefore\text{The diameter of the human eye is }2.4\text{ cm}.
\displaystyle \text{(iii) Ratio of surface areas is the square of the scale factor.}
\displaystyle \frac{\text{Surface area of model}}{\text{Surface area of human eye}}=k^2=36
\displaystyle \text{Surface area of human eye}=\frac{651.6}{36}=18.1\text{ cm}^2
\displaystyle \therefore\text{The external surface area of the human eye is }18.1\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 93: }\text{In the adjoining diagram }PQ,\ PR\text{ and }ST\text{ are the tangents to the}
\displaystyle \text{circle with centre }O\text{ and radius }7\text{ cm. Given }OP=25\text{ cm. Find:}
\displaystyle \text{(a) length of }ST\qquad\text{(b) value of }\angle OPQ,\text{ i.e. }\theta
\displaystyle \text{(c) }\angle QUR,\text{ in nearest degree. (Use mathematical tables)}
\displaystyle \text{Answer:}
\displaystyle \text{Since }PQ\text{ and }PR\text{ are tangents, }OQ\perp PQ\text{ and }OR\perp PR.
\displaystyle \text{Also, }OQ=OR=7\text{ cm and }OP=25\text{ cm}.
\displaystyle \text{In right-angled }\triangle OPQ,
\displaystyle PQ=\sqrt{OP^2-OQ^2}
\displaystyle =\sqrt{25^2-7^2}=\sqrt{576}=24\text{ cm}
\displaystyle \text{(a) Let }ST\text{ touch the circle at }V,\text{ where }V\text{ lies on }OP.
\displaystyle \therefore OV=7\text{ cm}
\displaystyle PV=OP-OV=25-7=18\text{ cm}
\displaystyle \text{Since }ST\parallel QR,\quad\triangle PST\sim\triangle PQR.
\displaystyle \frac{ST}{QR}=\frac{PV}{PO}=\frac{18}{25}\qquad ...(1)
\displaystyle \text{Area of }\triangle PQR=\frac{1}{2}(PQ)(OR)+\frac{1}{2}(PR)(OQ)
\displaystyle =\frac{1}{2}(24)(7)+\frac{1}{2}(24)(7)=168\text{ cm}^2
\displaystyle \text{Also, area of }\triangle PQR=\frac{1}{2}(QR)(OP)
\displaystyle 168=\frac{1}{2}(QR)(25)
\displaystyle QR=\frac{336}{25}\text{ cm}
\displaystyle \text{From (1),}\qquad ST=\frac{18}{25}\times\frac{336}{25}
\displaystyle =9.6768\text{ cm}
\displaystyle \text{(b) In }\triangle OPQ,
\displaystyle \sin\theta=\frac{OQ}{OP}=\frac{7}{25}=0.28
\displaystyle \therefore\theta=\sin^{-1}(0.28)\approx16^\circ.
\displaystyle \text{(c) In }\triangle OPQ,
\displaystyle \angle QOP=90^\circ-\theta\approx90^\circ-16.26^\circ=73.74^\circ
\displaystyle \text{By symmetry, }\angle POR=\angle QOP=73.74^\circ.
\displaystyle \therefore\angle QOR=2(73.74^\circ)=147.48^\circ
\displaystyle \angle QUR=\frac{1}{2}\angle QOR
\displaystyle =\frac{147.48^\circ}{2}=73.74^\circ\approx74^\circ
\displaystyle \therefore\angle QUR\approx74^\circ.
\displaystyle \\

\displaystyle \textbf{Question 94: }\text{Use ruler and compass to answer this question. Construct a triangle}
\displaystyle ABC\text{ where }AB=5.5\text{ cm},\ BC=4.5\text{ cm and }\angle ABC=135^\circ.
\displaystyle \text{Construct the circumcircle to the triangle }ABC.\text{ Measure and write down the length}
\displaystyle \text{of }AC.
\displaystyle \text{Answer:}
\displaystyle \text{Draw }AB=5.5\text{ cm}.
\displaystyle \text{At }B,\text{ construct }\angle ABC=135^\circ.
\displaystyle \text{With }B\text{ as centre and radius }4.5\text{ cm, mark point }C\text{ on the ray.}
\displaystyle \text{Join }AC\text{ to obtain }\triangle ABC.
\displaystyle \text{Draw the perpendicular bisectors of any two sides of }\triangle ABC.
\displaystyle \text{Let the perpendicular bisectors intersect at }O.
\displaystyle \text{With }O\text{ as centre and radius }OA,\text{ draw a circle through }A,\ B\text{ and }C.
\displaystyle \text{This is the required circumcircle of }\triangle ABC.
\displaystyle \text{On measuring, }AC\approx9.2\text{ cm}.
\displaystyle \therefore\text{The length of }AC\text{ is approximately }9.2\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 95: }\text{The curved surface area of a right circular cone is half of another}
\displaystyle \text{right circular cone. If the ratio of their slant heights is }2:1\text{ and that of their volumes}
\displaystyle \text{is }3:1,\text{ find ratio of their: (a) radii}\qquad\text{(b) heights}
\displaystyle \text{Answer:}
\displaystyle \text{Let the radii, heights and slant heights of the two cones be }r_1,h_1,l_1
\displaystyle \text{and }r_2,h_2,l_2\text{ respectively.}
\displaystyle \text{Given, }\frac{\text{CSA}_1}{\text{CSA}_2}=\frac{1}{2}\quad\text{and}\quad\frac{l_1}{l_2}=\frac{2}{1}.
\displaystyle \text{(a) Since curved surface area of a cone}=\pi rl,
\displaystyle \frac{\pi r_1l_1}{\pi r_2l_2}=\frac{1}{2}
\displaystyle \frac{r_1}{r_2}\times\frac{2}{1}=\frac{1}{2}
\displaystyle \frac{r_1}{r_2}=\frac{1}{4}
\displaystyle \therefore r_1:r_2=1:4.
\displaystyle \text{(b) Given, }\frac{V_1}{V_2}=\frac{3}{1}.
\displaystyle \frac{\frac{1}{3}\pi r_1^2h_1}{\frac{1}{3}\pi r_2^2h_2}=\frac{3}{1}
\displaystyle \left(\frac{r_1}{r_2}\right)^2\frac{h_1}{h_2}=3
\displaystyle \left(\frac{1}{4}\right)^2\frac{h_1}{h_2}=3
\displaystyle \frac{1}{16}\times\frac{h_1}{h_2}=3
\displaystyle \frac{h_1}{h_2}=48
\displaystyle \therefore h_1:h_2=48:1.
\displaystyle \therefore\text{Ratio of radii}=1:4\text{ and ratio of heights}=48:1.
\displaystyle \\

\displaystyle \textbf{Question 96: }\text{A cylindrical drum is unloaded from a truck by rolling it down along a}
\displaystyle \text{wooden plank. The length of the plank is }10\text{ m and it is making an angle of }10^\circ
\displaystyle \text{with the horizontal ground. Find the height from which the cylindrical drum was rolled down.}
\displaystyle \text{Give your answer correct to }3\text{ significant figures.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required height be }h\text{ m.}
\displaystyle \text{The plank forms the hypotenuse of a right-angled triangle.}
\displaystyle \sin10^\circ=\frac{h}{10}
\displaystyle h=10\sin10^\circ
\displaystyle =10(0.173648)
\displaystyle =1.73648\text{ m}
\displaystyle \therefore h\approx1.74\text{ m}\text{, correct to }3\text{ significant figures.}
\displaystyle \\

\displaystyle \textbf{Question 97: }\text{The data given below shows the marks of }12\text{ students in a test,}
\displaystyle \text{arranged in ascending order:}
\displaystyle 2,\ 3,\ 3,\ 3,\ 4,\ x,\ x+2,\ 8,\ p,\ q,\ 8,\ 9
\displaystyle \text{If the given value of the median and mode is }6\text{ and }8\text{ respectively, then find the}
\displaystyle \text{values of }x,\ p,\ q.
\displaystyle \text{Answer:}
\displaystyle \text{There are }12\text{ observations.}
\displaystyle \therefore\text{Median}=\frac{\text{6th observation}+\text{7th observation}}{2}
\displaystyle 6=\frac{x+(x+2)}{2}
\displaystyle 12=2x+2
\displaystyle 2x=10
\displaystyle x=5
\displaystyle \therefore x+2=7.
\displaystyle \text{The data becomes }2,\ 3,\ 3,\ 3,\ 4,\ 5,\ 7,\ 8,\ p,\ q,\ 8,\ 9.
\displaystyle \text{Since the observations are arranged in ascending order,}
\displaystyle 8\leq p\leq q\leq8.
\displaystyle \therefore p=8,\qquad q=8.
\displaystyle \text{Thus, }8\text{ occurs four times and is the mode.}
\displaystyle \therefore x=5,\qquad p=8,\qquad q=8.
\displaystyle \\

\displaystyle \textbf{Question 98: }\text{Solve the linear inequation, write down the solution set and}
\displaystyle \text{represent it on the real number line:}
\displaystyle 5(2-4x)>18-16x>22-20x,\quad x\in R
\displaystyle \text{Answer:}
\displaystyle 5(2-4x)>18-16x
\displaystyle 10-20x>18-16x
\displaystyle -8>4x
\displaystyle x<-2\qquad ...(1)
\displaystyle \text{Also,}\qquad18-16x>22-20x
\displaystyle 4x>4
\displaystyle x>1\qquad ...(2)
\displaystyle \text{From (1) and (2), }x<-2\text{ and }x>1\text{ must hold simultaneously.}
\displaystyle \text{This is not possible for any real value of }x.
\displaystyle \therefore\text{Solution set}=\varnothing.
\displaystyle \text{Hence, no point is represented on the real number line.}
\displaystyle \\

\displaystyle \textbf{Question 99: }\text{If a polynomial }x^3+2x^2-ax+b\text{ leaves a remainder }-6\text{ when}
\displaystyle \text{divided by }x+1\text{ and the same polynomial has }x-2\text{ as a factor, then find the}
\displaystyle \text{values of }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=x^3+2x^2-ax+b.
\displaystyle \text{Since the remainder on division by }x+1\text{ is }-6,
\displaystyle f(-1)=-6
\displaystyle (-1)^3+2(-1)^2-a(-1)+b=-6
\displaystyle -1+2+a+b=-6
\displaystyle a+b=-7\qquad ...(1)
\displaystyle \text{Since }x-2\text{ is a factor, by the factor theorem, }f(2)=0.
\displaystyle 2^3+2(2)^2-2a+b=0
\displaystyle 8+8-2a+b=0
\displaystyle -2a+b=-16\qquad ...(2)
\displaystyle \text{From (1),}\qquad b=-7-a.
\displaystyle \text{Substituting in (2),}
\displaystyle -2a-7-a=-16
\displaystyle -3a=-9
\displaystyle a=3
\displaystyle b=-7-3=-10
\displaystyle \therefore a=3,\qquad b=-10.
\displaystyle \\

\displaystyle \textbf{Question 100: }\text{If }A=\begin{bmatrix}-1&3\\2&0\end{bmatrix},\ B=\begin{bmatrix}1&-2\\0&3\end{bmatrix},
\displaystyle C=\begin{bmatrix}1&-4\end{bmatrix}\text{ and }D=\begin{bmatrix}4\\1\end{bmatrix}.
\displaystyle \text{(a) Is the product }AC\text{ possible? Justify your answer.}
\displaystyle \text{(b) Find the matrix }X,\text{ such that }X=AB+B^2-DC.
\displaystyle \text{Answer:}
\displaystyle \text{(a) The order of }A\text{ is }2\times2\text{ and the order of }C\text{ is }1\times2.
\displaystyle \text{The number of columns of }A\text{ is }2,\text{ while the number of rows of }C\text{ is }1.
\displaystyle \therefore AC\text{ is not possible.}
\displaystyle \text{(b) }X=AB+B^2-DC
\displaystyle AB=\begin{bmatrix}-1&3\\2&0\end{bmatrix}\begin{bmatrix}1&-2\\0&3\end{bmatrix}
\displaystyle =\begin{bmatrix}-1&11\\2&-4\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}1&-2\\0&3\end{bmatrix}\begin{bmatrix}1&-2\\0&3\end{bmatrix}
\displaystyle =\begin{bmatrix}1&-8\\0&9\end{bmatrix}
\displaystyle DC=\begin{bmatrix}4\\1\end{bmatrix}\begin{bmatrix}1&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}4&-16\\1&-4\end{bmatrix}
\displaystyle X=\begin{bmatrix}-1&11\\2&-4\end{bmatrix}+\begin{bmatrix}1&-8\\0&9\end{bmatrix} -\begin{bmatrix}4&-16\\1&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}0&3\\2&5\end{bmatrix}-\begin{bmatrix}4&-16\\1&-4\end{bmatrix}
\displaystyle =\begin{bmatrix}-4&19\\1&9\end{bmatrix}
\displaystyle \therefore X=\begin{bmatrix}-4&19\\1&9\end{bmatrix}.
\displaystyle \\

\displaystyle \textbf{Question 101: }\text{In the given figure (not drawn to scale), }BC\text{ is parallel to }EF,
\displaystyle CD\text{ is parallel to }FG,\ AE:EB=2:3,\ \angle BAD=70^\circ,\ \angle ACB=105^\circ,
\displaystyle \angle ADC=40^\circ\text{ and }AC\text{ is the bisector of }\angle BAD.
\displaystyle \text{(a) Prove }\triangle AEF\sim\triangle AGF.
\displaystyle \text{(b) Find: (i) }AG:AD\qquad\text{(ii) area of }\triangle ACB:\text{ area of }\triangle ACD
\displaystyle \text{(iii) area of quadrilateral }ABCD:\text{ area of }\triangle ACB.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AC\text{ bisects }\angle BAD=70^\circ,
\displaystyle \angle BAC=\angle CAD=35^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ABC=180^\circ-(35^\circ+105^\circ)=40^\circ.
\displaystyle \text{(a) Since }EF\parallel BC,
\displaystyle \angle AEF=\angle ABC=40^\circ.
\displaystyle \text{Since }FG\parallel CD,
\displaystyle \angle AGF=\angle ADC=40^\circ.
\displaystyle \text{Also, }\angle EAF=\angle GAF=35^\circ.
\displaystyle \therefore\triangle AEF\sim\triangle AGF\qquad\text{(AA similarity)}
\displaystyle \text{(b)(i) Since }EF\parallel BC,\quad\triangle AEF\sim\triangle ABC.
\displaystyle \therefore\frac{AF}{AC}=\frac{AE}{AB}
\displaystyle \text{Given }AE:EB=2:3,\quad AE:AB=2:5.
\displaystyle \therefore\frac{AF}{AC}=\frac{2}{5}.
\displaystyle \text{Also, since }FG\parallel CD,\quad\triangle AFG\sim\triangle ACD.
\displaystyle \therefore\frac{AG}{AD}=\frac{AF}{AC}=\frac{2}{5}
\displaystyle \therefore AG:AD=2:5.
\displaystyle \text{(ii) In }\triangle ACD,
\displaystyle \angle ACD=180^\circ-(35^\circ+40^\circ)=105^\circ.
\displaystyle \text{Thus, }\angle BAC=\angle CAD=35^\circ,
\displaystyle \angle ABC=\angle ADC=40^\circ\text{ and }\angle ACB=\angle ACD=105^\circ.
\displaystyle \therefore\triangle ABC\sim\triangle ACD.
\displaystyle \text{Since }AC\text{ corresponds to itself, the similarity ratio is }1:1.
\displaystyle \therefore\text{ar}(\triangle ACB):\text{ar}(\triangle ACD)=1:1.
\displaystyle \text{(iii) }\text{ar}(ABCD)=\text{ar}(\triangle ACB)+\text{ar}(\triangle ACD)
\displaystyle \text{Since the two triangular areas are equal,}
\displaystyle \text{ar}(ABCD)=2\text{ar}(\triangle ACB).
\displaystyle \therefore\text{ar}(ABCD):\text{ar}(\triangle ACB)=2:1.
\displaystyle \\

\displaystyle \textbf{Question 102: }\text{In the given figure }\angle ABC=70^\circ\text{ and }\angle ACB=50^\circ.
\displaystyle \text{Given }O\text{ is the centre of the circle and }PT\text{ is the tangent to the circle. Calculate:}
\displaystyle \text{(a) }\angle CBT\qquad\text{(b) }\angle BAT\qquad\text{(c) }\angle PBT\qquad\text{(d) }\angle APT
\displaystyle \text{Answer:}
\displaystyle \text{Since }C,\ O,\ T\text{ are collinear, }CT\text{ is a diameter of the circle.}
\displaystyle \text{(a) Angle in a semicircle is a right angle.}
\displaystyle \therefore\angle CBT=90^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC=180^\circ-(70^\circ+50^\circ)=60^\circ.
\displaystyle \text{Also, since }CT\text{ is a diameter, }\angle CAT=90^\circ.
\displaystyle \text{(b) }\angle BAT=\angle CAT-\angle CAB
\displaystyle =90^\circ-60^\circ=30^\circ
\displaystyle \therefore\angle BAT=30^\circ.
\displaystyle \text{At }B,\quad\angle ABT=\angle CBT-\angle CBA
\displaystyle =90^\circ-70^\circ=20^\circ
\displaystyle \text{(c) Since }P,\ A,\ B\text{ are collinear,}
\displaystyle \angle PBT=180^\circ-\angle ABT
\displaystyle =180^\circ-20^\circ=160^\circ
\displaystyle \therefore\angle PBT=160^\circ.
\displaystyle \text{Since }\angle ABT=20^\circ,\text{ angles subtended by chord }AT\text{ are equal.}
\displaystyle \therefore\angle ACT=\angle ABT=20^\circ.
\displaystyle \text{By the tangent-chord theorem, }\angle ATP=\angle ACT=20^\circ.
\displaystyle \text{Also, }\angle PAT=180^\circ-\angle BAT=150^\circ.
\displaystyle \text{(d) In }\triangle APT,
\displaystyle \angle APT=180^\circ-(150^\circ+20^\circ)=10^\circ
\displaystyle \therefore\angle APT=10^\circ.
\displaystyle \\

\displaystyle \textbf{Question 103: }\text{(Use a ruler and a compass for this question.)}
\displaystyle \text{(a) Construct a triangle }ABC\text{ such that }BC=8\text{ cm},\ AC=10\text{ cm and}
\displaystyle \angle ABC=90^\circ.
\displaystyle \text{(b) Construct an incircle to this triangle. Mark the centre as }I.
\displaystyle \text{(c) Measure and write the length of the in-radius.}
\displaystyle \text{(d) Measure and write the length of the tangents from vertex }C\text{ to the incircle.}
\displaystyle \text{(e) Mark points }P,\ Q\text{ and }R\text{ where the incircle touches sides }AB,\ BC\text{ and }AC
\displaystyle \text{respectively. Write the relationship between }\angle RIQ\text{ and }\angle QCR.
\displaystyle \text{Answer:}
\displaystyle \text{(a) Draw }BC=8\text{ cm}.
\displaystyle \text{At }B,\text{ construct a perpendicular to }BC.
\displaystyle \text{With }C\text{ as centre and radius }10\text{ cm, draw an arc cutting the perpendicular at }A.
\displaystyle \text{Join }AC\text{ to obtain }\triangle ABC.
\displaystyle \text{(b) Construct the bisectors of any two angles of }\triangle ABC.
\displaystyle \text{Let the angle bisectors intersect at }I.
\displaystyle \text{Draw }IP\perp AB.\text{ With }I\text{ as centre and radius }IP,\text{ draw the incircle.}
\displaystyle \text{(c) Since }\angle ABC=90^\circ,
\displaystyle AB=\sqrt{AC^2-BC^2}
\displaystyle =\sqrt{10^2-8^2}=\sqrt{36}=6\text{ cm}
\displaystyle \text{For a right-angled triangle, }r=\frac{AB+BC-AC}{2}
\displaystyle =\frac{6+8-10}{2}=2\text{ cm}
\displaystyle \therefore\text{The in-radius is }2\text{ cm}.
\displaystyle \text{(d) Semi-perimeter }s=\frac{6+8+10}{2}=12\text{ cm}
\displaystyle CQ=CR=s-AB=12-6=6\text{ cm}
\displaystyle \therefore\text{The length of each tangent from }C\text{ is }6\text{ cm}.
\displaystyle \text{(e) Since }IQ\perp BC\text{ and }IR\perp AC,
\displaystyle \angle IQC=\angle IRC=90^\circ.
\displaystyle \text{In quadrilateral }IQCR,
\displaystyle \angle RIQ+\angle QCR+90^\circ+90^\circ=360^\circ
\displaystyle \therefore\angle RIQ+\angle QCR=180^\circ.
\displaystyle \therefore\angle RIQ\text{ and }\angle QCR\text{ are supplementary angles.}
\displaystyle \\

\displaystyle \textbf{Question 104: }\text{The daily wages of workers in a construction unit were recorded}
\displaystyle \text{as follows:}
\displaystyle \begin{array}{c|cccccc} \text{Class Marks (Wages)}&425&475&525&575&625&675\\ \hline \text{No. of workers}&6&12&15&17&7&13 \end{array}
\displaystyle \text{Form a frequency distribution table with class intervals and find modal wage by}
\displaystyle \text{plotting a histogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Difference between two consecutive class marks}=475-425=50.
\displaystyle \therefore\text{Class width}=50.
\displaystyle \text{The required frequency distribution table is:}
\displaystyle \begin{array}{c|c} \text{Class Interval (Wages)}&\text{No. of workers}\\ \hline 400-450&6\\ 450-500&12\\ 500-550&15\\ 550-600&17\\ 600-650&7\\ 650-700&13 \end{array}
\displaystyle \text{Draw a histogram taking class intervals along the }x\text{-axis and frequencies along}
\displaystyle \text{the }y\text{-axis. The tallest rectangle corresponds to the class }550-600.
\displaystyle \therefore\text{The modal class is }550-600.
\displaystyle \text{Here, }l=550,\quad h=50,\quad f_1=17,\quad f_0=15,\quad f_2=7.
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =550+\frac{17-15}{2(17)-15-7}\times50
\displaystyle =550+\frac{2}{12}\times50
\displaystyle =550+\frac{25}{3}
\displaystyle =558.33
\displaystyle \therefore\text{The modal wage is approximately Rs. }558.33.
\displaystyle \\

\displaystyle \textbf{Question 105: }\text{A bag contains }13\text{ red cards, }13\text{ black cards and }13\text{ green}
\displaystyle \text{cards. Each set of cards are numbered }1\text{ to }13.\text{ From these cards, a card is drawn at}
\displaystyle \text{random. What is the probability that the card drawn is a:}
\displaystyle \text{(a) green card?}\qquad\text{(b) a card with an even number?}
\displaystyle \text{(c) a red or black card with a number which is a multiple of three?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=13+13+13=39.
\displaystyle \text{(a) Number of green cards}=13
\displaystyle P(\text{green card})=\frac{13}{39}=\frac{1}{3}
\displaystyle \therefore P(\text{green card})=\frac{1}{3}.
\displaystyle \text{(b) Even numbers from }1\text{ to }13\text{ are }2,\ 4,\ 6,\ 8,\ 10,\ 12.
\displaystyle \text{There are }6\text{ even-numbered cards of each colour.}
\displaystyle \text{Total even-numbered cards}=6\times3=18
\displaystyle P(\text{even number})=\frac{18}{39}=\frac{6}{13}
\displaystyle \therefore P(\text{even number})=\frac{6}{13}.
\displaystyle \text{(c) Multiples of }3\text{ from }1\text{ to }13\text{ are }3,\ 6,\ 9,\ 12.
\displaystyle \text{There are }4\text{ such cards among each colour.}
\displaystyle \text{Number of favourable red or black cards}=4+4=8
\displaystyle P(\text{red or black card with a multiple of }3)=\frac{8}{39}
\displaystyle \therefore P(\text{red or black card with a multiple of }3)=\frac{8}{39}.
\displaystyle \\

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