\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series PPQQB/2}  \hspace{3.0cm} \textbf{SET~1 } \hspace{4.0cm} \textbf{Q.P. Code }30/3/1
\displaystyle \text{Roll No.}

\displaystyle \textbf{NOTE}

\displaystyle \text{(I) Please check that this question paper contains 11 printed pages.}

\displaystyle \text{(II) Q.P. Code given on the right hand side of the question paper should be written on the}
\displaystyle \text{title page of the answer-book by the candidate.}

\displaystyle \text{(III) Please check that this question paper contains 14 questions.}

\displaystyle \text{(IV) Please write down the serial number of the question in the answer-book before}
\displaystyle \text{attempting it.}

\displaystyle \text{(V) 15 minute time has been allotted to read this question paper. The question paper}
\displaystyle \text{will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m. the students will read}
\displaystyle \text{the question paper only and will not write any answer on the answer-book during}
\displaystyle \text{this period.}

\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \text{Time allowed : 2 hours}                                                                 \displaystyle \text{Maximum marks : 40}

\displaystyle \textbf{General Instructions :}

\displaystyle \text{Read the following instructions very carefully and strictly follow them :}

\displaystyle \text{(i) This question paper contains 14 questions. All questions are compulsory.}

\displaystyle \text{(ii) This question paper is divided into three sections - Sections A, B and C.}

\displaystyle \text{(iii) Section A comprises of 6 questions (Q.no. 1 to 6) of 2 marks each. Internal}
\displaystyle \text{choice has been provided in two questions.}

\displaystyle \text{(iv) Section B comprises of 4 questions (Q.no. 7 to 10) of 3 marks each. Internal}
\displaystyle \text{choice has been provided in one question.}

\displaystyle \text{(v) Section C comprises of 4 questions (Q.no. 11 to 14) of 4 marks each. Internal}
\displaystyle \text{choice has been provided in one question. It also contains two case study based}
\displaystyle \text{questions.}

\displaystyle \text{(vi) Use of calculator is not permitted.}

\displaystyle \textbf{SECTION A}
\displaystyle \text{Question numbers 1 to 6 carry 2 marks each.}

\displaystyle \textbf{Question 1. }\text{(a) Solve the quadratic equation for }x:
\displaystyle x^2-2ax-(4b^2-a^2)=0

\displaystyle \textbf{OR}

\displaystyle \text{(b) If the quadratic equation }(1+a^2)x^2+2abx+(b^2-c^2)=0
\displaystyle \text{has equal and real roots, then prove that }b^2=c^2(1+a^2).
\displaystyle \text{Answer:}
\displaystyle \text{(a) }x^2-2ax-(4b^2-a^2)=0
\displaystyle x^2-2ax+a^2-4b^2=0
\displaystyle (x-a)^2-(2b)^2=0
\displaystyle (x-a-2b)(x-a+2b)=0
\displaystyle x=a+2b\quad \text{or}\quad x=a-2b
\displaystyle \therefore x=a\pm2b

\displaystyle \textbf{OR}

\displaystyle \text{(b) For equal and real roots, }D=0
\displaystyle [2ab]^2-4(1+a^2)(b^2-c^2)=0
\displaystyle 4a^2b^2-4(1+a^2)(b^2-c^2)=0
\displaystyle a^2b^2-(1+a^2)(b^2-c^2)=0
\displaystyle a^2b^2-b^2-a^2b^2+c^2+a^2c^2=0
\displaystyle -b^2+c^2(1+a^2)=0
\displaystyle \therefore b^2=c^2(1+a^2)
\\

\displaystyle \textbf{Question 2. }\text{Find the sum of first }20\text{ terms of an AP in which }d=5\text{ and }a_{20}=135.
\displaystyle \text{Answer:}
\displaystyle a_{20}=a+19d
\displaystyle 135=a+19(5)
\displaystyle 135=a+95
\displaystyle a=40
\displaystyle S_{20}=\frac{20}{2}(a+a_{20})
\displaystyle =10(40+135)
\displaystyle =10\times175
\displaystyle =1750
\displaystyle \therefore \text{Sum of first }20\text{ terms}=1750
\\

\displaystyle \textbf{Question 3. }\text{Find the mode of the given frequency distribution:}
\displaystyle \begin{array}{|c|c|}\hline \text{Class} & \text{Frequency}\\ \hline 15-25 & 6\\ \hline 25-35 & 11\\ \hline 35-45 & 22\\ \hline 45-55 & 23\\ \hline 55-65 & 14\\ \hline 65-75 & 5\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Highest frequency}=23
\displaystyle \therefore \text{Modal class}=45-55
\displaystyle l=45,\quad h=10,\quad f_1=23,\quad f_0=22,\quad f_2=14
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =45+\frac{23-22}{2(23)-22-14}\times10
\displaystyle =45+\frac{1}{46-36}\times10
\displaystyle =45+1
\displaystyle =46
\displaystyle \therefore \text{Mode}=46
\\

\displaystyle \textbf{Question 4. }\text{(a) }150\text{ spherical marbles, each of diameter }1.4\text{ cm, are dropped in a}
\displaystyle \text{cylindrical vessel of diameter }7\text{ cm containing some water, and are completely immersed}
\displaystyle \text{in water. Find the rise in the level of water in the cylindrical vessel.}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Three cubes of side }6\text{ cm each, are joined as shown in Figure }1.\text{ Find the total}
\displaystyle \text{surface area of the resulting cuboid.} \displaystyle \text{Answer:}
\displaystyle \text{(a) Radius of each marble}=\frac{1.4}{2}=0.7\text{ cm}
\displaystyle \text{Radius of cylindrical vessel}=\frac{7}{2}=3.5\text{ cm}
\displaystyle \text{Let the rise in water level be }h\text{ cm.}
\displaystyle \text{Volume of }150\text{ marbles}=\text{Volume of water displaced}
\displaystyle 150\times\frac{4}{3}\pi(0.7)^3=\pi(3.5)^2h
\displaystyle 150\times\frac{4}{3}\times0.343=12.25h
\displaystyle 68.6=12.25h
\displaystyle h=5.6
\displaystyle \therefore \text{Rise in water level}=5.6\text{ cm}

\displaystyle \textbf{OR}

\displaystyle \text{(b) Three cubes of side }6\text{ cm are joined to form a cuboid.}
\displaystyle \text{Length}=18\text{ cm},\quad \text{breadth}=6\text{ cm},\quad \text{height}=6\text{ cm}
\displaystyle \text{TSA of cuboid}=2(lb+bh+hl)
\displaystyle =2(18\times6+6\times6+6\times18)
\displaystyle =2(108+36+108)
\displaystyle =2\times252
\displaystyle =504\text{ cm}^2
\displaystyle \therefore \text{Total surface area of the resulting cuboid}=504\text{ cm}^2
\\

\displaystyle \textbf{Question 5. }\text{For what value of }n\text{, are the }n^{\text{th}}\text{ terms of the APs }9,7,5,\ldots\text{ and}
\displaystyle 15,12,9,\ldots\text{ the same?}
\displaystyle \text{Answer:}
\displaystyle \text{For first AP, }a=9,\quad d=7-9=-2
\displaystyle T_n=9+(n-1)(-2)
\displaystyle =9-2n+2=11-2n
\displaystyle \text{For second AP, }a=15,\quad d=12-15=-3
\displaystyle T_n=15+(n-1)(-3)
\displaystyle =15-3n+3=18-3n
\displaystyle \text{Since the }n^{\text{th}}\text{ terms are same,}
\displaystyle 11-2n=18-3n
\displaystyle n=7
\displaystyle \therefore \text{Required value of }n=7
\\

\displaystyle \textbf{Question 6. }\text{In Figure }2,\text{ }PQ\text{ and }PR\text{ are tangents to the circle centred at }O.\text{ If}
\displaystyle \angle OPR=45^\circ,\text{ then prove that }ORPQ\text{ is a square.} \displaystyle \text{Proof:}
\displaystyle \text{Since }PR\text{ is tangent at }R,\quad OR\perp PR
\displaystyle \therefore \angle ORP=90^\circ
\displaystyle \text{Given, }\angle OPR=45^\circ
\displaystyle \text{In }\triangle OPR,
\displaystyle \angle ROP=180^\circ-90^\circ-45^\circ=45^\circ
\displaystyle \therefore \angle OPR=\angle ROP
\displaystyle \therefore OR=PR \qquad (i)
\displaystyle \text{Similarly, }PQ\text{ is tangent at }Q,\quad OQ\perp PQ
\displaystyle \text{Tangents from an external point are equal, so }PQ=PR \qquad (ii)
\displaystyle \text{Radii of the same circle are equal, so }OQ=OR \qquad (iii)
\displaystyle \text{From }(i),(ii)\text{ and }(iii),
\displaystyle OR=RP=PQ=QO
\displaystyle \text{Also, }\angle ORP=90^\circ\text{ and }\angle OQP=90^\circ
\displaystyle \therefore \text{Quadrilateral }ORPQ\text{ has all sides equal and one angle }90^\circ.
\displaystyle \therefore ORPQ\text{ is a square.}
\\

\displaystyle \textbf{SECTION B}
\displaystyle \text{Question numbers 7 to 10 carry 3 marks each.}

\displaystyle \textbf{Question 7. }\text{(a) Draw a line segment }AB\text{ of length }8\text{ cm and locate a point }P\text{ on }AB
\displaystyle \text{such that }AP:PB=1:5.

\displaystyle \textbf{OR}

\displaystyle \text{(b) Draw a circle of radius }3\text{ cm. From a point }P\text{ lying outside the circle}
\displaystyle \text{at a distance of }6\text{ cm from its centre, construct two tangents }PA\text{ and }PB\text{ to the circle.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }AP:PB=1:5
\displaystyle \therefore AB\text{ is divided into }1+5=6\text{ equal parts.}
\displaystyle AP=\frac{1}{6}\times8=\frac{4}{3}\text{ cm}
\displaystyle PB=\frac{5}{6}\times8=\frac{20}{3}\text{ cm}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw a line segment }AB=8\text{ cm.}
\displaystyle \text{2. Draw a ray }AX\text{ making an acute angle with }AB.
\displaystyle \text{3. Mark }6\text{ equal points }A_1,A_2,A_3,A_4,A_5,A_6\text{ on }AX.
\displaystyle \text{4. Join }A_6B.
\displaystyle \text{5. Through }A_1,\text{ draw a line parallel to }A_6B\text{ meeting }AB\text{ at }P.
\displaystyle \therefore P\text{ divides }AB\text{ in the ratio }1:5.

\displaystyle \textbf{OR}

\displaystyle \text{(b) Steps of construction:}
\displaystyle \text{1. Draw a circle with centre }O\text{ and radius }3\text{ cm.}
\displaystyle \text{2. Mark a point }P\text{ such that }OP=6\text{ cm.}
\displaystyle \text{3. Join }OP.
\displaystyle \text{4. Draw the perpendicular bisector of }OP\text{ to get its midpoint }M.
\displaystyle \text{5. With centre }M\text{ and radius }MO,\text{ draw a circle cutting the given circle at }A\text{ and }B.
\displaystyle \text{6. Join }PA\text{ and }PB.
\displaystyle \therefore PA\text{ and }PB\text{ are the required tangents.}
\\

\displaystyle \textbf{Question 8. }\text{The tops of two poles of heights }20\text{ m and }28\text{ m are connected with a wire.}
\displaystyle \text{The wire is inclined to the horizontal at an angle of }30^\circ.\text{ Find the length of the wire}
\displaystyle \text{and the distance between the two poles.}
\displaystyle \text{Answer:}
\displaystyle \text{Difference in heights of poles}=28-20=8\text{ m}
\displaystyle \text{Let the length of the wire be }l\text{ m.}
\displaystyle \sin 30^\circ=\frac{8}{l}
\displaystyle \frac{1}{2}=\frac{8}{l}
\displaystyle l=16\text{ m}
\displaystyle \therefore \text{Length of wire}=16\text{ m}
\displaystyle \text{Let the distance between the poles be }x\text{ m.}
\displaystyle \cos 30^\circ=\frac{x}{16}
\displaystyle \frac{\sqrt{3}}{2}=\frac{x}{16}
\displaystyle x=8\sqrt{3}\text{ m}
\displaystyle \therefore \text{Distance between the two poles}=8\sqrt{3}\text{ m}
\\

\displaystyle \textbf{Question 9. }\text{The weights (in kg) of }50\text{ wild animals of a National Park were recorded}
\displaystyle \text{and the following data was obtained. Find the mean weight using assumed mean method.}
\displaystyle \begin{array}{|c|c|}\hline \text{Weight (in kg)} & \text{Number of animals}\\ \hline 100-110 & 4\\ \hline 110-120 & 12\\ \hline 120-130 & 23\\ \hline 130-140 & 8\\ \hline 140-150 & 3\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class} & f_i & x_i & d_i=x_i-125 & f_id_i\\ \hline 100-110 & 4 & 105 & -20 & -80\\ \hline 110-120 & 12 & 115 & -10 & -120\\ \hline 120-130 & 23 & 125 & 0 & 0\\ \hline 130-140 & 8 & 135 & 10 & 80\\ \hline 140-150 & 3 & 145 & 20 & 60\\ \hline \end{array}
\displaystyle A=125,\qquad \sum f_i=50,\qquad \sum f_id_i=-60
\displaystyle \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}
\displaystyle =125+\frac{-60}{50}
\displaystyle =125-1.2
\displaystyle =123.8
\displaystyle \therefore \text{Mean weight}=123.8\text{ kg}
\\

\displaystyle \textbf{Question 10. }\text{For the following frequency distribution, find the median:}
\displaystyle \begin{array}{|c|c|}\hline \text{Class} & \text{Frequency}\\ \hline 1400-1550 & 6\\ \hline 1550-1700 & 13\\ \hline 1700-1850 & 25\\ \hline 1850-2000 & 10\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class} & f & c.f.\\ \hline 1400-1550 & 6 & 6\\ \hline 1550-1700 & 13 & 19\\ \hline 1700-1850 & 25 & 44\\ \hline 1850-2000 & 10 & 54\\ \hline \end{array}
\displaystyle N=54,\qquad \frac{N}{2}=27
\displaystyle \text{Median class}=1700-1850
\displaystyle l=1700,\quad h=150,\quad f=25,\quad c.f.=19
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-c.f.}{f}\times h
\displaystyle =1700+\frac{27-19}{25}\times150
\displaystyle =1700+\frac{8}{25}\times150
\displaystyle =1700+48
\displaystyle =1748
\displaystyle \therefore \text{Median}=1748
\\

\displaystyle \textbf{SECTION C}
\displaystyle \text{Question numbers 11 to 14 carry 4 marks each.}

\displaystyle \textbf{Question 11. }\text{(a) In Figure }3,\text{ two circles with centres at }O\text{ and }O'\text{ of radii }2r\text{ and}
\displaystyle r\text{ respectively, touch each other internally at }A.\text{ A chord }AB\text{ of the bigger circle}
\displaystyle \text{meets the smaller circle at }C.\text{ Show that }C\text{ bisects }AB.

\displaystyle \textbf{OR}

\displaystyle \text{(b) In Figure }4,\text{ }O\text{ is centre of a circle of radius }5\text{ cm. }PA\text{ and }BC\text{ are}
\displaystyle \text{tangents to the circle at }A\text{ and }B\text{ respectively. If }OP=13\text{ cm, then find}
\displaystyle \text{the length of tangents }PA\text{ and }BC. \displaystyle \text{Answer:}
\displaystyle \text{(a) Draw }OM\perp AB\text{ and }O'N\perp AC.
\displaystyle \text{Since perpendicular from the centre to a chord bisects the chord,}
\displaystyle AM=MB=\frac{AB}{2}\quad \text{and}\quad AN=NC=\frac{AC}{2}
\displaystyle \text{Also, }OA=2r,\quad O'A=r
\displaystyle \therefore \frac{O'A}{OA}=\frac{r}{2r}=\frac{1}{2}
\displaystyle \text{In }\triangle AO'N\text{ and }\triangle AOM,
\displaystyle \angle ANO'=\angle AMO=90^\circ,\quad \angle O'AN=\angle OAM
\displaystyle \therefore \triangle AO'N\sim\triangle AOM
\displaystyle \therefore \frac{AN}{AM}=\frac{O'A}{OA}=\frac{1}{2}
\displaystyle AN=\frac{AM}{2}
\displaystyle \frac{AC}{2}=\frac{1}{2}\times\frac{AB}{2}
\displaystyle AC=\frac{AB}{2}
\displaystyle \therefore AC=CB
\displaystyle \therefore C\text{ bisects }AB.

\displaystyle \textbf{OR}

\displaystyle \text{(b) }OA=OB=5\text{ cm},\quad OP=13\text{ cm}
\displaystyle \text{Since radius is perpendicular to the tangent, }OA\perp PA
\displaystyle \text{In }\triangle OAP,
\displaystyle PA^2=OP^2-OA^2
\displaystyle =13^2-5^2
\displaystyle =169-25=144
\displaystyle PA=12\text{ cm}
\displaystyle \text{Let }BC=x\text{ cm.}
\displaystyle \text{Using similarity of right triangles in the figure,}
\displaystyle \frac{BC}{OB}=\frac{OB}{PA}
\displaystyle \frac{x}{5}=\frac{5}{12}
\displaystyle x=\frac{25}{12}
\displaystyle \therefore PA=12\text{ cm},\quad BC=\frac{25}{12}\text{ cm}
\\

\displaystyle \textbf{Question 12. }\text{A straight highway leads to the foot of a tower. A man standing at the top}
\displaystyle \text{of the tower observes a car at an angle of depression of }30^\circ,\text{ which is}
\displaystyle \text{approaching the foot of the tower with a uniform speed. Ten seconds later, the}
\displaystyle \text{angle of depression of the car is found to be }60^\circ.\text{ Find the time taken by the}
\displaystyle \text{car to reach the foot of the tower from this point.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower be }h\text{ m.}
\displaystyle \text{Let the distances of the car from the foot of the tower at the two observations}
\displaystyle \text{be }x\text{ m and }y\text{ m respectively.}
\displaystyle \text{Since angle of depression equals angle of elevation,}
\displaystyle \tan 30^\circ=\frac{h}{x}
\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{x}
\displaystyle x=h\sqrt{3}
\displaystyle \tan 60^\circ=\frac{h}{y}
\displaystyle \sqrt{3}=\frac{h}{y}
\displaystyle y=\frac{h}{\sqrt{3}}
\displaystyle \text{Distance travelled by the car in }10\text{ seconds}
\displaystyle =x-y
\displaystyle =h\sqrt{3}-\frac{h}{\sqrt{3}}
\displaystyle =\frac{2h}{\sqrt{3}}
\displaystyle \text{Speed of the car}=\frac{\frac{2h}{\sqrt{3}}}{10}
\displaystyle =\frac{h}{5\sqrt{3}}\text{ m/s}
\displaystyle \text{Remaining distance to the foot of the tower}=y=\frac{h}{\sqrt{3}}
\displaystyle \text{Time required}=\frac{\frac{h}{\sqrt{3}}}{\frac{h}{5\sqrt{3}}}
\displaystyle =5\text{ seconds}
\displaystyle \therefore \text{The car will take }5\text{ seconds to reach the foot of the tower.}
\\

\displaystyle \textbf{Question 13. }\textbf{Case Study - 1}
\displaystyle \text{In the picture given below, one can see a rectangular in-ground}
\displaystyle \text{swimming pool installed by a family in their backyard. There is a}
\displaystyle \text{concrete sidewalk around the pool of width }x\text{ m. The outside edges of the}
\displaystyle \text{sidewalk measure }7\text{ m and }12\text{ m. The area of the pool is }36\text{ sq. m.} \displaystyle \text{(a) Based on the information given above, form a quadratic equation}
\displaystyle \text{in terms of }x.
\displaystyle \text{(b) Find the width of the sidewalk around the pool.}
\displaystyle \text{Answer:}
\displaystyle \text{Outer dimensions of the pool with sidewalk are }12\text{ m and }7\text{ m.}
\displaystyle \text{Since the sidewalk is }x\text{ m wide on all sides,}
\displaystyle \text{Length of pool}=(12-2x)\text{ m}
\displaystyle \text{Breadth of pool}=(7-2x)\text{ m}
\displaystyle \text{Area of pool}=36\text{ sq. m}
\displaystyle (12-2x)(7-2x)=36
\displaystyle 84-24x-14x+4x^2=36
\displaystyle 4x^2-38x+48=0
\displaystyle 2x^2-19x+24=0
\displaystyle \therefore \text{Required quadratic equation is }2x^2-19x+24=0
\displaystyle \text{Solving }2x^2-19x+24=0
\displaystyle 2x^2-16x-3x+24=0
\displaystyle 2x(x-8)-3(x-8)=0
\displaystyle (2x-3)(x-8)=0
\displaystyle x=\frac{3}{2}\quad \text{or}\quad x=8
\displaystyle \text{Since }x=8\text{ m is not possible,}
\displaystyle x=\frac{3}{2}\text{ m}
\displaystyle \therefore \text{Width of the sidewalk}=1.5\text{ m}
\\

\displaystyle \textbf{Question 14. }\textbf{Case Study - 2}
\displaystyle \text{John planned a birthday party for his younger sister with his friends.}
\displaystyle \text{They decided to make some birthday caps by themselves and to buy a}
\displaystyle \text{cake from a bakery shop. For these two items, they decided the following}
\displaystyle \text{dimensions :}
\displaystyle \text{Cake : Cylindrical shape with diameter }24\text{ cm and height }14\text{ cm.}
\displaystyle \text{Cap : Conical shape with base circumference }44\text{ cm and height }24\text{ cm.} \displaystyle \text{Based on the above information, answer the following questions :}
\displaystyle \text{(a) How many square cm paper would be used to make }4\text{ such caps ?}
\displaystyle \text{(b) The bakery shop sells cakes by weight (}0.5\text{ kg, }1\text{ kg, }1.5\text{ kg, etc.). To}
\displaystyle \text{have the required dimensions, how much cake should they order, if}
\displaystyle 650\text{ cm}^{3}\text{ equals }100\text{ g of cake ?}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Base circumference of cap}=44\text{ cm}
\displaystyle 2\pi r=44
\displaystyle 2\times\frac{22}{7}\times r=44
\displaystyle r=7\text{ cm}
\displaystyle h=24\text{ cm}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{7^2+24^2}
\displaystyle =\sqrt{49+576}
\displaystyle =25\text{ cm}
\displaystyle \text{Paper required for one cap}=\pi rl
\displaystyle =\frac{22}{7}\times7\times25
\displaystyle =550\text{ cm}^2
\displaystyle \text{Paper required for }4\text{ caps}
\displaystyle =4\times550
\displaystyle =2200\text{ cm}^2
\displaystyle \therefore \text{Paper required}=2200\text{ cm}^2
\displaystyle \text{(b) Radius of cake}=\frac{24}{2}=12\text{ cm}
\displaystyle \text{Height}=14\text{ cm}
\displaystyle \text{Volume of cake}=\pi r^2h
\displaystyle =\frac{22}{7}\times12^2\times14
\displaystyle =6336\text{ cm}^3
\displaystyle 650\text{ cm}^3\equiv100\text{ g}
\displaystyle 6336\text{ cm}^3\equiv\frac{6336\times100}{650}\text{ g}
\displaystyle =974.77\text{ g}
\displaystyle \approx0.975\text{ kg}
\displaystyle \text{Since cakes are sold in }0.5\text{ kg, }1\text{ kg, }1.5\text{ kg, etc.,}
\displaystyle \therefore \text{they should order a }1\text{ kg cake.}
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.