Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2011)

Two and Half HourAnswers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.

The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions form Section A and any four questions from Section BAll working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the Answer. Omission of essential working will result in the loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables are provided.

SECTION A [40 Marks]

(Answer all questions from this Section.)

Question 1:

(a) Find the value of $k$ if $(x-2)$ is a factor of  $x^3+2x^2-kx+10$. Hence determine whether $(x+5)$ is also a factor.    [3]

(b) If $A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix}$  and $B = \begin{bmatrix} 4 \\ 2 \end{bmatrix}$ , is the product $AB$ possible ? Give a reason. If yes, find $AB$.     [3]

(c) Mr. Kumar borrowed $Rs. \ 15000$ for two years. The rate of interest  for the two successive years are $8\%$ and $10\%$ respectively. If the repays $Rs. \ 6200$ at the end of the first year, find the outstanding amount at the end of the second year.    [4]

(a)  Let $f(x) = x^3+2x^2-kx+10$.

Since given that $(x-2)$ is a factor $f(2) = 0$

Substituting the value of $x =2$ in the above function we get:

$f(2) = 0$

$f(2) = 8+8-2k+10=0$

$\Rightarrow k=13$

For $(x + 5)$ to be a factor $f(-5) = 0$

Substituting the value of $x =-5$ in the above function we get:

$f(-5) = (-5)^3+2(-5)^2-k(-5)+10 = -125+50+65+10=0$

Hence $(x+5$) is a factor of $x^3+2x^2-kx+10$

(b)  The order of matrix $A = 2 \times 2$ and the order of matrix  $B \ is \ 2 \times 1$.

Since the number of columns in $A$ is equal to the number of rows in  $B$, the product  $AB$ is possible.

$AB = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} . \begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 6+20 \\ 8-8 \end{bmatrix} = \begin{bmatrix} 26 \\ 0 \end{bmatrix}$

(c)   Principal $= 15000 \ Rs$

The rate of interest  for the two successive years are $8\%$ and $10\%$ respectively.

Formula: $A = P(1+\frac{r}{100})^n$

Therefore Amount after $1^{st}$ year $= 15000 \times (1+\frac{8}{100})^1 = 16200 \ Rs.$

Principal at the start of $2^{nd}$ year after repayment $= 16200 - 6200 = 10000 \ Rs.$

Amount outstanding at the end of second year $= 10000 \times (1+\frac{10}{100})^1 = 11000 \ Rs.$

$\\$

Question 2:

(a) From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random.What is the probability that the card drawn is;

(i) a face card (King, Jack, or Queen)

(i) an even number red card    [3]

(b) Solve the following equation:  $x -$ $\frac{18}{x}$ $= 6$  Give your answer correct to two significant figures.    [3]

(c) In the given figure $O$ is the center of the circle Tangent of $A$ and $B$ meet at $C$ if $\angle AOC = 30^o$, find (i) $\angle BCO$ (ii) $\angle AOB$ (iii) $\angle APB$    [4]

(a) Total number of cards $= 52$

Number of cards which are multiples of $3 = 12$

Total number of cards left $= 52 - 12 = 40$

(i) Number of face cards $= 12$

Probability (of a face card) $= \frac{12}{40} = 0.3$

(ii) Even numbered red cards $= 10$

Probability (of a even number red card) $= \frac{10}{40} = 0.25$

(b) Given $x -$ $\frac{18}{x}$ $= 6$

Simplifying: $x^2 - 6x - 18 = 0$

Compare with equation $ax^2 + bx + c = 0$, we get $a =1 b = -6$ and $c = -18$

We know, $x =$ $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$=$ $\frac{6 \pm \sqrt{36 + 72 }}{2}$

$=$ $\frac{6 \pm 6\sqrt{3}}{2}$ $= 3 \pm 3 \sqrt{3}$

Therefore $x = 8.196 \ or \ -2.196$

Answer correct to two significant figures: $x = 8.2 \ or \ -2.2$

(c)  Consider $\triangle AOC$ and $\triangle BOC$

$\angle OAC = \angle OBC = 90^o$

$OC$ is common

$AC = BC$  (two tangents drawn from a point on a circle are of equal lengths)

Therefore $\triangle AOC \cong \triangle BOC$ (RHS postulate)

(i) $\angle ACO = \angle BCO = 30^o$

(ii) $\angle AOC = 180^o - 90^o - 30^o = 60^o$

$\angle BOC = 180^o - 90^o - 30^o = 60^o$

$\therefore \angle AOB = \angle AOC + \angle BOC = 120^o$

(iii) $\angle AOB = 2 \angle APB$ (chord subtends twice the angle at the center than that it subtends on the circumference)

$\Rightarrow \angle AOB = 2 \angle APB$

$\Rightarrow \angle APB = 60^o$

$\\$

Question 3:

(a) Ahmed has a recurring deposit account in a bank. He deposits $Rs. 2500$ per month for $2$ years. If he gets $Rs. 66250$ at the time of maturity, find;

(i) The interest paid by the bank

(ii) The rate of interest    [3]

(b) Calculate the area of the shaded region, if the diameter of the semi circle is equal to $14 \ cm$.   (Take $\pi = \frac{22}{7}$)     [3]

(c) $ABC$ is a triangle and $G(4,3)$ is the central of the triangle. If $A=(1,3), B=(4,b)$ and $C=(a,1)$ find $'a'$ and $'b'$ find the length of side $BC$.       [4]

(a)   $P = Rs. \ 2500, \ no \ of \ months = 24, \\ \ rate = r\% \ Maturity Amount = Rs. 66250$

$Maturity \ Value = P \times n + P \times$ $\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$66250 =2500 \times 24 +2500 \times$ $\frac{24(24+1)}{2 \times 12} \times \frac{r}{100}$ $\Rightarrow r=10\%$

$Interest = P \times$ $\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$

$= 2500 \times$ $\frac{24(24+1)}{2 \times 12} \times \frac{10}{100}$ $= 6250$

(b) Area of shaded portion = Total area – area of the two quadrants

$= (Area \ of \ ACDE + \ Area \ of \ semi \ circle \ EFD) - \\ (Area \ of \ Quadrant \ ABE + Area \ of \ Quadrant \ BCD)$

$= (14 \times 7 + \frac{1}{2} \pi \times 7^2) - (\frac{1}{4} \times \pi \times 7^2 + \frac{1}{4} \times \pi \times 7^2 )$

$= 98 \ cm^2$

(c)  Since $G$ is the centroid

$4 =$ $\frac{1+4+a}{3}$ $\Rightarrow a=7$

$3=$ $\frac{3+b+1}{3}$ $\Rightarrow b = 5$

Therefore $B(4, 5) \ and \ C(7, 1)$

Therefore $BC=\sqrt{(7-4)^2+(1-5)^2} = \sqrt{25} = 5$ units.

$\\$

Question 4:

(a) Solve the following in equation and represent the solution set on the number line:

$2x-5 \le 5x + 4 \le 11 \ where \ x \in I$    [3]

(b) Evaluate without using trigonometric tables:

$2$ $(\frac{tan \ 35^o}{cot \ 55^o})^2$ $+$ $(\frac{cot \ 55^o}{tan \ 35^o})^2$ $- 3$ $(\frac{sec \ 40^o}{cosec \ 50^o})$    [3]

(c) A mathematics aptitude test of 50 students was recorded as follows:

 Marks 50-60 60-70 70-80 80-90 90-100 No. of students 4 8 14 19 5

Draw a histogram for the above data using a graph paper and locate the mode.   [4]

(a)  $2x-5 \leq 5x+4 < 11$

$2x-5 \leq 5x+4$ or $-9 \leq 3x$  or $-3 \leq x$

$5x+4 < 11$  or $5x < 7$  or  $x < \frac{7}{5}$

$-3 \leq x <\frac{7}{5}$

Therefore $x \in \{-3, -2, -1, 0, 1 \}$

(b) $2$ $(\frac{tan \ 35^o}{cot \ 55^o})^2$ $+$ $(\frac{cot \ 55^o}{tan \ 35^o})^2$ $- 3$ $(\frac{sec \ 40^o}{cosec \ 50^o})$

$= 2$ $(\frac{tan \ 35^o}{cot \ (90^o - 35^o)})^2$ $+$ $(\frac{cot \ 55^o}{tan \ (90^o - 55^o)})^2$ $- 3$ $(\frac{sec \ 40^o}{cosec \ (90^o - 40^o)})$

$= 2$ $(\frac{tan \ 35^o}{tan \ 35^o})^2$ $+$ $(\frac{cot \ 55^o}{cot \ 55^o})^2$ $- 3$ $(\frac{sec \ 40^o}{sec \ 40^o})$

$= 2 + 1 - 3 = 0$

(c)

$\\$

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5:

(a) A manufacturer sells a washing machine to a wholesaler for $Rs.\ 15000$. The wholesaler sells it to a trader at a profit of $Rs.\ 1200$ and the trader in turn sells it to a consumer at a profit of $Rs.\ 1800$. If the rate of VAT is $8\%$ find:

(i) The amount of VAT received by the state government on the sale of this machine from the manufacture and the wholesaler.

(ii) The amount that the consumer pays for the machine.    [3]

(b) A solid cone of radius $5 \ cm$ and height $8 \ cm$ is melted and made into small spheres of radius $0.5 \ cm$. Find the number of sphere formed.     [3]

(c) $ABCD$ is a parallelogram where $A(x, y), B (5, 8), C (4, 7)$ and $D (2, -4)$ Find

(i) Coordinates of $A$

(ii) Equation of diagonal $BD$     [4]

(a)  (i) Tax received by the manufacturer $=$ $\frac{8}{100}$ $\times 1500 = Rs. 1200$

For the trader the price  $= 15000+1200= Rs. \ 16200$

Tax paid by the trader $=$ $\frac{8}{100}$ $\times 16200 = Rs. \ 1296$

Therefore VAT received from wholesaler $= 1296-1200 = Rs. \ 96$

Price for the consumer $= 16200+1800 = Rs. \ 18000$

(ii) Tax paid by the consumer $=$ $\frac{8}{100}$ $\times 18000 = Rs. \ 1440$

Hence the total price paid by the consumer $= 18000+1440 = Rs. \ 19440$

(b)  Cone: Radius : $5 \ cm$ and Height $8 \ cm$

Sphere: $0.5 \ cm$

Number of sphere $= \frac{Volume \ of \ the \ cone}{Volume \ of \ the \ small sphere}$

$= \frac{\frac{1}{3} \pi (5)^2 \times 8}{\frac{4}{3} \pi (0.5)^2}$

$=$ $\frac{25 \times 8}{4 \times 0.125}$ $= 400$

(c)  (i) Mid point of $BD = (\frac{5+2}{2}, \frac{-4+8}{2})= (\frac{7}{2}, 2)$

Therefore we have $A(x, y), O(\frac{7}{2}, 2)$ and $C(4, 7)$

$O$ is the mid point of $AC$ as well  (diagonals of a parallelogram bisect each other)

Hence $\frac{x+4}{2} = \frac{7}{2}$ $\Rightarrow x = 3$

and $\frac{y+7}{2}$ $= 2 \Rightarrow y = -3$

Hence $A ( 3, -3)$

(ii) Equation of $BD$

$y - 8 =$ $\frac{-4-8}{2-5}$ $(x-5)$

$y-8 = 4(x-5)$

$y - 8 = 4x - 20$

$y + 12 = 4x$

$\\$

Question 6:

(a) Use a graph paper to answer the following (Take $1 \ cm =1 \ unit$ on both axes)

(i) Plot $A(4,4), B(4,-6)$ and $C(8,0)$ the vertices of a $\triangle ABC$

(ii) Reflect $ABC$ on the $y-axis$ and name it as $A'B'C'$

(iii) Write the coordinates of the image $A'B'$ and $C'$

(iv) Give a geometrical name for the figure $AA' C' B' BC$

(v) Identify the line of symmetry of $AA' C' B' BC'$    [5]

(b) Choudhury opened a Saving account at State Bank of India an 1st April 2007, The entries of one year as shows in his pass book are given below:the bank pays interest at the rate of 5% per annum, find the interest paid on 1st April 2008. Give your answer correct to the nearest rupees.      [5]

 Date Particulars Withdrawals (Rs) Deposit (Rs.) Balance (in Rs.) 1st April 2007 By cash – 8550 8550 12th April 2007 To Self 1200 – 7350 24th April 2007 By cash – 4550 11900 8th July 2007 By cheque – 1500 13400 10th Sep.2007 By cheque – 3500 16900 17th Sep. 2007 By cheque 2500 – 14400 11th Oct.2007 By cash – 800 15200 6th Jan 2008 To Self 2000 – 13200 9th March 2008 By cheque – 950 14150

(a)   (i) & (ii) Shown in the graph below.

(iii) $A'(-4, 4), C'(-8, 0)$ and $B'(-4, -6)$

(iv) The shape is that of a hexagon.

(v) Line of symmetry : $y-axis$

(b) Qualifying principal for various months:

 Month Principal (Rs.) April 7350 May 11900 June 11900 July 13400 August 13400 September 14400 October 14400 November 15200 December 15200 January 13200 February 13200 March 14150 Total 157700

$P = Rs. \ 157700 \ R = 5\% \ and \ T=$ $\frac{1}{12}$

$I = P \times R \times T = 157700 \times$ $\frac{5}{100} \times \frac{1}{12}$ $= Rs. \ 657.08 \ or \ Rs. \ 657$

$\\$

Question 7:

(a) Using component and dividend, find the value of x:

$\frac{\sqrt{3x+4}+ \sqrt{3x-5}}{\sqrt{3x+4}- \sqrt{3x-5}}$ $= 9$   [3]

(b) If  $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ , $B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}$ and $I$ is the identity matrix of the same order and $A^t$ is transpose of matrix $A$, find $A^t B +BI$.    [3]

(c) In the adjoining figure $ABC$ is a right angled triangle with $\angle BAC = 90^o$

(i) Prove $\triangle ADB \sim \triangle CDA$

(ii) If $BD=18\ cm, CD=8\ cm, \ find \ AD$

(iii) Find the ratio of the area of  $\triangle ADB$ is to the area  of  $\triangle CDA$    [4]

(a)  $\frac{\sqrt{3x+4}+ \sqrt{3x-5}}{\sqrt{3x+4}- \sqrt{3x-5}}$ $= 9$

Applying componendo and dividendo

$\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}$

$\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}$

Simplifying

$\frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}$

Square both sides

$\frac{3x+4}{3x-5} = \frac{25}{14}$

$42x+56 = 75x-125$

Simplifying we get $x = 7$

(b)   $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$

$A^t = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$

$A^t.B+BI$

$= \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} . \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}+ \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}$

(c)  (i)   Let $\angle DAB = \theta$

Therefore $\angle DAC = 90^o - \theta$

$\angle DBA = 90^o - \theta$

$\angle DCA = \theta$

Therefore $\triangle ADB \sim \triangle CDA$ (AAA postulate)

(ii) $\frac{CD}{AD}=\frac{AD}{BD}$

$\Rightarrow AD^2 = CD \times BD = 8 \times 18 = 144$

Therefore $AD = \sqrt{144} = 12$

(iii) $\frac{Area \ of \ \triangle ADB}{ Area \ of \ \triangle CDA} = \frac{\frac{1}{2} AD \times BD} {\frac{1}{2} AD \times CD} = \frac{BD}{CD} = \frac{18}{8}= \frac{9}{4}$

$\\$

Question 8:

(a) Using step division method, calculate the mean marks of the following distribution: State the modal class:     [5]

 Class Interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 Frequency 5 20 10 10 9 6 12 8

(b) Marks obtained by 200 students in an examination are given below:

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 11 10 20 28 37 40 29 14 6

Draw an ogive for the given distribution taking $2 \ cm=10$ marks on one axis and $2 \ cm=20$ students on the other axis. Using the graph, determine.

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is $40$.

(iii) If scoring $85$ and more marks is considered as grade one, find the number of students who secured grade on in the examination;     [5]

(a)

 C.I $f$ $x$ $d=x-67.5$ $u$ $f.u.$ 50-55 5 52.5 -15 -3 -15 55-60 20 57.5 -10 -2 -40 60-65 10 62.5 -5 -1 -10 65-70 10 67.5 0 0 0 70-75 9 72.5 5 1 9 75-80 6 77.5 10 2 12 80-85 12 82.5 15 3 36 85-90 8 87.5 20 4 32 $\Sigma f = 80$ $\Sigma fu = 24$

$A.M = 67.5$

(i) $\bar{x} = A.M. + \frac{\Sigma fu}{\Sigma f} \times i = 67.5 + \frac{24}{80} \times 5 = 69$

(ii) Modal class is $55-60$ (class with highest freq.)

(b)

 Class Interval Frequency Cumulative Frequency 0-10 5 5 10-20 11 16 20-30 10 20 30-40 20 46 40-50 28 74 50-60 37 111 60-70 40 151 70-80 29 180 80-90 14 194 90-100 6 200

(i) $n = 200$ (even)

Median $= (\frac{n}{2})^{th}$ observation $= (\frac{200}{2})^{th}$ observation $= 100^{th}$ observation $= 57$

(ii) Number of student who failed $= 46$

(iii) Number of students who secured grade one  $200 - 1888 = 12$

$\\$

Question 9:

(a) Parekh invested $Rs. \ 52000$ on $100$ shares at a discount of $Rs. \ 20$ paying $8\%$ dividend. At the end of one year he sells the shares at a premium of $Rs. \ 20$. Find;

(i) The annual dividend

(ii) The profit earned including hid dividend;

(b) Draw a circle of radius $3.5 \ cm$. mark a point $P$ outside the circle at a distance of $6 \ cm$. from the center. Construct two tangent from $P$ to the given circle. Measure and write down the length of one tangent.

(c) Prove that $(cosec\ A- sin\ A) (sec\ A-cos\ A) \ sec^2 A = tan\ A$

(a)   Nominal Value of the share $= 100 \ Rs.$

Market Value of the share $= 80 \ Rs.$

Number of shares bought $= \frac{52000}{80} = 650$

Dividend earned $= 650 \times 100 \times \frac{8}{100} = 5200 \ Rs.$

Sale proceeds $= 650 \times 120 = 78000 \ Rs.$

Profit $= (78000-52000)+5200 = 31200 \ Rs.$

(b)

Steps  of construction (diagram not to scale):

1. With the help of a ruler measure 3.5 cm in your compass and draw a circle of radius 3.5 cm.
2. The draw a point P, 6 cm away from the circle.
3. The next step is to bisect PP’
4. The with this as the center, O, draw a circle that goes through point Q and R on the original circles.
5. Join PQ and PR. These are the tangents.
6. Length of the tangents is 4.9 cm

(c) $(cosec\ A- sin\ A) (sec\ A-cos\ A) \ sec^2 A = tan\ A$

LHS $= (cosec\ A- sin\ A) (sec\ A-cos\ A) \ sec^2 A$

= $\frac{1 - sin^2 \ A}{sin \ A} . \frac{1 - cos^2 \ A}{cos \ A} . \frac{1}{cos^2 \ A}$

= $\frac{cos^2 \ A}{sin \ A}. \frac{sin^2 \ A}{cos \ A} . \frac{1}{cos^2 \ A}$

= $\frac{sin \ A}{cos \ A}$ $= tan \ A =$ RHS

Hence Proved.

$\\$

Question 10:

(a) $6$ is the mean proportion two numbers $x$ and $y$ and $48$ is the third proportion of $x$ and $y$. Find the numbers.

(b) In what period of time will $Rs.\ 12000$ yield $Rs.\ 3972$ as compound interest at $10\%$ per annum, if compounded on an yearly basis?

(c) A man observed the angle of elevation of the top of a building to be $30^o$. He walks towards it in a horizontal line through its base. On covering $60 \ m$ the angle of evaluation changes to $60^o$. Find the height of the building correct to the nearest meter.

(a)  Given  $6$ is the mean proportion between two numbers $x \ and \ y$

Therefore $\frac{x}{6}={6}{y} \Rightarrow xy=36 \Rightarrow x = \frac{36}{y}$  … … … … … … i)

Also given  $48$  is the third proportion to $x \ and \ y$

Therefore $\frac{x}{y}=\frac{y}{48} \Rightarrow y^2=48x$  … … … … … … ii)

Solving i) and ii)

$y^2 = 48 \frac{36}{y}$

$y^3 = 2^3 \times 6^3 \Rightarrow y = 12$

Hence $x = \frac{36}{12} = 3$

Hence the numbers are $3 \ and \ 12$.

(b) Given $P=12000 \ Rs.; A= (12000+3972)= 15972 \ Rs. ; \\ r=10\%; n=n$

$A=P(1+\frac{r}{100})^n \Rightarrow 15972 = 12000(1+\frac{10}{100})^n \\ \Rightarrow n = 3 \ years$

(c)  From $\triangle ABC$

$\frac{h}{x}$ $= tan \ 60^o$

$\Rightarrow x = \frac{h}{\sqrt{3}}$

Similarly, from $\triangle ADB$

$\frac{h}{60+x}$ $= tan \ 30^o$

$60+x = h \sqrt{3}$

Equating  for $x$ we get

$h \sqrt{3} - 60 = \frac{h}{\sqrt{3}}$

$\Rightarrow 3 h - 60 \sqrt{3} = h$

$\Rightarrow 2h = 60 \sqrt{3} \Rightarrow h = 30 \sqrt{3} = 51.96 \ m$

$\\$

Question 11:

(a) $\triangle ABC$  with $AB=10 \ cm, BC=8 \ cm$ and $AC=6 \ cm$ (not drawn to scale) Three circle are drawn touching each other with the vertices as their centers. Find the radii of the three circles.   [3]

(b) $Rs. \ 480$ is divided equally among $x$ children. If the number of children were $20$ more than each would have got $Rs. \ 12$ less. Find $x.$   [3]

(c) Give equation of line $L_1$ is $y=4$

(i) Write the slope of line $L_2$ if $L_2$ is the bisector of angle $O$.

(ii) Write the co-ordinates of point $P$.

(iii) Find the equation of $L_2$ [4]

(a)  Let the radius of the three circles $r_1, r_2 and r_3$

Given: $AB=10 \ cm, BC=8 \ cm$ and $AC=6 \ cm$

Therefore

$r_1 + r_2 = 10$ … … … … … (i)

$r_1 + r_3 = 6$ … … … … … (ii)

$r_2 + r_3 = 8$ … … … … … (iii)

Adding (i), (ii) and (iii) we get

$r_1 + r_2 + r_3 = 12$ … … … … … (iv)

Using (i) and (iv) we get $r_3 = 2 \ cm$

Using (ii) and (iv) we get $r_2 = 6 \ cm$

Using (iii) and (iv) we get $r_1 = 4 \ cm$

(b)  Let the number of children $= x$

Therefore

$\frac{480}{x} - \frac{480}{x+20}$ $= 12$

$480x+9600-480x= 12 (x^2+20x)$

$12x^2+240x-9600=0 \Rightarrow x = 20 \ or -40 \ (not \ possible)$

Hence the number of children is 20

(c)   (i) Slope of $L_2 = m = tan \ 45^o = 1$

(ii) Equation of line $L_2$. It passes through $(0, 0)$

$y - 0 = 1(x - 0) \Rightarrow y = x$

$P$ is the point of intersection of $L_1$ and $L_2$

Solving $L_1 (y = 4)$ and $L_2 (y = x)$ we get $x = 4$ and $y = 4$.

Hence $P (4, 4)$.

(iii) Equation of $L_2: y = x$

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