MATHEMATICS (ICSE 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]

Answer:

(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.

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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]

2011-1.jpg(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]

Answer:

(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

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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; 2011-2

(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]

Answer:

(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.

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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]

Answers:

(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 \}  

221.jpg

(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) 

2011-3

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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]

Answer:

(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

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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

 Answer:

(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

2011-4

(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

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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) sm2In 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]

Answer:

(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}

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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]

Answers:

(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

2011-8

(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 

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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 

Answer:

(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. 

2011-9(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.

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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. 

Answer:

(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)  2011-5.jpgFrom \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

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Question 11:

2011-6.jpg(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) 2011-7.jpgGive 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]

Answer:

(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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