MATHEMATICS (ICSE Paper 2013)

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) Given A = \begin{bmatrix}  2 & -6 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix}  -3 & 2 \\ 4 & 0 \end{bmatrix} \ and \  C = \begin{bmatrix}  4 & 0 \\ 0 & 2 \end{bmatrix} . Find the matrix X  such that A + 2X = 2B + C .   [3]

(b) At what rate \% p.a. will a sum of Rs. \ 4000 yield Rs. \ 1324 as compound interest in 3 years?   [3]

(c)   The median of the following observation 11,12,14 (x-2), (x+4), (x+9), 32, 38, 47 arranged in ascending order is 24 . Find the value of x and hence find the mean. [4]

Answers:

(a)  Given A = \begin{bmatrix}  2 & -6 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix}  -3 & 2 \\ 4 & 0 \end{bmatrix} \ and \  C = \begin{bmatrix}  4 & 0 \\ 0 & 2 \end{bmatrix}

Substituting these values in the given expression A + 2X = 2B + C we get,

\therefore  \begin{bmatrix}  2 & -6 \\ 2 & 0 \end{bmatrix} + 2X = 2 \begin{bmatrix}  -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix}  4 & 0 \\ 0 & 2 \end{bmatrix}

2X = \begin{bmatrix}  -6+4 & 4+0 \\ 8+0 & 0+2 \end{bmatrix} - \begin{bmatrix}  2 & -6 \\ 2 & 0 \end{bmatrix}

X = \frac{1}{2} \begin{bmatrix}  -4 & 10 \\ 6 & 2 \end{bmatrix}

X = \begin{bmatrix}  -2 & 5 \\ 3 & 1 \end{bmatrix}

(b)  Given: Principle =Rs.4000,\ C.I.=Rs.1324

Amount =P+C.I =4000+1324=Rs. \ 5324

Time =3 \ years

We know that; A=P(1+\frac{r}{100})^n 

5324=4000 (1+\frac{r}{100})^3

\frac{5324}{4000}=(1+\frac{r}{100})^3

\frac{1331}{1000}=(1+\frac{r}{100})^3

(\frac{11}{10})^3 =(1+\frac{r}{100})^3

Therefore, 1+\frac{r}{100}=\frac{11}{10}

\frac{r}{100}=\frac{11}{10}-1

\frac{r}{100}=\frac{1}{10}

r=\frac{100}{10}

r=10\% 

(c)  Given observation are 11,12,14, (x-2), (x+4), (x+9), 32, 38, 47 and mediam =24

Since n=9   which is odd, therefore

Median=\ \frac{n+1}{2}th\ term =\frac{9+1}{2}th\ term

24=5th\ term 

x+4=24 

\Rightarrow x=24-4 \Rightarrow x=20 

Therefore, 11,12,14, (20-2), (20+4), (20+9),32,38,47

=11,12,14,18,24,29,32,38,47

Now, Mean =\frac{\sum{x}}{n} 

=\ \frac{11+12+14+18+24+29+32+38+47}{9} =\frac{225}{9} =25 

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Question 2:

(a) What number must be added to each of the number 6, 15, 20 \ and \  43 to make them proportional?   [3]

(b) If (x-2) is a factor of the expression 2x^3 + ax^2 + bx - 14 and, when the expression is divided by (x - 3) , it leaves a remainder 52 , find the values of a \ and \  b  [3]

(c) Draw a histogram from the following frequency distribution and find the made from the graph:   [4]

Class 0-5 5-10 10-15 15-20 20-25 25-30
Frequency 2 5 18 14 8 5

Answers:

(a)  Let the number that must be added be x , then

The new number = (6 + x), (15 + x),(20 + x),(43 + x)

Since they are proportional,

(6+x) : (15+x) :: (20+x):(43+x)

(6+x)(43+x) = (15+x)(20+x)

\Rightarrow 258+6x+43x+x^2 = 300+20x+15x+x^2

\Rightarrow 49x - 35x = 300-258

\Rightarrow 14x = 42 \Rightarrow x= 3

(b) Let (x-2) is a factor of the given expression;

Since x-2=0 \Rightarrow x=2

2x^3 + ax^2 + bx - 14 = 0

In the given expression,  we substitute x = 2 2(2^3)+a(2^2)+b(2)-14 = 0 we get

16+4a+2b- 14 = 0

4a+2b+2 = 0

4a +2b = -2

2a+b = -1 … … … … … (i)

When given expression is divided by (x - 3)

x-3 = 0 \Rightarrow x=3

Similarly, in the given expression,  we substitute x = 3 2x^3 +ax^2+bx-14 = 52 we get

2 (3)^3 + a(3)^2 + b(3) - 66 = 0

54+9a+3b-66 = 0

9a+3b = 12

3a+b = 4    … … … … … (ii)

Solving equation (i) and (ii),

a = 5 \ and \  b = -11

(c)

16-1

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Question 3:

(a) Without using tables evaluate: 3  \  cos \  80^o. cosec \ 10^o + 2 \  sin \ 59^o sec \ 31^o    [3]

(b) In the given feature,16-2

\angle BAD=65^o \angle ABD=70^o \angle BDC=45^o

Prove:

(i) AC is the diameter of the circle

(ii) Find \angle ACB    [3]

(c) AB is a diameter of a circle with center C =(-2,5) , If A=(3,-7) , Find;

(i) The length of radius AC

(ii) The Coordinates of B    [4]

Answers:

(a) 3 \ {\mathrm{cos \ } 80{}^\circ \ cosec\ 10{}^\circ +2 \ {\mathrm{sin \ } 59{}^\circ \ sec \ 31{}^\circ \ }\ }

= 3 \ {\mathrm{cos \ } 80{}^\circ \ cosec\ \left(90{}^\circ -80{}^\circ \right)+2\ {\mathrm{sin \ } 59{}^\circ \ {\mathrm{sec \ } \left(90{}^\circ -59{}^\circ \right)\ }\ }\ }

= 3 \ cos \ 80^o sec \ 80^o + 2 \ sin \ 59^o cosec \ 59^o 

= 3 \ {\mathrm{cos \ } 80{}^\circ \ \times \frac{1}{{\mathrm{cos \ } 80{}^\circ \ }}+2 \ {\mathrm{sin \ } 59{}^\circ \times \frac{1}{sin \ 59{}^\circ }\ }\ }

= 3+2=5

(b) Given: \angle BAD=65^o ,\ \angle ABD=70^o ,\ \angle BDC=45^o

(i) Since  ABCD is a cyclic quadrilateral

In \triangle \ ABD

\angle BDA+\angle DAB+\angle ABD=180^o  ( sum property of a triangle)

\angle BDA=180^o -(65^o +70^o) =180^o -135^o =45^o

Now from \triangle ACD ,

\angle ADC=\angle ADB+\angle BDC =45^o +45^o=90^o

Hence \angle ADC makes right angle belongs in semi-circle therefore AC is a diameter of the circle.

(ii) \angle ACB = \angle ADB (Angles in the same segment of a circle)

Therefore \angle ABC = 45^o \ (since \  \angle ADB = 45^o)

(c)   (i) Length of the radius AC = \sqrt{(-2-3)^2+ (5+7)^2} = \sqrt{25+144} = 13

(ii) Let the point B be (x,y)

Given C is the mid-point of AB . Therefore

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

5 = \frac{-7+y}{2} \Rightarrow 10 = -7+y \Rightarrow y = 17

Hence, the co-ordinate of B (-7, 17)

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Question 4:

(a) Solve the following equation and calculate the answer correct to two decimal places. x^2-5x-10= 0 . [3]

(b)  In the given figure, AB and DE are perpendicular to BC  16-3

(i) Prove that \triangle ABC \sim \triangle DEC

(ii) If AB=6 \ cm, DE=4 \ cm \ and \  AC=15 \ cm , Calculate CD ,

(iii) Find the ratio of the area of \triangle ABC : area \ of \  \triangle DEC    [3]

(c)  Using graph paper, plot the point A(6,4) and B(0,4) .

(i) Reflect A and B in the origin to get the images A' and B' .

(ii) Write the co-ordinate of A'  and B'

(iii) State the geometrical name for the figure ABA'B' 

(iv) Find its perimeter   [4]

Answers:

(a)   Given : x^2 - 5x - 10 = 0

Comparing this expression with ax^2+bx+c = 0 , we get  a=1, b=5 \ and \  C=-10

\mathrm{D=}{\mathrm{b}}^{\mathrm{2}}\mathrm{-}\mathrm{4ac} \mathrm{=(-}{\mathrm{5)}}^{\mathrm{2}}\mathrm{-}\mathrm{4\times 1\times -10=65}

x=\frac{\mathrm{-}\mathrm{b\pm }\sqrt{\mathrm{D}}}{\mathrm{2a}} \mathrm{=}\frac{\mathrm{5\pm}\sqrt{\mathrm{65}}}{\mathrm{2\times 1}}\mathrm{=}\frac{\mathrm{5\pm 8.06}}{\mathrm{2}} \mathrm{=}\frac{\mathrm{5+8.06}}{\mathrm{2}}\mathrm{\ }\frac{\mathrm{5-8.06}}{\mathrm{2}} \ or \ \frac{\mathrm{13.06}}{\mathrm{2}}\mathrm{,\ -}\frac{\mathrm{3.06}}{\mathrm{2}}

x=6.53,\ -1.53

(b)   (i)  From \mathrm{\triangle }\mathrm{ABC\ and\ }\mathrm{\triangle }\mathrm{DEC,}

\mathrm{\angle ABC=\angle DEC=90{}^\circ } Givem

And \mathrm{\angle ACB=\ \angle }\mathrm{DCE=common}

\mathrm{\therefore \ \  }\mathrm{\triangle }\mathrm{ABC\ \sim DEC\ }\left(\mathrm{by\ AA\ similarity}\right)

(ii) In \mathrm{\triangle }\mathrm{ABC\ and\ }\mathrm{\triangle }\mathrm{DEC}

Since \mathrm{\triangle }\mathrm{ABC\sim }\mathrm{\triangle }\mathrm{DEC}

\mathrm{\therefore \  }\frac{\mathrm{AB}}{\mathrm{DE}}\mathrm{=}\frac{\mathrm{AC}}{\mathrm{CD}}

Given: AB =6 \ cm, DE=4 \ cm, AC=15 \ cm ,

\mathrm{\therefore  \ }\frac{\mathrm{6}}{\mathrm{4}}\mathrm{=}\frac{\mathrm{15}}{\mathrm{CD}}

\mathrm{\Rightarrow \ \ }\mathrm{6\times CD=15\times 4} \mathrm{ \ \ \Rightarrow \ \ }\mathrm{CD=}\frac{\mathrm{60}}{\mathrm{6}} \mathrm{ \ \ \Rightarrow  \ \ }\mathrm{CD=10 \ cm.}

(iii) Since $latex \triangle ABC \sim \triangle DEC,16-4

\frac{\mathrm{Area\ of\ }\mathrm{\triangle }\mathrm{ABC}}{\mathrm{Area\ of\ }\mathrm{\triangle }\mathrm{DEC}}\mathrm{=\ }\frac{{\mathrm{AB}}^{\mathrm{2}}}{{\mathrm{DE}}^{\mathrm{2}}} = \frac{{\mathrm{6}}^{\mathrm{2}}}{{\mathrm{4}}^{\mathrm{2}}}\mathrm{=}\frac{\mathrm{36}}{\mathrm{16}}\mathrm{=}\frac{\mathrm{9}}{\mathrm{4}}

\mathrm{\therefore \  }\mathrm{Area\ of\ }\mathrm{\triangle }\mathrm{ABC:Area\ of\ }\mathrm{\triangle }\mathrm{DEC=9:4}

(c)   (i) Please see graph

16-5

(ii) Reflection of A' and B' in the origin = A'(-6, -4) \ and \  B' (0,-4)

(iii) The geometrical name for the figure AB A'B' is a parallelogram

(iv) From the graph, AB=6 \ cm \ BB'=8 \ cm

In \triangle ABB' 

(AB')^2= AB^2+ (BB')^2 = 6^2+8^2= 36 + 64 = 100 

Therefore AB' = 10  = A'B  since ABA'B'  is a parallelogram

Perimeter of ABA'B' = A'B' +AB'+AB+A'B' = 6+10+6+10 =32 \ units 

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SECTION B [40 Marks]

(Answer any four questions in this Section.)


Question: 5

(a)   Solve the following inequation, write the solution set and represent it on the number line:   [3]

\mathrm{-}\frac{x}{\mathrm{3}}\mathrm{<}\frac{x}{\mathrm{2}}\mathrm{-}\mathrm{1}\frac{\mathrm{1}}{\mathrm{3}}\mathrm{<}\frac{\mathrm{1}}{\mathrm{6}} \ \ \ \  x \mathrm{\in }\mathrm{R}

(b)   Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8\% per annum and Mr. Britto gets Rs. \ 8088 from the bank after 3 years, find the value of his monthly installment.        [3]

(c) Salman buys 50 shares of face value Rs. \ 100 available at Rs.132 .

(i)  What is his investment?

(ii)  If the dividend is 7.5\% what will be his annual income?

(iii)  If he wants to increase his annual income by Rs. \ 150 . How many extra shares should he buy?      [4]

Answers:

(a) Given;  {-}\frac{ {x}}{ {3}} {\le }\frac{ {x}}{ {2}} {-} {1}\frac{ {1}}{ {3}} {<}\frac{ {1}}{ {6}}

Taking L.C.M. of 3, 2 and 6 is 6.

 {-}\frac{{x}}{ {3}} {\times 6} \  {\le } \  \frac{{x}}{ {2}} {\times 6-}\frac{ {4}}{ {3}} {\times 6 \ < \ }\frac{ {1}}{ {6}} {\times 6}

 {-}{2x} {\ \le \  }{3x-8<1}

 {\Rightarrow }{\ -2x} {\le }{3x-8\ and\ 3x-8<1}

\Rightarrow -2x \le 3x-8 \Rightarrow 8 \le 5x  \Rightarrow \frac{8}{5} \le x

Similarly, 3x-8 < 1 \Rightarrow 3x < 9 \Rightarrow x < 3 

 {\therefore } {\ The\ solution\ set\ is\ }\left(x: \frac{8}{5} \  {\le }x  {\le }3,\ x\  {\in } {R}\right)

16-12

(b) Let the monthly installment be Rs. \ x

Given: Maturity amount = Rs. \ 8,058, \ Time (n)=3 \ years =36 \ months, \ Rate (R)=8\% \ p.a.

{Principle\ for\ month=P\times }\frac{ {n}\left( {n+1}\right)}{ {2}} = \frac{ {x\ \times 36\times 37}}{ {2}} = {18\times 37x}

Interest = \frac{ {18\ \times 37x\times 8\times 1}}{ {100\times 12}} = \frac{ {444x}}{ {100}}

Actual sum deposited =36x

Maturity amount = Interest +Actual sum deposited

\Rightarrow {8088=}\frac{ {444x}}{ {100}} {+36x}

\Rightarrow 8088 = \frac{4044x}{100}

\Rightarrow {x=}\frac{ {8088\times 100}}{ {4044}} {=200}

Hence the monthly installment be Rs.200

(c) Number of shares =50

Face value of each share =Rs.100

Market value of each shares = Rs.132

Total face value =100 \times 50 = Rs.5000

(i) Total investment =132 \times 50 = Rs.6600

(ii) Rate of dividend =7.5\%

Annual income = \frac{ {5000\times 7.5}}{ {100}} {=Rs. \ 375}

(iii) Let extra share should he buy be x

Then total number of shares = 50+x

Total face value = Rs.100 \times (50+x)

Annual income = Rs.\frac{ {100\times }\left( {50+x}\right) {\times 7.5}}{ {100}} = \left( {50+x}\right) {\times 7.5}

(50+x) \times 7.5 = 375+150

50+x = \frac{525}{7.5} = 70

{x=70-50=20}

Hence the extra shares should be buy =20

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Question 6:

(a) Show that \sqrt {\frac{1-cos \ A}{1+cos \ A}} = \frac{sin \ A}{1+cos \ A}    [3]

(b) 16-6In the given circle with center O, \angle ABC=100^o \ \  \angle ACD=40^o and CT is a tangent to the circle at C . Find \angle ADC \ and \  \angle DCT . [3]

(c) Given below are the entries in a Saving Bank A/C pass book

Date Particular Withdrawal Deposit Balance
Feb. 8

Feb .18

April. 12

June.15

July. 8

B/F

To Self

By Cash

To Self

By Cash

 

Rs.4000

Rs.5000

Rs.2230

Rs.6000

Rs.8500

Calculate the interest for six months from February to July at 6\% \ p.a  [4]

Answers:

(a)  LHS = \sqrt{\frac{1-cos \ A}{1+cos \ A}}

= \sqrt{\frac{1-cos \ A}{1+cos \ A}} \times \sqrt{\frac{1+cos \ A}{1+cos \ A}}

= \sqrt{\frac{(1-cos \ A)(1+cos \ A)}{(1+cos \ A)(1+cos \ A)}}

= \sqrt{\frac{1-cos^2 \ A}{(1+cos \ A)^2}}

= \sqrt{\frac{sin^2 \ A}{(1+cos \ A)^2}}

= \frac{sin \ A}{1+cos \ A} = RHS. Hence Proved.

(b) Given: \angle ABC =100^o

We know that,16-7

\angle ABC+\angle ADC=180^o (sum of the opposite angles in a cyclic quadrilateral is 180^o )

100^o+\angle ADC=180^o

\angle ADC=180^o-100^o

\angle ADC=80^o

Join OA \ and \  OC , we have a isosceles \triangle OAC

{\therefore } {OA=OC} (Radius of the same circle)

{\angle AOC=2\times \angle ADC\ } (angle in a semi circle)

\Rightarrow {\angle AOC=2\times 80^o=160^o}

In \triangle AOC

{\angle AOC+\angle OAC+OCA=180^o}

{160^o+\angle OCA+\angle OCA=180^o}

Since \angle OCA = \angle OCA (in a triangle, angles opposite to equal sides are equal)

{2\angle OCA=20^o}

{\angle OCA=10^o}

{\angle OCA+\angle OCD=40^o}

{10^o+\angle OCD=40^o}

{\therefore } {\angle OCD=30^o}

Hence, \angle OCD+\angle DCT=\angle OCT 

Since {\angle } {OCT=90^o}

The tangent at a point to circle is perpendicular to the radius through the point to contact.

30^o+\angle DCT=90^o

{\angle DCT=60^o}

(c) 

Date Particular Withdrawal Deposit Balance
Feb. 8

Feb .18

April. 12

June.15

July. 8

B/F

To Self

By Cash

To Self

By Cash

Rs. 4000

Rs. 5000

Rs. 2230

Rs. 6000

Rs. 8500

Rs. 4500

Rs. 6730

Rs. 1730

Rs. 7730

Principle for the month of Feb = Rs. 4500

Principle for the month of March = Rs. 4500

Principle for the month of April = Rs. 4500

Principle for the month of May = Rs. 6730

Principle for the month of June = Rs. 1730

Principle for the month of July = Rs. 7730

Total principle from the month of Feb. to July = Rs. 29690

Time = \frac{1}{12}

Rate of interest = 6\%

Interest = \frac{P \times R \times T}{100} = \frac{29690 \times 6 \times 1}{100 \times 12} = Rs. \ 148.45

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Question 7:

(a) In \triangle ABC, A(3, 5), B(7, 8) \ and \  C(7, -10) . Find the equation of the equation of the median through A.   [3]

(b) A shopkeeper sells an article at the listed price of Rs. \ 1500 and the rate of VAT is 12% at each stage of sale. If the shopkeeper pays a VAT of Rs. \ 36 to the Government. What was the price, inclusive of Tax, at which the shopkeeper purchased the article from the wholesaler?   [3]

(c) 16-8In the figure given, from the top of a building AB=60 \ m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30^o and 60^o respectively.   [4]

Find:

(i) The horizontal distance between AB and CD

(ii) The height of the lamp paid.

Answers:

(a) Since D is midpoint of BC

The co-ordinate of D  = (\frac{7+1}{2} , \frac{8-10}{2}) = (4, -1)

Now equation of mediam AD ,

y-y_1 = \frac{y_2-y_1}{x_2-x_1} (x-x_1)

Here, x_1=3, y_1=5, x_2=4 \ and \  y_2=-1

y-5 = \frac{-1-5}{4-3} (x-3)

y-5 = \frac{-6}{-1} (x-3)

y-5 = -6x+18

y = -6x+18+5

6x+y-23 = 0

(b)  Given: List price of an article =Rs.1500 , Rate of VAT=12\%

VAT on the article =\frac{ {12}}{ {100}} {\times 1500,\ =Rs.180}

Let C.P. of this article be x , then

VAT=\frac{ 12}{ 100} \times x \ = \frac{ 12x}{ 100}

If the shopkeeper pays a VAT =Rs.36

Then, 180- \frac{ 12x}{ 100} =36

\frac{ 18000-12x}{ 100} =36

\Rightarrow 18000-12x=3600 

12x=18000-3600=14400 

x=Rs. \ 1200 

Therefore the price at which the shopkeeper purchased the article inclusive of sales tax

=1200+\frac{ 12}{ 100} {\times 1200} =1200+144 =Rs.1344

(c) Given; AB=60 m16-9

Since \angle PAC=60^o

\angle PAC=\angle BCA  (alternate angles, AP and BC are parallel)

(i)   Now in \triangle ABC 

tan \ 60^o =\frac{ AB}{ BC}

\sqrt{3}=\frac{60}{BC}

\Rightarrow {BC=}\frac{ {60}}{\sqrt{ {3}}} {\ \times }\frac{\sqrt{ {3}}}{\sqrt{ {3}}}

BC = \frac{60}{\sqrt{3}} = 20 \sqrt{3}

Hence the horizontal distance between AB \ and \  CD =20 \sqrt{3} \ m

(ii)  Let AE= x and proved above BC=20 \sqrt{3} 

Therefore BC = ED =20 \sqrt{3} 

In \triangle AED    tan \ 30^o =\frac{AE}{ED}

\frac{ 1}{\sqrt{3}} =\frac{ AE}{20 \sqrt{3}} 

AE=20 \ m 

EB=AB-AE 

EB=60-20=40 \ m 

Since EB=CD 

CD=40 \ m

Hence, the height of the lamp post =40 \ m

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Question 8:

(a) Find x \ and \  y if  \begin{bmatrix}  x & 3x \\ y & 4y \end{bmatrix} . \begin{bmatrix}  2  \\ 1 \end{bmatrix} = \begin{bmatrix}  5 \\ 12 \end{bmatrix}    [3]

(b) A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of comes recast. [3]

(c) Without solving the following quadratic equation, find the value of p for which the given equation has real and equal roots;

x^2+(p-3)x+p = 0     [4]

Answers:

(a)  Given \begin{bmatrix}  x & 3x \\ y & 4y \end{bmatrix} . \begin{bmatrix}  2  \\ 1 \end{bmatrix} = \begin{bmatrix}  5 \\ 12 \end{bmatrix}

\begin{bmatrix}  2x+3x  \\ 2y+ 4y \end{bmatrix} = \begin{bmatrix}  5 \\ 12 \end{bmatrix}

\begin{bmatrix}  5x  \\ 6y \end{bmatrix} = \begin{bmatrix}  5 \\ 12 \end{bmatrix}

Therefore 5x = 5 \Rightarrow x = 1

and 6y = 12 \Rightarrow y = 2

Hence x = 1 \ and \  y = 2

(b)  Radius of a solid sphere, r=15 \ cm

Volume of solid sphere =\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (15)^3 \ = 4500 \pi

Now, radius of right circular come =2.5 \ cm

Height h=8 \ cm

Volume of right circular cone = \frac{1}{2} \pi r^2 h = \frac{1}{2} \pi (2.5)^2 (8) =  \frac{50}{3} \pi

Therefore The number of cones = \frac{Volume \ of \ Sphere}{Volume \ of \  cone}  = \frac{4500 \pi}{\frac{50}{3} \pi} s=1$ = 270

(c)  Given equation x^2+(p-3)x + p = 0

Since roots are real and equal, b^2-4ac = 0 … … … … … (i)

Comparing the coefficient of a, b \ and \  c with equation ax^2+bx+c=0 , we get a =1 , b = p-3  \ and \  c= p

Substituting the values in (i) we get

(p-3)^2-4 \times 1 \times p = 0

p^2-10p+9 = 0

p^2-9p-p+9 = 0

p(p-9)-1(p-9)=0

(p-9)(p-1)=0 \Rightarrow p = 9 \ or \  1

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Question 9: 

(a) 16-10In the figure along side OAB is a quadrant of a circle, The radius OA=3.5 \ cm \ and \  OD=2 \ cm , Calculate the area of the shaded portion. (Take \pi = \frac{22}{7} )  [3]

(b) A box contain some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box.    [3]

(c) Find the mean of the following distribution by step deviation method:   [4]

Class Internal 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 10 6 8 12 5 9

Answers:

(a) Radius of quadrant OACB, r=3.5 \ cm

Area of quadrant OACB=\frac{1}{4} \pi r^2 = \frac{1}{4}. \frac{22}{7}. (3.5)^2 = 9.625 \ cm^2

Here, \angle AOD=90^o

base =3.5 \ cm and height =2 \ cm

Then area of \triangle AOD = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3.5 \times 2 = 3.5 \ cm^2

Area of shaded portion =Area of quadrant – Area of triangle =9.625-3.5 =6.125 \ cm^2

(b) Let the number of black balls be x , then

Total number of balls =30+x

Thus, the probability of blackballs =\frac{x}{30+x}

And the probability of white balls =\frac{30}{30+x}

Given, Probability of black ball = \frac{2}{5} . (probability \ of \ white \ ball)

Therefore \frac{x}{30+x} = \frac{2}{5}. \frac{30}{30+x}

\Rightarrow 5x = 60 \Rightarrow x = 12

Hence, the number of black balls =12 

(c) 

C.I Frequency

(f_i)

Mid-Value

(x)

 d_i = \frac{x-a}{h} (f_i. d_i)

20-30

30-40

40-50

50-60

60-70

70-80

10

6

8

12

5

9

25

35

45

55

65

75

-2

-1

0

1

2

3

-20

-6

0

12

10

27

 \Sigma f_i = 50  \Sigma f_i.d_i = 23

Here, a=45  and h=10 

Mean = a + \frac{\Sigma f_i.d_i }{\Sigma f_i}  = 45 + \frac{23}{50} . 10 = 49.6 

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Question:10

(a) Using a ruler and compasses only:

(i) Construct a \triangle ABC with the following data:

AB = 3.5 \ cm, BC = 6 \ cm \ and \  \angle ABC = 120^o

(ii) In the some diagram draw a circle with BC as diameter. Find a point P on the circumstance of the circle which is equidistant from AB and BC .

(iii) Measure \angle BCP        [4]

(b) The mark obtained by 120 students in a test are given below;

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Students 5 9 16 22 26 18 11 6 4 3

Draw an ogive for the given distribution on a graph sheet; Using suitable scale for ogive to estimate the following:

(i) The mediam

(ii) The number of students who obtained more than 75% marks in the test.

(iii) The number of students who did not pass the test if minimum marks required to pass is 40.   [6]

Answers:

(a) Steps of Construction:16-13

Using a ruler draw a line BC =6 cm

With the help of the point B , draw \angle ABC=120^o . You could do it by drawing an arc with B as the center. Then cut the arc twice using the same radius as set in the compass.

Given that the length of AB = 3.5 cm. Taking radius 3.5 cm cut BA=3.5 cm . This gives you point A.

Now join A to C with the help of a ruler.

We now need to draw a perpendicular bisector of BC. This can be done by taking a certain length in the compass and make arcs as shown in the diagram keeping point B and C as the center. Make sure that you keep the width of the compass same for all the four arcs.

The join the two points of intersection. This gives the perpendicular bisector. Draw perpendicular bisector MN of BC .

Draw a circle O as center and OC as radius.

Now draw angle bisector of  \angle ABC which intersects circle at P

Join BP and CP

Now, \angle BCP=30^o

(b) 

Marks No. of students Cumulative Frequency
0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

5

9

16

22

26

18

11

6

4

3

5

14

30

52

78

96

107

113

117

120

N=120

On the graph paper we plot the following points:

(10,5), (20,14), (30,30), (40,52), (50,78), (60,96), (70,107), (80,113), (90,117), (100,120)

16-11

(i) Mediam = (\frac{n}{2})^{th} \ term = \frac{120}{2} =60^{th} \ term 

From the graph 60th term =42

(ii) The number of students who obtained more than 75\% marks in test = 120-110 = 10

(iii) The number of students who did not pass the test if the minimum pass marks 40=52

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Question 11:

(a) In the figure given below the segment AB meets X-axis at A and Y-axis at B . The point P(-3,4) on AB divides it in the ration 2:3 find the coordinates of A and B .   [3]

(b) Using the properties of proportion solve for x , given \frac{x^4+1}{2x^2}= \frac{17}{8}    [3]

(c) A shopkeeper purchases a certain number of books for Rs. \  960 . If the cost per book was Rs. \ 8 less, the number of books that could be purchased for Rs. \ 960 would be 4 more. Write an equation, taking the original cost of each book to be Rs. \ x , and solve it to find the original cost of the books.  [4]

Answers:

(a) Let the co-ordinates of A \ and \ B \ be (x,0) \ and \ (0,y)

Since the co-ordinates of a point P(-3,4) \ on \ AB divides it in the ratio 2:3 it implies that AP:PB=2:3

By using section formula, we get

-3 = $latex  \frac{2 \times 0 + 3 \times x}{2+3} = \frac{3x}{5} &s=1$

\Rightarrow x = -5

Similarly, 4 =  \frac{2 \times y +3 \times 0}{2+3} = \frac{2y}{5}

\Rightarrow y = 10

Hence, the co-ordinate of A \ and \ B \ are (-5,0) \ and \ (0,10)

(b)  Given \frac{x^4+1}{2x^2}= \frac{17}{8}

By using componendo and dividendo, we get:

\frac{x^4+1+2x^2}{x^4+1-2x^2}= \frac{17+8}{17-8}

(\frac{x^2+1}{x^2-1})^2 = \frac{25}{9}

(\frac{x^2+1}{x^2-1})^2 = (\frac{5}{3})^2

Taking square root on both sides, we get

\frac{x^2+1}{x^2-1} = \frac{5}{3}

\Rightarrow  3x^2+3 = 5x^2-5

\Rightarrow 2x^2=8 

\Rightarrow x^2 = 4 \ or \  x = \pm 2

(c)  Given the original cost of each book be Rs.  \ x

Total cost  = Rs. \ 960

Therefore the number of books for Rs. \ 960 = \frac{960}{x}

If the cost per book was Rs. \ 8 less, (i.e. x-8 ) then

Number of books = \frac{960}{x-8}

According to question,

\frac{960}{x-8}=\frac{960}{x} + 4

\frac{960}{x-8}-\frac{960}{x} = 4

960 (\frac{x-x+8}{x(x-8)}) = 4

x^2-8x= 1920

x^2-8x-1920 = 0

x^2-48x+40x - 1920 = 0

x(x-48) +40(x-48) = 0

(x-48)(x+40) = 0 \Rightarrow x = 48 \ or \  -40 (not possible)

Hence the cost of the original book is Rs. \ 48

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