Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2012)

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 $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. Find  $A^2 - 5A + 7I$ [3]

(b) The monthly pocket money of Ravi and Sanjeev are in the ratio $5:7$. There expenditures are in the ratio $3:5$. If each saves $Rs. \ 80$ every month, find their monthly pocket money.   [3]

(c) Using the Reminder Theorem factories completely the following polynomial: $3x^2+2x^2-19x+6$    [4]

(a)  Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 - 5A + 7I$

$= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} . \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}- 5. \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7.\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$= \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-5+0 & 3-10+7 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$= 0$

(b) Let monthly pocket money be Ravi is $5x$ and Sanjeev is $7x$.

They both save $Rs. \ 80$ per month.

Therefore, their expenditure would be $(5x-80)$ and $(7x -80)$respectively.

Hence $\frac{5x-80}{7x-80} = \frac{3}{5}$

$\Rightarrow 25x - 400 = 21x -240$

$\Rightarrow x = 40 \ Rs.$

Ravi’s pocket money $= 5x = 200 \ Rs.$

Sanjeev’s pocket money $= 7x = 280 \ Rs.$

(c)  Given $f(x)=3x^3+ 2x^2-19x+6$

Try $x = 1, f(1)=3+2-19+6=0$

$f(1)=-3+2+19+6 \neq 0$, therefore (x-1) is not a factor of the given function.

Try $x = 2, f(2)=24+8-38+6=0$

Therefore  $(x-2)$ is a factor of $f(x)$

• $x-2 ) \overline {3x^3+ 2x^2-19x+6} (3x^2+8x-3$
•  $(-) \ \ \underline {3x^3-6x^2}$
•                           $8x^2-19x+6$
•                  $(-) \ \ \underline{8x^2-16x}$
•                                       ${ -3x+6}$
•                             $(-) \ \ \underline{ -3x+6}$
•                                          $\times$

To factories $3x ^2+8x-3$

$3x ^2+8x-3$ $= 3x ^2+9x-x-3$

$=3x(x+3)-1(x+3)$

$=(3x-1)(x+3)$

Hence, $3x ^3+ 2x ^2-19x+6=(x-2)(3x-1)(x+3)$

$\\$

Question 2:

(a) On what sum of money will the difference between the compound interest and simple interest for $2$ years be equal $Rs. \ 25$ if the rate of interest charged for both is $5 \%$ p.a.?   [3]

(b) $ABC$ is an is isosceles right-angled triangle with $\angle ABC = 90^o$. A semi-circle is drawn with $AC$ as the diameter. If $AB=BC=7 \ cm$ find the area of the shaded region. $( \pi = \frac{22}{7})$.    [3]

(c) Given a line segment $AB$ joining the points $A(-4, 6)$ and $B(8,-3)$ Find:

(i) the ratio in which $AB$ is divided by the $y-axis$

(ii) find the coordinates of the point of intersection.

(iii) the length of $AB$.    [4]

(a)  Let the sum be $x \ Rs.$

Simple Interest for 2 years $= x \times \frac{5}{100} \times 2 = 0.1x$

Amount Compound Interest

$A=P(1+\frac{r}{100})^n \Rightarrow A = x(1+\frac{5}{100})^2 \Rightarrow A = 1.1025x \ Rs.$

Given difference = 25 Rs.

Therefore $C.I. - S.I. = 25 \Rightarrow (1.1025x-x)-(0.1x) = 25 \Rightarrow x = 10000 \ Rs.$

(b)  $\triangle ABC$ is a right angled triangle. Therefore

$AC^2 = AB^2+BC^2 = (7)^2 + (7)^2 = 98$

$\Rightarrow AC = 7 \sqrt{2}$

Area of semi circle $= \frac{1}{2} \times \frac{22}{7} \times (\frac{7 \sqrt{2}}{2})^2 = 38.5 cm^2$

Area of $\triangle ABC = \frac{1}{2} \times 7 \times 7 = 24.5 cm^2$

Area of the shaded region = Area of the semi circle  – Area of $\triangle ABC = 38.5 - 24.5 = 14 cm^2$

(c)  Let the required ratio be $k: 1$ and the point of intersection $y-axis$   be $(0, y)$

Since $x =$ $\frac{kx_2+x_1}{k+1}$

$\Rightarrow 0 =$ $\frac{k \times (8)-4}{k+1}$

$\Rightarrow 8k-4=0$

$\Rightarrow k =$ $\frac{1}{2}$

$\Rightarrow m_1:m_2 = 1:2$

$y =$ $\frac{1 \times (-3)+2 \times (6)}{ 1+2}$ $= 3$

Therefore the point intersection is $= (0, 3)$

Length of $AB = \sqrt{(8-(-4))^2+(-3-6)^2} = \sqrt{144+81} = \sqrt{225} = 15 \ units$.

$\\$

Question 3:

(a) In the given figure $O$ is the central of the circle and $AB$ is a tangent at $B$. If $AB=15 \ cm$ and $AC =7.5 \ cm$. Calculate the radius of the circle.    [3]

(b) Evaluate without using trigonometric tables:

$cos^2 \ 26^o + cos \ 64^o.sin \ 26^o +$ $\frac{tan \ 36^o}{cot \ 54^o}$    [3]

(c) Marks obtained by $40$ students in a short assessment is given below, where $a \ and \ b$ are two missing data:

 Marks 5 6 7 8 9 No. of Students 6 A 16 13 B

If the mean of the distribution is $7.2$ find $a \ and \ b$.    [4]

(a)  Let the radius of the circle $= r$

Here we apply intercept theorem. Therefore:

$AC \times AD = AB^2$

$7 \times (7.5 + 2r ) = 15^2$

$2r = 30-7.5 = 22.5 \Rightarrow r = 11.25 \ cm$

(b)  $cos^2 \ 26^o + cos \ 64^o.sin \ 26^o +$ $\frac{tan \ 36^o}{cot \ 54^o}$

= $cos^2 \ 26^o + cos \ (90^o - 26^o).sin \ 26^o +$ $\frac{tan \ (90^o - 54^o)}{cot \ 54^o}$

= $cos^2 \ 26^o + sin^2 \ 26^o +$ $\frac{cot \ 54^o}{cot \ 54^o}$

$= 2$

(c)  Given, the total number of students $= 40$

Therefore $6 + a+ 16+ 13+ b = 40$

$\Rightarrow a + b = 5$ … … … (i)

Given mean $(\overline{x}) = \frac {\Sigma fx}{\Sigma f}$ $= 7.2$

Therefore $7.2 =$ $\frac{5 \times 6 + 6 \times a +7 \times 16 +8 \times 13 +9 \times b}{40}$

$246 + 6a + 9 b = 288$

$6a + 9 b = 42$

$2a + 3b = 14$… … … (ii)

Solving (i) and (ii) we get $a = 1$ and $b = 4$

$\\$

Question 4:

(a) Kiran deposited $Rs. 200$ per month for $36$ months in a Bank’s recurring deposit account. If the bank pays interest at the rate of $11 \%$ per annum, find the amount she gets on maturity.    [3]

(b) Two coins are tossed once; Find the probability of getting:

(i) $2$ heads

(ii) At least $1$ tail    [3]

(c) Using graph paper and taking $1 \ cm=1 \ unit$ along both $x-axis$ and $y-axis$;

(i) Plot the points $A(4,4)$ and $B (2,2)$

(ii) Reflect $A \ and \ B$ in the origin to get the images $A' \ and \ B'$respectively.

(iii) Write down the co-ordinates of $A' \ and \ B'$.

(iv) Give the geometrical name for the figure $ABA'B'$.

(v) Draw and name its lines of symmetry.    [4]

(a)  $P = Rs. \ 200, \ no \ of \ months = 36, \ r = 11\%$

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

$=200 \times 36 +200 \times \frac{36(36+1)}{2 \times 12} \times \frac{11}{100} = Rs. 8421$

(b) Let Heads – $H$ and Tails – $T$

If two coins are tossed once, then the total number of possibilities would be as shown:  Sample Space $(S) = {HH, HT, TH, TT}$

$n (S) = 4$ (i.e. there are 4 possible outcomes)

(i) Event: getting two heads $= {HH} \Rightarrow n(E) = 1$

Hence the probability $P(E) =$ $\frac{n(E)}{n(S)} = \frac{1}{4}$ $= 0.25$

(ii) Events : At least one tail  $= {HT, TH, TT} \Rightarrow n(E) = 3$

Hence the probability $P(E) =$ $\frac{n(E)}{n(S)} = \frac{3}{4}$ $= 0.75$

(c)   (i) Please refer to the graph shown below

(ii) Please refer to the graph shown below

(iii) Reflection of $A \ and \ B$ in the origin are $A'(4, -4) \ and \ B'(-2, -2)$ respectively.

(iv) Name of the geometrical figure in the graph show is Rhombus

(v) Two lines of symmetry: Both diagonal $AA' \ and \ BB'$

$\\$

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5:

(a) In the given figure, $AB$ is the diameter of a circle with center $O$. $\angle BCD = 130^o$. Find (i) $\angle DAB$ (ii) $\angle DBA$    [3]

(b) Given $A = \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}. X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$ Write (i) the order of the matrix $X$ (ii) the matrix $X$.    [3]

(c) A page from the savings Bank Account of Mr. Prateek is given below:    [4]

 Date Particular Withdrawal Deposit Balance Jan. 1st 2006 B/F – – 1,270 Jan. 7th 2006 By Cheque – 2310 3580 March 9th 2006 To Self 2000 – 1580 March 26th 2006 By Cash 6200 7780 June 10th 2006 To Cheque 4500 – 3280 July 15th 2006 By Clearing – 2630 5910 October 18th 2006 To Cheque 530 – 5380 October 27th 2006 To Self 2690 – 2690 November 3rd 2006 By Cash – 1500 4190 December 6th 2006 To Cheque 950 – 3240 December 23rd 2006 By Transfer – 2920 6160

If he receives $Rs. \ 198$ as interest on 1st January, 2007. Find the rate of interest paid by the bank.

(a)   (i) $\angle DAB = 180^o-\angle DCB$ $= 180^o-130^o = 50^o \ (ABCD$ is a cyclic quadrilateral)

(ii) In $\triangle ADB$

$\angle DAB + \angle ADB + \angle DBA = 180^o$

$\Rightarrow \angle DBA = 180^o-50^o-90^o = 140^o$

(b)   $\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

$A_{2 \times 2} . X_{p \times q} = B_{2 \times 1}$

Therefore $p = 2 \ and \ q = 1$. Hence the order of Matrix is $2 \times 1$

Let $X = \begin{bmatrix} x \\ y \end{bmatrix}$

Therefore

$\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

$\begin{bmatrix} 2x+y \\ -3x+4y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

Therefore

$2x+y = 7$

and $-3x+4y = 6$

Solving we get $x = 2 \ and \ y = 3$

Hence $X = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

(c)  Qualifying principal for various months:

 Month Principal (Rs.) January 3580 February 3580 March 1580 April 7780 May 7780 June 3280 July 3280 August 5910 September 5910 October 2690 November 4190 December 3240 Total 52800

$P = Rs. \ 52800 \ R = x\% \ and \ T= \frac{1}{12} I = Rs. \ 198$

$I = P \times R \times T \Rightarrow 52800 \times \frac{x}{100} \times \frac{1}{12} = 198 \Rightarrow x = 4.5 \%$

Therefore Rate $= 4.5\%$

$\\$

Question 6:

(a) The principal price of an article is $Rs. \ 60000$. The wholesaler allows a discount of $20 \%$ to the shopkeeper. The shopkeeper sells the article to the customer at the printed price. Sales tax (under VAT) is charged at the rate of $6 \%$ at every stage.    [4]

Find:

(i) The cost of the shopkeeper inclusive of tax

(ii) VAT paid by the shopkeeper to the Government.

(iii) The cost of the customer inclusive of tax.

(b) Solve the following inequation and represent the solution set on the number line:

$4x - 19 < \frac{3x}{5} - 2 \le \frac{-2}{5}+ x, x \in R$    [3]

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

$x^2 + 2 (m-1)x + (m+5) = 0$    [3]

(a)  Printed Price $= Rs. \ 60000$

Discounted price $= 60000 \times (1-\frac{20}{100}) = Rs. \ 48000$

Price charged by the wholesaler $= 48000 + \frac{6}{100} \times 48000 = Rs. \ 50880$

The cost to the shopkeeper inclusive of the tax $= Rs. \ 50880$

VAT paid by the shopkeeper $= Tax Charged - Tax Paid = \frac{6}{100} \times 60000 - \frac{6}{100} \times 48000 = Rs. \ 720$

The cost to the customer inclusive of the tax $= 60000+ 3600 = Rs. \ 63600$

(b)   $4x-19 < \frac{3x}{5}-2 \leq -\frac{2}{5}+x$

$4x-19 < \frac{3x}{5}-2$  or  $20x-95 < 3x-10$  or   $17x < 85$  or   $x < 5$

$\frac{3x}{5}-2 \leq -\frac{2}{5}+x$  or   $3x-10 \leq -2 +5x$  or  $-8 \leq 2x$  or  $-4 \leq x$

Therefore $\{x : -4 \leq x < 5, x \in R \}$

(c)   Comparing $x^2+2(m-1)x+(m+5)=0$ with $ax^2+bx+c=0$, we get $a = 1, b = 2(m-1) \ and \ c =(m+5)$

For roots to be equal, we should have $b^2-4ac = 0$

$(2(m-1))^2-4(1)(m+5)=0$

$4(m^2+1-2m)-4(m+5)=0$

$4m^2+4-8m-4m-20=0$

$4m^2-12m-16=0$

$m^2-3m-4=0$

$m^2-4m+m-4=0$

$m(m-4)+(m-4)=0$

$(m-4)(m+1)=0 \Rightarrow m = 4, -1$

$\\$

Question 7:

(a) A hollow sphere of internal and external radii $6 \ cm$ and $8 \ cm$ respectively is melted and recast into small cones of base radius $2 \ cm$ and height $8 \ cm$. Fins the number of cones.   [3]

(b) Solve the following equation and give your answer correct to $3$ significant figures: [3]

$5x^2-3x-4=0$

(c) As observed from the top of a $80 \ m$ tall lighthouse, the angles of depression of two ships on the same side of the light house in horizontal line with its base are $30^o$ and $40^o$ Find the distance between the two ships. Give your answer correct to the nearest meter.    [4]

(a)   Sphere: Internal radius $= 6 \ cm$, External radius $= 8 \ cm$

Cone: Radius $= 2 \ cm$, Height $= 8 \ cm$

$\frac{4}{3} \times \pi \times (8)^3 - \frac{4}{3} \times \pi \times (6)^3 = n \times \frac{1}{3} \times \pi \times (2)^2 \times 8$

$\Rightarrow n =$ $\frac{4 \times (8^3-6^3)}{2^2 \times 8}$ $= 37$

(b)  Given : $5x^2 - 3x - 4 = 0$

Comparing this expression with $ax^2+bx+c = 0$, we get  $a=5, b=-3 \ and \ C=-4$

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

$x =$ $\frac{3 \pm \sqrt{(-3)^2 - 4(5)(-4)}}{2(5)}$

$x =$ $\frac{3 \pm \sqrt{9 + 80}}{10}$

$x =$ $\frac{3 \pm \sqrt{89}}{10}$

$x =$ $\frac{3 \pm 9.43}{10}$

Therefore  $x =$ $\frac{3 + 9.43}{10}$ or $x =$ $\frac{3 - 9.43}{10}$

$x=1.243,\ -0.643$

(c)  In $\triangle ABC \ tan \ 50^o =$ $\frac{BC}{80}$

$\Rightarrow BC = 80 \times 1.1918$

$\Rightarrow BC = 95.34 \ m$

In $\triangle ABD, tan \ 60^o =$ $\frac{BD}{80}$

$\Rightarrow BD = 80 \sqrt{3} = 138.56 \ m$

Therefore $CD = BD - BC = 138.56 - 95.34 = 43.2 \ m$

$\\$

Question 8:

(a) A man invests $Rs. \ 9,600$ on $Rs. \ 100$ shares at $Rs. \ 80$.  If the company pays him $18 \ %$ dividend. Find:

(i) The number of shares he buys

(ii) His total dividend.

(iii) His percentage return on the shares.    [3]

(b)  In the given figure $\triangle ABC$ and $\triangle AMP$ are right angled at $B \ and \ M$ respectively.  Given $AC= 10 \ cm, AP = 15 \ cm$ and $PM = 12 \ cm$.

(i) Prove $\triangle ABC \sim \triangle AMP$

(ii) Find $AB \ and \ BC$    [3]

(c) If $x =$ $\frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt{a+1} - \sqrt{a-1}}$ using properties of proportion show that $x^2-2ax+1=0$    [4]

(a)  Investment $= Rs. \ 9600$ , Dividend $= 18\%$

Nominal Value of the share $= Rs. \ 100$

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

(i) Number of shares $=$ $\frac{Investment}{Market \ Value \ of \ the \ share} = \frac{9600}{80}$ $= 120$

(ii) Dividend $= 120 \times 100 \times \frac{18}{100} = Rs. \ 2160$

(iii) $\% Return$ $=$ $\frac{Dividend \ earned}{Investment \ made} = \frac{2160}{9600}$ $\times 100 = 22.5\%$

(b)  In  $\triangle ABC \ and \ \triangle AMP$

$\angle BAC = \angle PAM$ (common angle)

$\angle ABC =\angle PMA$ (right angles)

Therefore $\triangle ABC \sim \triangle AMP$ (AAA postulate)

$AM = \sqrt{AP^2-PM^2} = \sqrt{15^2-12^2} = \sqrt{121} = 11$

Since $\triangle ABC \sim \triangle AMP$

$\frac{AP}{AM}= \frac{BC}{PM}= \frac{AC}{AP}$

Given AC=10 cm, AP = 15 cm and PM= 12 cm

$\Rightarrow AB = \frac{10}{15} \times 11 = 7.33 \ cm$

$\Rightarrow BC = \frac{10}{15} \times 12 = 8 \ cm$

(c)  Given $x=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}$

Applying componendo and dividendo

$\frac{x+1}{x-1}=\frac{(\sqrt{a+1}+\sqrt{a-1})+(\sqrt{a+1}-\sqrt{a-1})}{(\sqrt{a+1}+\sqrt{a-1})-(\sqrt{a+1}-\sqrt{a-1})}$

Simplify

$\frac{x+1}{x-1}=\frac{\sqrt{a+1}}{\sqrt{a-1}}$

Now square both sides

$\frac{x^2+1+2x}{x^2-2x+1}=\frac{a+1}{a-1}$

Simplifying

$x^2+1 = 2ax$

or $x^2-2ax+1 = 0$

$\\$

Question 9:

(a) The line through $A (-2, 3)$ and $B (4, b)$ is perpendicular to the line $2x - 4y = 5$. Find the value of \$latex b.   [3]

(b) Prove that $\frac{tan^2 \ \theta}{(sec \ \theta -1)^2} = \frac{1 + cos \ \theta}{1 + cos \ \theta}$   [3]

(c) A car covers a distance of 400 km at a certain speed. Had the speed been 12 km per hours more, the line taken for the journey would have been 1 hour and 40 minutes less. Find the original speed of the care.   [4]

(a)  Slope of $AB = \frac{b-3}{4-(-2)} = \frac{b-3}{6}$

Given equation is $2x - 4y = 5$

$\Rightarrow 4y=2x-5$

$\Rightarrow y = \frac{1}{2}x-\frac{5}{4}$

$\Rightarrow Slope (m_2) = \frac{1}{2}$

Since they are perpendicular, $m_1 . m_2 =-1$

$\Rightarrow \frac{b-3}{6}. \frac{1}{2}=-1$

$\Rightarrow b=-9$

(b)  RHS $=$ $\frac{tan^2 \ \theta}{(sec \ \theta - 1)^2}$

$=$ $\frac{sin^2 \ \theta}{cos^2 \ \theta} \times \frac{cos^2 \ \theta}{(1-cos \ \theta)^2}$

$=$ $\frac{(1-cos \ \theta)(1 + cos \ \theta)}{(1-cos \ \theta)^2}$

$=$ $\frac{1+ cos \ \theta}{1- cos \ \theta} =$ LHS. Hence proved.

(c)  Let the speed of the car be $x \ km/hr$

Therefore $\frac{400}{x} = \frac{400}{x+12} + \frac{100}{60}$

$\frac{2400}{x} = \frac{2400}{x+12} +10$

$2400x+28800=10x^2+2520x$

$10x^2-120x-28800=0 \Rightarrow x = -60\ or \ 48 (ruled \ out)$

Therefore speed of the car is 48 km/hr.

$\\$

Question 10:

(a) Construct a $\triangle ABC$ in which base $BC =6 \ cm, AB=5.5 \ cm$ and $\angle ABC=120^o$ [3]

(i) Construct a circle circumscribing the $\triangle ABC$

(ii) Draw a cyclic quadrilateral $ABCD$ so that $D$ is equidistant from $B \ and \ C$.    [4]

(b) The following distribution represents the height of 160 students of a school

 Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 Students 12 20 30 38 24 16 12 8

Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm=20 students on the other axis. Using the graph, determine:

(i) The medium height

(ii) The interquartile range

(iii) The number of students whose height is above 172 cm.     [6]

(a) (i) Following Step to Constructions:

1. First, using a ruler, draw a line segment $BC = 6 \ cm$
2. Construct $\angle CBE=120^o$ This you can do it by using a compass. Draw an arc with B as the center. The using the same arc and  intersection point ‘x”, draw two arcs one at ‘y’ and then one at ‘z’. Craw a line through B and ‘z’. You will get 120^o angle.
3. The cut $BA = 5.5 \ cm$ from $BP$. This you can do by setting the compass for a width of 5.5 cm.
4. Then join $A \ to \ C$. So you get the \triangle ABC.
5. Now construct perpendicular bisectors of $AB$ and $BC$ intersecting at $O$. Join $AO$.
6. Taking O as the center and $OA$ as radius draw a circle, passing through $A, B$, and $C$

(ii)

1. Extend the right bisector of $BC$ intersecting the circle at $D$.
2. Join $A \ to\ D$ and $C \ to \ D$
3. $ABCD$ is required cyclic quadrilateral.

(b) Following table:

 Height F c.f. 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 12 20 30 38 24 16 12 8 12 32 62 100 124 140 152 160

(i) Mean $= 157.3$

(ii) Interquartile range $= Q_3 - Q_1 = 163.5 - 152 = 11.5$

(iii) No. of students above $172 \ cm = 160-144 = 16$

$\\$

Question 11:

(a) In $\triangle PQR, PQ = 24 \ cm, QR = 7 \ cm$ and $\angle PQR = 90^o$. Find the radius of the inscribed circle.    [3]

(b) Find the mode and medium of the following frequency distribution:    [3]

 X 10 11 12 13 14 15 f 1 4 7 5 9 3

(c) The line through $P(5,3)$ intersects $y-axis$ at $Q$. [4]

(i) Write the slope of the line

(ii) Write the equation of the line

(iii) Find the Co-ordinates of $Q$    [4]

(a)  Given $\triangle PQR$ is a right triangle. Therefore using Pythagoras theorem:

$PR^2 = PQ^2 + QR^2 = (24)^2 + (7)^2 = 625$

$\Rightarrow PR = \sqrt{625} =25$

Draw $OC \ and \ OB$ as shown. They will be perpendicular to the sides of the triangle.

$AOBQ$ is a square since the angels are $90^o$ and $QA = QB$ (Since the tangent to a circle from an exterior point are equal in length).

Therefore $QA = QB = x$

$AR = 7-x = RC$ (Since the tangent to a circle from an exterior point are equal in length).

$BP = 12-x = PC$ (Since the tangent to a circle from an exterior point are equal in length).

Since $PC +RC = PR$

$\Rightarrow 7-x+12-x = 25$

$\Rightarrow x = 3 \ cm$

(b)  Mode is the value of the highest frequency.

Therefore Mode $= 14$

For Median, first write the data in ascending order as follows:

$10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, \\ 13, 13, 13, 13,13, 14, 14, 14,14, 14, 14, 14, 14, 14,15, 15, 15$.

Since Median is the middle most value.

Median $= (\frac{N+1}{2})^{th} observation = (\frac{29+1}{2})^{th} observation = 15^{th} \ observation = 13$

(c)  Given points $P(-2, 0) \ and \ Q(0, y)$

Slope $= m = tan \ 45^o = 1$

Equation of line:

$y-3=1(x-5)$

$\Rightarrow y = x-2$

When $x=0, y = -2$

Hence the co-ordinates of $Q = (0, -2)$

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