Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2020)

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) Solve the following quadratic equation:

$x^2 - 7x + 3 = 0$

(b)  Given $A = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix}$

If $A^2= 3 I$, where $I$ is the identity matrix of Order $2$, find $x$ and $y$.           [3]

(c)   Using ruler and compass construct a triangle $ABC$ where $AB = 3$ cm, $BC = 4$ cm and $\angle ABC = 90^{\circ}$. Hence construct a circle circumscribing the triangle $ABC$. Measure and write down the radius of the circle.          [4]

(a) Given $x^2 - 7x + 3 = 0$

Comparing it with $ax^2 + bx + c = 0$ we get , $a = 1, b = -7$ and $c = 3$

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

$x =$ $\frac{7 \pm \sqrt{49-12}}{2}$

$\Rightarrow x=$ $\frac{7 \pm \sqrt{37}}{2}$

$\Rightarrow x =$ $\frac{7 \pm 6.08}{2}$

Therefore $x =$ $\frac{7+6.08}{2}$ $= 6.54$ or

$x =$ $\frac{7 - 6.08}{2}$ $= 0.46$

Hence roots are $6.54$ and $0.46$

(b) Given: $A = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix}$ and $A^2 = 3I$

$A^2 = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix} \times \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix}$

$\Rightarrow A^2 = \begin{bmatrix} x^2 + 3y & 3x+9 \\ xy+3y & 3y+9 \end{bmatrix}$

$3I = 3 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

Comparing we get

$\begin{bmatrix} x^2 + 3y & 3x+9 \\ xy+3y & 3y+9 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$\Rightarrow 3x + 9 = 0 \Rightarrow x = - 3$

Similarly, $3y + 9 = 3 \Rightarrow y = - 2$

(c)

$BC = 4$ cm and $AB = 3$ cm

Radius of the circle $= 2.5$ cm

$\\$

Question 2:

(a) Use factor theorem to factorize $6x^3 + 17x^2 + 4x - 12$ completely.     [3]

(b) Solve the following in-equation and represent the solution set on the number line.

$\frac{3x}{5}$ $< x+4 \leq$ $\frac{x}{5}$ $, x \in R$                                                                                            [3]

(c)  Draw a histogram for the given data using graph paper.                 [4]

 Weekly Wages (in Rs) No. of People 3000-4000 4 4000-5000 9 5000-6000 18 6000-7000 6 7000-8000 7 8000-9000 2 9000-10000 4

(a)  Let $f(x) = 6x^3 + 17x^2 + 4x - 12$

By trial and error, let $x = -2$

Therefore $f(-2) = 6(-8) + 17(4) + 4(-2) - 12 = -48 + 68 - 8 - 12 = 0$

Hence we can say that $(x+2)$ is a factor of $f(x)$

Now,

$\begin{array}{r l l} x+2 ) & \overline{6x^3+ 17x^2 + 4x - 12} & (6x^2+5x-6 \\ (-) & 6x^3+12x^2 & \\ \hline & \hspace{1.0cm} 5x^2 + 4x & \\ (-) & \hspace{1.0cm} 5x^2 + 10x & \\ \hline & \hspace{2.0cm} -6x-12 & \\ (-) & \hspace{2.0cm} -6x - 12 & \\ \hline & \hspace{3.0cm} 0 & \end{array}$

Now factorize

$6x^2+5x-6 = 6x^2 +9x - 4x - 6 = 3x( 2x+3) - 2 ( 2x+3) = (3x-2)(2x+3)$

Therefore complete factorization is $f(x) = (x+2)(3x-4)(2x+3)$

(b) $\frac{3x}{5}$ $< x+4 \leq$ $\frac{x}{5}$ $, x \in R$

First solve:  $\frac{3x}{5}$ $+ 2 < x+4$

$\Rightarrow 3x + 10 < 5x + 20$

$\Rightarrow -10 < 2x$

$\Rightarrow -5 < x$

Also solve $x+4 \leq$ $\frac{x}{2}$ $+ 5$

$\Rightarrow 2x + 8 \leq x + 10$

$\Rightarrow x \leq 2$

$\therefore -5 < x \leq 2, x \in R$

Hence the solution set $S = \{ x : x \in R, -5 < x \leq 2 \}$

(c)

$\\$

Question 3:

(a) In the figure given below, $O$ is the center of the circle and $AB$ is a diameter, if $AC = BD$ and $\angle ADC = 72^{\circ}$, find              [3]

i) $\angle ABC$

ii) $\angle BAD$

iii) $\angle ABD$

(b) Prove that

$\frac{\sin A}{1 + \cot A}$ $-$ $\frac{\cos A}{1+\tan A}$ $= \sin A - \cos A$                                            [3]

(c)  In what ratio is the line joining $P(5, 3)$ and $Q( -5, 3)$ is divided by y-axis. Also find the coordinates of the point of intersection.     [4]

(a)

i) We know, The angle which an arc of a circle subtends at the center is double that which it subtends on any point of the part of the circumference.

Therefore $\angle ABC =$ $\frac{1}{2}$ $\angle AOC =$ $\frac{1}{2}$ $(72) = 36^{\circ}$

ii) Given $AC = BD$

The chord of the same length subtend the same angle

$\therefore \angle BAD = 36^{\circ}$

iii) $\angle ABD = 90^{\circ}$  (semi circular angles are $90^{\circ}$)

In $\triangle ABD, \angle ABD = 180^{\circ} - 36^{\circ} - 90^{\circ} = 54^{\circ}$

(b)

LHS $=$ $\frac{\sin A}{1 + \cot A}$ $-$ $\frac{\cos A}{1 + \tan A}$

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

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

$=$ $\frac{(\sin A - \cos A )(\sin A + \cos A )}{\sin A + \cos A }$

$= \sin A - \cos A =$ RHS Hence proved.

(c) Let the point of intersection be $(0, b)$

Let the ratio of  in which the line segment gets divided be $k:1$

Using section formula

$x =$ $\frac{mx_2+nx_1}{m+n}$  and $y =$ $\frac{my_2+ny_1}{m+n}$

$\therefore 0 =$ $\frac{-5k+5}{k+1}$ $\Rightarrow -5k + 5 = 0 \Rightarrow k = 1$

Hence the ratio is $1:1$

Therefore $b =$ $\frac{3+3}{1+1}$ $= 3$

Therefore the coordinates of the point of intersection is $(0, 3)$

$\\$

Question 4:

(a) A solid spherical ball of radius $6$ cm is melted and recasted in $64$ identical spherical marbles. Find the radius of each marble.     [3]

(b) Each of the letters of the word $'AUTHORIZES'$ is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is

i) a vowel

ii) One of the first $9$ letters of the English alphabet which appears in a given word

iii) One of the last $9$ letters of the English alphabet which appears in a given word          [3]

(c)  Mr. Bedi visits the market and buys the following articles:

Medicines costing Rs. $950$, GST @ $5\%$

A pair of shoes costing Rs. $3000$, GST @ $18\%$

A laptop bag costing Rs $1000$ with a discount or $30\%$, GST @ $18\%$

i) Calculate the total amount of GST paid

ii) The total bill amount including GST paid by Mr. Bedi      [4]

(a) Let $R$ be the radius of spherical ball and $r$ be the radius of the spherical marble

Volume of spherical ball $=$ $\frac{4}{3}$ $\pi R^3 =$ $\frac{4}{3}$ $\pi (6)^3$

Volume of spherical marble $=$ $\frac{4}{3}$ $\pi r^3$

$\therefore$ $\frac{4}{3}$ $\pi (6)^3 = 64 \times$ $\frac{4}{3}$ $\pi r^3$

$r^3 =$ $\frac{6^3}{64}$ $=$ $\frac{6^3}{4^3}$

$\Rightarrow r =$ $\frac{6}{4}$ $=$ $\frac{3}{2}$ cm

(b) First 9 letters: $A, B, C, D, E, F, G, H, I$

Last 9 letters: $R,S, T, U, V, W, X, Y, Z$

Total outcomes $= 10$

i) No. of vowels $= 5$

Therefore probability $P(Vowels) =$ $\frac{5}{10}$ $=$ $\frac{1}{2}$

ii) There are $4$ probable outcomes

Therefore probability $P( \ first \ 9 \ letters\ ) =$ $\frac{4}{10}$ $=$ $\frac{2}{5}$

iii) There are $5$ probable outcomes

Therefore probability $P( \ last \ 9 \ letters \ ) =$ $\frac{5}{10}$ $=$ $\frac{1}{2}$

(c)

 Article Cost ( Rs.) Final Cost (Rs.) GST Rate GST ( Rs.) Final Price (Rs.) Medicines 950 950 5% 47.50 997.50 Shoes 3000 3000 18% 540 3540 Laptop Bag 1000 @ 30% discount 700 18% 126 829 Total (Rs.) 713.50 5363.50

Therefore

i) Calculate the total amount of GST paid $= 713.50$ Rs.

ii) The total bill amount including GST paid by Mr. Bedi $= 5363.50$ Rs.

$\\$

SECTION B [40 Marks]

(Attempt any four questions from this Section.)

Question 5:

(a) A company with $500$ shares of nominal value Rs. $120$ declares and annual dividend of $15\%$. Calculate:

i) the total amount of dividend paid by the company

ii) annual income of Mr. Sharma who holds $80$ shares of the company.

If the return percent of Mr. Sharma for his shares is $10\%$, find the market value of each share.              [3]

(b) The mean of the following data is $16$. Calculate the value of $f$  [3]

 Marks 5 10 15 20 25 No. of Students 3 7 $f$ 9 6

(c) The $4^{th}, 6^{th}$ and the last term of a geometric progression are $10, 40$ and $640$ respectively. If the common ratio is positive, find the first term, common ration and the number of terms of the series.     [4]

(a) Number of shares $= 500$

Nominal Value $= 120$ Rs.

Dividend $= 15\%$

i) Dividend $= n \times NV \times$ $\frac{div \%}{100}$ $= 500 \times 120 \times$ $\frac{15}{100}$ $= 9000$ Rs.

ii) $n = 80$ shares

Dividend $= 80 \times 120 \times$ $\frac{5}{100}$ $= 1440$ Rs.

Return $\% =$ $\frac{Annual \ Income }{Investment}$ $\times 100$

$\Rightarrow 10 =$ $\frac{1440}{I}$ $\times 100$

$\Rightarrow I =$ $\frac{1440 \times 100}{10}$ $= 14400$ Rs.

Therefore Market Value $=$ $\frac{14400}{80}$ $= 180$ Rs.

(b)

 Marks $( x)$ No. of Students $(f)$ $fx$ 5 3 15 10 7 70 15 $f$ $15f$ 20 9 180 25 6 150 $\Sigma f = 25 + f$ $\Sigma fx = 415 + 15f$

Given: Mean $= 16$

We know $\overline{x} =$ $\frac{\Sigma fx}{\Sigma f}$

$\Rightarrow 16 =$ $\frac{415+15f}{25+f}$

$\Rightarrow 400 + 16f = 415 + 15f$

$\Rightarrow f = 15$

(c) $4^{th}$ term $= 10$          $6^{th}$ term $= 40$          Last term $= 640$

We know, $t_n = ar^{n-1}$

$\therefore 10 = a r^{4-1}$     $\Rightarrow 10 = ar^3$   … … … … … i)

Similarly,  $40 = a r^{6-1}$     $\Rightarrow 40 = ar^5$     … … … … … ii)

Dividing ii) by i) we get

$\frac{ar^5}{ar^3}$ $=$ $\frac{40}{10}$

$\Rightarrow r^2 = 4$    $\Rightarrow r = 2$

Using i) we get $10 = a (2)^3$

$\Rightarrow a =$ $\frac{10}{8}$ $=$ $\frac{5}{4}$

We know that $t_n = a r^{n-1}$

$\therefore 640 =$ $\frac{5}{4}$ $2^{n-1}$

$\Rightarrow 2^{n-1} = \frac{640 \times 4}{5}$

$\Rightarrow 2^{n-1} = 512$

$\Rightarrow 2^{n-1} = 2^9$

$\Rightarrow n = 10$

$\\$

Question 6:

(a) If $A = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}$. Find:   $A^2 - 2AB + B^2$      [3]

(b) In the given figure $AB = 9$ cm, $PA = 7.5$ cm and $PC = 5$ cm

Chord $AD$ and $BC$ intersect at $P$.

i) Prove that $\triangle PAB \sim \triangle PCD$

ii) Find the length of $CD$

iii) Find area of $\triangle PAB :$ area of $\triangle PCD$    [3]

(c) From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be $45^{\circ}$ and $60^{\circ}$ respectively. If the height of the tower is $20$ m, Find:

i) the height of the cliff

ii) the distance between the cliff and the tower.    [4]

(a) Given

$A = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix}$

$A^2 = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix}$

$B^2 = \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix}$

$AB = \begin{bmatrix} 3 & 0 \\ 5 & 1 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix}$

$\therefore A^2 - 2AB + B^2 = \begin{bmatrix} 9 & 0 \\ 20 & 1 \end{bmatrix} - 2 \begin{bmatrix} -12 & 6 \\ -19 & 10 \end{bmatrix} + \begin{bmatrix} 18 & -8 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 51 & -20 \\ 54 & -17 \end{bmatrix}$

(b)

i) Consider $\triangle PAB$ and $\triangle PCD$

$\angle BPA = \angle DPC$ ( Vertically opposite angles)

$\angle ABC = \angle ADC$ (angles subtended by an arc on the circumference are equal)

$\therefore \triangle PAB \sim \triangle PCD$  (By AA similarity criterion)

ii) Since $\triangle PAB \sim \triangle PCD$

$\Rightarrow$ $\frac{PA}{PC}$ $=$ $\frac{AB}{CD}$ $=$ $\frac{PB}{PD}$

$\Rightarrow$ $\frac{7.5}{5}$ $=$ $\frac{9}{CD}$

$\Rightarrow CD =$ $\frac{5 \times 9}{7.5}$ $= 6$ cm

iii) $\frac{ar. of \triangle PAB}{ar. of \triangle PCD}$ $=$ $\frac{9^2}{6^2}$ $= \frac{9}{4}$

c)

In $\triangle ABC$

$\tan 60^{\circ} =$ $\frac{h}{x}$     $\Rightarrow \sqrt{3} =$ $\frac{h}{x}$     $\Rightarrow h = \sqrt{3} x$    … … … … … i)

In $\triangle CDE$

$\tan 45^{\circ} =$ $\frac{h-20}{x}$     $\Rightarrow 1 =$ $\frac{h-20}{x}$     $\Rightarrow x = h - 20$     … … … … … ii)

Substituting $x = \sqrt{3}x - 20$

$\Rightarrow ( \sqrt{3} -1 ) x = 20$

$\Rightarrow x =$ $\frac{20}{\sqrt{3} -1}$ $\times$ $\frac{\sqrt{3}+1}{\sqrt{3}+1}$ $= 10 ( 1.732+1) = 27.32$ m

$\therefore h = \sqrt{3} x = 1.732 \times 27.32 = 47.32$ m

i) the height of the cliff $= 47.32$ m

ii) the distance between the cliff and the tower $= 27.32$ m

$\\$

Question 7:

(a) Find the value of $'p'$ is the lines $5x - 3y + 2 = 0$ and $6x - py + 7 = 0$ are perpendicular to each other. Hence find the equation of a line passing through $(-2, -1)$ and parallel to $6x - py + 7 = 0$.     [3]

(b) Using properties of proportion find $x:y$ given $\frac{x^2 + 2x}{2x + 4}$ $=$ $\frac{y^2 + 3y}{3y + 9}$      [3]

(c) In the given figure, $TP$ and $TQ$ are two tangents to the circle with center $O$, touching at $A$ and $C$ respectively. If $\angle BCQ = 55^{\circ}$ and $\angle BAP = 60^{\circ}$, find:

i) $\angle OBA$ and $\angle OBC$

ii) $\angle AOC$

iii) $\angle ATC$                 [4]

(a) Given $5x - 3y + 2 = 0 \Rightarrow 3y = 5x + 2 \Rightarrow y =$ $\frac{5}{3}$ $x +$ $\frac{2}{3}$

Therefore slope $m_1 =$ $\frac{5}{3}$

Similarly, $6x - py + 7 = 0 \Rightarrow py = 6x + 7 \Rightarrow y =$ $\frac{6}{p}$ $x +$ $\frac{7}{p}$

Therefore slope $m_2 =$ $\frac{6}{p}$

We know $m_1 m_2 = - 1$

$\Rightarrow$ $\frac{5}{7}$ $\times$ $\frac{6}{p}$ $= -1 \Rightarrow p = -10$

Therefore slope $m_2 =$ $\frac{6}{-10}$ $= -$ $\frac{3}{5}$

Therefore equation of line

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

$\Rightarrow 5y + 5 = -3x - 6$

$\Rightarrow 3x + 5y + 11 = 0$

(b) $\frac{x^2 + 2x}{2x + 4}$ $=$ $\frac{y^2 + 3y}{3y + 9}$

Applying componendo and dividendo

$\frac{x^2 + 2x + 2x + 4}{x^2 + 2x - 2x - 4}$ $=$ $\frac{y^2 + 3y + 3y + 9}{y^2 + 3y -3y - 9}$

$\Rightarrow$ $\frac{x^2 + 4x + 4}{x^2 - 4}$ $=$ $\frac{y^2 + 6y + 9}{y^2 - 9}$

$\Rightarrow$ $\frac{(x+2)^2}{(x+2)(x-2)}$ $=$ $\frac{(y+3)^2}{(y+3)(y-3)}$

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

Applying componendo and dividendo

$\frac{x+2 + x - 2}{x+2 -x+2}$ $=$ $\frac{y+3 + y - 3}{y+3-y+3}$

$\Rightarrow$ $\frac{2x}{4}$ $=$ $\frac{2y}{6}$

$\Rightarrow$ $\frac{x}{y}$ $=$ $\frac{2}{3}$

$\therefore x : y = 2:3$

(c)

i) $\angle OAT = 90^{\circ}$

$\therefore \angle OAB = 90^{\circ} - 60^{\circ} = 30^{\circ}$

$\angle OCT = 90^{\circ}$

$\therefore \angle OCB = 90^{\circ} - 55^{\circ} = 35^{\circ}$

Since $OB = OC, \triangle OBC$ is an isosceles triangle

$\therefore \angle OBC = 35^{\circ}$

Similarly, $OA = OB, \triangle AOB$ is an isosceles triangle

$\therefore \angle OBA = 30^{\circ}$

ii) $\angle ABC = 30^{\circ} + 35^{\circ} = 65^{\circ}$

Angle subtended by an arc at the center is twice that subtended on the circumference.

$\therefore \angle AOC = 2\angle ABC = 2 \times 65^{\circ} = 130^{\circ}$

iii) $AOCT$ is a quadrilateral. We know that the sum of all the internal angles of a quadrilateral is $360^{\circ}$

Hence, $90^{\circ} + 130^{\circ} + 90^{\circ} + \angle ATC = 360^{\circ}$

$\Rightarrow \angle ATC = 50^{\circ}$

$\\$

Question 8:

(a) What must be added to the polynomial $2x^3 - 3x^2 - 8x$ , so that it leaves a remainder $10$ when divided by $2x+1$?      [3]

(b) Mr. Sonu has a recurring deposit account and deposits Rs. $750$ per month for $2$ years. If he gets Rs. $19125$ at the time of maturity, find the rate of interest.     [3]

(c) Use a graph paper for this.                             [4]

Take $1$ cm $= 1$ unit on both x and y axes.

i) Plot the following points on your graph sheet $A( -4, 0), B ( -3, 2), C ( 0, 4), D ( 4, 1)$ and $E(7, 3)$

ii) Reflect point $B, C, D$ and $E$ on the x axis and name then $B', C', D'$and $E'$ respectively.

iii) Join the points $A, B, C, D, E, E', D', C', B'$ and $A$ in order. Name the closed figure formed.

(a) Given : $2x^3 - 3x^2 - 8x$

When divided by $2x+1 \Rightarrow x = -$ $\frac{1}{2}$

$f(-$ $\frac{1}{2}$ $) = 2 ( -$ $\frac{1}{8}$ $) - 3 ($ $\frac{1}{4}$ $) - 8 (-$ $\frac{1}{2}$ $) + k$

$\Rightarrow 10 = -$ $\frac{1}{4}$ $-$ $\frac{3}{4}$ $+ 4 + k$

$\Rightarrow 10 = -1 + 4 + k$

$\Rightarrow k = 7$

(b) Given $P = 750$ Rs.    $n = 2$ years $= 24$ months    $MV = 19125$ Rs.

Interest $= MV - ( P \times n) = 19125 - 750 \times 24 = 1125$ Rs.

Interest $=$ $\frac{n(n+1)}{2}$ $\times$ $\frac{750}{12}$ $\times$ $\frac{r}{100}$

$1125 =$ $\frac{24 \times 25}{2}$ $\times$ $\frac{750}{12}$ $\times$ $\frac{r}{100}$

$\Rightarrow r =$ $\frac{45 \times 100}{750}$ $= 6\%$

(c)

The figure formed is a nonagon (nine edges). It is a FISH.

$\\$

Question 9:

(a) $40$ students enter for a game of shot put competition. The distance thrown ( in meters) is recorded below.                   [6]

 Distance in m 12-13 13-14 14-15 15-16 16-17 17-18 18-19 Number of Students 3 9 12 9 4 2 1

Use a graph paper to draw an ogive for the above distribution.

Use a scale of $2$ cm $= 1$ m on x-axis and $2$ cm $= 5$ students on the other axis.

Hence using your graph paper find:

i) the median

ii) Upper Quartile

iii) Number of students who cover a distance which is above $16$ $\frac{1}{2}$ m

(b) If $x =$ $\frac{\sqrt{2a+1}+\sqrt{2a-1}}{\sqrt{2a+1}- \sqrt{2a-1}}$, prove that $x^2 - 4ax + 1 = 0$       [4]

(a)

i) the median $= 14.7$

ii) Upper Quartile $= 15.6$

iii) Number of students who cover a distance which is above $16$ $\frac{1}{2}$ m $= 5$ students

(b)

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

Applying componendo and dividendo

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

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

Squaring both sides

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

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

Applying componendo and dividendo

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

$\Rightarrow$ $\frac{2(x^2+1)}{4x}$ $=$ $\frac{4a}{2}$

$\Rightarrow$ $\frac{x^2+1}{2x}$ $=$ $\frac{4a}{2}$

$x^2 + 1 = 4ax$

$x^2 - 4ax + 1 = 0$. Hence proved.

$\\$

Question 10:

(a) If the $6^{th}$ terms of an AP is equal to four time its first term and the sum of first six terms is $75$, find the first term and the common difference.  [3]

(b) The difference of two natural numbers us $7$ and their product is $450$. Find the numbers.      [3]

(c) Use ruler and compass for this question. Construct a circle of radius $4.5$ cm. Draw a chord $AB = 6$ cm.

i) Find the locus of points equidistant from $A$ and $B$. Mark the point where it meets the circle as $D$.

ii) Join $AD$ and find the locus of points which are equidistant from $AD$ and $AB$. mark the point where it meets the circle as $C$.

Join $BC$ and $CD$. Measure and write down the length of side $CD$ of quadrilateral $ABCD$.     [4]

(a) $6^{th} term = 4a$ and $S_6 = 75$

We know that $t_n = a + ( n-1) d$

$\Rightarrow 4a = a + 5 d$    $\Rightarrow 3a = 5d$   … … … … … i)

Also $S_n =$ $\frac{n}{2}$ $[ 2a + ( n-1) d ]$

$\Rightarrow 75 =$ $\frac{6}{2}$ $[ 2a + 5d ]$     $\Rightarrow 75 = 6a + 15d$    … … … … … ii)

Solving i) and ii)

$75 = 6a + 9a \Rightarrow a = 5$

Substituting in i)

$5d = 3a \Rightarrow d =$ $\frac{3}{5}$ $\times 5 = 3$

Therefore first term $(a) = 3$ and common difference $(d) = 3$

(b) Let the two natural numbers be $x$ and $y$

Therefore $x - y = 7$

Also $xy = 450 \Rightarrow y =$ $\frac{450}{x}$

Substituting $x -$ $\frac{450}{x}$ $= 7$

$\Rightarrow x^2 - 7x - 450 = 0$

$\Rightarrow x^2 - 25x + 18x - 450 = 0$

$\Rightarrow x( x - 25) + 18 ( x - 25) = 0$

$\Rightarrow (x-25) ( x + 18) = 0$

$\Rightarrow x = 25 \ or \ x = -18$ ( not a natural number).

Therefore $x = 25$. Substituting we get $y = 25- 7 = 18$

(c)

Step 1:Draw a circle of radius 3.5 cm.

Step 2: Then take a point A and draw an arc 6 cm long to intersect the circle at B. Join AB. That is the chord

Step 3: Locus of equidistant points for AB is the perpendicular bisector.  Draw perpendicular bisector.

Step 4: Mark point D. Join AD.

Step 5: Draw an angle bisector of angle DAB. This is the locus for equidistant point for point on line AD and AB.

Step 6: Measure CD.

Question 11:

(a) A model of a high rise building is made to a scale of $1:50$.

i) if the height of the model is $0.8$ m, find the height of the actual building.

ii) If the floor area of a flat in the building is $20 \ m^2$, find the floor are of that in the model.

(b) From a solid wooden cylinder of height $28$ cm and diameter $6$ cm. Two conical cavities are hollowed out. The diameters of the cone is also $6$ cm and height is $10.5$ cm.

Taking $\pi =$ $\frac{22}{7}$ , find the volume of the remaining solid.

(c) Prove the identity:

$\Big($ $\frac{1-\tan \theta}{1 - \cot \theta}$ $\Big)^2 = \tan^2 \theta$

(a) Scale $= 1:50$

$h_m = 0.8$ m        $A_b = 20 \ m^2$

i) $\frac{1}{50}$ $=$ $\frac{h_m}{h_b}$

$\Rightarrow h_b = h_m \times 50$

$\Rightarrow h_b = 0.8 \times 50= 40$ m

ii) $\frac{1^2}{50^2}$ $=$ $\frac{A_m}{A_b}$

$\Rightarrow$ $\frac{1}{2500}$ $=$ $\frac{A_m}{20}$

$\Rightarrow A_m =$ $\frac{1}{125}$ $\ m^2$  or $A_m = 0.008 \ m^2$

(b) Volume of a cylinder $= \pi r^2 H$

Volume of a cone $=$ $\frac{1}{3}$ $\pi r^2 h$

Given : $r = 3$ cm      $H = 28$ cm    $h = 10.5$ cm

Volume of cylinder $= \pi (3)^2 ( 28) = 252 \pi$

Volume of cones $= 2 [$ $\frac{1}{3}$ $\pi (3)^2 (10.5) ] = 63 \pi$

Therefore Volume of remaining solid $= 252 \pi - 63 \pi = 189 \times$ $\frac{22}{7}$ $= 594 \ cm^3$

(c) LHS $= \Big($ $\frac{1 - \tan \theta}{1 - \cot \theta}$ $\Big)^2$

$= \Big[$ $\frac{\cos \theta - \sin \theta }{\cos \theta}$ $\times$ $\frac{\sin \theta}{\sin \theta - \cos \theta}$ $\Big]^2$

$= \Big[$ $\frac{ - \sin \theta }{\cos \theta}$ $\Big]^2 = \tan^2 \theta =$ RHS. Hence proved.