Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2015)

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) A shopkeeper bought an article for Rs. 3450. He marks the price of the article 16% above the cost price. The rate of sales tax charged on the article is 10%. Find the:

(i) marked price of the article

(ii) price paid by a customer who buys the article. [3]

(b) Solve the following in equation and write the solution set:

$13x-5<15x+4<7x+12,\ \ x\in R\$

Represent the solution on a real number line. [3]

(c) Without using trigonometric tables evaluate:-

$\frac{{\mathrm{sin \ } 65{}^\circ \ }}{{\mathrm{cos\ } 25{}^\circ \ }}+\frac{{\mathrm{cos\ } \ 32{}^\circ \ }}{{\mathrm{sin \ } 58{}^\circ \ }}$ $\ - \ {\mathrm{sin \ } 28{}^\circ \ }.\ {\mathrm{sec\ } 62{}^\circ \ }+{\mathrm{cosec}^2 \ 30{}^\circ \ }$        [4]

(a)  (i) Cost price of the article $= Rs.3450$

Marked Price $=3450 + \frac{16}{100}\times 3450 = Rs. \ 4002$

(ii) Price paid by the customer $= Rs. 4002+\frac{10}{100}\times 4002 = Rs. \ 4402.20$

(b) Given

$13x-5<15x+4<7x+12$

$\Rightarrow 15x+4>13x-5\ \ and\ 15x+4<7x+12$

$\Rightarrow 15x-13x>-5-4 \ and \ 15x-7x<12-4$

$\Rightarrow 2x>-9 \ and \ 8x<8$

$\Rightarrow x>\frac{-9}{2} and x<1$

Solution: $\{ x: \frac{9}{2} < x< 1 , x \in R \}$

(c)   $\frac{{\mathrm{sin\ } 65{}^\circ \ }}{{\mathrm{cos\ } 25{}^\circ \ }}+\frac{{\mathrm{cos} \ 32{}^\circ \ }}{{\mathrm{sin\ } 58{}^\circ \ }}$ $-{\mathrm{sin} 28{}^\circ \ }.\ {\mathrm{sec \ } 62{}^\circ \ }+{\mathrm{cosec}^ 2 \ 30{}^\circ \ }$

$= \frac{{\mathrm{cos} (90{}^\circ -65{}^\circ )\ }}{{\mathrm{cos\ } 25{}^\circ \ }}+\frac{{\mathrm{sin\ (} 90{}^\circ -32{}^\circ )\ }}{{\mathrm{sin\ } 58{}^\circ \ }}$ $-{\mathrm{cos} (90{}^\circ -28{}^\circ )\times \frac{1}{{\mathrm{cos\ } 62{}^\circ \ }}+{(2)}^2\ }$

$= \frac{{\mathrm{cos} 25{}^\circ \ }}{{\mathrm{cos} 25{}^\circ \ }}+\frac{{\mathrm{sin\ } 58{}^\circ \ }}{{\mathrm{sin} 58{}^\circ \ }}$ $-{\mathrm{cos\ } 62{}^\circ \ }\times \frac{1}{cos\ 62{}^\circ }+4$

$= 1+1-1+4 = 5$

$\\$

Question 2

(a) If $A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix} ,\ find\ x\ and\ y\ when\ A^2=B$. [3]

(b) The present population of a town is $2,00,000$. It population increases by $10\%$ in the first year and $15\%$ in the second year. Find the population of the town at the end of the two years.   [3]

(c) Three verticals of a parallelogram $ABCD$ taken in order are $A(3,6), B(5,10) \ and \ C(3,2)$ find:

(i) the coordinates of the fourth vertex $D$.

(ii) length of diagonal $BD$

(iii) equation of side $AB$ of the parallelogram $ABCD$[4]

(a) $A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix}$

$A^2=B$

$\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

$\begin{bmatrix} 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

$\begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

Therefore $4x = 16 \Rightarrow x = 4$

And $1 = -y \Rightarrow y = -1$

(b) Given: $P =2,00,000$

Population after first year $= 200000\ [1+\frac{{10}}{100}] ^1 = 200000 \left[\frac{110}{100}\right]=220000$

Principle for the second year $= 220000$

Population at the end of second year $= 220000 \left[1+\frac{15}{100}\right]$

$= 220000\times \frac{115}{100} = 253000$

(c) Let the coordinate of $D \ be \ (x,y)$. In a parallelogram, the diagonals bisect each other.

Mid point of $AC = [\frac{3+3}{2},\ \frac{6+2}{2}] = (3,4)$

Mid point of $BD = [\frac{x+5\ }{2},\ \frac{y+10}{2}]$

(i) $3=\frac{x+5}{2} \Rightarrow x = 1$

$4=\ \frac{y+10}{2} \Rightarrow y = -2$

Coordinates of $D (1, -2)$

(ii) $BD = \sqrt{{(5-1}^2)+(10+{2)}^2}$

$= \sqrt{16+144} = 4 \sqrt{10} \ units$

(iii) Equation of AB:

$y-y_1$ $=\ \frac{y_{2-y_1}}{x_2-\ x_1}$ $\left(x-x_1\right)$

$y-6=$ $\ \frac{10-6}{5-3}$ $\left(x-3\right)$

$y-6=\ \frac{4}{2}\left(x-3\right)$

$y-6=2x-6$

$2x-y=0$

$\\$

Question 3

(a) In the given figure, ABCD is a square of side 21 cm, AC and BD are two diagonals of the square. Two semi circle are drawn with AD and BC as diameters. Find the area of the shaded region. (Taken $\pi =\frac{22}{7}$ )   [3]

(b) The marks obtained by 30 students in a class assessment of 5 marks is given below:

 Marks 0 1 2 3 4 5 No. of Students 1 3 6 10 5 5

Calculate the mean, mediam and mode of the above distribution. [3]

(c)   In the figure given below O is the center of the circle and SP is a tangent. If $\angle SRT =65{}^\circ ,\ find\ the\ value\ of\ x,\ y\ and\ z$  [4]

(a) Given: Side $=21\ cm$

Let Diagonal of the square $= \sqrt{2} \times \ side$

$\therefore AC=BD=21\sqrt{2}$

$\therefore AO=OC=BO=OD=\ \frac{21\sqrt{2}}{2}$

Area of  $\Delta AOD=Area\ of\ \Delta BOC=\ \frac{1}{2}\times \frac{21\sqrt{2}}{2}\times \frac{21\sqrt{2}}{2}=\frac{441}{4}{\ cm}^2$

Area of semicircle $= \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times \left(\frac{{21}^2}{2}\right)=\frac{693}{4}{\ cm}^2$

Area of shaded region = Area of 2 semicircles + Area of $\triangle AOD$$\triangle BOC$

$= 2\ \times \frac{693}{4}+\frac{441}{4}+\frac{441}{4}+\frac{2268}{4} = 567\ {cm}^2$

(b) Below

 $x$ $f$ $fx$ $cf$ 0 1 0 1 1 3 3 4 2 6 12 10 3 10 30 20 4 5 20 25 5 5 25 30 $\mathcal{E}f=30\ \ \ \ \ \ \ \mathcal{E}fx=90$

Mean $= \frac{\mathcal{E}fx}{\mathcal{E}f}=\frac{90}{30} = 3$

Median $= size \ of \left(\frac{N}{2}\right)th\ {obs}^n$

$= size \ of \left(\frac{30}{2}\right)th\ {obs}^n$

$= Size \ of 15 {}^{th} \ obs{}^{n} = 3$

Mode = 3 marks (as highest frequency is 10)

(c)  In $\Delta OSP,\ \ \angle OSR=90{}^\circ$ (Radius is always perpendicular to the tangent)

$\ In\ \Delta TSR$

$x+90{}^\circ +65{}^\circ =180{}^\circ$

$x=25{}^\circ$

$\angle SOQ=2\ \angle STR\ \ \left[Angle\ at\ centre=2\ \times angle\ at\ circumference \right]$

$y=2\times 25=50{}^\circ$

In  $\Delta OSP,\ \ 50{}^\circ +90{}^\circ +z=180{}^\circ$

$z=40{}^\circ$

$\\$

Question 4

(a) Katrina opened a recurring deposit account with a Nationalized Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is Rs. 1000 find the: [3]

(i) interest earned in 2 years

(ii) matured value

(b) Find the value of $k$ for which $x=3$ is a solution of the quadratic equation, $\left(k+2\right)x^2-kx+6=0$. Thus find the root of the equation.  [3]

(c) Construct a regular hexogon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown.  [4]

(a) (i) Given $P=Rs.1000,\ n=2\ years=\ 24\ months,\ r=6\%.$

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

$= 1000\times \frac{24\ \left(24+1\right)}{2}\times \frac{6}{12\times 100}$

$= 1000\times \frac{24\times 25}{2}\times \frac{6}{12\times 100}$

$=1500$

(ii) Total money deposit in 24 months = $Rs.\ 1000\times 24 = Rs. \ 24000$

$Matured\ Value=Total\ sum\ deposit+Interest =24000+1500 = Rs. \ 25500$

(b) $\left(k+2\right)x^2-kx+6=0$

$Putting\ x=3$  in given equation:

$\left(k+2\right)\times 9-k\times 3+6=0$

$9k+18-3k+6=0$

$6k=-24 \Rightarrow k = -4$

Putting $k= -4$ in given equation:

$-2x^2+4x+6=0$

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

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

$x\left(x-3\right)+1\left(x-3\right)=0$

$\left(x+1\right)\left(x-3\right)=0$

$x+1=0\ or\ x-3=0$

$x=-1\ or\ x=3$

$x=-1\$ is the other root of the given equation.

(c) Steps of Construction:

1. First draw a regular hexagon. The length of one side is 5 cm. You will get hexagon ABCDEF.
2. Take any two adjacent sides and draw perpendicular bisectors.
3. The point where these two bisectors intersect, is the center of the circle.
4. With O as the center, draw a circle which will pass through all the vertices of the hexagon.

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5

(a)  Use a graph paper for this question taking $1 \ cm = 1$ unit along both the $x \ and \ y \ axis$:

(i) Plot the points $A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) \ and \ F(0, -5)$

(ii) Reflect the points $B, C, D \ and \ E$ on the $y-axis$ and name them respectively as $B', C', D',U' \ and \ E'$.

(iv) Name the figure formed by $B C D E E'D'C'B'$.

(v) Name a line of symmetry for, the figure formed.  [5]

(b) Virat opened a Saving Bank account in a bank on 16th April 2010. His pass book shows the following entries:

 Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.) April 16, 2010 By Cash – 2500 2500 April 28th By Cheque – 3000 5500 May 9th To Cheque 850 – 4650 May 15th By Cash – 1600 6250 May 24th To Cash 1000 – 5250 June 4th To Cash 500 – 4750 June 30th By Cheque – 2400 7150 July 3rd To Cash – 1800 8950

Calculate the interest Virat earned at the end, of 31st July, 2010 at 4% per annum interest. What sum of money will he receives if he closes the account on 1st August, 2010? [5]

(a)   (i) Shown in the  diagram.

(ii) Shown in then diagram.

(iii) Coordinates : $B', (-2, 5), C', (-5 ,2), D', (-5 , -2),E', (-2, -5)$

(iv) Octagon

(v) $x-axis \ or \ y-axis$.

(b) Qualifying principal for various months:

 Month Principal (Rs.) April 0 May 4650 June 4750 July 8950 Total 18350

$P = Rs. \ 18350 \ R = 4\% \ and \ T= \frac{1}{12}$

$I = P \times R \times T = 18350 \times \frac{4}{100} \times \frac{1}{12} = Rs. \ 61.16$

Amount $= 8950+61.61= Rs. \ 9011.16$

$\\$

Question 6

(a)  If a, b, c are in continued proportion, prove that $(a + b + c) (a - b + c) = a^2 + b^2 + c^2$  [3]

(b)  In the given figure $ABC$  is a triangle and $BC$, is parallel to the $y-axis$. $AB$ and $AC$ intersects the $y-axis$ at $P$ and $Q$ respectively.

(i) Write the coordinates of $A$.

(ii) Find, the length of $AB \ and \ AC$.

(iii) Find the ratio in which $Q$ divides $AC$.

(iv) Find the equation, of the line $AC$ [4]

(c) Calculate the mean of the following distribution. [3]

 Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16

(a)  To prove:  $(a + b + c) (a - b + c) = a^2 + b^2 + c^2$

Given $a, b \ and\ c$ are in continued proportion

Therefore $\frac{a}{b} = \frac{b}{c} = k$

$\Rightarrow a = bk \ and \ b = ck$

This also implies $a = (ck)k = ck^2$

LHS $= (a + b + c) (a - b + c)$

$= (ck^2+ck+c)(ck^2-ck+c)$

$= c^2(k^2+k+1)(k^2-k+1)$

$= c^2 \{ (k^2+1)+k \} \{ (k^2+1)-k \}$

$=c^2 \{ (k^2+1)^2-k^2 \}$

$= c^2 \{ k^4+1+2k^2-k^2 \}$

$= c^2 \{ k^4+k^1+1 \}$

RHS $= a^2 + b^2 + c^2$

$= (ck^2)^2+(ck)^2+c^2$

$= c^2k^4+c^2k^2+c^2$

$=c^2(k^4+k^2+1)$

Hence LHS = RHS

Therefore Proved

(b)   (i) Coordinates of $A (4, 0)$

(ii) Coordinates of $A (4, 0) \ and \ B (-2, 3)$

Note: The distance between any two points $(x_1, y_1)$ and $(x_2, y_2)$ is $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Length $AB = \sqrt{(-2-4)^2+(3-0)^2} = \sqrt{36+9} = \sqrt{45} = 3 \sqrt{5} \ units$

Coordinates of $A (4, 0) \ and \ C (-2, -4)$

Length $AC = \sqrt{(-2-4)^2+(-4-0)^2} = \sqrt{36+16} = \sqrt{52} = 2 \sqrt{13} \ units$

(iii) Let the required ratio be  $k:1$

Let the coordinate of $P \ be \ (0, y)$, since it lies on the y axis.

Since  $x =$ $\frac{kx_2+x_1}{k+1}$

$\Rightarrow 0 =$ $\frac{k \times (-2) +4}{k+1}$

$\Rightarrow -2k+4=0$

$\Rightarrow k = 2$

$\Rightarrow m_1:m_2 = 2:1$

(iv) Equation of $AC$ through Coordinates of $A(4, 0) \ and \ C(-2, -4)$

Required  equation of the line: $(y-y_1)=m(x-x_1)$

$y-0=$ $(\frac{0+4}{4+2})$ $(x-4)$

$\Rightarrow 3y=2x-8$

$\Rightarrow \ or \ 3y-2x+8=0$

(c)

 Class Mid Value $x$ $f$ $fx$ 0-10 5 8 40 10-20 15 5 75 20-30 25 12 300 30-40 35 35 1225 40-50 45 24 1080 50-60 55 16 880 Total $\Sigma f=100$ $\Sigma fx=3600$

Mean $= \bar{x} = \frac{\mathcal{E}fx}{ \mathcal{E}f} = \frac{3600}{100}$ $= 36$

$\\$

Question 7

(a) The solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3]

(b) Find ‘a’ if the two polynomials ${ax}^3+{3x}^2-9 \ and \ {2x}^3+4x+a$, leaves the same remainder when divided by $(x+3)$. [3]

(c) Prove that  $\frac{{\mathrm{sin \ } \theta \ }}{1-{\mathrm{cot \ } \theta \ }}+\frac{{\mathrm{cos \ } \theta \ }}{1-{\mathrm{tan \ } \theta \ }}$ $={\mathrm{cos \ } \theta \ }+{\mathrm{sin \ } \theta }$ [4]

(a)  Volume of two spheres = Volume of cone

$\frac{4}{3}\pi \left(2^3\right)+\frac{4}{3}\pi \left(4^3\right)=\ \frac{1}{3}{\pi r}^2h$

$32+256=r^2\times 8$

$r^2 = \frac{288}{8}$

$r^2 = 36 \Rightarrow r = 6 \ cm$

(b)  Let $f(x) = {ax}^3+{3x}^2-9$ and $g(x) = {2x}^3+4x+a$

Putting $x = -3$ in both the expressions

$f(-3) = {a(-3)}^3+{3(-3)}^2-9$

$g(-3) ={2(-3)}^3+4(-3)+a$

Given $f(-3) = g(-3)$

Therefore ${a(-3)}^3+{3(-3)}^2-9 = {2(-3)}^3+4(-3)+a$

$-27a+27 -9 = -54-12+a$

$27a+a = 27-9+54+12$

$28a=84 \Rightarrow a = 3$

(c)  Given $\frac{{\mathrm{sin \ } \theta \ }}{1-{\mathrm{cot \ } \theta \ }}+\frac{{\mathrm{cos \ } \theta \ }}{1-{\mathrm{tan \ } \theta \ }}$ $={\mathrm{cos \ } \theta \ }+{\mathrm{sin \ } \theta }$

LHS $=\frac{{\mathrm{sin \ } \theta \ }}{1-\frac{\mathrm{cos \ } \theta \ }{\mathrm{sin \ } \theta \ }}+\frac{{\mathrm{cos \ } \theta \ }}{1-\frac{\mathrm{sin \ } \theta \ }{\mathrm{cos \ } \theta \ }}$

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

$=\frac{{\mathrm{sin}^2 \ \theta \ - \mathrm{cos}^2 \ \theta \ }}{{\mathrm{sin \ } \theta \ } - {\mathrm{cos \ } \theta \ }}$

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

$={\mathrm{cos \ } \theta \ }+{\mathrm{sin \ } \theta } =$ RHS

$\\$

Question 8

(a) $AB \ and \ CD$ are two chords of a circle intersecting at $P$. Prove that $AP \times PB =CP \times PD$. [3]

(b) A bag contain 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag, Find the probability that the ball drawn is:

(i) a green ball

(ii) a white or a red ball

(iii) is neither a green ball nor a white ball. [3]

(c) Rohit invested Rs. 9600 on Rs. 100 shares at Rs. 20 premium paying 8% dividend. Rohit sold the shares when the price rose to Rs. 160. He invested the proceeds (excluding dividend) in 10% Rs. 50 shares at Rs. 40. Find the:

(i) original Number of shares

(ii) sale proceeds

(iii) new number of shares

(iv) change in the two dividends. [4]

(a)   To prove: $AP\times PB=CP\times PD$

Construction : Join AD and CB.

$\angle A = \angle C$  (Angles of the same segment)

$\angle C = \angle B$  (Angles of the same segment)

Therefore $\Delta APD \sim \Delta \ BPC$   (AAA Postulate)

$\therefore \frac{AP}{CP}=\frac{PD}{PB}$

$\therefore AP \times PB=CP \times PD$

Hence Proved

(b)  Number of White Balls: $5$

Number of Red Balls: $6$

Number of Green Balls : $9$

Total number of Balls:  $5+6+9 = 20$

(i) Probability (Green) $= \frac{Favourable \ Cases}{Total \ number \ of \ possible \ cases} = \frac{9}{20}$

(ii) Probability (Red or White) $= \frac{5+6}{20} = \frac{11}{20}$

(iii) Probability (neither a green ball nor a white ball} $= \frac{6}{20} = \frac{3}{10}$

(c)   First Investment

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

Market Value of the share $= 120 \ Rs.$

Dividend $= 8 \%$

(i) Number of shares bought $= \frac{9600}{120} = 80$

(ii) Sale Price $= 160 \ Rs.$

Sale Proceed $= 80 \times 160 = 12800 \ Rs.$

Second Investment

Nominal Value of the share $= 50 \ Rs.$

Market Value of the share $= 40 \ Rs.$

Dividend $= 10\%$

(iii) Number of shares bought $= \frac{12800}{40} = 320$

Dividend on 1st Investment earned $= 80 \times \frac{8}{100} \times 100 = 640 \ Rs.$

Dividend on 2nd Investment earned $= 320 \times \frac{10}{100} \times 50 = 1600 \ Rs.$

Change in Dividend income $= 1600-640 = 960 \ Rs.$

$\\$

Question 9

(a) The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is $30^o$ and $24^o$ respectively.

Find the height of the two towers. Give your answer correct to 3 significant figures.   [4]

(b) The weight of 50 workers is given below:

 Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on  one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight.   [6]

(a)  Let AB and CD are two towers.

$BD=120 \ m=EC,\ \angle ACE=30{}^\circ ,\ \angle CBD=24{}^\circ$

In the right $\Delta CBD$

${\mathrm{tan} 24{}^\circ \ }=\ \frac{CD}{BD}=\frac{CD}{120}$

$CD=120\times 0.4452 =53.42 \ m$

In right angle $\Delta ACE$

${\mathrm{tan} 30{}^\circ {}^\circ =\ \frac{AE}{EC}=\frac{AE}{120}\ }$

$\mathrm{AE=120\times 0.5773} =69.28 \ m$

$AB=AE+EB$

$= AE+CD = 69.28+53.42 = 122.7 \ m.$

Height of $1{}^{st}$ tower is $122.7 \ m$ and $2{}^{nd}$ tower is $53.42 \ m$.

(b) Below the table

 Weight $f$ $cf$ 50-60 4 4 60-70 7 11 70-80 11 22 80-90 14 36 90-100 6 42 100-110 5 47 110-120 3 50 Total 50

(i)   Lower quartile= $\frac{N^{th}}{4}{ \ obs}^n$

$= \frac{{50}^{th}}{4}{ \ obs}^n = 12.5th { \ obs}^n = 72 \ kg$

Upper quartile = $\frac{3N}{4}\ th\ {obs}^n = \frac{150}{4}th\ {obs}^n = 37.5th { \ obs}^n = 92 \ kg$

(ii)  No. of over weight workers $= 50-39 = 11$

$\\$

Question 10

(a) A wholesale buys a TV from the manufacture for Rs. 25000. He marks the price of the TV 20% above his cost price and sell it to a retailer at 10% discount on the marked price. If the rate of VAT is 8%, find the: [3]

(i) marked Price

(ii) retailer’s cost price inclusive of tax

(iii) VAT paid by the wholesaler.

(b) If $A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$$B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix}$. Find AB-5C.   [3]

(c) ABC is a right angled triangle with $\angle ABC= 90^o$. D is any point on AB and DE is perpendicular to AC. Prove that:

(i) $\Delta ADE \ \sim \Delta ACB$

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

(iii) Find, area, of $\Delta ADE$: area of quadrilateral BCED.   [4]

(a)  (i) Cost price for wholesaler $= 25000 \ Rs.$

Market Price $= 25000+\frac{20}{100}\times 25000 = 30000 \ Rs.$

(ii) Discount $= 10\% \ of \ 30000= 3000 \ Rs.$

Cost Price for retailer $= Market Price -Discount$

$= \mathrm{30000 - 3000} = 27000 \ Rs.$

Cost Price inclusive Tax $= \mathrm{27000+}\frac{\mathrm{8}}{\mathrm{100}}\mathrm{\times 27000} = 29160 \ Rs.$

(iii) Cost price for wholesaler $= 25000 \ Rs.$

Sale price for wholesaler $= 27000 \ Rs.$

Profit for wholesalers $= 27000 - 25000= 2000 \ Rs.$

VAT $= \frac{8}{100}\times 2000 = 160 \ Rs.$

(b)  Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$,  $B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}$,  $C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix}$

$AB = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} . \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix}$

$5C = 5 . \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} = \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix}$

$AB - 5C= \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} - \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix} = \begin{bmatrix} 30 & 52 \\ 40 & -14 \end{bmatrix}$

(c)  To prove: $\Delta ADE\ \sim \Delta ACB$

(i) In $\Delta ADE\ and\ \Delta ACB$

$\angle A$ (is common)

$\angle AED = \angle ABC=90^o$

Therefore $\Delta ADE\ \sim \Delta ACB$ (AAA Postulate)

(ii) ${AC}^2={AB}^2+{BC}^2$

$169={AB}^2+25$

$AB = 12 cm$

Since $\ \Delta ADE \sim \Delta ACB$

Therefore $\ \frac{DE}{BC} = \frac{AD}{AC} = \frac{AE}{AB}$

Therefore $\ \frac{DE}{BC} = \frac{AE}{AB}$

$\frac{DE}{5} = \frac{4}{12}$ $\Rightarrow DE =$ $\frac{5}{3}$ cm

$\frac{AD}{AC} = \frac{AE}{AB}$

$\frac{AD}{13} = \frac{4}{12}$ $\Rightarrow AD =$ $\frac{13}{3}$ cm

(iii) $\frac{Ar.\ of\ \left(\Delta \ ABC\right)}{Ar.\ of\ \left(\Delta \ ADE\right)}=\ \frac{{AB}^2}{{AE}^2}=\frac{144}{16}=9$

$\Rightarrow \frac{Ar.\ of\ \left(\Delta ADE\right)+Ar.\ of\ \Delta (BCED)}{Ar.\ of\ (\Delta ADE)}$ $=9$

$\Rightarrow 1 + \frac{Ar.\ of\ \left(BCED\right)}{Ar.\ of\ \left(\Delta \ ADE\right)}$ $= 9$

$\Rightarrow \frac{Ar.\ of\ \left(\Delta \ ADE\right)}{Ar.\ of\ \left(BCED\right)}=\ \frac{1}{8}$

$\\$

Question 11

(a) Sum of two natural numbers is 8 and the difference of their reciprocal is $\frac{2}{15}$ find the numbers.  [3]

(b) Given $\frac{x^3+12x}{6x^2+8} = \frac{y^3+27y}{9y^2+27}$ using componendo and dividendo find $x:y$ [3]

(c) Construct the $\triangle ABC \ with \ AB=5.5 \ cm, AC=6 \ cm$, and $\angle BAC=105^o$. Hence

(i) Construct the locus of points equidistant from BA and BC

(ii) Construct the locus of points equidistant from B and C

(iii) Mark the point which satisfy he above two loci as P. Measure and write the length of PC.   [4]

(a)  Let 1st number be $x$ and the 2nd number be $(8-x)$

Given $\frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$

$\frac{8-x-x}{x\ \left(8-x\right)}=\frac{2}{15}$

$\frac{8-2x}{8x-x^2}=\frac{2}{15}$

$120-30x=16x-2x^2$

${2x}^2-46x+120=0$

$x^2-23x+60=0$

$x\ \left(x-20\right)-3\ \left(x-20\right)=0$

$\left(x-3\right)\left(x-20\right)=0 \Rightarrow x=3 \ or \ x=20 (not \ possible)$

Therefore the first number is 3 and the second number is 5.

(b) $\frac{x^3+12x}{{6x}^2+8}=\frac{y^3+27y}{{9y}^2+27}$

Applying componendo and dividendo

$\frac{x^3+12x+{6x}^2+8}{x^3+12x-{6x}^2-8}=\frac{y^2+27y+{9y}^2+27}{y^3+27y-{9y}^2-27}$

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

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

Again Applying componendo and dividendo

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

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

$\frac{x}{y}=\frac{4}{6}=\frac{2}{3}$

$x:y=2:3$

(c) Steps of Construction:

1. First draw a line AB=5.5 cm.
2. With the help of the point A, draw $\angle XAB =105^o$. This you can draw using a compass.
3. Take radius 6 cm, cut AC=6 cm and join C to B. This way you get CA and BC. This completes the triangle.
4. Then draw perpendicular bisector of BC and draw angle of bisector $\angle CBA$ both intersecting at P.
5. P is the required point PC=4.8 cm