2013 ICSE (Class 10) Board Paper Solution: Mathematics

MATHEMATICS (ICSE – Class X Board Paper 2013)

Two and Half Hour. Answers 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 B. All 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 $=\frac{\sum{x}}{n}$

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

$\\$

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)

$\\$

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,

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

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 $B (-7, 17)$

$\\$

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 $BC$

(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,

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

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

(c) (i) Please see graph

(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 $ABA'B' = A'B' +AB'+AB+A'B' = 6+10+6+10 =32 \ units$

$\\$

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.

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

${-}{2x} {\ \le \ }{3x-8<1}$

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

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

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

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

(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 $=20$

$\\$

Question 6:

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

(b) In 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,

$\angle ABC+\angle ADC=180^o$ (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 $= \frac{P \times R \times T}{100} = \frac{29690 \times 6 \times 1}{100 \times 12} = Rs. \ 148.45$

$\\$

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) In 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 m

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 $=40 \ m$

$\\$

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

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

$p^2-10p+9 = 0$

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

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

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

$\\$

Question 9:

(a) In 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

-2

-1

-20

-6

$\Sigma f_i = 50$

$\Sigma f_i.d_i = 23$

Here, $a=45$ and $h=10$

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

$\\$

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

107

113

117

120

N=120

On the graph paper we plot the following points:

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

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

From the graph 60^th 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 $40=52$

$\\$

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]