Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board 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:

$\displaystyle \text{(a) Given } A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}, B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} . \\ \\ \text{ Find the matrix } X \text{ such that } A + 2X = 2B + C . \hspace{7.0cm} [3]$

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

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

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

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

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

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

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

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

(b) Given: Principle $\displaystyle =\text{ Rs. } 4000,\ C.I.=\text{ Rs. } 1324$

Amount $\displaystyle =P+C.I =4000+1324=\text{ Rs. } 5324$

Time $\displaystyle =3 \text{ years }$

$\displaystyle \text{We know that; } A=P \Big(1+\frac{r}{100} \Big)^n$

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

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

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

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

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

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

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

$\displaystyle r=\frac{100}{10}$

$\displaystyle r=10\%$

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

Since $\displaystyle n=9$ which is odd, therefore

$\displaystyle \text{Median }=\ {\frac{n+1}{2}}^{th} \text{term } ={\frac{9+1}{2}}^{th} \text{term }$

$\displaystyle 24=5th\ \text{term}$

$\displaystyle x+4=24$

$\displaystyle \Rightarrow x=24-4 \Rightarrow x=20$

$\displaystyle \text{Therefore, } 11,12,14, (20-2), (20+4), (20+9),32,38,47$

$\displaystyle =11,12,14,18,24,29,32,38,47$

$\displaystyle \text{Now, Mean } =\frac{\sum{x}}{n}$

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

$\displaystyle \\$

Question 2:

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

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

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

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

Since they are proportional,

$\displaystyle (6+x) : (15+x) :: (20+x):(43+x)$

$\displaystyle (6+x)(43+x) = (15+x)(20+x)$

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

$\displaystyle \Rightarrow 49x - 35x = 300-258$

$\displaystyle \Rightarrow 14x = 42 \Rightarrow x= 3$

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

Since $\displaystyle x-2=0 \Rightarrow x=2$

$\displaystyle 2x^3 + ax^2 + bx - 14 = 0$

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

$\displaystyle 16+4a+2b- 14 = 0$

$\displaystyle 4a+2b+2 = 0$

$\displaystyle 4a +2b = -2$

$\displaystyle 2a+b = -1 \text{ ... ... ... ... ... (i)}$

When given expression is divided by $\displaystyle (x - 3)$

$\displaystyle x-3 = 0 \Rightarrow x=3$

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

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

$\displaystyle 54+9a+3b-66 = 0$

$\displaystyle 9a+3b = 12$

$\displaystyle 3a+b = 4 \text{ ... ... ... ... ... (ii)}$

Solving equation (i) and (ii),

$\displaystyle a = 5 \text{ and } b = -11$

(c)

$\displaystyle \\$

Question 3:

(a) Without using tables evaluate: $\displaystyle 3 \cos 80^\circ. \mathrm{cosec} 10^\circ + 2 \sin 59^\circ \sec 31^\circ$      [3]

(b) In the given feature,

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

Prove:

(i) AC is the diameter of the circle

(ii) Find $\displaystyle \angle ACB$                                                                                            [3]

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

(i) The length of radius $\displaystyle AC$

(ii) The Coordinates of $\displaystyle B$                                                                            [4]

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

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

$\displaystyle = 3 \cos \ 80^\circ \sec\ 80^\circ + 2 \sin\ 59^\circ \mathrm{cosec} 59^\circ$

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

$\displaystyle = 3+2=5$

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

(i) Since $\displaystyle ABCD$ is a cyclic quadrilateral

In $\displaystyle \triangle ABD$

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

$\displaystyle \angle BDA=180^\circ -(65^\circ +70^\circ) =180^\circ -135^\circ =45^\circ$

Now from $\displaystyle \triangle ACD$ ,

$\displaystyle \angle ADC=\angle ADB+\angle BDC =45^\circ +45^\circ=90^\circ$

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

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

Therefore $\displaystyle \angle ABC = 45^\circ (\text{since} \angle ADB = 45^\circ)$

(c)

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

(ii) Let the point $\displaystyle B$ be $\displaystyle (x,y)$

Given $\displaystyle C$ is the mid-point of $\displaystyle AB .$ Therefore

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

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

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

$\displaystyle \\$

Question 4:

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

(b) In the given figure, $\displaystyle AB \text{ and } DE$ are perpendicular to $\displaystyle BC$

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

(ii) If $\displaystyle AB=6 \text{ cm}, DE=4 \text{ cm} \text{ and } AC=15 \text{ cm}$ , Calculate $\displaystyle CD$

(iii) Find the ratio of the area of $\displaystyle \triangle ABC : \text{area of } \triangle DEC$                     [3]

(c) Using graph paper, plot the point $\displaystyle A(6,4) \text{ and } B(0,4) .$

(i) Reflect $\displaystyle A \text{ and } B$ in the origin to get the images $\displaystyle A' \text{ and } B' .$

(ii) Write the co-ordinate of $\displaystyle A' \text{ and } B'$

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

(iv) Find its perimeter                                                                                  [4]

(a) Given : $\displaystyle x^2 - 5x - 10 = 0$

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

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

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

$\displaystyle x=6.53,\ -1.53$

(b)

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

$\displaystyle \mathrm{\angle ABC=\angle DEC=90{}^\circ }$ Given

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

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

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

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

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

Given: $\displaystyle AB =6 \text{ cm}, DE=4 \text{ cm}, AC=15 \text{ cm},$

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

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

(iii) Since $\displaystyle \triangle ABC \sim \triangle DEC,$

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

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

(c)

(ii) Reflection of $\displaystyle A' \text{ and } B'$ in the origin $\displaystyle = A'(-6, -4) \text{ and } B' (0,-4)$

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

(iv) From the graph, $\displaystyle AB=6 \text{ cm } BB'=8 \text{ cm }$

In $\displaystyle \triangle ABB'$

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

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

Perimeter of $\displaystyle ABA'B' = A'B' +AB'+AB+A'B' = 6+10+6+10 =32 \text{ units }$

$\displaystyle \\$

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]

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

(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]

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

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

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

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

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

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

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

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

(b) Let the monthly installment $\displaystyle be \text{ Rs. } x$

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

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

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

Actual sum deposited $\displaystyle =36x$

Maturity amount = Interest +Actual sum deposited

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

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

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

Hence the monthly installment be $\displaystyle \text{ Rs. } 200$

(c) Number of shares $\displaystyle =50$

Face value of each share $\displaystyle =\text{ Rs. } 100$

Market value of each shares $\displaystyle = \text{ Rs. } 132$

Total face value $\displaystyle =100 \times 50 = \text{ Rs. } 5000$

(i) Total investment $\displaystyle =132 \times 50 = \text{ Rs. } 6600$

(ii) Rate of dividend $\displaystyle =7.5\%$

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

(iii) Let extra share should he buy be $\displaystyle x$

Then total number of shares $\displaystyle = 50+x$

Total face value $\displaystyle = \text{ Rs. } 100 \times (50+x)$

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

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

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

$\displaystyle {x=70-50=20}$

Hence the extra shares should be buy $\displaystyle =20$

$\displaystyle \\$

Question 6:

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

(b) In the given circle with center $\displaystyle O, \angle ABC=100^o \ \angle ACD=40^o \text{ and } CT$ is a tangent to the circle at $\displaystyle C .$ Find $\displaystyle \angle ADC \text{ and } \angle DCT .$                                                 [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 $\displaystyle 6\% p.a .$             [4]

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

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

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

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

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

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

(b) Given: $\displaystyle \angle ABC =100^{\circ}$

We know that,

$\displaystyle \angle ABC+\angle ADC=180^{\circ}$ (sum of the opposite angles in a cyclic quadrilateral is $\displaystyle 180^{\circ}$ )

$\displaystyle 100^{\circ}+\angle ADC=180^{\circ}$

$\displaystyle \angle ADC=180^{\circ}-100^{\circ}$

$\displaystyle \angle ADC=80^{\circ}$

Join $\displaystyle OA \text{ and } OC$ , we have a isosceles $\displaystyle \triangle OAC$

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

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

$\displaystyle \Rightarrow {\angle AOC=2\times 80^{\circ}=160^{\circ}}$

In $\displaystyle \triangle AOC$

$\displaystyle {\angle AOC+\angle OAC+OCA=180^{\circ}}$

$\displaystyle {160^{\circ}+\angle OCA+\angle OCA=180^{\circ}}$

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

$\displaystyle {2\angle OCA=20^{\circ}}$

$\displaystyle {\angle OCA=10^{\circ}}$

$\displaystyle {\angle OCA+\angle OCD=40^{\circ}}$

$\displaystyle {10^{\circ}+\angle OCD=40^{\circ}}$

$\displaystyle {\therefore } {\angle OCD=30^{\circ}}$

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

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

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

$\displaystyle 30^{\circ}+\angle DCT=90^{\circ}$

$\displaystyle {\angle DCT=60^{\circ}}$

(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

$\displaystyle \text{Time } = \frac{1}{12}$

$\displaystyle \text{Rate of interest } = 6\%$

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

$\displaystyle \\$

Question 7:

(a) In $\displaystyle \triangle ABC, A(3, 5), B(7, 8) \text{ and } C(7, -10) .$ 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 $\displaystyle AB=60 \text{ m}$ high, the angles of depression of the top and bottom of a vertical lamp post $\displaystyle CD$ are observed to be $\displaystyle 30^\circ \text{ and } 60^\circ$ respectively.                                                                                                [4]

Find:

(i) The horizontal distance between $\displaystyle AB \text{ and } CD$

(ii) The height of the lamp paid.

(a) Since $\displaystyle D \text{ is midpoint of } BC$

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

Now equation of mediam $\displaystyle AD$

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

Here, $\displaystyle x_1=3, y_1=5, x_2=4 \text{ and } y_2=-1$

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

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

$\displaystyle y-5 = -6x+18$

$\displaystyle y = -6x+18+5$

$\displaystyle 6x+y-23 = 0$

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

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

Let C.P. of this article be $\displaystyle x$ , then

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

If the shopkeeper pays a $\displaystyle VAT =\text{ Rs. } 36$

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

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

$\displaystyle \Rightarrow 18000-12x=3600$

$\displaystyle 12x=18000-3600=14400$

$\displaystyle x=\text{ Rs. } 1200$

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

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

(c) Given; AB=60 m

Since $\displaystyle \angle PAC=60^\circ$

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

(i) Now in $\displaystyle \triangle ABC$

$\displaystyle \tan 60^\circ =\frac{ AB}{ BC}$

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

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

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

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

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

Therefore $\displaystyle BC = ED =20 \sqrt{3}$

$\displaystyle \text{In } \triangle AED \tan 30^\circ =\frac{AE}{ED}$

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

$\displaystyle AE=20 \text{ m}$

$\displaystyle EB=AB-AE$

$\displaystyle EB=60-20=40 \text{ m}$

Since $\displaystyle EB=CD$

$\displaystyle CD=40 \text{ m}$

Hence, the height of the lamp post $\displaystyle =40 \text{ m}$

$\displaystyle \\$

Question 8:

$\displaystyle \text{(a) Find } x \text{ and } y \text{ if } \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix} . \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \hspace{6.5cm} [3]$

(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 $\displaystyle p$ for which the given equation has real and equal roots;                                            [4]

$\displaystyle x^2+(p-3)x+p = 0$

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

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

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

Therefore $\displaystyle 5x = 5 \Rightarrow x = 1$

and $\displaystyle 6y = 12 \Rightarrow y = 2$

Hence $\displaystyle x = 1 \text{ and } y = 2$

(b) Radius of a solid sphere, $\displaystyle r=15 \text{ cm}$

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

Now, radius of right circular come $\displaystyle =2.5 \text{ cm}$

Height $\displaystyle h=8 \text{ cm}$

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

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

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

Since roots are real and equal, $\displaystyle b^2-4ac = 0 \text{ ... ... ... ... ... (i) }$

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

Substituting the values in (i) we get

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

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

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

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

$\displaystyle (p-9)(p-1)=0 \Rightarrow p = 9 \text{ or } 1$

$\displaystyle \\$

Question 9:

(a) In the figure along side $\displaystyle OAB$ is a quadrant of a circle, The radius $\displaystyle OA=3.5 cm \text{ and } OD=2 \text{ cm} ,$ Calculate the area of the shaded portion. $\displaystyle \Big( \text{ Take } \pi = \frac{22}{7} \Big) \hspace{11.0cm} [3]$

(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

(a) Radius of quadrant $\displaystyle OACB, r=3.5 \text{ cm}$

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

$\displaystyle \text{Here, } \angle AOD=90^\circ$

$\displaystyle \text{base} =3.5 \text{ cm} \text{ and } \text{height} =2 \text{ cm}$

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

$\displaystyle \text{Area of shaded portion = Area of quadrant - Area of triangle} \\ \\ =9.625-3.5 =6.125 \text{ cm}^2$

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

Total number of balls $\displaystyle =30+x$

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

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

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

$\displaystyle \text{Therefore } \frac{x}{30+x} = \frac{2}{5} \cdot \frac{30}{30+x}$

$\displaystyle \Rightarrow 5x = 60 \Rightarrow x = 12$

Hence, the number of black balls $\displaystyle =12$

(c)

 C.I Frequency $\displaystyle (f_i)$$\displaystyle (f_i)$ Mid-Value $\displaystyle (x)$$\displaystyle (x)$ $\displaystyle d_i = \frac{x-a}{h}$$\displaystyle d_i = \frac{x-a}{h}$ $\displaystyle (f_i. d_i)$$\displaystyle (f_i. d_i)$ 20-30 30-40 40-50 50-60 60-70 70-80 106 8 12 5 9 25 35 45 55 65 75 -2 -1 0 1 2 3 -20 -6 0 12 10 27 $\displaystyle \Sigma f_i = 50$$\displaystyle \Sigma f_i = 50$ $\displaystyle \Sigma f_i \cdot d_i = 23$$\displaystyle \Sigma f_i \cdot d_i = 23$

Here, $\displaystyle a=45 \text{ and } h=10$

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

$\displaystyle \\$

Question:10

(a) Using a ruler and compasses only:

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

$\displaystyle AB = 3.5 \text{ cm}, BC = 6 \text{ cm} \text{ and } \angle ABC = 120^\circ$

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

(iii) Measure $\displaystyle \angle BCP$                                                                                                 [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]

(a) Steps of Construction:

Using a ruler draw a line $\displaystyle BC =6 \text{ cm}$

With the help of the point $\displaystyle B$ , draw $\displaystyle \angle ABC=120^o .$ 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 $AB = 3.5 \text{ cm}.$ Taking radius $\displaystyle 3.5 \text{ cm}$ cut $\displaystyle BA=3.5 \text{ cm} .$ This gives you point A.

Now join $\displaystyle A$ to $\displaystyle C$ 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 $\displaystyle MN$ of $\displaystyle BC .$

Draw a circle $\displaystyle O$ as center and $\displaystyle OC$ as radius.

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

Join $\displaystyle BP \text{ and } CP$

Now, $\displaystyle \angle BCP=30^o$

(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:

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

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

From the graph 60th term $\displaystyle =42$

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

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

$\displaystyle \\$

Question 11:

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

$\displaystyle \text{(b) Using the properties of proportion solve for } x , \text{ given } \frac{x^4+1}{2x^2}= \frac{17}{8} \hspace{2.0cm} [3]$

(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 $\displaystyle \text{ Rs. } x$ , and solve it to find the original cost of the books.                        [4]

(a) Let the co-ordinates of $\displaystyle A \text{ and } B \text{ be } (x,0) \text{ and } (0,y)$

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

By using section formula, we get

$\displaystyle -3 = \frac{2 \times 0 + 3 \times x}{2+3} = \frac{3x}{5}$

$\displaystyle \Rightarrow x = -5$

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

$\displaystyle \Rightarrow y = 10$

Hence, the co-ordinate of $\displaystyle A \text{ and } B \text{ are } (-5,0) \text{ and } (0,10)$

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

By using componendo and dividendo, we get:

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

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

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

Taking square root on both sides, we get

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

$\displaystyle \Rightarrow 3x^2+3 = 5x^2-5$

$\displaystyle \Rightarrow 2x^2=8$

$\displaystyle \Rightarrow x^2 = 4 \text{ or } x = \pm 2$

(c) Given the original cost of each book be $\displaystyle \text{ Rs. } x$

Total cost $\displaystyle = \text{ Rs. } 960$

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

$\displaystyle \text{If the cost per book was Rs. } 8 \text{ less, (i.e. x-8 ) then}$

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

According to question,

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

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

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

$\displaystyle x^2-8x= 1920$

$\displaystyle x^2-8x-1920 = 0$

$\displaystyle x^2-48x+40x - 1920 = 0$

$\displaystyle x(x-48) +40(x-48) = 0$

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

Hence the cost of the original book is $\displaystyle \text{ Rs. } 48$