Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2019)

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 in-equation and write down the solution set:

$11x - 4 < 15x + 4 \leq 13x + 14 , x \in W$

(b) A man invests Rs. $4500$ in shares of a company which is paying $7.5\%$ dividend. If Rs. $100$ shares are available at a discount of $10\%$. Find:

(i) Number of shares he purchases.

(ii) His annual income.

(c) In a class of $40$ students, marks obtained by the students in a class test (out of $10$) are given below: Calculate the following for the given distribution:

 Marks 1 2 3 4 5 6 7 8 9 10 Number of Students 1 2 3 3 6 10 5 4 3 3

(i) Median

(ii) Mode

(a) Given $11x - 4 < 15x + 4 \leq 13x + 14 , x \in W$

Case 1: $11x - 4 < 15x + 4$

$\Rightarrow -4x < 8$  $\Rightarrow -x < 2 \ or \ x > -2$

Case 2: $15x + 4 \leq 13x + 14$

$\Rightarrow 2x \leq 10$  $\Rightarrow x \leq 5$

$\therefore x \in [ -1, 0, 1, 2, 3, 4, 5]$

(b)  Rs. $100$ shares at a discount of $10\%$ will cost $= 100 - 10 = 90$ Rs.

i) Therefore Number of shares $=$ $\frac{4500}{90}$ $= 50$

ii) His annual income at $7.5\%$ dividend $=$ $\frac{7.5}{100}$ $\times 50 \times 100 = 337.5$ Rs.

(c)

 Marks $(x)$ No. of Students Cumulative Frequency 1 1 1 2 2 3 3 3 6 4 3 9 5 6 15 6 10 25 7 5 30 8 4 34 9 3 37 10 3 40 Total 40

i) Total number of Students $= 40$ which is even

Median $=$ $\frac{1}{2}$ $\Bigg[ \Big($ $\frac{n}{2}$ $\Big) \ term + \Big($ $\frac{n}{2}$ $\Big)^{th} \ term \Bigg]$

$=$ $\frac{1}{2}$ $\Bigg[ \Big($ $\frac{40}{2}$ $\Big) \ term + \Big($ $\frac{40}{2}$ $\Big)^{th} \ term \Bigg]$

$=$ $\frac{1}{2}$ $[ 20^{th}$ term $+ 21^{st}$ term $]$

$= \Big($ $\frac{41}{2}$ $\Big)^{th}$ term

$= 20.5^{th}$ term

Which is between $15$ and $25$. Therefore Median $= 6$

ii) Mode frequency of $6$ is the highest. Therefore Mode $= 3$

$\\$

Question 2:

(a) Using the factor theorem, show that $(x - 2)$ is a factor of $x^3 + x^2 - 4x - 4$ . Hence factorize the polynomial completely.

(b) Prove that:  $( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)=1$

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are $8$ and $14$ respectively. Find the:

(i) first term

(ii) common difference

(iii) sum of the first $20$ terms.

(a)  $f(x) = x^3 + x^2 - 4x - 4$

Let $x-2 = 0 \Rightarrow x = 2$

$\therefore f(2) = (2)^3 + (2)^2 - 4(2) - 4$

$\Rightarrow f(2) = 8 + 4 - 8 - 4 = 0$

$f(2) = 0$

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

$x-2 ) \overline{x^3 + x^2 - 4x - 4} ( x^2 + 3x + 2 \\ \hspace*{0.5cm} (-) \underline{x^3 - 2x^2 \hspace*{1.8cm}} \\ \hspace*{2.1cm} {3x^2 - 4x - 4} \\ \hspace*{1.5cm} (-) \underline{3x^2 - 6x \hspace*{0.8cm}} \\ \hspace*{3.1cm}2x-4 \\ \hspace*{2.5cm} (-) \underline {2x-4} \\ \hspace*{3.7cm}0$

Hence $x^3 + x^2 - 4x - 4 = (x-2)(x^2 + 3x + 2)$

$\Rightarrow x^3 + x^2 - 4x - 4 = (x-2)(x+2)(x+1)$

(b)   To prove: $( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)=1$

LHS $= ( \mathrm{cosec} \theta - \sin \theta )(\sec \theta - \cos \theta)(\tan \theta+\cot \theta)$

$= \Big($ $\frac{1}{\sin \theta}$ $- \sin \theta \Big) \Big($ $\frac{1}{\cos \theta}$ $- \cos \theta\Big) \Big($ $\frac{\sin \theta}{\cos \theta}$ $+$ $\frac{\cos \theta}{\sin \theta}$ $\Big)$

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

$= \Big($ $\frac{\cos^2 \theta}{\sin \theta}$ $\Big) . \Big($ $\frac{\sin^2 \theta}{\cos \theta}$ $\Big) . \Big($ $\frac{1}{\sin \theta . \cos \theta}$ $\Big)$

$= 1 =$ RHS. Hence proved.

(c)  Let the first term of the sequence is $a$ and the common difference is $d$.

$a_4 = a + 3d = 8$   … … … … … i)

$a_6 = a + 5d = 14$    … … … … … ii)

Solving i) and ii) we get $d = 3$, and $a = -1$

Therefore

i) First term $(a) = -1$

ii) Common difference $(d) = 3$

iii) Sum of the first $20$ terms $= S_n =$ $\frac{n}{2}$ $\Big[ 2a + (n-1)d \Big]$

$=$ $\frac{20}{2}$ $\Big[ -2 + (20-1)d \Big] = 10 \times 55 = 550$

$\\$

Question 3:

(a) Simplify: $\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}$

(b) $M$ and $N$ are two points on the $X$ axis and $Y$ axis respectively. $P (3, 2)$ divides the line segment $MN$ in the ratio $2 : 3$. Find:

(i) the coordinates of $M$ and $N$

(ii) slope of the line $MN$.

(c) A solid metallic sphere of radius $6$ cm is melted and made into a solid cylinder of height $32$ cm. Find the:

(ii) curved surface area of the cylinder     [Take $\pi = 3.1$ ]

(a)  Given $\sin A \begin{bmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{bmatrix} + \cos A \begin{bmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{bmatrix}$

$\Rightarrow$ $\begin{bmatrix} \sin^2 A & - \sin A \cos A \\ \sin A \cos A & \sin^2 A \end{bmatrix} + \begin{bmatrix} \cos^2 A & \cos A \sin A \\ - \cos A \sin A & \cos^2 A \end{bmatrix}$

$\Rightarrow$ $\begin{bmatrix} \sin^2 A + \cos^2 A & - \sin A \cos A + \cos A \sin A \\ \sin A \cos A - \cos A \sin A & \sin^2 A + \cos^2 A \end{bmatrix}$

$\Rightarrow$ $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

(b) Let coordinates of $M$ is $(x_1, 0)$ and $N$ is $(0, y_2)$. Point $P$ divides $MN$ in $2 : 3$ ratio

$\therefore 3 =$ $\frac{3x_1+ 2 \times 0}{3+2}$ $\Rightarrow 3x_1 = 15 \Rightarrow x_1 = 5$

and $2 =$ $\frac{3 \times 0 + 2 \times y_2}{3+2}$ $\Rightarrow 3y_2 = 10 \Rightarrow y_2 = 5$

Therefore i) The coordinates of $M$ is $(5, 0)$ and $N(0, 5)$

ii) Slope of line $MN =$ $\frac{0-5}{5-0}$ $= - 1$

(c)  Let the radius of the sphere is $r_1$ and radius of cylinder is $r_2$ and the height of the cylinder is $h$.

Therefore  Volume of sphere $=$ Volume of cylinder

$\Rightarrow$ $\frac{4}{3}$ $\pi {r_1}^3 = \pi {r_2}^2 h$

$\Rightarrow$ $\frac{4}{3}$ $( {6}^3) = {r_2}^2 (32)$

$\Rightarrow {r_2}^2 =$ $\frac{12 \times 6}{8}$ $= 9$

$\Rightarrow r_2 = 3$

Therefore Radius of the cylinder is $3$ cm

Curved surface area $= 2 \pi r h = 2 \times 3.1 \times 3 \times 32 = 595.2 \ cm^2$

$\\$

Question 4:

(a) The following numbers, $K + 3, K + 2, 3K - 7$ and $2K - 3$ are in proportion. Find $K$.

(b) Solve for $x$ the quadratic equation $x^2 - 4x - 8 = 0$. Give your answer correct to three significant figures

(c) Use ruler and compass only for answering this question.

Draw a circle of radius $4$ cm. Mark the center as $O$. Mark a point $P$ outside the circle at a distance of $7$ cm from the center. Construct two tangents to the circle from the external point $P$.

Measure and write down the length of any one tangent.

(a)  Given $K + 3, K + 2, 3K - 7$ and $2K - 3$ are in proportion.

$\Rightarrow$ $\frac{K+3}{K+2}$ $=$ $\frac{3K-7}{2K-3}$

$\Rightarrow (K+3)(2K-3) = (3K-7)(K+2)$

$\Rightarrow 2K^2 + 6K - 3K - 9 = 3K^2 - 7K + 6K - 14$

$\Rightarrow 2K^2 + 3K - 9 = 3K^2 - K - 14$

$\Rightarrow K^2 - 4K-5 = 0$

$\Rightarrow K^2 - 5K + K -5 = 0$

$\Rightarrow K(K-5) + (K-5) = 0$

$\Rightarrow (K-5)(K+1) = 0$

$\Rightarrow K = 5 \ or \ K = -1$

(b)  Given $x^2 - 4x - 8 = 0$

Comparing the above equation by $ax^2 + bx + c = 0$

$a =1, b = -4, c = -8$

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

$x =$ $\frac{4 \pm \sqrt{(-4)^2 - 4(1)(-8)} }{2(1)}$

$x =$ $\frac{4 \pm \sqrt{16+32} }{2}$

$x =$ $\frac{ 4 \pm \sqrt{48} }{2}$

$x =$ $\frac{4 \pm 6.928}{2}$

$x = 5.464$ or $x = 1.464$

(c)  i. Draw a line segment $OP = 7$ cm

ii. With center $O$ and radius $4$ cm, draw a circle.

iii. Draw the mid point of $OP$.

iv. With center $M$ and diameter $OP$, draw a circle which intersect the circle at $T$ and $S$.

v. Joint $PT$ and $PS$

$PT$ and $PS$ are the required tangent on measuring the length of $PT = PS = 5.74$ cm

SECTION B (40 Marks)
Attempt any four questions from this Section

Question 5:

(a) There are $25$ discs numbered $1$ to $25$. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by $2$ and $3$ both.

(iii) a number less than $16$.

(b) Rekha opened a recurring deposit account for $20$ months. The rate of interest is $9\%$ per annum and Rekha receives Rs. $441$ as interest at the time of maturity. Find the amount Rekha deposited each month.

(c) Use a graph sheet for this question. Take $1$ cm $= 1$ unit along both $x$ and $y$ axis.

(i) Plot the following points: $A(0,5), B(3,0), C(1,0)$ and $D(1,-5)$

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

(iii) Write down the coordinates of $B', C'$ and $D'$.

(iv) Join the points $A, B, C, D, D', C', B', A$ in order and give a name to the closed figure $ABCDD'C'B'$.

(a) Total number of cases $= 25$

(i) an odd number $\{ 1,3,5,7,9,11,13,15,17,19,21,23,25 \}$

Therefore Probability of an odd number $=$ $\frac{13}{25}$

ii) Divisible by $2$ and $3$ both $\{ 6,12,18,24 \}$

Probability of the number divisible by $2$ and $3$ both $=$ $\frac{4}{25}$

iii) Probability a number less than $16 =$ $\frac{15}{25}$ $=$ $\frac{3}{5}$

(b)  Let the monthly installment i.e $P =$ Rs. $x$

Since $n = 20$ months and $r = 9\%$

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

$=$ $\frac{x \times 20(20+1)}{2 \times 12}$ $\times$ $\frac{9}{100}$

$=$ $\frac{x \times 20 \times 21 }{2 \times 12}$ $\times$ $\frac{9}{100}$

$=$ $\frac{x \times 5 \times 7 }{2 }$ $\times$ $\frac{9}{100}$

Interest  $= 1.575x$ Rs.

$\Rightarrow 441 = 1.575x \Rightarrow x = 280$ Rs.

(c)  Please refer to the graph below for answers. the shape of the figure is “Arrow Head”.

$\\$

Question 6:

(a) In the given figure, $\angle PQR = \angle PST = 90^o, PQ = 5$ cm and $PS = 2$ cm.

(i) Prove that $\triangle PQR \sim \triangle PST$.

(ii) Find Area of $\triangle PQR :$ Area of quadrilateral $SRQT$.

(b) The first and last term of a Geometrical Progression (G.P.) are $3$ and $96$ respectively. If the common ratio is $2$, find:

(i) $'n'$ the number of terms of the G.P.

(ii) Sum of the $n$ terms.

(c) A hemispherical and a conical hole is scooped out of a solid wooden cylinder. Find the volume of the remaining solid where the measurements are as follows:

The height of the solid cylinder is $7$ cm, radius of each of hemisphere, cone and cylinder is $3$ cm. Height of cone is $3$ cm. Give your answer correct to the nearest whole number. Take $\pi =$ $\frac{22}{7}$.

(a)

i)    To prove $\triangle PQR \sim \triangle PST$

Consider $\triangle PQR$ and $\triangle PST$

$\angle PQR = \angle PST = 90^o$     (Given)

$\angle P$ is common

$\therefore \triangle PQR \sim \triangle PST$   (By AA criterion)

ii)   $\frac{ Ar \triangle PQR}{ Ar \triangle PST }$ $=$ $\frac{ \frac{1}{2} \times PQ \times QR}{\frac{1}{2} \times PS \times ST}$ $=$ $\frac{5}{2}$ $\times$ $\frac{5}{2}$ $=$ $\frac{25}{4}$                      $\Big[ \because$ $\frac{PQ}{PS}$ $=$ $\frac{QR}{ST}$ $\Big]$

Taking the reciprocals on both sides

$\frac{ Ar \triangle PST }{ Ar \triangle PQR}$ $=$ $\frac{4}{25}$

Now deducting both sides by $1$

$1 -$ $\frac{ Ar \triangle PST }{ Ar \triangle PQR}$ $= 1 -$ $\frac{4}{25}$

$\Rightarrow$ $\frac{ Ar \triangle PQR - Ar \triangle PST }{ Ar \triangle PQR}$ $=$ $\frac{25 - 4}{25}$

$\Rightarrow$ $\frac{ Ar \ of \ Quadrilateral \ QRST }{ Ar \triangle PQR}$ $=$ $\frac{21}{25}$

(b) Given first term $(a) = 3$ and last term $(l) = 96$

i)    $r = 2$

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

$\Rightarrow 96 = 3 \times 2^{n-1}$

$\Rightarrow 32 = 2^{n-1}$

$\Rightarrow 2^5 = 2^{n-1}$

$\therefore n = 6$

ii) Sum of the $n$ terms $= (S_n) =$ $\frac{a r^{n-1}}{r-1}$ $=$ $\frac{(3) 2^{6-1}}{2-1}$ $= 3 \times 63 = 189$

(c) Required volume $=$ Volume of cylinder $-$ Volume of hemisphere $-$ Volume of cone

Volume of cone $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\times \pi \times (3)^2 \times 3 = 9 \pi \ cm^3$

Volume of hemisphere $=$ $\frac{2}{3}$ $\pi r^3 =$ $\frac{2}{3}$ $\times \pi \times 3^3 = 18 \pi \ cm^3$

Volume of cylinder $= \pi r^2 h = \pi \times 3^3 \times 7 = 63 \pi \ cm^3$

Therefore required volume $= 63 \pi - 18\pi - 9 \pi = 36 \pi = 36 \times$ $\frac{22}{7}$ $= 113.14 \ cm^3$

$\\$

Question 7:

(a) In the given figure $AC$ is a tangent to the circle with center $O$. If $\angle ADB = 55^o$, find $x$ and $y$. Give reasons for your answers.

(b) The model of a building is constructed with the scale factor $1 : 30$.

(i) If the height of the model is $80$ cm, find the actual height of the building in meters.

(ii) If the actual volume of a tank at the top of the building is $27 \ m^3$, find the volume of the tank on the top of the model.

(c) Given $\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$, where $M$ is a matrix and $I$ is unit matrix of order $2x^2$.

(i) State the order of matrix $M$.

(ii) Find the matrix $M$.

(a) In $\triangle ABD, \angle BAD = 90^o$

$\therefore \angle ABD+ \angle BAD+ \angle ADB = 180^o$  [Sum of angles in a triangle]

$\therefore \angle ABD + 90^o + 55^o= 180^o$

$\therefore \angle ABD= 35^o$

Also $\angle AOE = 2\angle ABE = 2\angle ABD$

$\Rightarrow y = 70^o$

In $\triangle AOC,$

$\angle OAC + x + y = 180^o$

$\therefore 90^o + x + 70^o=180^o$

$\therefore 90^o + x + 70^o=180^o$

$\Rightarrow x = 20^o$

(b)

i) If the scale factor is $1 : 30$, then actual height will be $30$ times the height of the model.

The height of the model is $80$ cm

Therefore actual height $= 80$ cm $\times 30 = 2400$ cm $=24 \ m$

ii) Now, actual volume of a tank will be $(30)^3$ times the volume of a tank in the model.

Therefore Volume of tank in model $\times (30)^3 =$ Actual volume of a tank

Therefore volume of tank $=$ $\frac{27}{30^3}$ $\ m^3 = 0.001 \ m^3 = 1000 \ cm^3$

(c) Given

i) $\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ $M = 6I$

$\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ $M = 6$ $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ $M =$ $\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

Therefore $M$ has the order of $2 \times 2$

ii) Let us assume $M =$ $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\therefore$ $\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ = $\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

$\Rightarrow$ $\begin{bmatrix} 4a+2c & 4b+2d \\ -a + c & -b+d \end{bmatrix}$ = $\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

$\Rightarrow -a+c = 0 \Rightarrow a = c$

Similarly, $4b+2d = 0 \Rightarrow d = - 2b$

Now, $4a + 2c = 6 \Rightarrow 4a + 2a = 6 \Rightarrow 6a = 6 \Rightarrow a = 1$

also, $-b + d = 6$ $\Rightarrow -b + (-2b) = 6$ $\Rightarrow -3b =6$ $\Rightarrow b= -2$

$\therefore a =1, b =- 2, c =1, d = 4$

$\\$

Question 8:

(a) The sum of the first three terms of an Arithmetic Progression (A.P.) is $42$ and the product of the first and third term is $52$. Find the first term and the common difference.

(b) The vertices of a $\triangle ABC$ are $A(3, 8), B(-1, 2)$ and $C(6, -6)$. Find:

(i) Slope of $BC$.

(ii) Equation of a line perpendicular to $BC$ and passing through $A$.

(c) Using ruler and a compass only construct a semi-circle with diameter $BC = 7$ cm. Locate a point $A$ on the circumference of the semicircle such that $A$ is equidistant from $B$ and $C$. Complete the cyclic quadrilateral $ABCD$, such that $D$ is equidistant from $AB$ and $BC$. Measure $\angle ADC$ and write it down.

(a)  Let the three times terms of an A.P. be $(a - d ),a, (a + d)$

Sum: $42= (a - d)+ a + (a + d) \Rightarrow 42 = 3a \Rightarrow a = 14$

Also $(a -d) (a + d) = 52$

$\Rightarrow a^2 - d^2 = 52$

$d^2 = a^2 - 52 = 196 - 52 = 144$

$\Rightarrow d = \pm 12$

When $d = 12, a = 14- 12 = 2$

When $d = - 12, a = 14 + 12 = 26$

Hence the two APs are $2, 14, 26, \cdots$ and $26, 14, 2, \cdots$

(b) Given $A(3, 8), B(-1, 2), C(6, -6)$

i) Slope  of BC $=$ $\frac{y_2 - y_1}{x_2 - x_1}$ $=$ $\frac{-6-2}{6-(-1)}$ $= -$ $\frac{8}{7}$

ii) Slope of line perpendicular to $BC =$ $\frac{-1}{-\frac{8}{7}}$ $=$ $\frac{7}{8}$

Therefore required line is $y - y_1 = m(x-x_1)$

$\Rightarrow y - 8 =$ $\frac{7}{8}$ $(x-3)$

$\Rightarrow 8y - 64 = 7x - 21$

$\Rightarrow 7x - 8y + 43 = 0$

(c)

i. Draw a line segment $BC = 7$ cm

ii. Taking mid point of $BC$ as center $O$ , draw a semi-circle with radius $= 3.5$ cm

iii. Now, the semicircle circumscribes the $\triangle ABC$

iv. Draw angle bisector of $\angle ABC$ and make it intersect the semi-circle at $D$.

v. Measure the angle $\angle DBC$ which comes out to be $22.5^o$

$\\$

Question 9:

(a) The data on the number of patients attending a hospital in a month are given below. Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as $45$. Give your answer correct to $2$ decimal places.

 Number of Patients 10-20 20-30 30-40 40-50 50-60 60-70 Number of Days 5 2 7 9 2 5

(b) Using properties of proportion solve for $x$, given $\frac{\sqrt{5x} + \sqrt{2x-6}}{\sqrt{5x} + \sqrt{2x-6}}$ $= 4$

(c) Sachin invests Rs. $8500$ in $10\%$, Rs. $100$ shares at Rs. $170$. He sells the shares when the price of each share rises by Rs. $30$. He invests the proceeds in $12\%$ Rs. $100$ shares at Rs. $125$. Find:

(i) the sale proceeds.

(ii) the number of Rs. $125$ shares he buys.

(iii) the change in his annual income.

(a)

 Number of Patients Number of Days $(f_i)$ Mid Value $(x_i)$ Assumed Mean $A = 45$  $\therefore d = x_i - A$ $f_i \times d_i$ 10-20 5 15 -30 -150 20-30 2 25 -20 -40 30-40 7 35 -10 -70 40-50 9 45 0 0 50-60 2 55 10 20 60-70 5 65 20 100 Total 30 -140

Mean $= A +$ $\frac{ \Sigma f_id_i }{f_i}$ $= 45 +$ $\frac{-140}{30}$ $= 45 -$ $\frac{14}{3}$ $=$ $\frac{121}{3}$ $= 40.33$

(b) Given $\frac{\sqrt{5x} + \sqrt{2x-6}}{\sqrt{5x} + \sqrt{2x-6}}$ $= 4$

Using componendo and dividendo on both sides

$\frac{(\sqrt{5x} + \sqrt{2x-6})+ (\sqrt{5x} + \sqrt{2x-6})}{\sqrt{5x} + \sqrt{2x-6}) - (\sqrt{5x} + \sqrt{2x-6})}$ $=$ $\frac{4+1}{4-1}$

$\frac{\sqrt{5x}}{\sqrt{2x-6}}$ $=$ $\frac{5}{3}$

On squaring both sides

$\frac{5x}{2x-6}$ $=$ $\frac{25}{9}$

$\Rightarrow 45x = 50x -150$

$\Rightarrow 150 = 5x$

$\Rightarrow x = 30$

(c)

i) Market value of shares $= 170$ Rs.

Therefore the number of shares bought $=$ $\frac{8500}{170}$ $= 50$

Total face value of $50$ shares $= 50 \times 100 = 5000$ Rs.

Income form these shares $=$ $\frac{10}{100}$ $\times 5000 = 500$ Rs.

Selling price of the shares $= 170 + 30 = 200$ Rs.

Sale proceed $= 50 \times 200 = 10000$ Rs.

ii) Market value of the shares bought $= 125$ Rs.

Therefore number of Rs. $125$ shares bought $=$ $\frac{10000}{125}$ $= 80$

Total face value of $80$ shares $= 80 \times 100 = 8000$ Rs.

iii) Income from Rs. $125$ shares $=$ $\frac{12}{100}$ $\times 8000 = 960$ Rs.

Therefore increase in income $= 960 - 500 = 460$ Rs.

$\\$

Question 10:

(a) Use graph paper for this question. The marks obtained by $120$ students in an English 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 Number of Students 5 6 16 22 26 18 11 6 4 3

Draw the ogive and hence, estimate:

(i) the median marks.

(ii) the number of students who did not pass the test if the pass percentage was $50$.

(iii) the upper quartile marks.

(b) A man observes the angle of elevation of the top of the tower to be $45^o$. He walks towards it in a horizontal line through its base. On covering $20$ m the angle of elevation changes to $60^o$. Find the height of the tower correct to $2$ significant figures.

(a)

 Class Interval Frequency Cumulative Frequency 0-10 5 5 10-20 9 14 20-30 16 30 30-40 22 52 40-50 26 78 50-60 18 96 60-70 11 107 70-80 6 113 80-90 4 117 90-100 3 120

Here $n = 120$, therefore $\frac{n}{2}$ $= 60$

i) Median $= 43$

ii) Number of students who failed $= 78$

iii) The upper Quartile marks $=$ $\frac{3}{4}$ $n^{th}$ term $= 90^{th}$ term

(b) Let the height of the tower be $h$ m.

In $\triangle ADC, \tan 45^o =$ $\frac{h}{20+x}$

$\Rightarrow 1 =$ $\frac{h}{20+x}$

$\Rightarrow h = 20 + x$   … … … … … i)

In $\triangle BDC, \tan 60^o =$ $\frac{h}{x}$

$\sqrt{3} =$ $\frac{h}{x}$

$\Rightarrow x =$ $\frac{h}{\sqrt{3}}$   … … … … … ii)

Using i) and ii), we get

$h = 20 +$ $\frac{h}{\sqrt{3}}$

$\Rightarrow h -$ $\frac{h}{\sqrt{3}}$ $= 20$

$\Rightarrow h \Big($ $\frac{\sqrt{3} -1 }{\sqrt{3}}$ $\Big) = 20$

$\Rightarrow h = 47.32$ m

Therefore Height of the tower $= 47.32$ m

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Question 11:

(a) Using the Remainder Theorem find the remainders obtained when $x^3 + (kx + 8)x + k$ is divided by $x + 1$ and $x - 2$. Hence find $k$ if the sum of the two remainders is $1$.

(b) The product of two consecutive natural numbers which are multiples of $3$ is equal to $810$. Find the two numbers.

(c) In the given figure, $ABCDE$ is a pentagon inscribed in a circle such that $AC$ is a diameter and side $BC \parallel AE$. If $\angle BAC = 50^o$, find giving reasons:

(i) $\angle ACB$

(ii) $\angle EDC$

(iii) $\angle BEC$

Hence prove that $BE$ is also a diameter.

(a) Remainder theorem:

Dividend $=$ Divisors $\times$ Quotient $+$ Remainder

$\therefore Let f (x) = x^3 + (kx + 8) x + k = x^3 + kx^2 + 8x + k$

Dividing $f (x)$ by $x + 1$ gives remainder as $R_1$

$\therefore f(-1) = R_1$

Also $f(2) = R_2$

$\therefore f(-1) = (-1)^3 + k(-1)^2 + 8(-1) + k = -1 + k - 8 + k = 2k-9 = R_1$

$f(2) = (2)^3 + k (2)^2 + 8 \times 2 + k = 8 + 4k + 16 + k = 5k + 24 = R_2$

Sum of remainder $= R_1 + R_2 = 1$

$\therefore 2k-9 + 5k + 24 = 1$

$\Rightarrow 7k + 15 = 1$

$\Rightarrow k = -2$

(b)    Let the numbers be $3x, 3(x+1)$

Therefore $(3x) \times 3(x+1) = 810$

$9(x^2 + x) = 810$

$x^2 + x = 90$

$x^2 + x - 90 = 0$

$x^2 + 10x - 9 x - 90 = 0$

$x(x+10) - 9 (x+10) = 0$

$(x+10)(x-9) = 0$

$x = 9$ or $x = -10$

But $x \neq -10$, because $3x$ must be natural number.

Therefore $x = 9$

Therefore the numbers are $27, 30$

(c)  Let $\angle ACB = x, \angle EDC = y$ and $\angle BEC = z$

We know $\angle ABC = 90^o$     Angle in a semi-circle

i) In $\triangle ABC, \angle ABC + \angle BAC + \angle ACB = 180^o$

$\Rightarrow 90^o + 50^o + \angle ACB = 180^o$

$\Rightarrow \angle ACB = 40^o= x$

ii) We know $\angle EAC = \angle ACB$

Alternate interior angles since $AC$ is transversal to parallel  lines  $AE$ and $BC$

$\therefore \angle EAC = 40^o$

Also $\angle EAC + \angle EDC = 180^o$

$\therefore 40^o + \angle EDC = 180^o$

$\Rightarrow \angle EDC = 140^o = y$

iii)  $\angle EBC + \angle EDC = 180^o$

$\Rightarrow \angle EBC + 140^o = 180^o$

$\Rightarrow \angle EBC = 40^o$

Therefore in $\triangle EBC$

$\angle BEC + \angle ECB + \angle EBC =180^o$

$\angle BEC + 90^o + 40^o =180^o$

$\angle BEC = 50^o$

Also in $\triangle EAB$

$\angle EAB = \angle EAC + \angle BAC = 40^o + 50^o$

$\angle EAB = 90^o$

We know, if an angle of a triangle in a circle is $90^o$ then, the hypotenuse must be the diameter of the circle. Hence, $BE$ is the diameter of the circle.