Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2016)

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) Using remainder theorem, find the value of $k$ if on dividing $2x^3+3x^2-kx+5$ by $x-2$, leaves a remainder of $7$[3]

(b) Given $A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix}$  and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$   and $A^2=9A+mI$. Find $m$[4]

(c) The mean of the following numbers is $68$. Find the value of $x: 45, 52, 60, x, 69, 70, 26, 81 \ and \ 94$. Hence estimate the medium. [3]

(a)    Let $f(x)=2x^3+3x^2-kx+5$

By remainder theorem, when  $f(x)$ is divided by  $(x-2)$ means  $x-2=0$, $\Rightarrow x=2$, then the remainder is  $7$.

Therefore $2(2)^3+3(2)^2-k(2)+5 = 7$

$\Rightarrow 16+12-2k+5=7$

$\Rightarrow 2k = 26$

$\Rightarrow k = 13$

$\\$

(b)     Given $A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$   and $A^2=9A+mI$.

LHS  $= A^2 = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix}$

$= \begin{bmatrix} 4 & 0 \\ -2-7 & 0+49 \end{bmatrix}$

$= \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}$

RHS $= 9A+mI= 9 \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix}+ \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}$

$= \begin{bmatrix} 18+m & 0 \\ -9 & 63+m \end{bmatrix}$

Since LHS = RHS

$\begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}= \begin{bmatrix} 18+m & 0 \\ -9 & 63+m \end{bmatrix}$

$\Rightarrow 4= 18+m \Rightarrow m = -14$

Also $\Rightarrow 49 = 63+m \Rightarrow m = -14$

Hence the value of $m = -14$

$\\$

(c)   $Mean$ $= \frac{Sum \ of \ Variates}{Number \ of \ Variates}$

$\Rightarrow 68$ $= \frac{45+52+60+x+69+70+26+81+94}{9}$

$\Rightarrow 612=497+x$

$\Rightarrow x = 115$

Now arrange the numbers in ascending order we get

$26, 45, 52, 60, 69, 70, 81, 94, 115$

Number of terms $(n) = 9$ (odd number of terms)

Median $= Value \ of \ the \ (\frac{n+1}{2})^th$terms

$= Value \ of \ the (\frac{9+1}{2})^{th}$ term

$= Value \ of \ the \ (5)^{th} = 69$ (median)

$\\$

Question 2

(a) The slope of a line joining $P(6, k) \ and \ Q(1-3k, 3) \ is$ $\frac{1}{2}$. Find: i)  $k$ ii) Midpoint of $PQ$ using the value of $k$ found in (i). [3]

(b)  Without using trigonometrical tables evaluate: $cosec^2 \ 57^o-tan^2 \ 33^o+cos 44^o \ cosec \ 46^o-\sqrt{2} cos \ 45^o-tan^2 \ 60^o$ [4]

(c) A certain number of metallic cones, each of radius $2 \ cm$ and height $B \ cm$ are melted and, recast into a solid, sphere of radius $6 \ cm$. Find, the number of cones. [3]

(a)  i)     Let $P(6, k)=(x_1, y_1) \ and \ Q(1-3k, 3)= (x_2, y_2)$

Given: Slope of $PQ = \frac{1}{2}$

Formula for slope of $PQ$ $= \frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow \frac{3-k}{1-3k-6}= \frac{1}{2}$

$\Rightarrow \frac{3-k}{-3k-5}= \frac{1}{2}$

$\Rightarrow 6-2k=-3k-5$

$\Rightarrow k = -11$

ii)    Let $P(6, k)=(x_1, y_1) \ and \ Q(1-3k, 3)= (x_2, y_2)$

Given: Slope of $PQ = \frac{1}{2}$

Formula for slope of $PQ$ $= \frac{y_2-y_1}{x_2-x_1}$

$\Rightarrow \frac{3-k}{1-3k-6}= \frac{1}{2}$

$\Rightarrow \frac{3-k}{-3k-5}= \frac{1}{2}$

$\Rightarrow 6-2k=-3k-5$

$\Rightarrow k = -11$

$\\$

(b)   Given:

$Cosec^2 \ 57^o-tan^2 \ 33^o+cos 44^o \ cosec \ 46^o-\sqrt{2} cos \ 45^o-tan^2 \ 60^o$

$= Cosec^2 \ (90-33)^o-tan^2 \ 33^o+cos (90-46)^o \ cosec \ 46^o-\sqrt{2} cos \ 45^o-tan^2 \ 60^o$

$= Sec^2 \ 33^o-tan^2 \ 33^o+sin \ 46^o . \frac{1}{ sin \ 46^o } -\sqrt{2} . \frac{1}{\sqrt{2}}-(\sqrt{3})^2$

$= 1+1 -1-3 = -2$

$\\$

(c)  Let number of cones $= n$

Volume of cones = Volume of sphere [Provide Formulas]

$\frac{1}{3} \pi r^2 h \times n = \frac{4}{3} \pi r^3$

$\frac{1}{3} \pi 2^2 \times 3 \times n = \frac{4}{3} \pi 6^3$

$\Rightarrow n$ $= \frac{\frac{4}{3} \pi 6^3}{\frac{1}{3} \pi 2^2 \times 3}$

$\Rightarrow n = \frac{216}{3}$ $= 72$

Therefore the number of cones needed is $72$

$\\$

Question 3

(a) Solve the following inequation, write the solution set and represent it on number line. [3]

$-3x(x-7) \geq 15-7x \ge \frac{x+1}{3}, x \in R$

(b) In the given figure below, $AD$ is the diameter. $O$ is the center of the circle. $AD$ is parallel to  $BC$ and  $\angle CBD = 32^o$. Find:  i) $\angle OBD$  ii) $\angle AOB$  iii) $\angle BED$ [4]

(c) If $(3a+2b): (5a+3b) =18:29$$Find \ a:b$ [3]

(a)    $-3x(x-7) \geq 15-7x \ge$  $\frac{x+1}{3}$

$-3x+21 \geq 15-7x \ge$  $\frac{x+1}{3}$

Therefore we have:

$-3x+21 \geq 15-7x$

$\Rightarrow 4x \geq -6$

$\Rightarrow x \geq -\frac{3}{2}$ … … … … … i)

Also we have

$15-7x \ge$  $\frac{x+1}{3}$

$\Rightarrow 45-21x \ge x+1$

$\Rightarrow 44 \ge 22x$

$\Rightarrow 2 \ge x$… … … … … i)

Combining i) and ii) we get

Solution Set $=\{x: -\frac{3}{2} \geq x \ge 2 \ and \ x \in R \}$

$\\$

(b)    Given  $AD \parallel BC$

Therefore $\angle ADB = \angle DBC = 32^o$  (alternate angles)

Since $OB=OD$  (radius of the same circle)

Therefore  $\angle OBD = 32^o$  (angles opposite to equal side of the triangle are equal)

$\angle AOB =2 \angle ODB = 2 \times 32^o$  (angle at the center is twice that subtended at the circumference)

In $\triangle ABC$

$\angle ABD = 90^o, \angle ADB = 32^o$

$\angle BAD = 180-90-32 = 58^o$   (sum of the angles of a triangle is $180^o$)

$\angle BAD = \angle BED$  (angle in the same segment)

Therefore $\angle BED = 58^o$

$\\$

(c)    $(3a+2b):(5a+3b)= 18:29$

$\frac{3a+2b}{5a+3b}=\frac{18}{29}$

Cross multiplying

$\Rightarrow 87a+58b=90a+54b$

$3a=4b$

Or $\frac{a}{b}=\frac{4}{3}$

$\\$

Question 4

(a) A game of number has cards marked with $11, 12, 13, \ldots 40.$ A card is drawn at random. Find the probability that the number  on the card  drawn is:   i) A perfect square ii) Divisible by $7$ [3]

(b) Use graph paper for, this question. (Take $2 cm = 7$ unit along both $x \ and \ y \ axis$).  Plot the point  $O (0, 0), A(-4, 4), B(-3, 0) \ and \ C (0, -3)$[4]

Reflect points $A \ and \ B$ on the y axis and name them $A' \ and \ B'$ respectively.  i) Write down their coordinates.  ii) Name the figure $OABCB'A'$  iii) State the line of symmetry of this figure

(c) Mr. Lalit invested $Rs. 5000$ at a certain rate of interest, compounded annually for two years. At the end of first year it amounts $Rs. 5325$. Calculate:  i) The rate of interest.  ii) The amount at the end of second year, to the nearest rupee. [3]

(a)   Total number of all possible outcomes $= 30$

Formula used: $Probability$ $= \frac{number \ of \ favorable \ outcome}{total \ number \ of \ possible \ outcome}$

i)    The cards with perfect squares are: $16, 25 \ and \ 36$

The number of favorable outcomes $= 3$

$P (A \ perfect \ square) = \frac{3}{30}=\frac{1}{10}=0.10$

ii)   The cards with numbers divisible by $7$ are:: $14, 21, 28 \ and \ 35$

Therefore the number of favorable outcomes $= 4$

$P (Divisible \ by \ 7) = \frac{4}{30} = \frac{2}{15}$

$\\$

(b)  i) $A'(4, 4) \ and \ B'(3, 0)$

iii) $y-axis$ is the line of symmetry

$\\$

(c)        Given: Principal $= Rs. \ 5000$, Time $= 2 \ years$, After one year amount = $\ 5325 \ Rs.$

i)    We know that Amount $(A) = P + I$

For $1^{st} \ year: 5325=5000+\frac{PRT}{100}$

$\Rightarrow R= \frac{325}{50} = 6.5\%$

ii)   Amount (A) at the end of $2^{nd} \ year$

Formula for compound interest: $A = P (1+\frac{r}{100})^n$

Given  $P = Rs. \ 5000, r=6.5\% \ (calculated \ in \ (i)) \ and \ n = 2$

Therefore  $A = 5000 (1+\frac{6.5}{100})^2$

$\Rightarrow A = 5000 \times (\frac{106.5}{100})^2 = 5671.125 \ Rs.$

Hence Amount at the end of 2 years $= Rs. \ 5671$

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5

(a) Solve the quadratic equation $x^2-3(x+3) =0$. Give answer correct to three significant figures. [3]

(b) A page form the saving bank account of Mrs. Ravi is given below:

 Date Particulars Withdrawal (Rs.) Deposit (Rs.) Balance (Rs.) April 3rd 2006 B / F – – 6,000 April 7th By Cash – 2,300 8,300 April 15th By Cheque – 3,500 11,800 May 20th To Self 4,200 – 7,600 June 10th By Cash – 5,800 13,400 June 15th To Self 3,100 – 10,300 August 13th By Cheque – 1,000 11,300 August 25th To Self 7,400 – 3,900 September 6th 2006 By Cash – 2,000 5,900

She closed the account an $30th \ September, \ 2006$. Calculate the interest Mrs. Ravi earned, at the end of $30th \ September, \ 2006 \ at \ 4.5\%$ per annunm interest. Hence find the amount she receives on closing the account. [4]

(c)  In what time will $Rs. 1500$ yield $Rs. 1996.50$ as compound interest at $15\%$ per annum compounded annually ? [3]

(a)  Given $x^2-3(x+3) =0$

$\Rightarrow x^2-3x-9=0$

Comparing $x^2-3x-9=0$ with $ax^2+bx+c=0$, we get $a = 1, b = -3 \ and \ c =-9$

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

Therefore $x$ $= \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}$

$\Rightarrow x$ $= \frac{3 \pm \sqrt{45}}{2}$

Solving we get $x = 4.85, -1.85$

$\\$

(b)  Qualifying principal for various months:

 Month Qualifying Principal (Rs.) April 8300 May 7600 June 10300 July 10300 August 3900 Total 40400

$P = Rs. \ 40400 R = 4.5\% \ and \ T= \frac{1}{12}$

$I = P \times R \times T = 40400 \times \frac{4.5}{100} \times \frac{1}{12} = Rs. \ 151.50$

Amount received on 30th September (on closing the account)

$Amount = Last \ Balance + Interest$

$Amount = 5900+151.50 = 6051.50 \ Rs.$

$\\$

(c)  Given $P=1500 \ Rs.; A= 1996.50 \ Rs. ; r=10\%; n=n$

$A=P(1+\frac{r}{100})^n \Rightarrow 1996.50 = 1500(1+\frac{10}{100})^n$

$(\frac{11}{10})^3= (\frac{11}{10})^n$

$\Rightarrow n = 3 \ years$

$\\$

Question 6

(a) Construct a regular hexagon of side $5 \ cm$. Hence construct all its lines of symmetry and name them. [3]

(b) In the given figure $PQRS$ is a cyclic quadrilateral $PQ \ and \ RS$ produced meet at point $T$.  i) Prove $\triangle TPS \sim \triangle TRQ$  ii) Find. $SP \ if \ TP = 18 \ cm,$ $\ RQ = 4 \ cm \ and, \ TR = 6 \ cm$  iii) Find area, of quadrilateral $PORS$ if area of  $\triangle PTS=27 \ cm^2$ [4]

(c)  Given matrix $A = \begin{bmatrix} 4 \ sin\ 30^o & cos \ 0^o \\ cos \ 0^o & 4 \ sin\ 30^o \end{bmatrix}$  and $B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

If, $AX=B$: i) Write the order of matrix $X$. ii) Find the matrix $X$[3]

(a)     Step 1: Given side of the hexagon is $5 \ cm$. Construct the hexagon as follows:

• First draw a line using a ruler of length 5 cm. Mark it AB.
• The using a compass, make an arc of $5 \ cm$
• Using compass, draw 120 degree angle and cut the line into 5 cm lengths using the compass. Continue this until the hexagon is completed
• One the hexagon is completed, draw perpendicular bisectors of each of the arms of the hexagon. You will get mid point for each of the arms of hexagon as marked $(U, V, W, X, Y, Z)$
• Now join the mid points of the opposite sides to get lines $(UV, WX, \ and \ YZ)$
• Now draw the diagonals passing through the center to get lines $(AD, BE \ and \ CF)$

The six lines of symmetry $(UV, WX, YZ, AD, BE \ and \ CF).$

$\\$

(b)  (i) Consider $\triangle TPS$ and $\triangle TRQ$

Since exterior angle of cyclic quadrilateral is equal to the interior opposite angle, we get

$\angle PST = \angle RQT$

$\angle SPQ = \angle QRT$

$\angle T$ is common

Therefore $\triangle TPS \sim \triangle TRQ$

(ii) Since $\triangle TPS \sim \triangle TRQ$

$\frac{TP}{TR}=\frac{PS}{RQ}$

$\Rightarrow PS = RQ \times \frac{TP}{TR} = 4 \times \frac{18}{6} = 12 \ cm$

(iii) Since $\triangle TPS \sim \triangle TRQ$

$\frac{Area \ of \ \triangle TPS}{Area \ of \ \triangle TRQ}=\frac{TP^2}{TR^2}$

$\Rightarrow Area \ of \ \triangle TRQ = Area \ of \ \triangle TPS \times \frac{TR^2}{TP^2} = 27 \times \frac{6^2}{18^2} = 3 \ cm^2$

Therefore Area of quadrilateral $PQRS = Area \ of \ \triangle TPS - Area \ of \ \triangle TQR = 27-3 = 24 \ cm ^2$

$\\$

(c)    Given matrix $A = \begin{bmatrix} 4 \ sin\ 30^o & cos \ 0^o \\ cos \ 0^o & 4 \ sin\ 30^o \end{bmatrix}$  and $B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

If, $AX=B$:

(i)    From the given matrix, order of $A = 2 \times 2$ and order of  $B = 2 \times 1$

We know

$A_{m \times n} \times X_{m \times n} = B_{m \times n}$

or $A_{2 \times 2} \times X_{m \times n} = B_{2 \times 1}$

Hence $m = 2 and n = 1$

Therefore the order of $X \ is \ 2 \times 1$

(ii)  Let $X = \begin{bmatrix} a \\ b \end{bmatrix}$

Therefore

$\begin{bmatrix} 4 \ sin\ 30^o & cos \ 0^o \\ cos \ 0^o & 4 \ sin\ 30^o \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

$\begin{bmatrix} 4a \ sin\ 30^o +b cos \ 0^o \\ a \ cos \ 0^o + 4b \ sin\ 30^o \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

$4a \ sin\ 30^o +b cos \ 0^o=4$

$\Rightarrow 2a+b=4$ … … … … … i)

$a \ cos \ 0^o + 4b \ sin\ 30^o = 5$

$\Rightarrow a+2b=5$ … … … … … ii)

Solving i) and ii) we get $a = 1 \ and \ b = 2$

Therefore $X = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

$\\$

Question 7

(a) An aero plane is at a height of $1500$  meters finds that two ships are sailing towards it in the same direction. The angle of depression as observed from the aero plane are $45^o \ and \ 35^o$  respectively.  Find the distance between the two ships. [4]

(b) The table shows the distribution of the scores obtained by $160$  shooters in a shooting competition.  Use a graph sheet and draw an ogive of the distribution.

(Take $2 cm = 10$  scores on the $x \ axis$  and $2 cm = 20$  shooters on the $y-axis$).

 Scores 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No of shooters 9 13 20 26 30 22 15 10 8 7

Use your graph to estimate the following:  i) The median.  ii) The interquartile range.  iii) The number of shooters who obtained a score of more than $85\%$ [6]

(a)     In the adjacent diagram, let $D$ be the air plane and $A \ and \ B$ be the two ships.

Given: $CD=1500 \ m$

$\angle EDA = 30^o$

$\angle EDB = 45^o$

Consider $\triangle DBC$

$tan \ 45^o = \frac{DC}{BC} = \frac{1500}{BC} \Rightarrow BC = 1500 \ m$

Consider $\triangle ADC$

$tan \ 30^o = \frac{DC}{AC} = \frac{1500}{AB+BC} \Rightarrow AB+BC = 1500 \sqrt{3}$

Therefore $AB = 1500 \sqrt{3}-1500 = 1098 \ m$.

Hence, distance between two ships $= 1098 \ m$.

$\\$

(b)

 Scores No of Shooters Cumulative Frequency (c.f.) 0-10 9 9 10-20 13 22 20-30 20 42 30-40 26 68 40-50 30 98 50-60 22 120 60-70 15 135 70-80 10 145 80-90 8 153 90-100 7 160

(i)    $N = 160$ (which is even)

Median $= (\frac{N}{2})^{th} \ Term = 80^{th} \ Term$

From the graph, Median $= 43.5$

(ii) Lower Quartile $= (\frac{N}{4})^{th} \ Term = 40^{th} \ Term = 28.5$

Upper Quartile $= (\frac{3N}{4})^{th} \ Term = 120^{th} \ Term = 60$

Inter Quartile range $60-28.5 = 31.5$

(iii) The number of shooter who obtained a score of more $85\% = 160-150 = 10$ (using the graph)

$\\$

Question 8

(a) If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$  show that  $\frac{x^3}{a^3}+ \frac{y^3}{b^3}+ \frac{z^3}{c^3} = \frac{3xyz}{abc}$ [3]

(b) Draw, a line $AB = 5 \ cm$. Mark a point $C \ on \ AB$ such that $AC = 3 \ cm$. Using a ruler and compass only, construct:

(i) A circle of radius $2.5 \ cm$, passing through $A \ and \ C$.

(ii) Construct two tangents to the circle from the external point $B$. Measure and record the length of the tangents. [4]

(c) A line $AB$ meets the $x-axis$ at $A$ and $y-axis$ at $B$.  $P(4, -1)$ divides $AB$ in the ratio $1 : 2.$

(i) Find the coordinates of $A \ and \ B$.

(ii) Find the equation of the line through $P$ and perpendicular to $AB$. [3]

(a)   Let $\frac{x}{a}=\frac{y}{b}=\frac{z}{c} = k$

Therefore $= ak, y =bk, z=ck$

LHS $= \frac{x^3}{a^3}+ \frac{y^3}{b^3}+ \frac{z^3}{c^3}$

$= \frac{(ak)^3}{a^3}+ \frac{(bk)^3}{b^3}+ \frac{(ck)^3}{c^3}$

$= k^3+k^3+k^3 = 3k^3$

Since  $k =\frac{x}{a}, k =\frac{y}{b}, k =\frac{z}{c}$

Substituting

$= 3$ $\frac{x}{a} \frac{y}{b} \frac{z}{c} =$ RHS Hence proved.

$\\$

(b)

1. First draw line $AB = 5 \ cm$. This you can just draw by using a ruler.
2. From $AB$ cut $AC = 3 \ cm$. This can be done by using a compass. Set the compass at 3 cm and then make this arc cutting AB at C.
3. Draw perpendicular bisector of $AC$.
4. Taking $A$ as center and $AO = 2.5 \ cm$  radius, draw an arc to meet the perpendicular bisector of $AC \ at \ O$.
5. Taking $O$ as the center and $AO = 2.5 \ cm$ as radius, draw a circle, which passing through $A \ and \ C$
6. Join $BO$.
7. Draw right bisector of $BO$, which cuts $BO \ at \ P$.
8. Taking $P$ as center and $OP$ as radius, draw a circle which cuts old Circle at $R$ and  $T$.
9. Join $BR \ and \ BT$.

These are required tangents.

(ii)   Length of tangent $= 3 \ cm$.

$\\$

(c)

(i)    Let coordinates of $A \ and \ B$ be $(a, 0) \ and \ (0, b)$

The coordinate of a point $P (4, -1) \ on \ AB$ divides it in the ratio $1 : 2$

$AP: PB = 1:2$

If a point divides two points $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1:m_2$, then the coordinates of the point at

$x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$

$y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$

Therefore

$4 = \frac{1 \times 0+2 \times a}{1+2} \Rightarrow a = 6$

$-1 = \frac{1 \times b+2 \times 0}{1+2} \Rightarrow b = -3$

Therefore coordinates are $A(6,0) \ and \ B(0, -3)$

(ii)  Slope of AB $= \frac{y_2-y_1}{x_2-x_1}= \frac{-3-0}{0-6}= \frac{1}{2}$

Slope of perpendicular bisector $= \frac{-1}{ \frac{1}{2}} = -2$

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

$y-(-1)=-2(x-4)$

$\Rightarrow y+1=-2x+8$

$\Rightarrow 2x+y=7$

$\\$

Question 9

(a) A dealer buys an article at a discount of $30\%$ from the wholesaler, the marked price being $Rs. \ 6,000$. The dealer sells it to a shopkeeper at a discount of $10\%$ on the marked price. If the rate of VAT is $6\%$, find:

(i) The price paid by the shopkeeper including the tax.

(ii) The VAT paid by the dealer. [3]

(b)  The given figure represents a kite with a, circular and a Semicircular motifs stuck on it. The radius of circle is $2.5 \ cm$ and the semicircle is $2 \ cm$ . If diagonals $AC \ and \ BD$ are the lengths $12 \ cm \ and \ 8 \ cm$ respectively, find the area of the :

(c)  A model of a ship is made to a scale $1 : 300$ .

(i) The length of the model of the ship is $2 m$ . Calculate the length of the ship.

(ii) The area of the deck ship is $180,000 m^2$ . Calculate the area of the deck of the model.

(iii) The volume of the model is $6.5 m^3$ . Calculate the volume of the ship. [3]

(a) Given: Market Price $= Rs. 6,000$ Rate of discount  $=30\%$

Therefore the cost price of the article for the dealer

$= 6000 - \frac{30}{100} \times 6000$

$= 6000-1800 = 4200 \ Rs.$

Sales tax paid by the dealer to the wholesaler

$= \frac{6}{100} \times 4200 = 252 \ Rs.$

(i)   Discount rate for shop keeper $= 10\%$

The cost price for the shopkeeper

$= 6000-\frac{10}{100} \times 6000$

$= 6000 - 600 = 5400$

The price paid by the shopkeeper including the tax

$= 5400 + \frac{6}{100} \times 5400$

$= 5400 + 324 = 5724 \ Rs.$

(ii) VAT paid by the dealer = Tax charged – Tax paid

$= 324 - 252 = 72 \ Rs.$

$\\$

(b)

(i)    Given: Radius of circle $= 2.5 \ cm$ Radius of semicircle $= 2 \ cm$.

Area of shaded part = Area of semicircle + Area of circle

$= \frac{1}{2} \pi r^2 + \pi r^2$

$= \frac{1}{2} \pi (2)^2+\pi (2.5)^2$

$= 2\pi + 6.24 \pi$

$= 8.25 \pi$

$= 25.92 cm^{2}$

(ii) Let kite $ABCD$ be a quadrilateral.

Area of quad $= \frac{1}{2} \times \ the \ product \ of \ the \ diagonals$

Area of kite $= \frac{1}{2} BD x AC = \frac{1}{2} \times 8 \times 12 = 48 \ cm^{2}$

Area of the unshaded part = Area of kite – Area of shaded part $= 48-26 = 22 \ cm^{2}$

$\\$

(c)  $Scale \ factor =$ $\frac{length \ of \ the \ model}{length \ of \ the \ ship}$

Length of the ship $= 2 \times 300 = 600 \ m$

$(Scale factor)^2 =$  $\frac{area \ of \ the \ model}{area \ of \ the \ ship}$

Area of the model $= \frac{180000}{90000} = 2 m^2$

$(Scale factor)^3 =$  $\frac{Volume \ of \ the \ model}{Volume \ of \ the \ ship}$

Volume of the ship $= 6.5 \times (300)^3 = 175500000 cm^3$

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Question 10

(a) Mohan has a recurring deposit account in a bank for $2$  years at $6\%$  p.a. simple interest. If he gets $Rs. \ 1200$  as interest at the time of maturity, find:

(i) the monthly installment

(ii) the amount of maturity [3]

(b) The histogram below represents the scores obtained by $25$  students in a Mathematics mental test. Use the data to:

(i) Frame a frequency distribution, table.

(ii) To calculate mean.

(iii) To determine the Modal class. [4]

(c) A bus covers a distance of  $240$  km at a uniform speed. Due to heavy rain its speed gets reduced by $10 \ km/hr$  and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be $x \ km/hr$ , form an equation and solve it to evaluate $x$ . [3]

(a) $n =$ the number of months for which the money is deposited $= 2 x 12 = 24$

$Rate = 6\%, \ Interest = 1200 \ Rs.$

(i)   Let the monthly instalment be $x$

Using the formula:

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

$1200 = x \times \frac{24(24+1)}{2 \times 12} \times \frac{6}{100}$

$1200 = x \times \frac{24 \times 25 }{2 \times 12} \times \frac{6}{100}$

$\Rightarrow x = \frac{2 \times 1200}{3} = 800$

Therefore the monthly instalment $= 800 \ Rs.$

(ii) The amount at maturity = Total money deposited + interest

$= 24 \times 800 + 1200 = 20400 \ Rs.$

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(b)   (i) Frequency Distribution Table

 C.I Frequency 0-10 2 10-20 5 20-30 8 30-40 4 40-50 6

(ii) Mean

 C.I $f$$f$ $x$$x$ $f.x$$f.x$ 0-10 2 5 10 10-20 5 15 75 20-30 8 25 200 30-40 4 35 140 40-50 6 45 270 $\Sigma f = 25$$\Sigma f = 25$ $\Sigma f.x = 695$$\Sigma f.x = 695$

Mean $\frac{\Sigma fx}{\Sigma f} = \frac{695}{25}= 27.8$

(iii) Model Class = 20-30 &s=0\$

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(c)   Let the speed of the bus for onward journey $= x \ km/hr$

Distance $= 240 \ km$

Time taken for onward journey $= \frac{240}{x} \ hrs.$

Due to rain the return speed of the bus $= (x-10) \ km/hr$

Time taken for the return journey $= \frac{240}{x-10} \ hrs$

Given that the return time is $2 \ hours$ more, we have

$\frac{240}{x-10} - \frac{240}{x} = 2$

$\frac{240x - 240x + 2400}{x(x-10)} = 2$

$2x^2 - 20x = 2400$

$x^2-10x-1200 = 0$

$x^2 - 40x +30x -1200 = 0$

$x(x-40)+30(x-40)=0$

$(x-40)(x+30) = 0$

$\Rightarrow x = 40 or -30 \ (not \ possible)$

Hence $x = 40 \ km/hr$

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Question 11

(a) Prove that $\frac{cos \ A}{1+sin \ A}$ $+tan \ A = sec \ A$ [3]

(b) Use ruler and compasses only for the following question. All construction lines and arcs much be clearly shown.

(i) Construct a $\triangle ABC$ in which $BC = 6.5 \ cm \ \angle ABC =60^o, AB=5 \ cm$ .

(ii) Construct the locus of points at a distance of $3.5 cm$ from $A$.

(iii) Construct the locus of points equidistant from $AC \ and \ BC$ .

(iv) Mark $2$ points $X \ and \ Y$ which are at a distance of $3.5 \ cm$ from $A$ and also equidistant from $AC \ and \ BC$ . Measure $XY$. [4]

(c) Ashok invested $Rs. 26,400$ on $12\%, Rs. \ 25$ shares of a company. If he receives a  dividend of $Rs. 2,475$ find the :

(i) number of shares he bought.

(ii) Market value of each share. [3]

(a) To prove $\frac{cos \ A}{1+sin \ A}$ $+tan \ A = sec \ A$

$= \frac{cos \ A}{1+sin \ A} + \frac{sin \ A}{cos \ A}$

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

$= \frac{1+ sin \ A}{ (1+ sin \ A) cos \ A}$

$= \frac{1}{cos \ A}$

= RHS Hence Proved

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(b)

1. Construct $\triangle ABC$. Draw a line BC which is 6.5  cm long. The using your compass, make an angle of $60^o$ at point B of line BC.
2. The taking the compass, make a cut a 5 cm mark on the side of the angle to mark A. AB is 5 cm long.
3. Taking $A$ as center and radius is $3.5$, draw a circle. This has been told in the question. Which is required locus.
4. Draw an angle bisector of $C$Angle bisector of $C$ cut above circle at two points i.e., $X and Y$These points are required points.
5. Measure the length of $XY = 5 cm$.

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(c)

(i)  Let number of shares be $x$

Face Value  of x shares $= 25 x$

Dividend $= \frac{25x \times 12}{100}= 3x$

Therefore $3x = 2475 \Rightarrow x = 825$

(ii) Market value $= \frac{26400}{825} = 32 \ Rs.$

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