Other Solved Mathematics Board Papers

MATHEMATICS (ICSE – Class X Board Paper 2010)

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 inequation and. represent the, solution set on the number line.

$-3 < - \frac{1}{2} - \frac{2x}{3} \le \frac{5}{6}, x \in R$     [3]

(b)  Tarun bought an article for $Rs. \ 8000$ and, spent $Rs. \ 1000$ for transportation. He marked the article at $Rs. \ 11700$ and, sold it to a customer. If the customer had. To pay $10\%$ sales tax, find

(i) the customer’s price

(ii) Tarun’s profit percent     [3]

(c)  Mr. Gupta opened a recurring deposit account in a bank. He deposited $Rs. \ 2500$ per month for two years. At the time of maturity he got $Rs. \ 67500$. Find. :

(i) the total interest earned by Mr Gupta.

(ii) the rate of interest per annum     [4]

(a)  $-3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}$

$-3 < -\frac{1}{2}-\frac{2x}{3}$

$-18 < -3 -4x$

$4x < 15$  or  $x < \frac{15}{4}$

$-\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}$  or  $-3-4x \leq 5$

$-8 \leq 4x$  or  $-2 \leq x$

Therefore $\{ x: -2 \leq x < \frac{15}{4}, x \in R \}$

(b)  Cost Price  $= Rs. 8000$

Overheads $= Rs. 1000$

Listed Price $= Rs. 11700$

Sales Tax rate $= 10\%$

(i) Customer Price $= 11700+11700 \times \frac{10}{100} = Rs. 12870$

(ii) Profit $=$ $\frac{11700-8000-1000}{9000}$ $= 30\%$

(c)  $P = Rs. \ 2500, \ no \ of \ months = 24, \ \\ \\ rate = r\% \ Maturity Amount = Rs. 67500$

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

$67500 =2500 \times 24 +2500 \times$ $\frac{24(24+1)}{2 \times 12} \times \frac{r}{100}$

$r =$ $\frac{(67500-2500 \times 24) \times (2 \times 12) \times 100}{2500 \times 24 \times 25}$ $\Rightarrow r=12\%$

$Interest =2500 \times$ $\frac{24(24+1)}{2 \times 12} \times \frac{12}{100}$ $= Rs. \ 7500$

$\\$

Question 2:

(a) Given $A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}$,   $B = \begin{bmatrix} 6 \\ 1 \end{bmatrix}$, $C = \begin{bmatrix} -4 \\ 5 \end{bmatrix}$ and $D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$. Find $AB+2C-4D$.     [3]

(b)  Nikita invests $Rs. \ 6000$ for two years at a certain rate of interest compounded annually, At the end of first year it amounts to $Rs. \ 6720$. Calculate;

(i) The rate of interest;

(ii) The amount at the end of the second year.     [3]

(c)  $A \ and \ B$ are two points on the $x-axis \ and \ y-axis$ respectively. $P (2, -3)$ is the mid-point of $AB$ . Find the

(i) Co-ordinates of $A \ and \ B$

(ii) Slope of line $AB$

(iii) Equation of line $AB$     [4]

(a) $AB+2C-4D$

$= \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}. \begin{bmatrix} 6 \\ 1 \end{bmatrix} + 2 \begin{bmatrix} -4 \\ 5 \end{bmatrix} - 4 \begin{bmatrix} 2 \\ 2 \end{bmatrix}$

$= \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}$

$= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

(b)  Compound Interest for 1 year

$P=6000\ Rs.; \ r=x\%; Compounded \ yearly \ n=1 \ year$

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{x}{100})^{1}$

Given $6000(1+\frac{x}{100})^{1}=6720 \Rightarrow x= 12\%$

Amount at the end of second year

$A=P(1+\frac{r}{100})^{1} \Rightarrow A= 6000(1+\frac{12}{100})^{2} = 7526.40 \ Rs.$

(c) Let $A(x,0)$ and $B(0, y)$

$P(2, -3)$ is the mid point

(i) Therefore  $2 = \frac{0+x}{2} \Rightarrow x = 4$

$-3 = \frac{y+0}{2} \Rightarrow y = -6$

Hence $A(4,0)$ and $B(0, -6)$

(ii) Slope of $AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}$

(iii) Equation of $AB$

$y = 0 = \frac{3}{2} (x-4)$

$2y = 3x-12$

$\\$

Question 3:

(a) Cards marked with numbers 1, 2, 3, 4 … 20 are well shuffled and a card is drawn at random. What is the probability that the number of the cards is

(i) a prime number

(ii) divisible by 3

(iii) a perfect square     [3]

(b)  Without using trigonometric tables evaluate :

$\frac{sin \ 35^o cos \ 55^o + cos \ 35^o sin \ 55^o}{cosec^2 \ 10^o - tan^2 \ 80^o}$     [3]

(c)  Use graph paper for this question $A (0, 3), B (3, -2) \ and \ O (0, 0)$ are the vertices of $\triangle ABO$.

(i) Plot $D$ the reflection of $B$ in the $y-axis$, and write its co-ordinates.

(ii) Give the geometrical name of the figure $ABOD$.

(iii) Write the equation of the line of symmetry of the line $ABOD$     [4]

(a)   Given: Cards marked with numbers 1,2, … , 20

$n(S) = 20$

(i) Prime Numbers $= 2, 3, 5, 7, 11, 13, 17, 19$

$n(E) = 8$

P (Prime number) = P(A) = $\frac{ n(E) }{ n(S) } = \frac{8}{20}$ $= 0.4$

(ii) No. divided by $3 = 3, 6, 9, 12, 15, 18$

$n(E) = 6$

P (no. divided by 3) =  P(A) = $\frac{ n(E) }{ n(S) } = \frac{6}{20}$ $= 0.3$

(iii) No. perfect square $= 1, 4, 9, 16$

$n(E) = 4$

P (Perfect square) = P(A) = $\frac{ n(E) }{ n(S) } = \frac{4}{20}$ $= 0.2$

(b)  Given: $\frac{sin \ 35^o cos \ 55^o + cos \ 35^o sin \ 55^o}{cosec^2 \ 10^o - tan^2 \ 80^o}$

$=$ $\frac{sin \ 35^o cos \ (90^o- 35^o) + cos \ 35^o sin \ (90^o- 35^o)}{cosec^2 \ 10^o - tan^2 \ (90^o - 10^o)}$

$=$ $\frac{sin^2 \ 35^o+cos^2 \ 35^o}{1+ cot^2 \ 10^o - cot^2 \ 10^o}$

$=$ $\frac{1}{1}$ $= 1$

(c)

(i)  Plotted above

(ii) The coordinates of $D (-3, -2)$

(iv) $x = 0$ is the equation of line of symmetry

$\\$

Question 4:

(a) When divided by $(x-3)$ the polynomials $x^3-px^2+x+6$ and $2x^3-x^2-(p+3)x-6$ leave the same remainder. Find the value of $p$.     [3]

(b) In the figure, given below, $AB$ and $CD$ are two parallel chords and $O$ is the center. If the radius of the circle is $15\ cm$, find the distance $MN$ between the two chords of lengths $24\ cm$ and $18\ cm$ respectively.      [3]

(c)  The distribution given below shows the marks obtained by $25$ students in an aptitude test. Find the mean, median and mode of the distribution.     [4]

 Marks obtained 5 6 7 8 9 10 No. of students 3 9 6 4 2 1

(a)  When $x=3$

Remainder1 $= (3)^3-p(3)^2+(3)+6$

$= 27-9p+9$

$= 36-9p$

Remainder2 $= 2(3)^3-(3)^2-(p+3)(3)-6$

$=54-9-3p-9-6$

$=30-39$

Given Remainder1 = Remainder 2

$36-9p=30-3p$

$6=6p \Rightarrow p =1$

(b) We know that perpendicular drawn from the center of the circle will bisect the chord.

Therefore $MO^2 = 15^2-12^2 \\ = 225 - 144 = 81 \Rightarrow MO = 81 \ cm$

$ON^2 = 15^2-9^2 = 225 - 81 = 12 \\ \Rightarrow ON = 12 \ cm$

Hence $MN = 9 + 12 = 21 \ cm$

(c)

 $x_i$ $f_i$ $x_if_i$ $cf$ 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32 22 9 2 18 24 10 1 10 25 $\Sigma = 25$ $\Sigma x_if_i = 171$

Mean = $\frac{\Sigma x_if_i}{N} = \frac{171}{25}$ $= 6.84$

Since $N = 25$ is odd,

Median = $(\frac{n+1}{2})^2 \ term = 13^{th} \ term = 7$

Mode $= 6$ (maximum frequency)

$\\$

Section B [40 Marks]

Answer any four questions in this section.

Question 5:

(a) Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal.$px^2-4x+3=0$     [3]

(b) Rohit borrows $Rs. \ 86000$ from Arun for $2$ years at $5\%$ per annum simple interest. He immediately lends his money to Akshay at $5\%$ compounded interest annually for the same period. Calculate Rohits profit at the end of two years.     [3]

(c) Mrs. Kapoor opened a Saving Bank Account in State Bank of India on 9th January 2008. Her passbook entries for the year 2008 are given below:

 Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.) Jan. 9, 2008 By Cash – 10,000 10,000 Feb. 12, 2008 By Cash – 15,500 25,500 April 6, 2008 To Cheque 3,500 – 22,000 April 30, 2008 To Self 2,000 – 20,000 July 16, 2008 By Cheque – 6,500 26,500 Aug. 4, 2008 To Self 5,500 – 21,000 Aug. 20, 2008 To Cheque 1,200 – 19,800 Dec. 12, 2008 By Cash – 1,700 21,500

Mrs. Kapoor closed the account on 31st December 2008. If the bank pays interest at 4% per annum, find the interest he receives on closing the account. Give your answer correct to the nearest rupee.     [4]

(a)  Comparing $px^2-4x+3=0$ with $ax^2+bx+c=0$, we get $a = p, b = -4 \ and \ c =3$

For roots to be equal, we should have $b^2-4ac = 0$

$(-4)^2-4(p)(3)=0$

$16-12p=0$

$p=\frac{4}{3}$

(b)  Simple Interest for 2 years

$S.I. = P \times \frac{r}{100} \times 2 \ Rs.$

$S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \ Rs.$

Compound Interest for 2 years

$P=8600\ Rs.; \ r=5\%; Compounded \ yearly \ n=2 \ year$

$A=P(1+\frac{r}{100})^{2} \Rightarrow A= 86000(1+\frac{5}{100})^{2} \Rightarrow A = 9481.50 \ Rs.$

Gain $=(94815-86000)-8600 = 215 \ Rs.$

(c)  Qualifying principal for various months:

 Month Principal (Rs.) January 10000 February 10000 March 25500 April 20000 May 20000 June 20000 July 20000 August 19800 September 19800 October 19800 November 19800 Total 204700

$P = Rs. \ 204700 \ \ R = 4.0\% \ and \ T= \frac{1}{12}$

$I = P \times R \times T = 204700 \times \frac{4}{100} \times \frac{1}{12} = Rs. \ 682.33 \ or \ Rs. 682$

$\\$

Question 6:

(a) A manufacturer marks an article at $Rs. \ 5000$. He sells this article to a wholesaler at a discount of $25 \%$ on the marked price and the wholesaler sells it to a retailer at a discount of $15 \%$ on its marked price. If the retailer sells the article without any discount and at each stage the sales-tax is $8 \ %$, calculate the amount of VAT paid by:

i. The Wholesaler

ii. The Retailer     [3]

(b) In the following figure O is the center of the circle and AB is a tangent to it at  point B. $\angle BDC = 65^o$.  Find $\angle BAO$.     [3]

(c) A doorway is decorated as shown in the figure. There are four semi-circles. $BC$, the diameter of the larger Semi-circles. $BC$ the diameter of the larger semi-circle is of length $84 \ cm$. Center of the three equal semi-circles lie on $BC$. $ABC$ is an isosceles triangle with $AB = AC$. If $BO = OC$, find the area of the shaded, region.     [4]

(a)  Market Price of the article for wholesaler $= Rs. \ 5000$

Cost Price of the article for wholesaler $= \frac{100-25}{100} \times 5000 =Rs. \ 3750$

Amount of tax paid by the wholesaler $= \frac{8}{100} \times 3750 = Rs. \ 300$

Cost Price of the article for retailer $= \frac{100-15}{100} \times 5000 =Rs. \ 4250$

Amount of tax paid by the retailer $= \frac{8}{100} \times 4250 = Rs. \ 340$

Selling price of the article for retailer $= Rs. \ 5000$

Amount of tax paid by the end customer $= \frac{8}{100} \times 5000 = Rs. \ 400$

Vat paid by retailer $= 400 - 340 = Rs. 60$

Vat paid by wholesaler $= 340 - 300 = Rs. 40$

(b) $AB$ is a tangent to the circle.

$\Rightarrow \angle ABO = 90^o$

$\angle BDC = 65^o$ (given)

$\angle BCD = 90^o-65^o = 25^o$ (angle at the center)

$\angle BOE = 2 \times 25^o = 50^o$

$\angle BAO = 90^o - \angle BOE$

$\therefore \angle BAO = 90^o - 50^o = 40^o$

(c)  Let $AB = AC = x \ cm$

As angle in semi-circle is $90^o$

$\therefore \ \angle A = 90^o$

In right angled $\triangle ABC$, by Pythagoras theorem, we get

$AB^2+AC^2 = BC^2$

$x^2+x^2 = 84^2$

$2x^2 = 84 \times 84$

$x^2 = 84 \times 42$

Also, Area of $\triangle ABC = \frac{1}{2} \times AB \times AC$

$= \frac{1}{2} \times 84 \times 42 = 1764 \ cm^2$

Diameter of semicircle $(2r) = 84 \ cm$

$Radius(r) = \frac{1}{2} \times 84 = 42 \ cm$

Area of semi-circle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 42 \times 42 = 2772 \ cm^2$

Diameter of each (three equal) semi-circles $= \frac{1}{3} \times 84 = 28 \ cm$

Radius of the 3 equal semi-circles $= \frac{1}{2} \times 28 = 14 \ cm$

Therefore Area of three equal semi circles $= \frac{1}{3} \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 = 924 \ cm^2$

Area of shaded region = Area of semi-circles + Area of three equal circles – Area of $\triangle ABC$

$= 2772 + 924 - 1764 = 3696 - 1764 = 1932 \ cm^2$

$\\$

Question 7:

(a)  Use ruler and compasses only for this question :

(i) Construct $\triangle ABC$, where $AB = 3.5 \ cm, \ BC = 6 \ cm$ and $\angle ABC = 60^o$.

(ii) Construct the locus of points inside the triangle which are equidistant from $BA$ and $BC$.

(iii) Construct the locus of points inside the triangle which are equidistant from $B$ and $C$.

(iv) Mark the point $P$ which is equidistant from $AB, \ BC$ and also equidistant from $B$ and $C$. Measure and record the length of $B$.     [3]

(b) The equation of a line is $3x + 4y - 7 = 0$ . Find:

(i) Slope of the line.

(ii) The equation of a line perpendicular to the given line and passing through the intersection of the lines $x -y + 2 = 0$ and $3x + y- 10 = 0$     [3]

(c) The mean of the following distribution is 52 and the frequency of class interval 30-40 is $f$. Find $f$     [4]

 Class Interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 5 3 $f$ 7 2 6 13

(a)

(i) Step 1: Draw a line $BC \ of \ length = 6 cm$. The with $B$ as the vertex, draw an angle of $60^o$ using a compass.

Step 2: Draw $BA$ of length $3.5 \ cm$. Join $A \ and \ C$. That gives is the $\triangle ABC$.

(ii) Points which are equidistant from $BA$ and $BC$ are on the angle bisector of $\angle ABC$. Therefore draw angle bisector of $\angle ABC$

(iii) Points which are on the perpendicular bisector of $BC$ are equidistant from point $B$ and $C$. Draw a perpendicular bisector of $BC$.

(iv) The point where the angle bisector and the perpendicular bisector intersect is the point that is equidistant from $AB$ and $BC$ but also equidistant form points $B$ and $C$.

(b) (i) Slope $= -\frac{3}{4}$

(ii) Slope of perpendicular $= \frac{4}{3}$

For point of intersection solve $x -y + 2 = 0$ and $3x + y- 10 = 0$

$y = 4$ and $x = 2$

Therefore intersection $= (2, 4)$

Therefore equation of line

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

$3y-12 = 4x-8$

$3y = 4x + 4$

(c)

 Interval Frequency $(f_i)$ $x-i$ $d_i = x_i - A$ $f_id_i$ 10-20 5 15 -30 -150 20-30 3 25 -20 -60 30-40 F 35 -10 -10f 40-50 7 45 (A) 0 0 50-60 2 55 10 20 60-70 6 65 20 120 70-80 13 75 30 390 $36+f$ $\Sigma f_id_i= 320 - 10f$

$Mean = A +$ $\frac{\Sigma f_id_i}{N}$

$\Rightarrow 52 = 45 +$ $\frac{320-10f}{36+f}$

$\Rightarrow 7 =$ $\frac{320-10f}{36+f}$

$\Rightarrow 252 + 7f = 320 - 10f$

$\Rightarrow 17f = 68$

$\Rightarrow f = 4$

$\\$

Question 8:

(a) Use the remainder theorem to factorize the following expression: $2x^3+x^2-13x+6$.     [3]

(b) If $x, y \ and \ z$ are in continued proportion, prove that: $\frac{(x+y)^2}{(y+z)^2} =\frac{x}{y}$     [3]

(c) From the top of a light house $100 \ m$ high the angles of depression of two ships on opposite sides of it are $48^o$ and $36^o$ respectively. Find the distance between the two ships to the nearest meter.     [4]

(a) Let $x =2$

Remainder $= 2(2)^3+(2)^2-13(2)+6 = 16+4-26+6=0$

Hence $(x-2)$ is a factor of  $2x^3+x^2-13x+6$

• $x-2 ) \overline {2x^3+x^2-13x+6} (2x^2+5x-3$
•  $(-) \ \ \underline {2x^3-4x^2}$
•                   $5x^2-13x+6$
•          $(-) \ \ \underline{5x^2-10x}$
•                              $-3x + 6$
•                      $(-) \ \ \underline{ -3x+6}$
•                                      $\times$

$2x^3+x^2-13x+6 = (x-2)(2x^2+5x-3)$

$= (x-2)(2x^2+6x-x-3)$

$= (x-2)[2x(x+3)-(x+3)]$

$= (x-2)(x+3)(2x-1)$

Hence $2x^3+x^2-13x+6 = (x-2)(x+3)(2x-1)$

(b)   If $x, y \ and \ z$ are in continued proportion, then

$\frac{x}{y}=\frac{y}{z}$ $\Rightarrow x =$ $\frac{y^2}{z}$

Applying componendo and dividendo

$\frac{x+y}{x-y}=\frac{y+z}{y-z}$

$\Rightarrow \frac{x+y}{y+z}=\frac{x-y}{y-z}$

Squaring both sides

$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=(\frac{x-y}{y-z})^2$

$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=(\frac{x-y}{y-z})^2$

Substituting

$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=(\frac{\frac{y^2}{z}-y}{y-z})^2$

$\Rightarrow \frac{(x+y)^2}{(y+z)^2}=(\frac{y^2-yz}{z(y-z)})^2= \frac{y^2}{z^2} = \frac{zx}{z^2}=\frac{x}{z}$

(c)   In $\triangle ABD: tan \ 48^o = \frac{AD}{BD}$

$\Rightarrow 1.1 = \frac{100}{BD}$

$\Rightarrow BD = 90.09 \ m$

In $\triangle ACD tan \ 36^o = \frac{AD}{CD}$

$\Rightarrow 0.7265 = \frac{100}{CD}$

$\Rightarrow DC = 137.64 \ m$

Therefore $BC = BD + CD = 90.09 + 137.64 = 227.73 \ m$

$\\$

Question 9:

(a) Evaluate  $\begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}$     [3]

(b) In the given figure, $ABC$ is a triangle with $\angle EDB = \angle ACB$. Prove that $\triangle ABC \sim \triangle EBD$. If $BE=6 \ cm, EC = 4 \ cm, BD = 5 \ cm$ and area of $\triangle BED = 9 cm^2$. Calculate the:

(i) length of $AB$

(ii) area of $\triangle ABC$      [3]

(c) Vivek invests $Rs. \ 4500$  in $8\%$ , $Rs. \ 10$  shares at $Rs. \ 15$ . He sells the shares when the price rises to $Rs. \ 30$ , and invests the proceeds in $12\% \ Rs. \ 100$  shares at $Rs. \ 125$ . Calculate; i) The sale proceeds  ii) The number of $Rs. \ 125$  shares he buys;  iii) The change in his annual income from dividend.     [4]

(a)  $\begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}$

$= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}$

$= \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}$

(b)  Consider $\triangle ABC \ and \ \triangle EBD$

$\angle EDB = \angle ACB$ (given)

$\angle DBE = \angle ABC$ (common)

Therefore $\angle DEB = \angle BAC$

$\triangle ABC \sim \triangle EBD$   (AAA postulate)

(i) Given $BE=6\ cm, EC=4\ cm, BD=5\ cm$

$\frac{AB}{ EB} = \frac{BC}{ BD} = \frac{AC }{ED}$

$AB = \frac{BE+EC }{5} \times 6 = 2 \ cm$

(ii)  $\frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle EBD}=\frac{AB^2}{EB^2}=\frac{144}{36}$

Area of  $\triangle ABC = \frac{144}{ 36} \times 9 = 36 \ cm^2$

(c)  First Investment

Let the amount invested $= 4500 \ Rs.$

Nominal Value of the share $= 10 \ Rs.$

Market Value of the share $= 15 \ Rs.$

Dividend earned $= 8\%$

Number of shares bought $= \frac{4500}{15} = 300$

Sale Proceed $= 300 \times 30 = 9000 \ Rs.$

Dividend earned $= 300 \times 10 \times \frac{8}{100} = 240 \ Rs.$

Second Investment

Therefore  the amount invested $= 9000 \ Rs.$

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

Market Value of the share $= 125 \ Rs.$

Dividend earned $= 12\%$

Number of shares bought $= \frac{9000}{125} = 72$

Dividend earned $= 72 \times 100 \times \frac{12}{100} = 720 \ Rs.$

Hence the change in income $= 720-240 = 480 \ Rs.$

$\\$

Question 10:

(a) A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.     [4]

(b)  The monthly income of a group of 320 employees in a company is given below:

 Monthly Income No. of Employees 6000-7000 20 7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5

Draw an ogive of the given. distribution on a graph sheet taking $2 \ cm = Rs. \ 1000$ on one axis and $2 \ cm = 50 \ employees$ on the other axis. From the graph determine:

(i) the median wage

(ii) the number of employees whose income is below $Rs. \ 8500$

(iii) If the salary of a senior employee is above $Rs. \ 11500$ find the number of senior employees in the company.

(iv) The upper quartile     [6]

(a)  Let the two parts be $x \ and \ y$

Given $x^2+y^2 = 20$

Also $y^2 = 8x$

Substituting it back $x^2+8x-20 = 0$

$\Rightarrow x = 2 or -10$   (ignore this as the number is positive)

Therefore the larger part is $y^ = 16 \Rightarrow y = 4$

Hence the number is $2+4 = 6$

(b)

 Monthly Income No. of Employees Cumulative Frequency 6000-7000 20 20 7000-8000 45 65 8000-9000 65 130 9000-10000 95 225 10000-11000 60 285 11000-12000 30 315 12000-13000 5 320

Here n (no. of employees) $= 320$ (even)

(i) Median $= \frac{1}{2} \{ \frac{n}{2} + (\frac{n}{2} + 1) \} = \frac{1}{2} (160+161) = 160.5$

Required median $= Rs. \ 9800$ (from graph)

(ii) Number of employees whose income is below $Rs. \ 8500 = 95$ approx

(iii) Number of senior employees in the company $= 320 - 305 = 15$

(iv) Upper Quartile $= (\frac{3n}{4})^{th} = \frac{3 \times 320}{4} = 240^{th}$term

Upper Quartile $= Rs. \ 10200$

$\\$

Question 11:

(a) Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.      [3]

(b) A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.     [3]

(c) Given $x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$ . Use componendo and dividendo to prove that: $x^2=\frac{2a^2 x}{x^2+1}$.     [4]

(a)  Given side of the hexagon is $4 \ cm$. 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 $4 \ cm$
• 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)  Given: Diameter of hemispherical bowl $= 7.2 \ cm$

Radius of hemispherical bowl $= 3.6 \ cm$

Volume of hemispherical bowl $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (3.6)^3 = 97.76 \ cm^3$

Volume of cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (4.8)^2 \times h = 24.14h \ cm^3$

Volume of cone \$latex = Volume of hemi-sperical bowl

$\therefore 24.14h = 97.76 \Rightarrow h = 4.05 \ cm$

(c)  Given $x=$ $\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$

Applying componendo and dividendo

$\frac{x+1}{x-1}=\frac{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})+(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}{(\sqrt{a^2+b^2}+\sqrt{a^2-b^2})-(\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 })}$

Simplifying

$\frac{x+1}{x-1}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2-b^2 }}$

Square both sides

$\frac{x^2+1+2x}{x^2-2x+1}=\frac{a^2+b^2}{a^2-b^2}$

Applying componendo and dividendo

$\frac{x^2+1+2x+x^2-2x+1}{x^2+1+2x-x^2+2x-1}=\frac{a^2+b^2+a^2-b^2}{a^2+b^2-a^2+b^2}$

$\frac{2(x^2+1)}{4x}=\frac{2a^2}{2b^2}$

$\frac{x^2+1}{2x}=\frac{a^2}{b^2}$

Simplifying

$b^2 =$ $\frac{2a^2x}{x^2+1}$

$\\$