2010 ICSE (Class 10) Board Paper Solution: Mathematics

MATHEMATICS (ICSE – Class X Board Paper 2010)

Two and Half Hour. Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.

The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions form Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the Answer. Omission of essential working will result in the loss of marks.

The intended marks for questions or parts of questions are given in brackets [ ].

Mathematical tables are provided.

SECTION A [40 Marks]

(Answer all questions from this Section.)

Question 1:

(a) Solve the following inequation and. represent the, solution set on the number line.

$\displaystyle -3 < - \frac{1}{2} - \frac{2x}{3} \le \frac{5}{6}, x \in R \hspace{5.0cm} [3]$

(b) Tarun bought an article for $\displaystyle \text{ Rs. } 8000$ and, spent $\displaystyle \text{ Rs. } 1000$ for transportation. He marked the article at $\displaystyle \text{ Rs. } 11700$ and, sold it to a customer. If the customer had. To pay $\displaystyle 10\%$ 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 $\displaystyle \text{ Rs. } 2500$ per month for two years. At the time of maturity he got $\displaystyle \text{ Rs. } 67500$ . Find. :

(i) the total interest earned by Mr Gupta.

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

Answers:

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

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

$\displaystyle -18 < -3 -4x$

$\displaystyle 4x < 15 \text{ or } x < \frac{15}{4}$

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

$\displaystyle -8 \leq 4x \text{ or } -2 \leq x$

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

(b) Cost Price $\displaystyle =\text{ Rs. } 8000$

Overheads $\displaystyle =\text{ Rs. } 1000$

Listed Price $\displaystyle =\text{ Rs. } 11700$

Sales Tax rate $\displaystyle = 10\%$

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

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

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

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

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

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

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

$\displaystyle \\$

Question 2:

$\displaystyle \text{(a) Given } A = \begin{bmatrix} 3 & -2 -1 & 4 \end{bmatrix} , B = \begin{bmatrix} 6 1 \end{bmatrix} , C = \begin{bmatrix} -4 5 \end{bmatrix} \text{ and } D = \begin{bmatrix} 2 2 \end{bmatrix} . \\ \\ \text{ Find } AB+2C-4D \hspace{10.0cm} [3]$

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

(i) The rate of interest;

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

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

(i) Co-ordinates of $\displaystyle A \text{ and } B$

(ii) Slope of line $\displaystyle AB$

(iii) Equation of line $\displaystyle AB$ [4]

Answers:

(a) $\displaystyle AB+2C-4D$

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

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

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

(b) Compound Interest for 1 year

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

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

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

Amount at the end of second year

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

$\displaystyle \text{(c) Let } A(x,0) \text{ and } B(0, y)$

$\displaystyle P(2, -3) \text{ is the mid point }$

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

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

$\displaystyle \text{Hence } A(4,0) \text{ and } B(0, -6)$

$\displaystyle \text{Slope of } AB = \frac{-6-0}{0-4} = \frac{-6}{-4} = \frac{3}{2}$

(iii) Equation of $\displaystyle AB$

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

$\displaystyle 2y = 3x-12$

$\displaystyle \\$

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 :

$\displaystyle \frac{\sin 35^{\circ} \cos 55^{\circ} + \cos 35^{\circ} \sin 55^{\circ}}{\mathrm{cosec}^2 10^{\circ} - \tan^2 80^{\circ}} \hspace{8.0cm} [3]$

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

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

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

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

Answers:

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

$\displaystyle n(S) = 20$

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

$\displaystyle n(E) = 8$

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

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

$\displaystyle n(E) = 6$

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

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

$\displaystyle n(E) = 4$

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

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

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

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

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

(c)

(i) Plotted above

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

(iii) Arrow Head

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

$\displaystyle \\$

Question 4:

(a) When divided by $\displaystyle (x-3)$ the polynomials $\displaystyle x^3-px^2+x+6 \text{ and } 2x^3-x^2-(p+3)x-6$ leave the same remainder. Find the value of $\displaystyle p \hspace{10.0cm} [3]$

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

$\displaystyle \left| \begin{array}{c | c | c | c | c | c |c } \hline \text{Marks Obtained} \hspace{1.0cm} & 5 \hspace{1.0cm} & 6 \hspace{1.0cm} & 7 \hspace{1.0cm} & 8 \hspace{1.0cm} & 9 \hspace{1.0cm} &10 \hspace{1.0cm} \\ \hline \text{No. of Students} \hspace{1.0cm} & 3 \hspace{1.0cm} & 9 \hspace{1.0cm} & 6 \hspace{1.0cm} & 4 \hspace{1.0cm} & 2 \hspace{1.0cm} &1 \hspace{1.0cm} \\ \hline \end{array} \right|$

Answers:

(a) When $\displaystyle x=3$

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

$\displaystyle = 27-9p+9$

$\displaystyle = 36-9p$

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

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

$\displaystyle =30-39$

Given Remainder1 = Remainder 2

$\displaystyle 36-9p=30-3p$

$\displaystyle 6=6p \Rightarrow p =1$

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

$\displaystyle \text{Therefore } MO^2 = 15^2-12^2 = 225 - 144 = 81 \Rightarrow MO = 81 cm$

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

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

(c)

$\displaystyle \left| \begin{array}{c | c | c | c } \hline x_i \hspace{1.0cm} & f_i & x_if_i & cf \\ \hline 5 \hspace{1.0cm} & 3 \hspace{1.0cm} & 15 \hspace{1.0cm} & 3 \hspace{1.0cm} \\ \hline 6 \hspace{1.0cm} & 9 \hspace{1.0cm} & 54 \hspace{1.0cm} & 12 \hspace{1.0cm} \\ \hline 7 \hspace{1.0cm} & 6 \hspace{1.0cm} & 42 \hspace{1.0cm} & 18 \hspace{1.0cm} \\ \hline 8 \hspace{1.0cm} & 4 \hspace{1.0cm} & 32 \hspace{1.0cm} & 22 \hspace{1.0cm} \\ \hline 9 \hspace{1.0cm} & 2 \hspace{1.0cm} & 18 \hspace{1.0cm} & 24 \hspace{1.0cm} \\ \hline 10 \hspace{1.0cm} & 1 \hspace{1.0cm} & 10 \hspace{1.0cm} & 25 \hspace{1.0cm} \\ \hline \ \hspace{1.0cm} & \Sigma = 25 \hspace{1.0cm} & \Sigma x_if_i = 171 \hspace{1.0cm} & \ \hspace{1.0cm} \\ \hline \end{array} \right|$

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

Since $\displaystyle N = 25$ is odd,

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

Mode $\displaystyle = 6$ (maximum frequency)

$\displaystyle \\$

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. $\displaystyle px^2-4x+3=0$ [3]

(b) Rohit borrows Rs. 86000 from Arun for2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohit’s 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:

$\displaystyle \left| \begin{array}{c | c | c | c | c } \hline \text{Date} \hspace{1.0cm} & \text{Particulars} & \text{ Withdrawls ( Rs.) } & \text{ Deposits ( Rs. ) } & \text{ Balance ( Rs. ) } \\ \hline \text{Jan. 9, 2008 } \hspace{1.0cm} & \text{By Cash } \hspace{1.0cm} & - \hspace{1.0cm} & 10000 \hspace{1.0cm} & 10000 \hspace{1.0cm} \\ \hline \text{Feb. 12, 2008 } \hspace{1.0cm} & \text{By Cash } \hspace{1.0cm} & - \hspace{1.0cm} & 15500 \hspace{1.0cm} & 25500 \hspace{1.0cm} \\ \hline \text{April 6, 2008 } \hspace{1.0cm} & \text{To Cheque } \hspace{1.0cm} & 3500 \hspace{1.0cm} & - \hspace{1.0cm} & 22000 \hspace{1.0cm} \\ \hline \text{April 30, 2008 } \hspace{1.0cm} & \text{To Self } \hspace{1.0cm} & 2000 \hspace{1.0cm} & - \hspace{1.0cm} & 20000 \hspace{1.0cm} \\ \hline \text{July 16, 2008 } \hspace{1.0cm} & \text{To Cheque } \hspace{1.0cm} & - \hspace{1.0cm} & 6500 \hspace{1.0cm} & 26500 \hspace{1.0cm} \\ \hline \text{ Aug. 4, 2008} \hspace{1.0cm} & \text{To Self } \hspace{1.0cm} & 5500 \hspace{1.0cm} & - \hspace{1.0cm} & 21000 \hspace{1.0cm} \\ \hline \text{Aug. 20, 2008 } \hspace{1.0cm} & \text{To Cheque } \hspace{1.0cm} & 1200 \hspace{1.0cm} & - \hspace{1.0cm} & 19800 \hspace{1.0cm} \\ \hline \text{Dec. 12, 2008 } \hspace{1.0cm} & \text{By Cash } \hspace{1.0cm} & - \hspace{1.0cm} & 1700 \hspace{1.0cm} & 21500 \hspace{1.0cm} \\ \hline \end{array} \right|$

Answers:

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 $\displaystyle px^2-4x+3=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = p, b = -4 \text{ and } c =3$

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

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

$\displaystyle 16-12p=0$

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

(b) Simple Interest for 2 years

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

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

Compound Interest for 2 years

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

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

Gain $\displaystyle =(94815-86000)-8600 = 215 \text{ Rs. }$

(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

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

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

$\displaystyle \\$

Question 6:

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

The Wholesaler
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. $\displaystyle \angle BDC = 65^{\circ}$ . Find $\displaystyle \angle BAO \hspace{10.0cm} [3]$

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

Answers:

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

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

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

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

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

$\displaystyle \text{Selling price of the article for retailer } =\text{ Rs. } 5000$

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

$\displaystyle \text{Vat paid by retailer } = 400 - 340 =\text{ Rs. } 60$

$\displaystyle \text{Vat paid by wholesaler } = 340 - 300 =\text{ Rs. } 40$

(b) $\displaystyle AB$ is a tangent to the circle.

$\displaystyle \Rightarrow \angle ABO = 90^{\circ}$

$\displaystyle \angle BDC = 65^{\circ}$ (given)

$\displaystyle \angle BCD = 90^{\circ}-65^{\circ} = 25^{\circ}$ (angle at the center)

$\displaystyle \angle BOE = 2 \times 25^{\circ} = 50^{\circ}$

$\displaystyle \angle BAO = 90^{\circ} - \angle BOE$

$\displaystyle \therefore \angle BAO = 90^{\circ} - 50^{\circ} = 40^{\circ}$

(c) Let $\displaystyle AB = AC = x \text{ cm }$

As angle in semi-circle is $\displaystyle 90^{\circ}$

$\displaystyle \therefore \angle A = 90^{\circ}$

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

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

$\displaystyle x^2+x^2 = 84^2$

$\displaystyle 2x^2 = 84 \times 84$

$\displaystyle x^2 = 84 \times 42$

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

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

Diameter of semicircle $\displaystyle (2r) = 84 \text{ cm }$

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

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

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

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

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

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

$\displaystyle = 2772 + 924 - 1764 = 3696 - 1764 = 1932 \text{ cm }^2$

$\displaystyle \\$

Question 7:

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

(i) Construct $\displaystyle \triangle ABC$ , where $\displaystyle AB = 3.5 \text{ cm }, BC = 6 \text{ cm } \text{ and } \angle ABC = 60^{\circ} .$

(ii) Construct the locus of points inside the triangle which are equidistant from $\displaystyle BA \text{ and } BC .$

(iii) Construct the locus of points inside the triangle which are equidistant from $\displaystyle B \text{ and } C .$

(iv) Mark the point $\displaystyle P$ which is equidistant from $\displaystyle AB, BC$ and also equidistant from $\displaystyle B \text{ and } C$ . Measure and record the length of $\displaystyle B \hspace{4.0cm} [3]$

(b) The equation of a line is $\displaystyle 3x + 4y - 7 = 0 .$ 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 $\displaystyle x -y + 2 = 0 \text{ and } 3x + y- 10 = 0$ [3]

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

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

Answers:

(a)

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

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

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

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

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

$\displaystyle \text{(b) (i) Slope } = -\frac{3}{4}$

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

For point of intersection solve $\displaystyle x -y + 2 = 0 \text{ and } 3x + y- 10 = 0$

$\displaystyle y = 4 \text{ and } x = 2$

Therefore intersection $\displaystyle = (2, 4)$

Therefore equation of line

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

$\displaystyle 3y-12 = 4x-8$

$\displaystyle 3y = 4x + 4$

(c)

Interval	Frequency $\displaystyle (f_i)$	$\displaystyle x-i$	$\displaystyle d_i = x_i - A$	$\displaystyle f_id_i$
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
	$\displaystyle 36+f$			$\displaystyle \Sigma f_id_i= 320 - 10f$

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

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

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

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

$\displaystyle \Rightarrow 17f = 68$

$\displaystyle \Rightarrow f = 4$

$\displaystyle \\$

Question 8:

(a) Use the remainder theorem to factorize the following expression: $\displaystyle 2x^3+x^2-13x+6 \hspace{9.0cm} [3]$

(b) If $\displaystyle x, y \text{ and } z$ are in continued proportion, prove that: $\displaystyle \frac{(x+y)^2}{(y+z)^2} =\frac{x}{y} \hspace{1.0cm} [3]$

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

Answers:

(a) Let $\displaystyle x =2$

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

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

$\displaystyle \begin{array}{r l l } x-2 ) & \overline {2x^3+x^2-13x+6} & (2x^2+5x-3 \\ (-) & \underline {2x^3-4x^2} & \\ & \hspace{1.0cm} 5x^2-13x+6 & \\(-)& \hspace{1.0cm} \underline{5x^2-10x} & \\ & \hspace{2.0cm} -3x + 6 & \\ (-) & \hspace{2.0cm} \underline{ -3x+6} & \\ & \hspace{3.0cm} \times & \end{array}$

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

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

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

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

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

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

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

Applying componendo and dividendo

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

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

Squaring both sides

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

Substituting

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

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

(c) $\displaystyle \text{In } \triangle ABD: \tan 48^{\circ} = \frac{AD}{BD}$ 2010-5

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

$\displaystyle \Rightarrow BD = 90.09 \text{ m}$

$\displaystyle \text{In } \triangle ACD \tan 36^{\circ} = \frac{AD}{CD}$

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

$\displaystyle \Rightarrow DC = 137.64 \text{ m}$

$\displaystyle \text{Therefore } BC = BD + CD = 90.09 + 137.64 = 227.73 \text{ m}$

$\displaystyle \\$

Question 9:

(a) Evaluate $\displaystyle \begin{bmatrix} 4\sin{30^{\circ}} & 2\cos{60^{\circ}} \sin{90^{\circ}} & 2\cos{0^{\circ}} \end{bmatrix} . \begin{bmatrix} 4 & 5 5 & 4 \end{bmatrix} \hspace{3.0cm} [3]$

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

(i) length of $\displaystyle AB$

(ii) area of $\displaystyle \triangle ABC \hspace{10.0cm} [3]$

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

Answers:

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

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

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

(b) Consider $\displaystyle \triangle ABC \text{ and } \triangle EBD$

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

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

$\displaystyle \text{Therefore } \angle DEB = \angle BAC$

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

(i) $\displaystyle \text{Given } BE=6 \text{ cm }, EC=4 \text{ cm }, BD=5 \text{ cm }$

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

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

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

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

(c) First Investment

Let the amount invested $\displaystyle = 4500 \text{ Rs. }$

Nominal Value of the share $\displaystyle = 10 \text{ Rs. }$

Market Value of the share $\displaystyle = 15 \text{ Rs. }$

Dividend earned $\displaystyle = 8\%$

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

Sale Proceed $\displaystyle = 300 \times 30 = 9000 \text{ Rs. }$

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

Second Investment

Therefore the amount invested $\displaystyle = 9000 \text{ Rs. }$

Nominal Value of the share $\displaystyle = 100 \text{ Rs. }$

Market Value of the share $\displaystyle = 125 \text{ Rs. }$

Dividend earned $\displaystyle = 12\%$

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

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

Hence the change in income $\displaystyle = 720-240 = 480 \text{ Rs. }$

$\displaystyle \\$

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 $\displaystyle 2 \text{ cm } =\text{ Rs. } 1000$ on one axis and $\displaystyle 2 \text{ cm } = 50 \text{ employees }$ on the other axis. From the graph determine:

(i) the median wage

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

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

(iv) The upper quartile [6]

Answers:

(a) Let the two parts be $\displaystyle x \text{ and } y$

$\displaystyle \text{Given } x^2+y^2 = 20$

Also $\displaystyle y^2 = 8x$

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

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

Therefore the larger part is $\displaystyle y^2 = 16 \Rightarrow y = 4$

Hence the number is $\displaystyle 2+4 = 6$

(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) $\displaystyle = 320$ (even)

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

Required median $\displaystyle =\text{ Rs. } 9800$ (from graph)

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

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

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

Upper Quartile $\displaystyle =\text{ Rs. } 10200$

$\displaystyle \\$

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) $\displaystyle \text{Given } x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2 }-\sqrt{a^2-b^2 }}$ . Use componendo and dividendo to prove that: $\displaystyle x^2=\frac{2a^2 x}{x^2+1} \hspace{1.0cm} [4]$

Answers:

(a) Given side of the hexagon is $\displaystyle 4 \text{ cm }$ . 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 $\displaystyle 4 \text{ cm }$
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 $\displaystyle (U, V, W, X, Y, Z)$
Now join the mid points of the opposite sides to get lines $\displaystyle (UV, WX, \text{ and } YZ)$
Now draw the diagonals passing through the center to get lines $\displaystyle (AD, BE \text{ and } CF)$

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

(b) Given: Diameter of hemispherical bowl $\displaystyle = 7.2 \text{ cm }$

Radius of hemispherical bowl $\displaystyle = 3.6 \text{ cm }$

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

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

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

$\displaystyle \therefore 24.14h = 97.76 \Rightarrow h = 4.05 \text{ cm }$

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

Applying componendo and dividendo

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