2015 ICSE (Class 10) Board Paper Solution: Mathematics

MATHEMATICS (ICSE – Class X Board Paper 2015)

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) A shopkeeper bought an article for Rs. 3450. He marks the price of the article 16% above the cost price. The rate of sales tax charged on the article is 10%. Find the:

(i) marked price of the article

(ii) price paid by a customer who buys the article. [3]

(b) Solve the following in equation and write the solution set:

$\displaystyle 13x-5<15x+4<7x+12,\ x\in R\$

Represent the solution on a real number line. [3]

$\displaystyle \frac{{\mathrm{sin } 65{}^\circ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{cos\ } 32{}^\circ }}{{\mathrm{sin } 58{}^\circ }} - {\mathrm{sin } 28{}^\circ }.\ {\mathrm{sec\ } 62{}^\circ }+{\mathrm{cosec}^2 30{}^\circ } \hspace{5.5cm} [3]$

Answers:

(a)

(i) Cost price of the article $\displaystyle = \text{ Rs. }3450$

$\displaystyle \text{Marked Price } =3450 + \frac{16}{100}\times 3450 = \text{ Rs. } 4002$

$\displaystyle \text{(ii) Price paid by the customer } = \text{ Rs. } 4002+\frac{10}{100}\times 4002 = \text{ Rs. } 4402.20$

(b) Given

$\displaystyle 13x-5<15x+4<7x+12$

$\displaystyle \Rightarrow 15x+4>13x-5 \text{ and } 15x+4<7x+12$

$\displaystyle \Rightarrow 15x-13x>-5-4 \text{ and } 15x-7x<12-4$

$\displaystyle \Rightarrow 2x>-9 \text{ and } 8x<8$

$\displaystyle \Rightarrow x>\frac{-9}{2} \text{ and } x<1$

$\displaystyle \text{Solution: } \{ x: \frac{9}{2} < x< 1 , x \in R \}$

$\displaystyle \text{(c)} \frac{{\mathrm{sin\ } 65{}^\circ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{cos} 32{}^\circ }}{{\mathrm{sin\ } 58{}^\circ }} -{\mathrm{sin} 28{}^\circ }.\ {\mathrm{sec } 62{}^\circ }+{\mathrm{cosec}^ 2 30{}^\circ }$

$\displaystyle = \frac{{\mathrm{cos} (90{}^\circ -65{}^\circ )\ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{sin\ (} 90{}^\circ -32{}^\circ )\ }}{{\mathrm{sin\ } 58{}^\circ }} -{\mathrm{cos} (90{}^\circ -28{}^\circ )\times \frac{1}{{\mathrm{cos\ } 62{}^\circ }}+{(2)}^2\ }$

$\displaystyle = \frac{{\mathrm{cos} 25{}^\circ }}{{\mathrm{cos} 25{}^\circ }}+\frac{{\mathrm{sin\ } 58{}^\circ }}{{\mathrm{sin} 58{}^\circ }} -{\mathrm{cos\ } 62{}^\circ }\times \frac{1}{cos\ 62{}^\circ }+4$

$\displaystyle = 1+1-1+4 = 5$

$\displaystyle \\$

Question 2

$\displaystyle \text{(a) If } A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix} , \text{ find } x \text{ and } y \text{ when } A^2=B . \hspace{3.0cm} [3]$

(b) The present population of a town is $\displaystyle 2,00,000 .$ It population increases by $\displaystyle 10\%$ in the first year and $\displaystyle 15\%$ in the second year. Find the population of the town at the end of the two years. [3]

(c) Three verticals of a parallelogram $\displaystyle ABCD$ taken in order are $\displaystyle A(3,6), B(5,10) \text{ and } C(3,2)$ find:

(i) the coordinates of the fourth vertex $\displaystyle D .$

(ii) length of diagonal $\displaystyle BD$

(iii) equation of side $\displaystyle AB$ of the parallelogram $\displaystyle ABCD .$ [4]

Answers:

$\displaystyle \text{(a) } A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix}$

$\displaystyle A^2=B$

$\displaystyle \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

$\displaystyle \begin{bmatrix} 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

$\displaystyle \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$

Therefore $\displaystyle 4x = 16 \Rightarrow x = 4$

And $\displaystyle 1 = -y \Rightarrow y = -1$

$\displaystyle \text{(b) Given: } P =2,00,000$

$\displaystyle \text{Population after first year } = 200000\ \Big[1+\frac{{10}}{100} \Big] ^1 = 200000 \left[\frac{110}{100}\right]=220000$

Principle for the second year $\displaystyle = 220000$

$\displaystyle \text{Population at the end of second year } = 220000 \left[1+\frac{15}{100}\right]$

$\displaystyle = 220000\times \frac{115}{100} = 253000$

$\displaystyle \text{Mid point of } AC = \Big[\frac{3+3}{2},\ \frac{6+2}{2} \Big] = (3,4)$

$\displaystyle \text{Mid point of } BD = \Big[\frac{x+5\ }{2},\ \frac{y+10}{2} \Big]$

$\displaystyle \text{(i) } 3=\frac{x+5}{2} \Rightarrow x = 1$

$\displaystyle 4=\ \frac{y+10}{2} \Rightarrow y = -2$

Coordinates of $\displaystyle D (1, -2)$

(ii) $\displaystyle BD = \sqrt{{(5-1}^2)+(10+{2)}^2}$

$\displaystyle = \sqrt{16+144} = 4 \sqrt{10} \text{ units }$

(iii) Equation of AB:

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

$\displaystyle y-6= \frac{10-6}{5-3} \left(x-3\right)$

$\displaystyle y-6=\ \frac{4}{2}\left(x-3\right)$

$\displaystyle y-6=2x-6$

$\displaystyle 2x-y=0$

$\displaystyle \\$

Question 3

(a) In the given figure, ABCD is a square of side 21 cm, AC and BD are two diagonals of the square. Two semi circle are drawn with AD and BC as diameters. Find the area of the shaded region. $\displaystyle \text{(Take } \pi =\frac{22}{7} ) \hspace{3.0cm} [3]$

(b) The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks	0	1	2	3	4	5
No. of Students	1	3	6	10	5	5

Calculate the mean, mediam and mode of the above distribution. [3]

(c) In the figure given below O is the center of the circle and SP is a tangent. If $\displaystyle \angle SRT =65{}^\circ , \text{ find the value of } x,\ y \text{ and } z$ [4]

Answer:

(a) Given: Side $\displaystyle =21 \text{ cm}$

$\displaystyle \text{Let Diagonal of the square } = \sqrt{2} \times \text{ side }$

$\displaystyle \therefore AC=BD=21\sqrt{2}$

$\displaystyle \therefore AO=OC=BO=OD=\ \frac{21\sqrt{2}}{2}$

$\displaystyle \text{Area of } \Delta AOD= \text{ Area of } \Delta BOC=\ \frac{1}{2}\times \frac{21\sqrt{2}}{2}\times \frac{21\sqrt{2}}{2}=\frac{441}{4} \text{ cm}^2$

$\displaystyle \text{Area of semicircle } = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times \left(\frac{{21}^2}{2}\right)=\frac{693}{4} \text{ cm}^2$

Area of shaded region = Area of 2 semicircles + Area of $\displaystyle \triangle AOD$ + $\displaystyle \triangle BOC$

$\displaystyle = 2\ \times \frac{693}{4}+\frac{441}{4}+\frac{441}{4}+\frac{2268}{4} = 567 \text{ cm}^2$

(b) Below

$\displaystyle x$	$\displaystyle f$	$\displaystyle fx$	$\displaystyle cf$
0	1	0	1
1	3	3	4
2	6	12	10
3	10	30	20
4	5	20	25
5	5	25	30
	$\displaystyle \Sigma f=30$ $\displaystyle \Sigma fx=90$

$\displaystyle \text{Mean } = \frac{\Sigma fx}{\Sigma f}=\frac{90}{30} = 3$

$\displaystyle \text{Median} = \text{size of } \left(\frac{N}{2}\right)th\ {obs}^n$

$\displaystyle = \text{size of } \left(\frac{30}{2}\right)th\ {obs}^n$

$\displaystyle = \text{Size of } 15 {}^{th} obs{}^{n} = 3$

Mode = 3 marks (as highest frequency is 10)

$\displaystyle In\ \Delta TSR$

$\displaystyle x+90{}^\circ +65{}^\circ =180{}^\circ$

$\displaystyle x=25{}^\circ$

$\displaystyle \angle SOQ=2\ \angle STR\ \left[Angle\ at\ centre=2\ \times angle\ at\ circumference \right]$

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

In $\displaystyle \Delta OSP,\ 50{}^\circ +90{}^\circ +z=180{}^\circ$

$\displaystyle z=40{}^\circ$

$\displaystyle \\$

Question 4

(a) Katrina opened a recurring deposit account with a Nationalized Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is Rs. 1000 find the: [3]

(i) interest earned in 2 years

(ii) matured value

(b) Find the value of $\displaystyle k$ for which $\displaystyle x=3$ is a solution of the quadratic equation, $\displaystyle \left(k+2\right)x^2-kx+6=0 .$ Thus find the root of the equation. [3]

(c) Construct a regular hexogon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown. [4]

Answer:

(a)

(i) Given $\displaystyle P= \text{ Rs. }1000,\ n=2\ \text{years} =\ 24\ \text{months},\ r=6\%.$

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

$\displaystyle = 1000\times \frac{24\ \left(24+1\right)}{2}\times \frac{6}{12\times 100}$

$\displaystyle = 1000\times \frac{24\times 25}{2}\times \frac{6}{12\times 100}$

$\displaystyle =1500$

$\displaystyle \text{(ii) Total money deposit in 24 months = } \text{Rs. }\ 1000\times 24 = \text{ Rs. } 24000$

$\displaystyle \text{Matured Value = Total sum deposit + Interest} =24000+1500 = \text{ Rs. } 25500$

(b) $\displaystyle \left(k+2\right)x^2-kx+6=0$

Putting $\displaystyle x=3$ in given equation:

$\displaystyle \left(k+2\right)\times 9-k\times 3+6=0$

$\displaystyle 9k+18-3k+6=0$

$\displaystyle 6k=-24 \Rightarrow k = -4$

Putting $\displaystyle k= -4$ in given equation:

$\displaystyle -2x^2+4x+6=0$

$\displaystyle x^2-2x-3=0$

$\displaystyle x^2-3x+x-3=0$

$\displaystyle x\left(x-3\right)+1\left(x-3\right)=0$

$\displaystyle \left(x+1\right)\left(x-3\right)=0$

$\displaystyle x+1=0\ or\ x-3=0$

$\displaystyle x=-1\ or\ x=3$

$\displaystyle x=-1\$ is the other root of the given equation.

First draw a regular hexagon. The length of one side is 5 cm. You will get hexagon ABCDEF.
Take any two adjacent sides and draw perpendicular bisectors.
The point where these two bisectors intersect, is the center of the circle.
With O as the center, draw a circle which will pass through all the vertices of the hexagon.

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5

(a) Use a graph paper for this question taking $\displaystyle 1 cm = 1$ unit along both the $\displaystyle x \text{ and } y axis$ :

(i) Plot the points $\displaystyle A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) \text{ and } F(0, -5)$

(ii) Reflect the points $\displaystyle B, C, D \text{ and } E$ on the $\displaystyle y-axis$ and name them respectively as $\displaystyle B', C', D',U' \text{ and } E' .$

(iv) Name the figure formed by $\displaystyle B C D E E'D'C'B' .$

(v) Name a line of symmetry for, the figure formed. [5]

(b) Virat opened a Saving Bank account in a bank on 16th April 2010. His pass book shows the following entries:

Date	Particulars	Withdrawals (Rs.)	Deposits (Rs.)	Balance (Rs.)
April 16, 2010	By Cash	–	2500	2500
April 28th	By Cheque	–	3000	5500
May 9th	To Cheque	850	–	4650
May 15th	By Cash	–	1600	6250
May 24th	To Cash	1000	–	5250
June 4th	To Cash	500	–	4750
June 30th	By Cheque	–	2400	7150
July 3rd	To Cash	–	1800	8950

Calculate the interest Virat earned at the end, of 31st July, 2010 at 4% per annum interest. What sum of money will he receives if he closes the account on 1st August, 2010? [5]

Answer:

(a)

(i) Shown in the diagram.

(ii) Shown in then diagram.

(iii) Coordinates : $\displaystyle B', (-2, 5), C', (-5 ,2), D', (-5 , -2),E', (-2, -5)$

(iv) Octagon

(v) $\displaystyle x-axis \text{ or } y-axis.$

(b) Qualifying principal for various months:

Month	Principal (Rs.)
April	0
May	4650
June	4750
July	8950
Total	18350

$\displaystyle P = \text{ Rs. } 18350 R = 4\% \text{ and } T= \frac{1}{12}$

$\displaystyle I = P \times R \times T = 18350 \times \frac{4}{100} \times \frac{1}{12} = \text{ Rs. } 61.16$

Amount $\displaystyle = 8950+61.61= \text{ Rs. } 9011.16$

$\displaystyle \\$

Question 6

(a) If a, b, c are in continued proportion, prove that $\displaystyle (a + b + c) (a - b + c) = a^2 + b^2 + c^2$ [3]

(b) In the given figure $\displaystyle ABC$ is a triangle and $\displaystyle BC$ , is parallel to the $\displaystyle y-axis .$ $\displaystyle AB \text{ and } AC$ intersects the $\displaystyle y-axis$ at $\displaystyle P \text{ and } Q$ respectively.

(i) Write the coordinates of $\displaystyle A .$

(ii) Find, the length of $\displaystyle AB \text{ and } AC .$

(iii) Find the ratio in which $\displaystyle Q$ divides $\displaystyle AC .$

(iv) Find the equation, of the line $\displaystyle AC$ [4]

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	8	5	12	35	24	16

Answer:

(a) To prove: $\displaystyle (a + b + c) (a - b + c) = a^2 + b^2 + c^2$

Given $\displaystyle a, b \text{ and } c$ are in continued proportion

$\displaystyle \text{Therefore } \frac{a}{b} = \frac{b}{c} = k$

$\displaystyle \Rightarrow a = bk \text{ and } b = ck$

This also implies $\displaystyle a = (ck)k = ck^2$

LHS $\displaystyle = (a + b + c) (a - b + c)$

$\displaystyle = (ck^2+ck+c)(ck^2-ck+c)$

$\displaystyle = c^2(k^2+k+1)(k^2-k+1)$

$\displaystyle = c^2 \{ (k^2+1)+k \} \{ (k^2+1)-k \}$

$\displaystyle =c^2 \{ (k^2+1)^2-k^2 \}$

$\displaystyle = c^2 \{ k^4+1+2k^2-k^2 \}$

$\displaystyle = c^2 \{ k^4+k^1+1 \}$

RHS $\displaystyle = a^2 + b^2 + c^2$

$\displaystyle = (ck^2)^2+(ck)^2+c^2$

$\displaystyle = c^2k^4+c^2k^2+c^2$

$\displaystyle =c^2(k^4+k^2+1)$

Hence LHS = RHS

Therefore Proved

(b)

(i) Coordinates of $\displaystyle A (4, 0)$

(ii) Coordinates of $\displaystyle A (4, 0) \text{ and } B (-2, 3)$

Note: The distance between any two points $\displaystyle (x_1, y_1) \text{ and } (x_2, y_2)$ is $\displaystyle = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Length $\displaystyle AB = \sqrt{(-2-4)^2+(3-0)^2} = \sqrt{36+9} = \sqrt{45} = 3 \sqrt{5} \text{ units }$

Coordinates of $\displaystyle A (4, 0) \text{ and } C (-2, -4)$

Length $\displaystyle AC = \sqrt{(-2-4)^2+(-4-0)^2} = \sqrt{36+16} = \sqrt{52} = 2 \sqrt{13} \text{ units }$

(iii) Let the required ratio be $\displaystyle k:1$

Let the coordinate of $\displaystyle P \text{ be } (0, y)$ , since it lies on the y axis.

$\displaystyle \text{Since } x = \frac{kx_2+x_1}{k+1}$

$\displaystyle \Rightarrow 0 = \frac{k \times (-2) +4}{k+1}$

$\displaystyle \Rightarrow -2k+4=0$

$\displaystyle \Rightarrow k = 2$

$\displaystyle \Rightarrow m_1:m_2 = 2:1$

(iv) Equation of $\displaystyle AC$ through Coordinates of $\displaystyle A(4, 0) \text{ and } C(-2, -4)$

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

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

$\displaystyle \Rightarrow 3y=2x-8$

$\displaystyle \Rightarrow or 3y-2x+8=0$

(c)

Class	Mid Value $\displaystyle x$	$\displaystyle f$	$\displaystyle fx$
0-10	5	8	40
10-20	15	5	75
20-30	25	12	300
30-40	35	35	1225
40-50	45	24	1080
50-60	55	16	880
Total		$\displaystyle \Sigma f=100$	$\displaystyle \Sigma fx=3600$

$\displaystyle \text{Mean } = \bar{x} = \frac{\Sigma fx}{ \Sigma f} = \frac{3600}{100} = 36$

$\displaystyle \\$

Question 7

(a) The solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3]

(b) Find ‘a’ if the two polynomials $\displaystyle {ax}^3+{3x}^2-9 \text{ and } {2x}^3+4x+a$ , leaves the same remainder when divided by $\displaystyle (x+3) .$ [3]

$\displaystyle \text{(c) Prove that } \frac{{\mathrm{sin } \theta }}{1-{\mathrm{cot } \theta }}+\frac{{\mathrm{cos } \theta }}{1-{\mathrm{tan } \theta }} ={\mathrm{cos } \theta }+{\mathrm{sin } \theta } \hspace{6.0cm} [4]$

Answers:

(a) Volume of two spheres = Volume of cone

$\displaystyle \frac{4}{3}\pi \left(2^3\right)+\frac{4}{3}\pi \left(4^3\right)=\ \frac{1}{3}{\pi r}^2h$

$\displaystyle 32+256=r^2\times 8$

$\displaystyle r^2 = \frac{288}{8}$

$\displaystyle r^2 = 36 \Rightarrow r = 6 cm$

(b) Let $\displaystyle f(x) = {ax}^3+{3x}^2-9 \text{ and } g(x) = {2x}^3+4x+a$

Putting $\displaystyle x = -3$ in both the expressions

$\displaystyle f(-3) = {a(-3)}^3+{3(-3)}^2-9$

$\displaystyle g(-3) ={2(-3)}^3+4(-3)+a$

Given $\displaystyle f(-3) = g(-3)$

Therefore $\displaystyle {a(-3)}^3+{3(-3)}^2-9 = {2(-3)}^3+4(-3)+a$

$\displaystyle -27a+27 -9 = -54-12+a$

$\displaystyle 27a+a = 27-9+54+12$

$\displaystyle 28a=84 \Rightarrow a = 3$

$\displaystyle \text{(c) Given } \frac{{\mathrm{sin } \theta }}{1-{\mathrm{cot } \theta }}+\frac{{\mathrm{cos } \theta }}{1-{\mathrm{tan } \theta }} ={\mathrm{cos } \theta }+{\mathrm{sin } \theta }$

$\displaystyle \text{LHS } =\frac{{\mathrm{sin } \theta }}{1-\frac{\mathrm{cos } \theta }{\mathrm{sin } \theta }}+\frac{{\mathrm{cos } \theta }}{1-\frac{\mathrm{sin } \theta }{\mathrm{cos } \theta }}$

$\displaystyle =\frac{{\mathrm{sin}^2 \theta }}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }} + \frac{{\mathrm{cos}^2 \theta }}{{\mathrm{cos } \theta } - {\mathrm{sin } \theta }}$

$\displaystyle =\frac{{\mathrm{sin}^2 \theta - \mathrm{cos}^2 \theta }}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }}$

$\displaystyle =\frac{{(\mathrm{sin } \theta - \mathrm{cos } \theta )(\mathrm{sin } \theta + \mathrm{cos } \theta )}}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }}$

$\displaystyle ={\mathrm{cos } \theta }+{\mathrm{sin } \theta } =$ RHS

$\displaystyle \\$

Question 8

(a) $\displaystyle AB \text{ and } CD$ are two chords of a circle intersecting at $\displaystyle P .$ Prove that $\displaystyle AP \times PB =CP \times PD .$ [3]

(b) A bag contain 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag, Find the probability that the ball drawn is:

(i) a green ball

(ii) a white or a red ball

(iii) is neither a green ball nor a white ball. [3]

(c) Rohit invested Rs. 9600 on Rs. 100 shares at Rs. 20 premium paying 8% dividend. Rohit sold the shares when the price rose to Rs. 160. He invested the proceeds (excluding dividend) in 10% Rs. 50 shares at Rs. 40. Find the:

(i) original Number of shares

(ii) sale proceeds

(iii) new number of shares

(iv) change in the two dividends. [4]

Answers:

(a) To prove: $\displaystyle AP\times PB=CP\times PD$

Construction : Join AD and CB.

$\displaystyle \angle A = \angle C$ (Angles of the same segment)

$\displaystyle \angle C = \angle B$ (Angles of the same segment)

Therefore $\displaystyle \Delta APD \sim \Delta BPC$ (AAA Postulate)

$\displaystyle \therefore \frac{AP}{CP}=\frac{PD}{PB}$

$\displaystyle \therefore AP \times PB=CP \times PD$

Hence Proved

(b) Number of White Balls: $\displaystyle 5$

Number of Red Balls: $\displaystyle 6$

Number of Green Balls : $\displaystyle 9$

Total number of Balls: $\displaystyle 5+6+9 = 20$

$\displaystyle \text{(i) Probability (Green) } = \frac{\text{Favorable Cases}}{\text{Total number of possible cases}} = \frac{9}{20}$

$\displaystyle \text{(ii) Probability (Red or White) } = \frac{5+6}{20} = \frac{11}{20}$

$\displaystyle \text{(iii) Probability (neither a green ball nor a white ball) } = \frac{6}{20} = \frac{3}{10}$

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

Market Value of the share $\displaystyle = 120 \text{ Rs. }$

Dividend $\displaystyle = 8 \%$

$\displaystyle \text{(i) Number of shares bought } = \frac{9600}{120} = 80$

(ii) Sale Price $\displaystyle = 160 \text{ Rs. }$

Sale Proceed $\displaystyle = 80 \times 160 = 12800 \text{ Rs. }$

Second Investment

Nominal Value of the share $\displaystyle = 50 \text{ Rs. }$

Market Value of the share $\displaystyle = 40 \text{ Rs. }$

Dividend $\displaystyle = 10\%$

$\displaystyle \text{(iii) Number of shares bought }= \frac{12800}{40} = 320$

$\displaystyle \text{Dividend on 1st Investment earned } = 80 \times \frac{8}{100} \times 100 = 640 \text{ Rs. }$

$\displaystyle \text{Dividend on 2nd Investment earned } = 320 \times \frac{10}{100} \times 50 = 1600 \text{ Rs. }$

$\displaystyle \text{Change in Dividend income } = 1600-640 = 960 \text{ Rs. }$

$\displaystyle \\$

Question 9

(a) The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is $\displaystyle 30^{\circ} \text{ and } 24^{\circ}$ respectively.

Find the height of the two towers. Give your answer correct to 3 significant figures. [4]

(b) The weight of 50 workers is given below:

Weight in kg	50-60	60-70	70-80	80-90	90-100	100-110	110-120
No. of workers	4	7	11	14	6	5	3

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight. [6]

Answers:

(a) Let AB and CD are two towers.

$\displaystyle BD=120 \text{ m}=EC,\ \angle ACE=30{}^\circ ,\ \angle CBD=24{}^\circ$

In the right $\displaystyle \Delta CBD$

$\displaystyle {\tan 24{}^\circ }=\ \frac{CD}{BD}=\frac{CD}{120}$

$\displaystyle CD=120\times 0.4452 =53.42 \text{ m}$

In right angle $\displaystyle \Delta ACE$

$\displaystyle {\tan 30{}^\circ =\ \frac{AE}{EC}=\frac{AE}{120}\ }$

$\displaystyle \mathrm{AE=120\times 0.5773} =69.28 \text{ m}$

$\displaystyle AB=AE+EB$

$\displaystyle = AE+CD = 69.28+53.42 = 122.7 \text{ m}$

Height of $\displaystyle 1{}^{st}$ tower is $\displaystyle 122.7 m \text{ and } 2{}^{nd}$ tower is $\displaystyle 53.42 \text{ m}$

(b) Below the table

Weight	$\displaystyle f$	$\displaystyle cf$
50-60	4	4
60-70	7	11
70-80	11	22
80-90	14	36
90-100	6	42
100-110	5	47
110-120	3	50
Total	50

$\displaystyle \text{(i) Lower quartile }= \frac{N^{th}}{4}{ obs}^n$

$\displaystyle = \frac{{50}^{th}}{4}{ obs}^n = 12.5^{th} { obs}^n = 72 kg$

$\displaystyle \text{Upper quartile } = \frac{3N}{4}^{th} \ {obs}^n = \frac{150}{4}^{th}\ {obs}^n = 37.5^{th} { obs}^n = 92 kg$

(ii) No. of over weight workers $\displaystyle = 50-39 = 11$

$\displaystyle \\$

Question 10

(a) A wholesale buys a TV from the manufacture for Rs. 25000. He marks the price of the TV 20% above his cost price and sell it to a retailer at 10% discount on the marked price. If the rate of VAT is 8%, find the: [3]

(i) marked Price

(ii) retailer’s cost price inclusive of tax

(iii) VAT paid by the wholesaler.

$\displaystyle \text{(b) If } A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$ , $\displaystyle B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} . \text{ Find} AB-5C. \hspace{1.0cm} [3]$

(c) ABC is a right angled triangle with $\displaystyle \angle ABC= 90^{\circ} .$ D is any point on AB and DE is perpendicular to AC. Prove that:

(i) $\displaystyle \Delta ADE \sim \Delta ACB$

(ii) If $\displaystyle AC = 13 \text{ cm}, BC = 5 \text{ cm} \text{ and } AE = 4 \text{ cm}. \text{ Find } DE \text{ and } AD.$

(iii) Find, area, of $\displaystyle \Delta ADE$ : area of quadrilateral BCED. [4]

Answers:

(a)

(i) Cost price for wholesaler $\displaystyle = 25000 \text{ Rs. }$

$\displaystyle \text{Market Price } = 25000+\frac{20}{100}\times 25000 = 30000 \text{ Rs. }$

(ii) Discount $\displaystyle = 10\% \text{ of } 30000= 3000 \text{ Rs. }$

Cost Price for retailer $\displaystyle = \text{ Market Price -Discount }$

$\displaystyle = \mathrm{30000 - 3000} = 27000 \text{ Rs. }$

$\displaystyle \text{Cost Price inclusive Tax } = \mathrm{27000+}\frac{\mathrm{8}}{\mathrm{100}}\mathrm{\times 27000} = 29160 \text{ Rs. }$

(iii) Cost price for wholesaler $\displaystyle = 25000 \text{ Rs. }$

Sale price for wholesaler $\displaystyle = 27000 \text{ Rs. }$

Profit for wholesalers $\displaystyle = 27000 - 25000= 2000 \text{ Rs. }$

$\displaystyle \text{VAT } = \frac{8}{100}\times 2000 = 160 \text{ Rs. }$

$\displaystyle \text{(b) Given } A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$ , $\displaystyle B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}$ , $\displaystyle C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix}$

$\displaystyle AB = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} . \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix}$

$\displaystyle 5C = 5 . \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} = \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix}$

$\displaystyle AB - 5C= \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} - \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix} = \begin{bmatrix} 30 & 52 \\ 40 & -14 \end{bmatrix}$

(i) In $\displaystyle \Delta ADE\ and\ \Delta ACB$

$\displaystyle \angle A$ (is common)

$\displaystyle \angle AED = \angle ABC=90^{\circ}$

Therefore $\displaystyle \Delta ADE\ \sim \Delta ACB$ (AAA Postulate)

(ii) $\displaystyle {AC}^2={AB}^2+{BC}^2$

$\displaystyle 169={AB}^2+25$

$\displaystyle AB = 12 cm$

Since $\displaystyle \Delta ADE \sim \Delta ACB$

$\displaystyle \text{Therefore } \frac{DE}{BC} = \frac{AD}{AC} = \frac{AE}{AB}$

$\displaystyle \text{Therefore } \frac{DE}{BC} = \frac{AE}{AB}$

$\displaystyle \frac{DE}{5} = \frac{4}{12} \Rightarrow DE = \frac{5}{3}$ cm

$\displaystyle \frac{AD}{AC} = \frac{AE}{AB}$

$\displaystyle \frac{AD}{13} = \frac{4}{12} \Rightarrow AD = \frac{13}{3}$ cm

$\displaystyle \text{(iii) } \frac{Ar.\ of\ \left(\Delta ABC\right)}{Ar.\ of\ \left(\Delta ADE\right)}=\ \frac{{AB}^2}{{AE}^2}=\frac{144}{16}=9$

$\displaystyle \Rightarrow \frac{Ar.\ of\ \left(\Delta ADE\right)+Ar.\ of\ \Delta (BCED)}{Ar.\ of\ (\Delta ADE)} =9$

$\displaystyle \Rightarrow 1 + \frac{Ar.\ of\ \left(BCED\right)}{Ar.\ of\ \left(\Delta ADE\right)} = 9$

$\displaystyle \Rightarrow \frac{Ar.\ of\ \left(\Delta ADE\right)}{Ar.\ of\ \left(BCED\right)}=\ \frac{1}{8}$

$\displaystyle \\$

Question 11

(a) Sum of two natural numbers is 8 and the difference of their reciprocal is $\displaystyle \frac{2}{15}$ find the numbers. [3]

$\displaystyle \text{(b) Given } \frac{x^3+12x}{6x^2+8} = \frac{y^3+27y}{9y^2+27} \text{ using componendo and dividendo find } x:y$ [3]

$\displaystyle \text{(c) Construct the } \triangle ABC \text{ with } AB=5.5 \text{ cm}, AC=6 \text{ cm} , \text{ and } \angle BAC=105^{\circ} . \text{ Hence }$

(i) Construct the locus of points equidistant from BA and BC

(ii) Construct the locus of points equidistant from B and C

(iii) Mark the point which satisfy he above two loci as P. Measure and write the length of PC. [4]

Answers:

(a) Let 1st number be $\displaystyle x$ and the 2nd number be $\displaystyle (8-x)$

$\displaystyle \text{Given } \frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$

$\displaystyle \frac{8-x-x}{x\ \left(8-x\right)}=\frac{2}{15}$

$\displaystyle \frac{8-2x}{8x-x^2}=\frac{2}{15}$

$\displaystyle 120-30x=16x-2x^2$

$\displaystyle {2x}^2-46x+120=0$

$\displaystyle x^2-23x+60=0$

$\displaystyle x\ \left(x-20\right)-3\ \left(x-20\right)=0$

$\displaystyle \left(x-3\right)\left(x-20\right)=0 \Rightarrow x=3 \text{ or } x=20 \text{ (not possible) }$

Therefore the first number is 3 and the second number is 5.

$\displaystyle \text{(b) }\frac{x^3+12x}{{6x}^2+8}=\frac{y^3+27y}{{9y}^2+27}$

Applying componendo and dividendo

$\displaystyle \frac{x^3+12x+{6x}^2+8}{x^3+12x-{6x}^2-8}=\frac{y^2+27y+{9y}^2+27}{y^3+27y-{9y}^2-27}$

$\displaystyle \frac{(x+{2)}^3}{(x-2)^3}=\frac{{\left(y+3\right)}^3}{{\left(y-3\right)}^3}$

$\displaystyle \frac{x+2}{x-2}=\frac{y+3}{y-3}$

Again Applying componendo and dividendo

$\displaystyle \frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}$

$\displaystyle \frac{2x}{4}=\frac{2y}{6}$

$\displaystyle \frac{x}{y}=\frac{4}{6}=\frac{2}{3}$

$\displaystyle x:y=2:3$

First draw a line AB=5.5 cm.
With the help of the point A, draw $\displaystyle \angle XAB =105^{\circ} .$ This you can draw using a compass.
Take radius 6 cm, cut AC=6 cm and join C to B. This way you get CA and BC. This completes the triangle.
Then draw perpendicular bisector of BC and draw angle of bisector $\displaystyle \angle CBA$ both intersecting at P.
P is the required point PC=4.8 cm

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

2015 ICSE (Class 10) Board Paper Solution: Mathematics

Like this:

Related

Leave a Reply Cancel reply

Share this:

Like this:

Related

Leave a Reply Cancel reply