2016 ICSE (Class 10) Board Paper Solution: Mathematics

MATHEMATICS (ICSE – Class X Board Paper 2016)

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

$\displaystyle \text{(b) Given } A = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \text{ and } A^2=9A+mI . \text{ Find } m . \hspace{2.0cm} [4]$

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

Answers:

(a) Let $\displaystyle f(x)=2x^3+3x^2-kx+5$

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

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

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

$\displaystyle \Rightarrow 2k = 26$

$\displaystyle \Rightarrow k = 13$

$\displaystyle \\$

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

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

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

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

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

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

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

Since LHS = RHS

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

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

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

Hence the value of $\displaystyle m = -14$

$\displaystyle \\$

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

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

$\displaystyle \Rightarrow 612=497+x$

$\displaystyle \Rightarrow x = 115$

Now arrange the numbers in ascending order we get

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

$\displaystyle \text{Number of terms } (n) = 9 \text{(odd number of terms)}$

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

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

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

$\displaystyle \\$

Question 2

$\displaystyle \text{(a) The slope of a line joining } P(6, k) \text{ and } Q(1-3k, 3) \text{ is } \frac{1}{2} . \\ \\ \text{ Find: i) } k \text{ ii) Midpoint of } PQ \text{ using the value of } k \text{ found in (i). } \hspace{2.0cm} [3]$

(b) Without using trigonometrical tables evaluate:

$\displaystyle \mathrm{cosec} ^2 57^{\circ} -\tan^2 33^{\circ} +\cos 44^{\circ} \mathrm{cosec} 46^{\circ} -\sqrt{2} \cos 45^{\circ} -\tan^2 60^{\circ} \hspace{3.0cm} [4]$

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

Answers:

(a)

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

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

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

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

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

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

$\displaystyle \Rightarrow k = -11$

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

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

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

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

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

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

$\displaystyle \Rightarrow k = -11$

$\displaystyle \\$

(b) Given:

$\displaystyle \mathrm{cosec} ^2 57^{\circ} -\tan^2 33^{\circ} +\cos 44^{\circ} \mathrm{cosec} 46^{\circ} -\sqrt{2} \cos 45^{\circ} -\tan^2 60^{\circ}$

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

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

$\displaystyle = 1+1 -1-3 = -2$

$\displaystyle \\$

Volume of cones = Volume of sphere [Provide Formulas]

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

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

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

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

Therefore the number of cones needed is $\displaystyle 72$

$\displaystyle \\$

Question 3

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

$\displaystyle -3x(x-7) \geq 15-7x \ge \frac{x+1}{3}, x \in R \hspace{5.0cm} [3]$

icse-1 (b) In the given figure below, AD is the diameter. O is the center of the circle. AD is parallel to $\displaystyle BC \text{ and } \angle CBD = 32^{\circ} .$ Find: i) $\displaystyle \angle OBD$ ii) $\displaystyle \angle AOB$ iii) $\displaystyle \angle BED$ [4]

$\displaystyle \text{(c) If } (3a+2b): (5a+3b) =18:29 . \text{ Find } a:b$ [3]

Answers:

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

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

Therefore we have:

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

$\displaystyle \Rightarrow 4x \geq -6$

$\displaystyle \Rightarrow x \geq - \frac{3}{2} \text{ ... ... ... ... ... i) }$

Also we have

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

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

$\displaystyle \Rightarrow 44 \ge 22x$

$\displaystyle \Rightarrow 2 \ge x \text{ ... ... ... ... ... ii) }$

Combining i) and ii) we get

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

$\displaystyle \\$

(b) Given $\displaystyle AD \parallel BC$

Therefore $\displaystyle \angle ADB = \angle DBC = 32^{\circ}$ (alternate angles)

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

Therefore $\displaystyle \angle OBD = 32^{\circ}$ (angles opposite to equal side of the triangle are equal)

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

In $\displaystyle \triangle ABC$

$\displaystyle \angle ABD = 90^{\circ} , \angle ADB = 32^{\circ}$

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

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

Therefore $\displaystyle \angle BED = 58^{\circ}$

$\displaystyle \\$

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

Cross multiplying

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

$\displaystyle 3a=4b$

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

$\displaystyle \\$

Question 4

(a) A game of number has cards marked with $\displaystyle 11, 12, 13, \ldots 40.$ 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 $\displaystyle 2 \text{ cm } = 7$ unit along both $\displaystyle x \text{ and } y axis$ ). Plot the point $\displaystyle O (0, 0), A(-4, 4), B(-3, 0) \text{ and } C (0, -3) .$ [4]

Reflect points $\displaystyle A \text{ and } B$ on the y axis and name them $\displaystyle A' \text{ and } B'$ respectively. i) Write down their coordinates. ii) Name the figure $\displaystyle OABCB'A'$ 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]

Answers:

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

$\displaystyle \text{Formula used: Probability } = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

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

The number of favorable outcomes $\displaystyle = 3$

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

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

Therefore the number of favorable outcomes $\displaystyle = 4$

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

$\displaystyle \\$

(b)

i) $\displaystyle A'(4, 4) \text{ and } B'(3, 0)$

ii) Arrow Head

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \text{Hence Amount at the end of 2 years } = \text{ Rs. } 5671$

SECTION B [40 Marks]

(Answer any four questions in this Section.)

Question 5

(a) Solve the quadratic equation $\displaystyle x^2-3(x+3) =0 .$ 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]

Answers:

(a) Given $\displaystyle x^2-3(x+3) =0$

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

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

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

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

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

Solving we get $\displaystyle x = 4.85, -1.85$

$\displaystyle \\$

(b) Qualifying principal for various months:

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

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

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

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

$\displaystyle \text{Amount = Last Balance + Interest }$

$\displaystyle \text{Amount } = 5900+151.50 = 6051.50 \text{ Rs. }$

$\displaystyle \\$

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

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

$\displaystyle \Rightarrow n = 3 \text{ years }$

$\displaystyle \\$

Question 6

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

icse-3 (b) In the given figure $\displaystyle PQRS$ is a cyclic quadrilateral $\displaystyle PQ \text{ and } RS$ produced meet at point $\displaystyle T .$

i) Prove $\displaystyle \triangle TPS \sim \triangle TRQ$

ii) Find. $\displaystyle SP if TP = 18 \text{ cm}, RQ = 4 \text{ cm and }, TR = 6 \text{ cm }$

iii) Find area, of quadrilateral $\displaystyle PORS$ if area of $\displaystyle \triangle PTS=27 \text{ cm}^2$ [4]

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

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

Answers:

(a)

icse-22 Step 1: Given side of the hexagon is $\displaystyle 5 \text{ cm} .$ 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 $\displaystyle 5 \text{ cm }$
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 $\displaystyle (UV, WX, YZ, AD, BE \text{ and } CF).$

$\displaystyle \\$

(b)

(i) Consider $\displaystyle \triangle TPS \text{ and } \triangle TRQ$

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

$\displaystyle \angle PST = \angle RQT$

$\displaystyle \angle SPQ = \angle QRT$

$\displaystyle \angle T$ is common

Therefore $\displaystyle \triangle TPS \sim \triangle TRQ$

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

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

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

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

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

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

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

$\displaystyle \\$

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

If, $\displaystyle AX=B$ :

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

We know

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

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

Hence $\displaystyle m = 2 \text{ and } n = 1$

Therefore the order of $\displaystyle X \text{ is } 2 \times 1$

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

Therefore

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

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

$\displaystyle 4a \sin \ 30^{\circ} +b \cos 0^{\circ} =4$

$\displaystyle \Rightarrow 2a+b=4 \text{ ... ... ... ... ... i) }$

$\displaystyle a \cos 0^{\circ} + 4b \sin \ 30^{\circ} = 5$

$\displaystyle \Rightarrow a+2b=5 \text{ ... ... ... ... ... ii) }$

Solving i) and ii) we get $\displaystyle a = 1 \text{ and } b = 2$

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

$\displaystyle \\$

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 $\displaystyle 45^{\circ} \text{ and } 35^{\circ}$ 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 $\displaystyle 2 \text{ cm } = 10$ scores on the $\displaystyle x-axis \text{ and } 2 \text{ cm } = 20$ shooters on the $\displaystyle y-axis$ ).

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]

Answer:

icse-4 (a) In the adjacent diagram, let $\displaystyle D$ be the air plane and $\displaystyle A \text{ and } B$ be the two ships.

Given: $\displaystyle CD=1500 \text{ m}$

$\displaystyle \angle EDA = 30^{\circ}$

$\displaystyle \angle EDB = 45^{\circ}$

Consider $\displaystyle \triangle DBC$

$\displaystyle \tan 45^{\circ} = \frac{DC}{BC} = \frac{1500}{BC} \Rightarrow BC = 1500 \text{ m}$

Consider $\displaystyle \triangle ADC$

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

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

Hence, distance between two ships $\displaystyle = 1098 \text{ m}$

$\displaystyle \\$

(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) $\displaystyle N = 160$ (which is even)

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

From the graph, Median $\displaystyle = 43.5$

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

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

Inter Quartile range $\displaystyle 60-28.5 = 31.5$

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

$\displaystyle \\$

Question 8

$\displaystyle \text{(a) If } \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \text{ show that } \frac{x^3}{a^3}+ \frac{y^3}{b^3}+ \frac{z^3}{c^3} = \frac{3xyz}{abc} \hspace{6.0cm} [3]$

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

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

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

icse-6 $\displaystyle \text{(c) A line } AB \text{ meets the x-axis at } A \text{ and y-axis at } B. \ P(4, -1) \\ \\ \text{ divides } AB \text{ in the ratio } 1 : 2.$

(i) Find the coordinates of $\displaystyle A \text{ and } B .$

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

Answers:

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

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

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

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

$\displaystyle = k^3+k^3+k^3 = 3k^3$

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

Substituting

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

$\displaystyle \\$

(b) icse-21

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

These are required tangents.

(ii) Length of tangent $\displaystyle = 3 \text{ cm} .$

$\displaystyle \\$

(c)

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

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

$\displaystyle AP: PB = 1:2$

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

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

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

Therefore

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

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

Therefore coordinates are $\displaystyle A(6,0) \text{ and } B(0, -3)$

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

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

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

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

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

$\displaystyle \Rightarrow 2x+y=7$

$\displaystyle \\$

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]

icse-7 (b) The given figure represents a kite with a, circular and a Semicircular motifs stuck on it. The radius of circle is $\displaystyle 2.5 \text{ cm }$ and the semicircle is $\displaystyle 2 \text{ cm} .$ If diagonals $\displaystyle AC \text{ and } BD$ are the lengths $\displaystyle 12 \text{ cm } \text{ and } 8 \text{ cm }$ respectively, find the area of the :

(i) shaded, part. Give your answer correct to the nearest whole number.

(ii) unshaded part. [4]

(i) The length of the model of the ship is $\displaystyle 2 \text{ m} .$ Calculate the length of the ship.

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

(iii) The volume of the model is $\displaystyle 6.5 \text{ m}^3 .$ Calculate the volume of the ship. [3]

Answers

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

Therefore the cost price of the article for the dealer

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

$\displaystyle = 6000-1800 = 4200 \text{ Rs. }$

Sales tax paid by the dealer to the wholesaler

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

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

The cost price for the shopkeeper

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

$\displaystyle = 6000 - 600 = 5400$

The price paid by the shopkeeper including the tax

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

$\displaystyle = 5400 + 324 = 5724 \text{ Rs. }$

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

$\displaystyle = 324 - 252 = 72 \text{ Rs. }$

$\displaystyle \\$

(b)

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

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

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

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

$\displaystyle = 2\pi + 6.24 \pi$

$\displaystyle = 8.25 \pi$

$\displaystyle = 25.92 \text{ cm}^{2}$

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

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

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

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

$\displaystyle \\$

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

Length of the ship $\displaystyle = 2 \times 300 = 600 \text{ m}$

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

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

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

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

$\displaystyle \\$

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]

icse-8

(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 $\displaystyle x \text{ km/hr},$ form an equation and solve it to evaluate $\displaystyle x .$ [3]

Answers

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

$\displaystyle \text{ Rate } = 6\%, \text{ Interest } = 1200 \text{ Rs. }$

(i) Let the monthly instalment be $\displaystyle x$

Using the formula:

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

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

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

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

Therefore the monthly instalment $\displaystyle = 800 \text{ Rs. }$

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

$\displaystyle = 24 \times 800 + 1200 = 20400 \text{ Rs. }$

$\displaystyle \\$

(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	$\displaystyle f$	$\displaystyle x$	$\displaystyle f.x$
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
	$\displaystyle \Sigma f = 25$		$\displaystyle \Sigma f.x = 695$

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

(iii) Model Class = 20-30

$\displaystyle \\$

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

$\displaystyle \text{Distance } = 240 \text{ km}$

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

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

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

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

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

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

$\displaystyle 2x^2 - 20x = 2400$

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

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

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

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

$\displaystyle \Rightarrow x = 40 \text{ or } -30 ( \text{not possible})$

Hence $\displaystyle x = 40 \text{ km/hr}$

$\displaystyle \\$

Question 11

$\displaystyle \text{(a) Prove that } \frac{\cos A}{1+\sin A} +\tan A = \sec A \hspace{7.0cm} [3]$

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

(i) Construct a $\displaystyle \triangle ABC$ in which $\displaystyle BC = 6.5 \text{ cm } \angle ABC =60^{\circ} , AB=5 \text{ cm} .$

(ii) Construct the locus of points at a distance of $\displaystyle 3.5 \text{ cm }$ from $\displaystyle A .$

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

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

(i) number of shares he bought.

(ii) Market value of each share. [3]

Answers

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

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

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

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

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

= RHS Hence Proved

$\displaystyle \\$

(b) icse-201

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