MATHEMATICS (ICSE 2016)

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

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

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

Mathematical tables are provided.


SECTION A [40 Marks]

(Answer all questions from this Section.)


Question 1

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

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

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

Answers:

(a)    Let  f(x)=2x^3+3x^2-kx+5

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

Therefore  2(2)^3+3(2)^2-k(2)+5 = 7

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

\Rightarrow 2k = 26

\Rightarrow k = 13

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

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

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

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

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

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

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

Since LHS = RHS

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

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

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

Hence the value of  m = -14

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

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

 \Rightarrow 612=497+x

 \Rightarrow x = 115

Now arrange the numbers in ascending order we get

 26, 45, 52, 60, 69, 70, 81, 94, 115

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

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

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

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

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Question 2

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

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

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

Answers:

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

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

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

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

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

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

\Rightarrow k = -11

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

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

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

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

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

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

\Rightarrow k = -11

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(b)   Given:

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

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

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

= 1+1 -1-3 = -2

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(c)  Let number of cones = n

Volume of cones = Volume of sphere [Provide Formulas]

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

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

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

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

Therefore the number of cones needed is 72

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Question 3

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

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

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

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

Answers:

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

-3x+21 \geq 15-7x \ge  \frac{x+1}{3}

Therefore we have:

-3x+21 \geq 15-7x

\Rightarrow 4x \geq -6

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

Also we have

15-7x \ge  \frac{x+1}{3}

\Rightarrow 45-21x \ge x+1

\Rightarrow 44 \ge 22x

\Rightarrow 2 \ge x … … … … … i)

Combining i) and ii) we get

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

icse 9

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(b)    Given  AD \parallel BC

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

Since OB=OD  (radius of the same circle)

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

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

In \triangle ABC

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

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

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

Therefore \angle BED = 58^o

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(c)    (3a+2b):(5a+3b)= 18:29

\frac{3a+2b}{5a+3b}=\frac{18}{29}  

Cross multiplying

\Rightarrow 87a+58b=90a+54b  

3a=4b  

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

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Question 4

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

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

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

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

Answers:

(a)   Total number of all possible outcomes = 30

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

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

The number of favorable outcomes = 3

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

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

Therefore the number of favorable outcomes = 4

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

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(b)  i) A'(4, 4) \ and \ B'(3, 0)

ii) Arrow Head

iii) y-axis is the line of symmetry

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

i)    We know that Amount (A) = P + I

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

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

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

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

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

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

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

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


SECTION B [40 Marks]

(Answer any four questions in this Section.)


Question 5

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

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

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

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

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

Answers:

(a)  Given x^2-3(x+3) =0

\Rightarrow  x^2-3x-9=0 

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

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

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

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

Solving we get x = 4.85, -1.85

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(b)  Qualifying principal for various months:

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

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

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

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

Amount = Last \ Balance + Interest

Amount = 5900+151.50 = 6051.50 \ Rs.  

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

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

(\frac{11}{10})^3= (\frac{11}{10})^n

\Rightarrow n = 3 \ years 

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Question 6

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

(b) icse 3In the given figure PQRS is a cyclic quadrilateral PQ \ and \ RS produced meet at point T .  i) Prove \triangle TPS \sim \triangle TRQ  ii) Find. SP \ if \ TP = 18 \ cm,  \ RQ = 4 \ cm \ and, \ TR = 6 \ cm  iii) Find area, of quadrilateral PORS if area of  \triangle PTS=27 \ cm^2  [4]

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

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

Answers:

(a)     Step 1: Given side of the hexagon is 5 \ cm  . Construct the hexagon as follows:icse 22

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

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

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

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

\angle PST = \angle RQT 

\angle SPQ = \angle QRT 

\angle T  is common

Therefore \triangle TPS \sim \triangle TRQ 

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

\frac{TP}{TR}=\frac{PS}{RQ} 

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

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

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

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

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

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

If, AX=B :

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

We know

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

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

Hence m = 2 and n = 1  

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

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

Therefore

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

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

 4a \ sin\ 30^o +b cos \ 0^o=4  

\Rightarrow 2a+b=4  … … … … … i)

a \ cos \ 0^o +  4b \ sin\ 30^o = 5 

\Rightarrow a+2b=5    … … … … … ii)

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

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

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Question 7

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

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

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

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

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

Answer:

(a)     In the adjacent diagram, let D be the air plane and A \ and \  B  be the two ships.icse 4.jpg

Given: CD=1500 \ m

\angle EDA = 30^o

\angle EDB = 45^o

Consider \triangle DBC

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

Consider \triangle ADC

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

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

Hence, distance between two ships = 1098 \ m .

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(b) 

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

(i)    N = 160 (which is even)

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

From the graph, Median = 43.5

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

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

Inter Quartile range 60-28.5 = 31.5

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

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Question 8

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

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

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

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

(c) icse 6.jpgA line AB  meets the x-axis  at A  and y-axis at B .  P(4, -1)  divides AB in the ratio 1 : 2.

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

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

Answers:

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

Therefore = ak, y =bk, z=ck 

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

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

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

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

Substituting

= 3 $latex  \frac{x}{a} \frac{y}{b} \frac{z}{c} = &s=2$ RHS Hence proved.

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(b) 

  1. First draw line AB = 5   \ cm . This you can just draw by using a ruler.icse 21
  2. From AB cut AC = 3   \ cm . This can be done by using a compass. Set the compass at 3 cm and then make this arc cutting AB at C.
  3. Draw perpendicular bisector of AC .
  4. Taking A as center and AO = 2.5   \ cm   radius, draw an arc to meet the perpendicular bisector of AC \ at \ O .
  5. Taking O as the center and AO = 2.5   \ cm  as radius, draw a circle, which passing through A \ and \ C
  6. Join BO .
  7. Draw right bisector of BO , which cuts BO \ at \ P .
  8. Taking P as center and OP as radius, draw a circle which cuts old Circle at R and  T .
  9. Join BR \ and \ BT .

These are required tangents.

(ii)   Length of tangent = 3  \ cm .

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(c) 

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

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

AP: PB = 1:2

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

x = \frac{m_1x_2+m_2x_1}{m_1+m_2}  

y = \frac{m_1y_2+m_2y_1}{m_1+m_2}  

Therefore

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

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

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

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

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

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

 y-(-1)=-2(x-4)

 \Rightarrow y+1=-2x+8

 \Rightarrow 2x+y=7

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Question 9

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

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

(ii) The VAT paid by the dealer. [3]

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

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

(ii) unshaded part. [4]

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

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

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

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

Answers

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

Therefore the cost price of the article for the dealer

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

= 6000-1800 = 4200 \ Rs.

Sales tax paid by the dealer to the wholesaler

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

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

The cost price for the shopkeeper

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

= 6000 - 600 = 5400

The price paid by the shopkeeper including the tax

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

= 5400 + 324 = 5724 \ Rs.

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

= 324 - 252 = 72 \ Rs.

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(b)

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

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

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

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

= 2\pi + 6.24 \pi

= 8.25 \pi

= 25.92 cm^{2}

(ii) Let kite ABCD be a quadrilateral.

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

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

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

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

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

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

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

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

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

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Question 10

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

(i) the monthly installmenticse 8

(ii) the amount of maturity [3]

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

(i) Frame a frequency distribution, table.

(ii) To calculate mean.

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

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

Answers

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

Rate = 6\%,  \ Interest = 1200 \ Rs.

(i)   Let the monthly instalment be x

Using the formula:

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

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

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

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

Therefore the monthly instalment = 800 \ Rs.

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

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

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(b)   (i) Frequency Distribution Table

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

(ii) Mean

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

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

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

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

Distance = 240 \ km

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

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

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

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

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

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

2x^2 - 20x = 2400

x^2-10x-1200 = 0

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

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

(x-40)(x+30) = 0  

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

Hence x = 40 \ km/hr

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Question 11

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

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

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

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

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

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

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

(i) number of shares he bought.

(ii) Market value of each share. [3]

Answers

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

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

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

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

= \frac{1}{cos \ A}

= RHS Hence Proved

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(b)

  1. Construct \triangle ABC . Draw a line BC which is 6.5  cm long. The using your compass, make an angle of 60^o at point B of line BC. icse 201
  2. The taking the compass, make a cut a 5 cm mark on the side of the angle to mark A. AB is 5 cm long.
  3. Taking A as center and radius is 3.5 , draw a circle. This has been told in the question. Which is required locus.
  4. Draw an angle bisector of C Angle bisector of C cut above circle at two points i.e., X and Y These points are required points.
  5. Measure the length of XY = 5 cm .

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(c)

(i)  Let number of shares be x

Face Value  of x shares = 25 x

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

Therefore 3x = 2475 \Rightarrow x = 825

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

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