MATHEMATICS (ICSE 2014)

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) Ranbir borrows Rs. 20,000 at 12\% per annum compound interest. If he repays Rs. 8400  at the end of the first year and Rs.9680 at the end of the second year, find the amount of loan outstanding at the beginning of the third year.  [3]

(b) Find the value of x , which satisfy the in equation -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2, x \in W . Graph the solution set on the number line.   [3]

(c) A die has 6 faces marked by the given numbers as shown below:  1, 2, 3, -1, -2, -3 . The die is thrown once. What is the probability of getting?

(i) A positive integer

(ii) An integer greater than -3

(iii) The smallest integer   [4]

Answer.

(a)  Given: Principal for the first year (P)= Rs. \ 20000,  \ Rate (r)=12\%

We known that A=P( 1+\frac{1}{100})^n     [Reference Link]

Amount after the 1st year = 20000 (1+\frac{12}{100})^1 = 20000 (\frac{112}{100}) = Rs. \ 22400

Money repaid at the end of 1st year Rs. \ 8400

Principle for the 2nd year 22400 - 8400 = Rs. \ 14000

Amount after 2nd year 14000 ( 1+\frac{12}{100} )^1= Rs. \ 15680

Money repaid at the end of the second year = Rs. \ 9680

The loan amount outstanding at the beginning of the third year  15680-9680 = Rs. \ 6000

(b)   Given -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2 

\Rightarrow - \frac{17}{6} < \frac{3-4x}{6} \le 2 

Multiplying throughout by 6

- 17 <3-4x \le 12 

-17<3-4x \ and \  3-4x \le 12 

\Rightarrow 4x<3+17 \ \ \ \ \ \ \ \ \ \  \Rightarrow 3-12 \le 4x 

\Rightarrow 4x < 20 \ \ \ \ \ \ \ \ \ \ \ \ \  \Rightarrow - 9 \le 4x

\Rightarrow x<5 \ \ \ \ \ \ \ \ \ \ \ \ \ \  \Rightarrow - \frac{9}{4}<x 

Therefore (5>x\ge \frac{-9}{4}) 

Hence the solution set is \{x \in W, - \frac{9}{4} \le x<5 \}

Therefore the values of x are \{0,1,2,3,4 \}

The graph of the solution set is shown by dots on the number line.

icse 14 3

(c) No. of sample space n(S) =6

A positive integer = \{1,2,3\}

No. of favorable cases n(E) = 3

Probability = \frac{n(E)}{n(S)}=\frac{3}{6}= \frac{1}{2} 

An integer greater than -3 = \lbrace 1,2,3,-1,-2\rbrace  

No. of favorable cases n(E)=5

Probability = \frac{n(E)}{n(S)}=\frac{5}{6} 

Smallest integer = -3

Probability of smallest integer = \frac{n(E)}{n(S)}=\frac{1}{6}

\\

Question 2:

(a) Find x,y if   \begin{bmatrix}  -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix}  -1 \\ 2x  \end{bmatrix} + 3 \begin{bmatrix}  -2  \\  1 \end{bmatrix} = 2 \begin{bmatrix}  y  \\  3 \end{bmatrix}    [3]

(b) Sharukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for 1\frac{1}{2} years. If he received Rs. 15,084 at the time of maturity. Find the rate of interest per annum.   [3]

(c) Calculate the ratio in which the line joining A(-4,2) \ and \ B(3,6)  is divided by point P(x,3) Also find (i) x (ii) Length of AP .   [4] 

Answer:

(a) Given \begin{bmatrix}  -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix}  -1 \\ 2x  \end{bmatrix} + 3 \begin{bmatrix}  -2  \\  1 \end{bmatrix} = 2 \begin{bmatrix}  y  \\  3 \end{bmatrix}

\Rightarrow \begin{bmatrix}  -2 \times -1+0 \times 2x  \\ 3 \times -  1 + 1 \times 2x \end{bmatrix} + \begin{bmatrix}  -6  \\  3 \end{bmatrix} = \begin{bmatrix}  2y  \\  6 \end{bmatrix}

\Rightarrow \begin{bmatrix}  2  \\  -3+2x \end{bmatrix} + \begin{bmatrix}  -6  \\  3 \end{bmatrix} = \begin{bmatrix}  2y  \\  6 \end{bmatrix}

\Rightarrow \begin{bmatrix}  2-6  \\  -3+2x+3 \end{bmatrix} = \begin{bmatrix}  2y  \\  6 \end{bmatrix}

\Rightarrow \begin{bmatrix}  -4  \\  2x \end{bmatrix} = \begin{bmatrix}  2y  \\  6 \end{bmatrix}

2y = 4  \ or \  y = -2 \ and \ 2x = 6 \ or \ x = 3 

(b)   Here, Principal (P) = money deposited per month =Rs. \ 800

N= Time for which the money is deposited = 1\frac{1}{2} \ years =18 \ months

Let the rate of interest be r\% per annum, then

Interest=P\times \frac{n(n+1)}{2\times 12}\times \frac{r}{100}    [Reference Link]

= 800 \times \frac{18\times 19}{2\times 12}\times \frac{r}{100}  = 114r \ Rs.

Total money deposited = 18 \times 800= Rs. \ 14400

Since money deposited +Interest = Maturity value

14400+114r=15084 

114r=15084-14400 

114r=684 

r=\frac{684}{114}=6 

Hence rate of interest = 6\% p.a. 

(c)    Let P(x, 3)  divide the line segment joining the points A (-4, 2)   and B (3, 6)  in the ratio k :1 

Coordinate of P is:                                                           [Reference Link]

( \frac{m_1. x_2+m_2.x_1}{m_1+m_2},  \frac{m_1. y_2+m_2.y_1}{m_1+m_2} = \frac{3k-4}{k+1}, \frac{6k+2}{k+1})

But Coordinate of P \ is \ (x,3) 

\frac{6k+2}{k+1}=3 

6k+2=3k+3 

3k=1 \Rightarrow k=\frac{1}{3} 

The required ratio is \frac{1}{3}:1 \ i.e. \ 1:3 , (Divides Internally)

(i)   Therefore x= \frac{3k-4}{k+1} 

Substituting k =\frac{1}{3} 

x= \frac{3\times \frac{1}{3}-4}{\frac{1}{3}+1}= \frac{1-4}{\frac{1+3}{3}} = \frac{-3}{\frac{4}{3}}= \frac{-9}{4} 

(ii) Coordinate of P  is (\frac{-9}{4}, 3)

Length of AP = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} 

= \sqrt{(-\frac{9}{4}+4)^2+(3-2)^2} 

= \sqrt{(\frac{-9+16}{4})^2+(3-2)^2} = \sqrt{\frac{49}{16}+1}

= \sqrt{\frac{49+16}{16}}= \sqrt{\frac{65}{16}}=\sqrt{\frac{65}{4}}

\\

Question 3:

(a) Without using trigonometric tables, evaluate

sin^2 \ 34^o + sin^2 \ 56^o + 2 \ tan \ 18^o tan \ 72^o  - cot ^2 \ 30^o   [3]

(b) Using the remainder and factor theorem, factor the following polynomial: x^{3}+10x^{2}-37x+26    [3]

(c) icse 14 2In the figure given below ABCD  is a rectangle AB=14 \ cm, \ BC=7\ cm . From the rectangle a quarter circle BFEC  and a semicircle DGE  are removed Calculate the area of the remaining piece of the rectangle (Take \pi = \frac{22}{7} )   [4]

Answers:

(a) Given sin^2 \ 34^o + sin^2 \ 56^o + 2 \ tan \ 18^o tan \ 72^o  - cot ^2 \ 30^o

= sin^2 \ 34^o + sin^2 \ (90^o - 34^o) + 2 \ tan \ 18^o \ tan \ (90^o-18^o)  - cot ^2 \ 30^o

= sin^2 \ 34^o + cos^2 \  34^o + 2 \ tan \ 18^o \ cot \ 18^o  - (\sqrt{3})^2

= 1+ 2 \ tan \ 18^o \times \frac{1}{tan \ 18^o}-3

= 1+2-3 = 0 

(b)  Let f(x)=x^3+10x^2-37x+26

Putting x =1 , we get

f(1)=1+10-37+26=0

By factor theorem x-1 is actor of f(x)

  • x-1 ) \overline {x^3+10x^2-37x+26} (x^2+11x-26
  •  (-) \ \  \underline {x^3-x^2}  
  •                           11x^2-37x+26
  •                  (-) \ \   \underline{11x^2-37x}
  •                                       { -26x+26}
  •                             (-) \ \   \underline{ -26x+26}
  •                                          \times

On dividing x^{3}+10x^{2}-37x+26 by (x-1) , we get x^{2}+11x-26 as the quotient and remainder = 0

Therefore the other factor of f(x) are the factor of x^{2}+11x-26

Now, x^{2}+11x-26

= x^{2}+13x-2x-26

= x(x+13)-2(x+13)

= (x+13)(x-2)

Hence x^{3}+10x^{2}-37x+26=(x-1)(x-2)(x+13)

(c) Area of rectangle ABCD= 14 \times  7 = 98cm

Area of quarter circle BFEC=\frac{1}{4}\pi (7^{2})=\frac{49}{4}\pi

Area of semicircle DGE= \frac{1}{2}\pi (\frac{7}{2})^2=\frac{1}{2}\times \frac{49}{4}\pi 

Area of remaining piece of rectangle = 98-[\frac{49}{4}\pi +\frac{1}{2}\times \frac{49}{4}\pi ]

= 98-\frac{49}{4}\pi \lbrack 1+\frac{1}{2}\rbrack 

= 98-\frac{49}{4}\times \frac{22}{7}\times \frac{3}{2}=98-\frac{231}{4}

= 98-57.75 = 40.25 cm^2

\\

Question 4: 

(a) The number 6,8,10,12,13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x [3]

(b) In the figure, \angle DBC=58^o. BD is a diameter of the circle. Calculate: [3]icse 14 1.jpg

(i) \angle BDC

(ii) \angle BEC

(iii) \angle BAC

(c) Using graph paper to answer the following questions. (Take 2 \ cm = 1 unit on both axis)

(i) Plot the points A (-4,2) and B (2,4)

(ii) A' is the image of A when reflected in the y-axis . Plot it on the graph paper and write the coordinates of A'

(iii) B' is the image of B when reflected in the line AA' write the coordinates of B'

(iv) Write the geometric name of the figure ABA'B'

(v) Name a line of symmetry of the figure formed.    [4]

Answers:

(a) Arrange numbers in ascending order are 6,8,10,12,13, x.

Mean = \frac{6+8+10+12+13+x}{6}=\frac{49+x}{6}

No. of terms (n) = 6 \ (even)

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

Median = \frac{ (\frac{6}{2})^{th} \ term + (\frac{6}{2}+1)^{th} \ term }{2} = \frac{3^{rd}+4^{th}}{2} = \frac{10+12}{2} = \frac{22}{2} = 11

According to given condition

\frac{49+x}{6}=11 \Rightarrow 49+x=66 

or x=17

(b) In \triangle BCD , \angle DBC = 58^o \ (given) 

(i) \angle BCD = 90^o  (angle in the semi circle)

\therefore \angle DBC + \angle BCD + \angle BDC = 180^o 

58^o+90^o+\angle BDC = 180^o 

\Rightarrow \angle BDC = 180^o-148^o = 32^o 

(ii) \angle BEC + \angle BDC = 180^o  (cyclic quadrilateral)

\angle BEC = 180^o-32^o = 148^o 

(iii) \angle BAC = \angle BDC  (angles in the same segment)

\angle BAC = 32^o 

(c) As shown in the graph below:

(i) Coordinate of A' = (4,2) 

(ii) Coordinate of B' = (2, 0) 

(iii) Geometric name = kite 

(iv) AA'  is the symmetric line.

sy5

 


SECTION B [40 Marks]

(Answer any four questions in this Section.)


Question 5:

(a) A shopkeeper bought a washing machine at a discount of 20 \% from a wholesaler, the printed price of the washing machine being Rs. \ 18,000 . The shopkeeper sells it to a consumer at a discount of 10 \%  on the printed price. If the rate of sales tax is 8 \%  find:

(i) the VAT paid by the shopkeeper

(ii) the total amount that the consumer pays for the washing machine.    [3]

(b) If \frac{x^2+y^2}{x^2-y^2} = \frac{17}{8} , then find the value of

(i) x:y

(ii) \frac{x^{3}+y^{3}}{x^{3}-y^{3}}    [3]

(c) In \triangle ABC, \angle ABC = \angle DAC. AB = 8 \ cm, AC = 4 \ cm, AD = 5 \ cm icse 14 10

(i) Prove that \triangle ACD is similar to \triangle BCA

(ii) Find BC \ and \  CD

(iii) Find area of \triangle ACD : \ area \ of  \ \triangle ABC    [4]

Answers:

(a)  Given: Printed price of washing machine = Rs.\ 18000

Rate of discount =20\% 

(i)   Amount of discount to shopkeeper = \frac{20}{100}\times 18000=Rs.\ 3600

Shopkeeper’s Price = 18000-3600 = Rs.\ 14400

Sales Tax paid by shopkeeper = \frac{8}{100}\times 14400=Rs.\ 1152

Discount for consumer = \frac{10}{100}\times 18000=Rs.\ 1800

Price of consumer =18000-1800= Rs.\ 16200

Tax charged by the shopkeeper = \frac{8}{100}\times 16200=Rs.\ 1296

Since, Tax paid by the shopkeeper = Rs.\ 1152

VAT paid by the shopkeeper=Tax charged – Tax Paid =1296-1152=Rs.\ 144

(ii) Total amount paid by the consumer for washing machine = 1620+1296 = Rs. \ 17496

(b) Given: \frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{17}{8}

(i) Applying componendo and dividend

\frac{(x^{2}+y^{2}) +(x^{2}-y^{2})}{(x^{2}+y^{2})-(x^{2}-y^{2})}=\frac{17+8}{17-8}

\Rightarrow \frac{2x^{2}}{2y^{2}}=\frac{25}{9}= \frac{x^{2}}{y^{2}}=\frac{25}{90}

\Rightarrow \frac{x}{y}=\frac{5}{3}

\Rightarrow x:y=5:3

(ii) Cubing both sides we get

\frac{x^{3}}{y^{3}}=\frac{5^{3}}{3^{3}}=\frac{125}{27}

Applying componendo and dividend

\frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{125+27}{125-27}

\Rightarrow \frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{152}{98}

\Rightarrow \frac{x^{3}+y^{3}}{x^{3}-y^{3}}=\frac{76}{49}

(c)  (i) In \triangle ACD and \triangle BCA

\angle C (common angle)icse 14 9

\angle ABC = \angle CAD (given)

Therefore \triangle ACD \sim \triangle BCA (AAA Postulate)

(ii) Since \triangle ACD \sim \triangle BCA 

\Rightarrow  \frac{AC}{BC}=\frac{CD}{CA}=\frac{AD}{BA} (corresponding sides are proportional)

\frac{4}{BC}=\frac{CD}{4}=\frac{5}{8}

\Rightarrow  \frac{4}{BC}=\frac{5}{8} \ and \  \frac{CD}{4}=\frac{5}{8}

\Rightarrow  BC=\frac{4\times 8}{5}=\frac{32}{5}=6.4 \ cm and CD=\frac{5}{8}\times  4=\frac{5}{2}=2.5 \ cm 

(iii) Since \Delta ACD \sim \Delta ABC

\frac{area \ (\Delta ACD)}{area \ (\Delta ABC)}=\frac{AC^{2}}{AB^{2}} = \frac{4^{2}}{8^{2}}=\frac{16}{64}=\frac{1}{4}

area \ (\Delta ACD ) \ : \ area \ (\Delta ABC)=1:4

Question 6:

(a) Find the value of 'a' for which the following points A (a,3), B(2,1) and C (5,a) are collinear. Hence find the equation of the line.   [3]

(b) Salman invests a sum of money in Rs.\ 50 shares, paying 15\% dividend quoted at 20\% premium. If his annual dividend is Rs.\ 600 , calculate;

(i) the number of shares he bought

(ii) his total investment

(iii) the rate of return on his investment   [3]

(c) The surface area of a solid metallic sphere is 2464 \ cm^2 . it is melted and recast into solid right circular cones of radius 3.5 \ cm and height 7 \ cm . calculate:

(i) the radius of the sphere.

(ii) the number of cones recast. (Take \pi = \frac{22}{7}  [4]

Answers:

(a)  Given: A (a,3), B(2,1) and C(5,a) are collinear.

Slope \ of \ AB= Slope \ of \ BC

\frac{1-3}{2-a}=\frac{a-1}{5-1}

\frac{-2}{2-a}=\frac{a-1}{3}

\Rightarrow \ \ \  -6=(2-a)(a-1)

\Rightarrow \ \ \ -6=2a-2-a^{2}+a

\Rightarrow \ \ \ a^{2}-3a-4=0

\Rightarrow \ \ \ a^{2}-4a+a-4=0

\Rightarrow \ \ \ (a-4)(a+1)=0

a=4, -1

Rejecting a=-1

Slope of BC = \frac{a-1}{5-2}=\frac{4-1}{3}=\frac{3}{3} =1

Equation \ of \ BC; (y-1)=1 (x-2)

y-1=x-2

x-y=1

(b) Nominal value of share = Rs. \ 50

Dividend on 1 share = \frac{15}{100}\times 50=Rs. \ 7.50

Total dividend of Salman =Rs. \ 600

(i) No. of shares Salman bought = \frac{600}{7.50}=80

(ii) Premium on 1 share =\frac{20}{100}\times 50=Rs. \ 10

Market value of 1 share = 50+10=Rs. \ 60

Total investment for 80 shares = 80 \times 60 = Rs. \ 4800

(iii) Rate of return= \frac{600}{4800}\times 100=12.5\%

(c)  (i) Let the radius sphere = r \ cm

Surface area of sphere = 4\pi r^{2}=2464 \ cm^{2}

r^{2}=\frac{2464}{4\pi }

r^{2}=\frac{2464\times 7}{4\times 22}=196

r=14\ cm

(ii) Volume of sphere = \frac{4}{3}\pi r^{2}=\frac{4}{3}\pi (14)^{3}

Volume of cone = \frac{1}{3}\pi ^{2}h=\frac{1}{3}\pi (3.5)^2 \times 7

No. of cones recast = \frac{Volume \ of \ sphere}{Volume \ of \ cone}

= \frac{\frac{4\pi }{3}(14)^{3}}{\frac{1}{3}\pi (3.5)2\times 7} = \frac{4\times 14\times 14\times 14}{3.5\times 3.5\times 7}=\frac{3200}{25} = 128 

Question 7:

(a) Calculate the mean of the distribution given below using the short cut method;     [3]

Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80
No. of students 2 6 10 12 9 7 4

(b) icse 14 8In the figure given below, diameter AB \ and \  CD of a circle meet at P. PT  is a tangent to the circle at T. CD=7.8 \ cm, \ PD=5 \ cm  \ PB=4 \ cm find:

(i) AB

(ii) the length of tangent PT  [3]

(c) Let A = \begin{bmatrix}  2 & 1 \\ 0 & -2  \end{bmatrix} B = \begin{bmatrix}  4 & 1 \\ -3 & -2  \end{bmatrix} and C = \begin{bmatrix}  -3 & 2 \\ -1 & 4  \end{bmatrix} . Find A^2+AC-5B  [3]

Answers:

(a) Table as follows

Marks \\ (C.I)

f Mean Value

x

A=45.5

d=x-A

f \times d

11-20

21-30

31-40

41-50

51-60

61-70

71-80

2

6

10

12

9

7

4

15-5

25-5

35-5

45-5

55-5

65-5

75-5

-30

-20

-10

0

10

20

30

-60

-120

-100

0

90

140

120

\Sigma f=50  \Sigma fd=70

 

Mean = A + \frac{ \Sigma fd}{\Sigma f} = 45.5+\frac{70}{50} = 45.5 + 1.4 = 46.9

(b)  (i) Since chord CD and tangent at point T intersect each other at P .

PC \times PD=PT^{2} … … … … (i)

Since chord AB and tangent at point T interested each other at P ,

PA\times PB= PT^{2}   … … … … (ii)

From (i) \& (ii) PC\times PD=PA\times PB

Given; CD=7.8 \ cm, PD=5 \ cm, PB=4 \ cm

PA=PB+AB=4+AB \\ PC=PD+CD=5+7.8=12.8 \ cm

Putting these values in eq. (3)

12.8\times 5=(4+AB)\times 4

\Rightarrow 4+AB= \frac{12.8\times 5}{4}

\Rightarrow 4+AB=16

\Rightarrow AB=12 \ cm

Hence, AB = 12 \ cm

(ii) From  (i), PT^{2}=PA\times PB=12.8\times 5

PT^{2}=64

Length of tangent  PT=8 \ cm=

(c)  A^2+AC-5B

= \begin{bmatrix}  2 & 1 \\ 0 & -2  \end{bmatrix} .  \begin{bmatrix}  2 & 1 \\ 0 & -2  \end{bmatrix} +  \begin{bmatrix}  2 & 1 \\ 0 & -2  \end{bmatrix} .  \begin{bmatrix}  -3 & 2 \\ -1 & 4  \end{bmatrix} - 5. \begin{bmatrix}  4 & 1 \\ -3 & -2  \end{bmatrix}

= \begin{bmatrix}  4 & 0 \\ 0 & 4  \end{bmatrix} + \begin{bmatrix}  -7 & 8 \\ 2 & -8  \end{bmatrix} - \begin{bmatrix}  20 & 5 \\ -15 & -10  \end{bmatrix} 

= \begin{bmatrix}  -23 & 3 \\ 17 & 6  \end{bmatrix}  

Question 8:

(a) The compound interest, calculate yearly, on a certain sum of money for the second year is Rs. \ 1320 and for the third year is Rs. \ 1452 . Calculate the rate of interest and the original sum of money.   [3]

(b) Construct a \triangle ABC with BC=6.5 \ cm, AB=5.5 \ cm, AC=5 \ cm , Construct the in circle of the triangle measure and record the radius of the in circle.  [3]

(c) Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below;

Pocket expenses (in Rs.) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Number of students (frequency) 10 14 28 42 50 30 14 12

Draw a histogram representing the above distribution and estimate the mode from the graph.   [4]

Answers:

(a)  Compound Interest for the third year = Rs. \ 1452

Compound Interest for the second year = Rs. \ 1320

Simple Interest on Rs. \ 1320 for one year = 1452-1320=Rs. \ 132

Rate of interest = \frac{132\times 100}{1320}=10\%

Let the original money be Rs. \ 'P'

Amount after 2 year – amount after one year = Compound Interest for second year.

\Rightarrow P (1+\frac{10}{100})^2-P(1+\frac{10}{100})=1320

\Rightarrow P[(\frac{110}{100})^2-(\frac{110}{100})]=1320

\Rightarrow P\lbrack (\frac{11}{10})^2-\frac{11}{10}\rbrack =1320

\Rightarrow (\frac{121}{100}-\frac{11}{10})=Rs \ 1320

\Rightarrow  P\times \frac{11}{100}=Rs.1320

\Rightarrow  P= \frac{1320\times 100}{11}=Rs. \ 12,000

Rate of interest 10\%

Original sum of money = Rs. \ 12,000

(b) Steps of construction:icse 14 12.jpg

  1. Construct a \triangle ABC   with the given data:
    1. Draw a line BC of 6.5 cm length using a ruler
    2. The make an arc of 5.5 cm from B and similarly, make an arc of 5 cm from C
    3. The place where the two arcs intersect, join that point to B and C and complete the triangle.
  2. Draw the internal bisectors of \angle B \ and \  \angle C . Let these bisectors cut at O
    1. Draw an arc from point B so that it cuts the two sides of the angle ABC
    2. From the point of intersections, draw two arcs
    3. Join the point B and the point of intersection and draw a line. This line bisects the angle ABC.
    4. Repeat the above 3 steps for point C.
    5. The two bisectors will intersect at the point O which is the center of the in-circle.
  3. Taking O as center and touching the side of the circle as the radius, Draw a in-circle which touches all the sides of the \triangle ABC
  4. From O draw a perpendicular to side BC which cut at N .
  5. Measure ON which is required radius of the in circle. ON=1.5 \ cm .

(c)

icse 14 7

Question: 9

(a) If (x-9): (3x+6) is the duplicate ratio of 4:9 , find the value of x .   [3]

(b) Solve for x using the quadratic formula. Write your answer correct to two significant figures. (x-1)^2-3x+4=0    [3]

(c) A page from the saving bank account of Priyanka is given below:   [4]

Date Particular Amount

Withdrawal

Amount

Deposited

Balance
03/04/2006

05/04/2006

18/04/2006

25/05/2006

30/05/2006

20/07/2006

10/09/2006

19/09/2006

B/F

By cash

By Cheque

To Cheque

By Cash

By Self

By Cash

To Cheque

5,000.00

4,000.00

1,000.00

2,000.00

6,000.00

3,000.00

2,000.00

4,000.00

6,000.00

12,000.00

7,000.00

10,000.00

6,000.00

8,000.00

7,000.00

If the interest earned by Priyanka for the period of ending September, 2006 is Rs.175 , find the rate of interest.

Answers:

(a)   Given: (x-9): (3x+6) is the duplicate ratio of  4:9

\Rightarrow \frac{x-9}{3x+6}=(\frac{4}{9})^2

\Rightarrow \frac{x-9}{3x+6}=(\frac{16}{81})

\Rightarrow 81x-729=48x+96

\Rightarrow 81x-48x=96+729

\Rightarrow 33x=825

\Rightarrow x=\frac{825}{33}=25

(b)  Given: (x-1)^2-3x+4=0

\Rightarrow x^{2}+1-2x-3x+4=0

\Rightarrow x^{2}-5x+5=0

Comparing x^{2}-5x+5=0 with ax^{2}+bx+c=0 , we get a=1, b=-5, C=5

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

x= \frac{-(-5)\pm \sqrt{(-5)^{2}-4(1)(5)}}{2 \times 1}

= \frac{5\pm \sqrt{25-20}}{2} =\frac{5\pm \sqrt{5}}{2}

= \frac{5\pm 2.236}{2}= \frac{5+2.236}{2}   and  \frac{5-2.236}{2}

= \frac{7.236}{2}   and  \frac{2.764}{2}

=3.618  \ and \   1.382

(c) Qualifying principal for various months:

Month Principal (Rs.)
April 6000
May 7000
June 10000
July 6000
August 6000
September 7000
Total for 1 month 42000

Now, P =Rs. \ 42,000, I=Rs. \ 175, \ T=\frac{1}{12}years, Rate= R

Interest = \frac{P\times R\times T}{100}

175 = \frac{42,000\times R\times 1}{100\times 12}

R = \frac{175\times 100\times 12}{42,000\times 1} = \frac{2100}{420}=5\%

Question: 10

(a) A two digit number is such that the product of its digits is 6, If 9 is added to the number, the digits interchange their places. Find the number. [4]

(b) The number obtained by 100 students in a Mathematics test are given below; [6]

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No. of Students 3 7 12 17 23 14 9 6 5 4

Draw an ogive for the distribution on a graph sheet. (Use a scale of 2 cm = 10 units on both axis)

Use the ogive to estimate the:

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the if the pass percentage was 35

Answers:

(a)  Let the required two digit number be 10x+y

Given: xy=6 and 10x+y+9=10y+x

10x-x+y-10y+9=0

\Rightarrow \ \ \ \ 9x-9y+9=0

\Rightarrow \ \ \ \ x-y+1=0

\Rightarrow \ \ \ \ y=x+1

Since xy = 6 (given)

\Rightarrow \ \ \ \ x(x+1)=6

\Rightarrow \ \ \ \ x^{2}+x-6=0

\Rightarrow \ \ \ \ x^2 +3x-2x-6=0

\Rightarrow \ \ \ \ (x+3)(x-2)=0

x=-3 \ or  \ 2

x=-3  (not possible)

When x=2, y=x+1=2+1=3

The required two digit number = 10x+y = 10 \times 2+3 = 23

(b) 

Marks No. of Students Cumulative Frequency (c.f)
0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

8

7

12

17

23

14

9

6

5

4

3

10

22

39

62

76

85

91

96

100

On the graph paper, we plot the following points:

(10, 3), (20, 10), (30, 22), (40,39), (50, 62), (60,76), (70,85),  \\ (80, 91), (90, 96),(100, 100)

icse 14 11

(i) Medium = (\frac{n}{2})^{th} term = \frac{100}{2}=50^{th} term

From the graph 50^{th} term=44

(ii) Lower quartile = (\frac{n}{4})^{th} term = \frac{100}{4}=25^{th} term

From the graph 25^{th} term =32

(iii) The number of students who obtained more than 85\% marks in test = 100-94 = 6 students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 = 30 students.

Question 11:

(a) In the figure given below, O is the center of the circle, AB and CD are two chords of the circle. OM \perp AB and ON \perp CD .icse 14 6

AB=24 \ cm, OM=5 \ cm, ON=12 \ cm Find the:

(i) radius of the circle

(ii) length of chord CD    [3]

(b) Prove the identity:

(sin \  \theta + cos \ \theta) (tan \ \theta + cot \ \theta) = sec \ \theta + cosec \ \theta    [3]

(c) An airplane at an altitude of 250 \ m observed of depression of two boats on the opposite banks of a river to be 45^o and 60^o respectively. Find the answer correct to the nearest whole number.   [4]

Answers:

(a)  Given AB=24\ cm; OM=5 \ cm, ON=12 \ cm

OM \perp AB icse 14 5

M is mid point of AB

AM=12 \ cm

 (i) Let radius of circle = r

From \triangle AMO

AO^{2}=AM^{2}+OM^{2}

r^{2}=12^{2}+5^{2}=144+25=169

r=13 \ cm

(ii) Now from \triangle CNO; CO^{2}=ON^{2}+CN^{2}

r^{2}=(12^{2})+CN^{2}

13^{2}-12^{2}=CN^{2}

169-144=CN^{2}

\Rightarrow CN^{2}=25

\Rightarrow CN=5

As ON \perp CD, N  is mid point of CD

\therefore CD=2 CN=2 \times 5=10 \ cm

(b)   To prove: (sin \  \theta + cos \ \theta) (tan \ \theta + cot \ \theta) = sec \ \theta + cosec \ \theta

LHS = (sin \  \theta + cos \  \theta)(tan \  \theta+ cot  \  \theta)

= (sin \  \theta + cos \  \theta) (\frac{sin \  \theta}{cos \  \theta} + \frac{cos \  \theta}{sin \  \theta})

= (sin \  \theta + cos \  \theta) (\frac{sin^2 \  \theta+ cos^2 \  \theta}{sin \  \theta . cos \  \theta})

= (sin \  \theta + cos \  \theta) (\frac{1}{sin \  \theta . cos \  \theta})

=  (\frac{sin \  \theta}{sin \  \theta . cos \  \theta}) + (\frac{cos \  \theta}{sin \  \theta . cos \  \theta})

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

= sec \  \theta+cosec \  \theta = RHS

Hence Proved.

(c)  Let AD =250 \ m height of airplaneicse 14 4

Two boats are at B \ and \  C .

Let BD =x \ and \  DC=y as shown in the diagram.

From \triangle ADB; \frac{x}{250}=\cot 45^o

\frac{x}{250}=1   \Rightarrow x=250 \ m

From \triangle ADC; \frac{y}{250}=\cot 60^o

\frac{y}{250}=\frac{1}{\sqrt{3}}

Width of river BC = BD+DC = x + y

= 250 +\frac{250}{\sqrt{3}}

= 250 (1+\frac{1}{\sqrt{3}})=250 (\frac{\sqrt{3}+1}{\sqrt{3}})

= 250 (\frac{1.732+1}{1.732})=250 (\frac{2.732}{1.732})

= 250 \times 1.577

= 394.25 \ m = 394 \ m

 

Advertisements