Mathematics

Class X

Board Paper – 2012

Time: 2½ hour Total Mark

1. Answer to this paper must be written on the paper provided separately.

2. You will NOT be allowed to write during the first 15 minutes. This time is to be spent in reading the

question paper.

3. The time given at the head of this paper is the time allowed for writing the answers.

4. This question paper is divided into two Sections. Attempt all questions from Section A and any four

questions from Section B.

5. Intended marks for questions or parts of questions are given in brackets along the questions.

6. All working, including rough work, must be clearly shown and should be done on the same sheet as

the rest of the answer. Omission of essential working will result in loss of marks.

7. Mathematical tables are provided

Solution

Section - A (40 Marks)

Sol. 1

(a)

10

01,

21

13IA

21

13

21

132A =

41

23

23

19=

35

58

10

017

21

135

35

58752 IAA

70

07

105

515

35

58

70

07

70

07

000

00

(b) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x

respectively.

Let their expenditures be 3y and 5y respectively.

Ravi’s savings = 5x – 3y

Sanjeev’s savings = 7x – 5y

By the given information,

5x – 3y = 80 … (1)

7x – 5y = 80 … (2)

From (1) and (2), we have:

5x – 3y = 7x – 5y

x = y

From equation (1),

5x – 3x = 80

2x = 80

x = 40

Hence,

Monthly pocket money of Ravi = 5 x 40 = Rs.200

Monthly pocket money of Sanjeev = 7 x 40 = Rs.280

(c) Let p(x) = 3x3+2x2-19x+6

P(2)=3(2)3+2(2)2-19(2)+6 = 24 +8 – 38 + 6 = 0

Using remainder theorem, (x-2) is a factor of p(x).

P(-3) = 3(-3)3 + 2(-3)2 - 19(-3) + 6= -81 + 18 + 57 + 6 = 0

Using remainder theorem, (x+3) is a factor of p(x).

Thus, (x-2)(x-3)=x2+x-6 is a factor of p(x).

Dividing p(x) by x2- 5x + 6, we have:

P(x)= (x2 + x - 6)(3x - 1) = (x-2)(x + 3)(3x - 1)

Sol.2

(a) We aknow that:

CI-SI (for 2 years) = SI for 1 year on the SI of first year

Let the principal be P.

R = 5% p.a.

Interest for the first year = 20100

15

100

PPTRP

Interest on Rs.40010020

151

20

PPyearfor

P

It is given that CI-SI=Rs.25

25.400

RsP

000,10.RsP

Thus, the sum of money is Rs.10,000.

(b) Using Pythagoras theorem,

AC 22222 77 BCAB

AC = 7 2

Radius of semi-circle = 72

27

Area of the shaded region = Area of semi-circle – Area of triangle ABC

Area of semi-circle = 2

2

2

2

77

4

49

8

98

2

27

2

1

2

1cmr

Area of triangle ABC = 2

2

4977

2

1cmcmcm

Area of the shaded region = 2222 142

28

2

49

2

77cmcmcmcm

(c) (i) Let the line segment AB be divided by the point C in the ratio k: 1.

Using section formula, the coordinates of point C are:

1

63,

1

48

k

k

k

k

( Since, C lies on y-axis, coordinates of C are (0, y). On comparing, we have:

01

48

k

k

048 k

2

1 k

Thus, the required ratio in which AB is divided by the y-axis is 1: 2.

(ii) The point of intersection of AB and the y-axis is C.

The coordinates of point C are

1

63.

1

48

k

k

k

k

=

12

1

62

13

,

12

1

42

18

=

12

1

62

3

,

12

1

44

=

2

212

123

,0

=(0,3)

iii) using distance formula,

AB= (8 − 4)2 + (−3− 6)2 = (12)2 + (−3− 6)2

= 144 + 81 = 225 = 15 units.

Sol.3

(a) Let OD = OC = x cm(radius of same circle)

Since, ACD is a secant and AB is a tangle to the given circle, we have,

AC × AD = AB2

(7.5)(7.5+2X)=152

2251525.56 x

75.16815 x

25.11 x

Thus, the radius of the circle is 11.25 cm.

(b)

54cot

36tan26sin64cos26cos2

= cos)3690cot(

36tan26sin)2690cos(262

=

36tan

36tan26sin.26sin26cos2

= tan)90cot(,sin)90cos(

= cos 126sin26 22

= 1+1 1sincos 22 = 2

C)

Marks(x) Number of students(f) fx

5 6 30

6 A 6a

7 16 112

8 13 104

9 b 9b

baf 35 bafx 96246

It is given that the number of students is 40.

35 + a + b = 40

05 ba ……..(1)

Mean = f

fx

= 2.735

96246

ba

bs (Given)

246+6a+9b=7.2(35+a+b)

246+6a+9b=252+7.2a+7.2b

0=252-246+7.2a-6a+7.2b+9b

6+1.2a-1.8b=0

10+2a-3b=0 …….(2)

Solving equations (1) and (2), we have:

5a-5=0

1a

From(1), we have:

b=4

Hence, the values of a and b are 1 and 4 respectively.

Sol.4

(a) Given that Recurring deposit per month = Rs 200

Period = 36 months

Rate of interest = 11%

Money deposited = Monthly value x No. of months

= 200x36=Rs 7200 -----(1)

Total principal fro 1 month = Rs 200,33,1.2

)136)(36(200Rs

Interest = Rs 1221.10012

13320011Rs

------- (2)

Hence, Maturity Amount = (1) + (2)

= Rs (1, 33,200 + 1,221) = Rs 1, 34,421

∴ The amount Kiran gets on maturity = Rs 1, 34, 421

(b) Let S be the event when two coins are tossed

Sample space, S = {(H, H), (H, T), (T, H), (T, T)}

Total number of outcomes = 4

(i) Let A be the event of getting 2 heads.

The outcomes favoring A is {(H, H)}

P (A) =4

1

(c) Taking the unit 1 cm = 1 unit the points are plotted as below

(i) – (ii)

(iii) The image of A'= 4,4

B 2,2'

(iv) Now, AB, A'B, AB' and A'B' are equal in length.

So, ABA'B' is a rhombus.

(v) The lines of symmetry are the diagonals of the rhombus ABA'B',

i.e. AA' and BB'.

Section - A (40 Marks)

Attempt any for question for this section

Sol.5

(a)

Given: AB is the diameter of the circle with centre O, ,130 BCD

To find: DBADAB ,

(i)Clearly ABCD is a cyclic quadrilateral.

We know, sum of pair of opposite angles of a cyclic quadrilateral is equal to 180 .

180 DCBDAB

180130 DAB

50130180 DAB

(ii)Consider triangle DAB

Here, 90ADB

since angle in a semi circle is a right angle.

So, by angle sum property of a triangle,

180 ADBDBADAB

50 18090 DBA

40140180 DBA

(b)

Given

6

7

43

12

(i) Let the order of X be a x b

.2243

12

X

126

7

ba

a=2 and b=1

The order of the matrix X= a x b = 2x1

(ii) let X =

y

x

6

7

43

12

y

x

6

7

43

2

yx

yx

2x + y = 7 ---------- (1)

-3x + 4y = 6 ---------- (2)

Multiplying (1) by 4 and subtracting from (2) ,

-3x – 8x = -22

-11x = -22

x = 211

22

(1) gives,

Y = 7-2x = 7 -2(2)=7-4=3

The matrix X=

3

2

y

x

(c) Principal for January = Rs 3580

Principal for February = Rs 3580

Principal for March = Rs 7780

Principal for April = Rs 7780

Principal for May = Rs 7780

Principal for June = Rs 3280

Principal for July = Rs 5910

Principal for August = Rs 5910

Principal for September = Rs 5910

Principal for October = Rs 2690

Principal for November = Rs 4190

Principal for December = Rs 6160

Total equivalent principal for 1 month = Rs 64550

Now, P=Rs.64550, I=Rs. 198

yearT12

1

Rate % = %100

TP

I

= %68.3164550

12100198

Sol.6

(a) The List price of the article = Rs 60,000

Discount to the Shopkeeper = 20%

Thus, Cost price of the article to the shopkeeper

= 000,60100

20100

= 000,60100

80

= Rs.48, 000

Since the shopkeeper sells the article to the customer at the printed price.

So, C.P to the customer = Rs 60,000

Now, VAT= 6 % at every stage.

(i)The cost to the shopkeeper inclusive of tax = Rs 48000+ 6% VAT on it.

= 48000 48000100

6

= 48000+2880

= Rs 50880

(ii) VAT paid by the shopkeeper to the Government is the difference of the

VAT to the customer and the shopkeeper

= Rs 3600- Rs2880

= Rs720

(iii) The Cost to the customer inclusive of Tax =Rs 60,000+6% of 60,000

= Rs 60,000 60000100

6

= Rs 60,000+3600

= Rs 63600

(b) The Given inequation is

4x -19< ,5

22

5

3

xxR

(4x-19)<5

52

5

103 xx

5(4x-19) <3x-10<-2+5x

20x – 95< 3x-10 x52

Solving 20x – 95<3x-10

17x < 85

x<5

Solving 3x -10 -2 +5x

4 x 4x

So, the solution is 54 x

Representation on the Number line

(c) Given Quadratic Equation is

0)5()1(22 mxmx

Here a = 1, b = 2(m-1) and c = (m+5)

Discriminant is given by D .422 acb For Real and equal roots, D=0

b 042 ac

0)5(4)1(22

mm

m 05212 mm

m 0432 m

Factorising, we get (m+1)(m-4)=0

m=-1 or m=4

Sol:7

(a) Given a hollow sphere of internal and external radii 6cm and 8cm resp.

is melted and recast into small cones of base radius 2cm and height 8cm.

So, Volume remains the same.

Let the external radii be R =8cm

And the internal radii be r=6cm

Thus, radius of the Hollow sphere =(R-r) = (8-6) cm =2 cm Let ‘n’ be the number of cones.

Then,

hrnrR 23

3

1)(

3

4

4(R-r) hnr23

4×8 =n×(2)2×h

n=1

Thus, there is only 1 cone. (b) Given quadratic equation is

5x2 -3x - 4 =0

Comparing it with ax 2 + bx + c=0, we get

a=5, b=-3, c=-4

x=a

acbb

2

42

=

52

454332

= 10

51693

= 10

893

= 10

433.93

So, roots are 10

433.93

10

433.93 and

= 1.2433 and -0.6433

So, the roots correct to 3 significant figures: 1.24 and -0.64

(c)

Let AB be the light house and C and D be the two ships.

The angles of Depression of the 2 ships are 30 40and

So, 30ADB

40ACB

Let the distance between the ships be CD=x m.

Also, Let BD= y m.

In ABC,

tan 40xy

80

y- x = )1(352.958390.0

80

Also, from ,ABD

tan 30y

80

y=80 m3 = mm 56.138732.180

So from (1), we get

138.56 – x = 95.352

x=138.56-95.352 cm=43.208m

Sol.8

(a)

(i)Market value = Rs 80

Sum invested = Rs 9600

Number of shares = 12080

9600

(i) Nominal value (face value) = Number of shares x face value

= 120 x Rs 100 = Rs 12000

Annual income (dividend) =Dividend% x Nominal value

= 12000100

18

= 2160

(ii) Percentage return = %100Investment

Income

= %5.22%1009600

2160

(b)

In ABC and AMP

AMPABC (each 90

)

PACBAC (common)

ABC AMP (By AA – similarity)

Since the triangles are similar, we have

AP

AC

MP

BC

AM

AB

15

10

12

BC

AM

AB

Taking, cmBCBC

815

1012

15

10

12

Now using Pythagoros theorem in triangle ABC,

AB 222 ACBC

36810 222 AB

cmAB 6

Hence, AB= 6 cm and BC= 8 cm.

(c)

Consider, x=11

11

aa

aa

11

11

1

aa

aax

By using componando and dividendo, we have

1111

11)11(

1

1

aaaa

aaaa

x

x

1

1

12

12

1

1

a

a

a

a

x

x

Squaring both sides, we get

1

1

1

12

2

2

a

a

x

x

1

1

12

122

2

a

a

xx

xx

Again using componando and dividendo, we get

11

11

1212

12122

22

aa

aa

xxxx

xxxx

2

2

4

22 2 a

x

x

12

12 a

x

x

x ax212

x 0122 ax

Sol.9

(a) Slope of the line through A(-2,3) and B(4,b) = ).(6

3

24

31

12

12 mbb

xx

yy

Slope of the line 2x – 4y = 5 is = ).(2

1

4

22m

B

A

Since the lines are perpendicular therefore, m 1 12 m

12

1

6

3

b

b - 3= -12

b=-12 + 3 = -9

(c) LHS = 2

2

2

2

1sec

1sec

1sec

tan

=

21sec

1sec1sec

= 1sec

1sec

=

cos

cos1cos

cos1

1cos

1

1cos

1

=

cos1

cos1

= RHS

Hence, LHS = RHS Proved.

(c) Let the speed of car = x km/h

Distance covered = 400 km

Time taken = .400tan

hrsxSpeed

ceDis

If the speed is increased by 12 km/h the time taken is = Ashrs

xSpeed

ceDis.

12

400tan

per given in question,

hxx

112

400400

40min = 1

60

40

3

5

3

21

12

400400

xx

4003

5

12

11

xx

240

1

4003

5

)12(

12

xx

xx

2880122 xx

02880122 xx

( 06048 xx

60,48 x

Hence, the speed is 48 km/h.

Sol.10

(a) Steps of construction:

1) Draw a line segment BC of length 6 cm. At B, draw a ray BX making an angle of

1200 with BC.

2) With B as centre and radius 5.5 cm, draw an arc to cut the ray BX at A. Triangle

ABC will be obtained

3) Draw the perpendicular bisectors of AB and BC to meet at point O.

4) With O as centre and radius OA, draw a circle. The circle will circumscribe ∆ABC.

5) Draw the angle bisector of ABC.

6) The angle bisector of ABC and perpendicular bisector of line segment BC will

intersect at point D. Point D will be equidistant from points B and C.

7) Join AD and DC to obtain the required cyclic quadrilateral ABCD.

(b) Taking scale as 2cm=5 cm on x-axis and 2cm=20 students on the y- axis.

Ogive for the distribution.

Height (in cm) Number of students (f) Cumulative Frequency (c.f)

140-145 12 12

145-150 20 32

150-155 30 62

155-160 38 100

160-165 24 124

165-170 16 140

170-175 12 152

175-180 8 160

N=160

(i) Median Height = 2

Nth observation

=2

160th observation

= 80th observation

=157cm (Approx)

(ii)Lower Quartile, Q1 = 4

Nth Observation

=4

160th observation

=40th Observation = 152cm

Upper quartile, Q3 =4

3Nth Observation

= 4

1603th Observation

= 120th Observation = 164 cm

So, Inter Quartile Range, Q3 – Q1 = 164 – 152 = 12cm

(ii)The number of students whose height is more than 172cm = 160 – 142 =18 students (approx)

Sol. 11

(a) Since the triangle is right angle,

PR = 22 247 = 57649 = 625 =25cm

Now Perimeter of triangle (P) = 7 + 24 + 25 = 56 cm

Area of triangle (A) = 2

1×PQ×QR = 84724

2

1 cm2

Therefore, x = radius of the inscribed circle = cmp

A3

56

8422

(b)

X f Cf

10 1 1

11 4 5

12 7 12

13 5 17

14 9 26

15 3 29

N =29

Here, N = 29 5.142

29

2

N

We find that the cf just greater than 5.142

N is 17 and the value of x

corresponding to 17 is 13.

Therefore, Median = 13

To find mode: From the table it is clear that the value 14 has maximum

frequency 9. Hence mode is 14.

(c) (i) Since the line make 450 with positive direction of x-axis,

Therefore slope (m) = tan450 = 1

(ii) Equation of line, using point-slope is:

y – y1 m(x – x1 )

y – 3 1 (x – 5 )

y – 3 x – 5

y = x – 2

Which is the required equation of line.

(iii) Since Q is a point on y-axis, its x-coordinate is zero.

Putting x = 0 in the equation of line, we get y = –2

Hence the point Q is (0, –2).