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)
 3 1
1
A 
,I  

 1 2
0
 3 1  3
A2  

 1 2  1
0
1
1  9  1 3  2  8 5
=
=
2  3  2  1  4  5 3
 8 5  3 1
1 0
A2  5 A  7 I  
 5
 7



 5 3  1 2
0 1
 8 5  15 5  7 0




 5 3  5 10 0 7
 7 0  7 0



 0  7 0 7
0 0

0
0 0
(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 =
Interest on Rs.
P  R  T P  5 1
P


100
100
20
P
P  5 1
P
for 1 year 

20
20  100 400
It is given that CI-SI=Rs.25
P
 Rs.25
400
 P  Rs.10,000

Thus, the sum of money is Rs.10,000.
(b) Using Pythagoras theorem,
AC 2  AB 2  BC 2  7 2  7 2
 AC = 7 2
Radius of semi-circle = 7
7 2
2
Area of the shaded region = Area of semi-circle – Area of triangle ABC
2
1
1 7 2 
98 49 77 2
 


cm
Area of semi-circle = r 2   

2
2  2 
8
4
2
1
49
Area of triangle ABC =  7cm  7cm  cm2
2
2
77
49
28
Area of the shaded region = cm2  cm2  cm2  14cm2
2
2
2
(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:
 8k  4  3k  6 
,


k 1 
 k 1
( Since, C lies on y-axis, coordinates of C are (0, y). On comparing, we have:
8k  4
0
k 1
 8k  4  0
1
k 
2
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.
 8k  4    3k  6 
.

 k 1   k 1 
The coordinates of point C are 
1
1


 8  4  3  6 
2
2

,
=
1
 1 1
 1 

2
 2

3


 6
44

, 2
=
1
 1 1 1 


2
2

  3  12 


2


= 0,
1 2 



2


=(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
 56.25  15x  225
 15x  168.75
 x  11.25
Thus, the radius of the circle is 11.25 cm.
(b) cos2 26   cos 64  sin 26  
tan 36 
cot 54 
= cos 2 26  cos(90  26 ) sin 26 
= cos2 26   sin 26  . sin 26  
tan 36
cot(90  36 )
tan 36 
tan 36 

=  cos(90   )  sin  , cot(90   )  tan 

= cos 2 26  sin 2 26  1
= 1+1
 cos    sin   1
2
2
=2
C)
Marks(x)
5
6
7
8
9
Number of students(f)
6
A
16
13
b
f  35  a  b
It is given that the number of students is 40.
 35 + a + b = 40
……..(1)
a b50
fx
Mean =
f
246  6s  9b
(Given)
 7.2
35  a  b
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
 a 1
From(1), we have:
b=4
Hence, the values of a and b are 1 and 4 respectively.
=
fx
30
6a
112
104
9b
fx  246  6a  9b
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)
200  (36)(36  1)
Total principal fro 1 month = Rs
 Rs.1,33,200
2
11  133200
Interest = Rs
------- (2)
 Rs.1221
12  100
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)}
1
 P (A) =
4
(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, BCD  130  ,
To find: DAB, DBA
(i)Clearly ABCD is a cyclic quadrilateral.
We know, sum of pair of opposite angles of a cyclic quadrilateral is equal to 180  .

DAB  DCB  180


DAB  130  180
DAB  180  130  50
(ii)Consider triangle DAB
Here, ADB  90
since angle in a semi circle is a right angle.
So, by angle sum property of a triangle,
DAB  DBA  ADB  180


50  DBA  90  180
DBA  180  140  40
(b)
 2 1
7
 
Given 

 3 4
6
(i)
Let the order of X be a x b
 2 1
7 
 3 4 2  2. X ab  6
  21




a=2 and b=1
 The order of the matrix X= a x b = 2x1
 x
(ii) let X =  
 y
 2 1  x  7

  3 4  y   6 

   
 2 x  y  7

 3x  4 y   6

  

2x + y = 7
-3x + 4y = 6
---------- (1)
---------- (2)
Multiplying (1) by 4 and subtracting from (2) ,
-3x – 8x = -22
 -11x = -22
 22
x=

2
 11

(1) gives,
Y = 7-2x = 7 -2(2)=7-4=3
 x  2

The matrix X=     
 y  3
(c) Principal for January
Principal for February
= Rs 3580
= 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
1
T
year
12
I  100

Rate % =
%
P T
198  100  12
=
 3.68%
64550  1
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
100  20
=
 60,000
100
80
=
 60,000
100
= 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.
6
= 48000 
 48000
100
= 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
6
= Rs 60,000 
 60000
100
= Rs 60,000+3600
= Rs 63600
(b) The Given inequation is
3x
2
4x -19<  2 
 , x  R
5
5
3x  10  2  5 x
 (4x-19)<

5
5
 5(4x-19) <3x-10<-2+5x
 20x – 95< 3x-10  2  5x
Solving 20x – 95<3x-10
 17x < 85
 x<5
Solving 3x -10  -2 +5x
 x4
 x  4
So, the solution is  4  x  5
Representation on the Number line
(c) Given Quadratic Equation is
x 2  2(m  1) x  (m  5)  0
Here a = 1, b = 2(m-1) and c = (m+5)
Discriminant is given by D 2  b 2  4ac.
For Real and equal roots, D=0
 b 2 4ac  0
 2(m  1)2  4(m  5)  0
 m 2 1  2m  m  5  0
 m 2 3m  4  0
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,
4
1
 ( R  r )3  n r 2h
3
3
3
2
 4(R-r)  nr h
 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=
=
 b  b 2  4ac
2a
3
 32  4  5  4
25
=
3  9  16  5
10
=
3  89
10
=
3  9.433
10
So, roots are
3  9.433
3  9.433
and
10
10
= 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  and 40

So, ADB  30
ACB  40
Let the distance between the ships be CD=x m.
Also, Let BD= y m.
In  ABC,

tan 40 
80
yx
80
 95.352      (1)
0.8390
Also, from ABD,
 y- x =

tan 30 
80
y
 y=80 3m = 801.732m  138.56m
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
9600
Number of shares =
 120
80
(i)
Nominal value (face value) = Number of shares x face value
= 120 x Rs 100 = Rs 12000
Annual income (dividend) =Dividend% x Nominal value
18
 12000
=
100
= 2160
Income
 100%
(ii) Percentage return =
Investment
2160
 100%  22.5%
=
9600
(b)
In ABC and AMP
ABC  AMP
BAC  PAC

(each 90 )
(common)
 ABC  AMP
(By AA – similarity)
Since the triangles are similar, we have
AB
BC AC


AM MP AP
AB BC 10



AM
12 15
BC 10
12  10
Taking,
  BC 
 8cm
12 15
15
Now using Pythagoros theorem in triangle ABC,
AB 2  BC 2  AC 2
 AB2  102  82  36
 AB  6cm
Hence, AB= 6 cm and BC= 8 cm.
(c)
Consider, x=
a 1  a 1
a 1  a 1
x
a 1 a 1

1
a 1  a 1
By using componando and dividendo, we have


 a 1 
  a 1 
x  1 ( a  1  a  1) 

x 1
a 1  a 1 


a  1
a 1
x 1 2 a 1
a 1


x 1 2 a 1
a 1
Squaring both sides, we get

x  12

x  12




a  1
a 1
2
x2  2x  1 a  1

 2
x  2x  1 a  1
Again using componando and dividendo, we get
x

x

2
 
 


 2 x  1  x 2  2 x  1 a  1  a  1

 2 x  1  x 2  2 x  1 a  1  a  1
2 x 2  2 2a

4x
2
x2  1 a

2x
1
 x 2 1  2ax

 x 2 2ax  1  0
Sol.9
(a) Slope of the line through A(-2,3) and B(4,b) =
y2  y1 b  3 b  3


( m1 ).
x2  x1 4  2
6
 A 2 1

 ( m2 ).
B
4 2
Since the lines are perpendicular therefore, m 1  m2  1
Slope of the line 2x – 4y = 5 is =
b3 1
  1
6
2
 b - 3= -12
 b=-12 + 3 = -9

tan 2 
sec2   1

sec  12 sec  12
sec  1sec  1
=
sec  12
(c) LHS =
=
sec  1
sec  1
1
1  cos
1
= cos
 cos
1
1  cos
1
cos
cos
1  cos
1  cos
= RHS
Hence, LHS = RHS Proved.
(c) Let the speed of car = x km/h
Distance covered = 400 km
Dis tan ce 400

hrs.
Time taken =
Speed
x
=
Dis tan ce
400

hrs. As
x  12
If the speed is increased by 12 km/h the time taken is = Speed
per given in question,
400 400
40

 1h 40min = 1
x
x  12
60
400 400
2 5


1 
x
x  12
3 3
1
x
 400  

1  5

x  12  3
x  12  x
5
1


x( x  12 ) 3  400 240
 x2  12x  2880
 x2  12x  2880  0
 ( x  48 x  60   0
 x  48,60
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)
140-145
145-150
150-155
155-160
160-165
165-170
170-175
175-180
Number of students (f)
12
20
30
38
24
16
12
8
N=160
N
th observation
2
160
=
th observation
2
= 80th observation
(i) Median Height =
=157cm (Approx)
Cumulative Frequency (c.f)
12
32
62
100
124
140
152
160
N
th Observation
4
160
=
th observation
4
=40th Observation = 152cm
(ii)Lower Quartile, Q1 =
3N
th Observation
4
3 160
=
th Observation
4
= 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)
Upper quartile, Q3 =
Sol. 11
(a) Since the triangle is right angle,
PR = 7 2  242 = 49  576 = 625 =25cm
Now Perimeter of triangle (P) = 7 + 24 + 25 = 56 cm
1
1
Area of triangle (A) = ×PQ×QR =  24  7  84 cm2
2
2
2 A 2  84

 3 cm
Therefore, x = radius of the inscribed circle =
p
56
(b)
X
10
11
12
13
14
15
f
1
4
7
5
9
3
N =29
Here, N = 29 
Cf
1
5
12
17
26
29
N 29

 14.5
2
2
We find that the cf just greater than
N
 14.5 is 17 and the value of x
2
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).