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).