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DESIGN OF THE SAMPLE QUESTION PAPERS

MATHEMATICS-CLASS X
Time : 3 Hours

Max
...


Weightage to Learning Outcomes

S
...

1
...

3
...

2
...
No
...

2
...

4
...

6
...


Learning Outcomes
Algebra
Commerical Mathematics
Mensuration
Trigonometry
Geometry
Statistics
Coordinate Geometry

Marks
26
12
10
10
22
12
8
Total : 100

3
...
No
...

2
...

4
...
No
...

Short answer type (SA I)

No
...
Time
3-5 minutes

2
...


Long answer type (LA)

Upto 8 Credit Points

8-10 minutes

136

These ranges of steps and time requirements for the answers are, however, suggestive
...
As the total time is calculated on
the basis of the number of questions required to be answered and the length of their anticipated
answers, it would, therefore, be advisable for the candidates to budget their time properly by
cutting out the superfluous lengths and be within the expected limits
...


Scheme of Options

All questions are compulsory i
...
there is no overall choice in the question paper
...
These choices have been given from within the
same topic and in questions which test higher mental abilities of students
...


Weightage to difficulty level of questions

S
...

1
...


Average

70%

3
...
As such, the assessment
in respect of each question will be made by the paper setter on the basis of general anticipation
from the group as whole taking the examination
...

Based on the above design, there are two separate sample papers along with their Blue Prints
as well as questionwise analysis
...

Note : Though weightages to content/subject units, objectives and forms of questions etc
...


137

138
-

Sub-Total

Total

G
...
Maths
Instalments
Income Tax

Objective
ü
Form of ü
questions
Content Unit
Algebra
Linear Eqns
Polynomials
Rational Exp
...
Prog
...


All questions are compulsory
...


The question paper consists of 25 questions divided into three sections A, B and C
...


3
...
However, internal choice has been provided in two questions
of three marks each, two questions of four marks each and two questions of six marks
each
...


In question on construction, the drawing should be neat and exactly as per the given
measurements
...


Use of calculators is not permitted
...

SECTION A

Q1
...


Reduce the following rational expression to its lowest terms :
x2 + 3x + 9
x3 — 27
÷ (x2 + 3x — 10)
x2 — 25

Q3
...


A suit is available for Rs
...
500 cash down payment followed by 3
monthly instalments of Rs
...
Find the rate of interest charged under the
instalment scheme
...


A loan has to be returned in two equal annual instalments
...
1682, find the sum
borrowed and the total interest paid
...


If (x — 2) is a factor of x2 + ax + b and a + b = 1, find the values of a and b
...


Using quadratic formula, solve the following equation for x :
abx2 + (b 2 —ac) x — bc = 0
OR
The sum of the squares of two positive integers is 208
...


Q8
...
P
...
is 132 more than its 54th term ?
OR
Derive the formula for the sum of first n terms of an A
...
whose first term is 'a' and the
common difference is 'd'

Q9
...
+199

Q10
...

SECTION B
Q11
...
Find graphically, the vertices of the triangle formed by the x-axes and the lines
2x — y + 8 = 0
8x + 3y — 24 = 0
Q13
...
Draw its
incircle and measure its radius
...
The total surface area of a closed right circular cylinder is 6512 cm², and the
22
circumference of its base is 88 cm
...
Prove the identity :
(1 + Cotθ - Cosecθ) (1 + tanθ + secθ) = 2
...
Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right
triangle
...

Q17
...
Hence find the value of p
...
Compute the missing frequencies 'f 1' and 'f 2' in the following data if the mean is
9
166 26 and the sum of observations is 52
...
An unbiased dice is tossed
i) Write the sample space of the experiment
ii) Find the probability of getting a number greater than 4
iii) Find the probability of getting a prime number
...
The pie chart (as shown in the
figure) represents the amount
spent on different sports by a
sports club in a year
...
1,08,000/-,
find the amount spent on each sport
...
Prove that the angle subtended by an arc of a circle at its center is double the angle
subtended by it at any point on the remaining part of the circle
...

Q22
...

Using the above, prove that the area of an equilateral triangle described on the side of
a square is half the area of the equilateral triangle described on its diagonal
...
From the top of a tower 60m
...
Find the height of the building
...
5km above the ground is observed at
a certain point on earth to subtend an angle of 60°
...
Calculate the speed of the aeroplane in km/h
...
A solid toy is in the form of a hemisphere surmounted by a right circular cone
...
14)
OR
A bucket of height 8cm
...

Calculate :
i) the height of the cone of which the bucket is a part
ii) the volume of water which can be filled in the bucket
...
Anil's total annual salary excluding HRA is Rs
...
He contributes Rs
...
P
...
How much he should invest in N
...
C
...

Find the amount which should be deducted per month towards tax from his salary
...
30,000/- in case the total
annual income is up to Rs
...

(ii) Rs
...
100,001 to Rs
...


b)

Rate of income Tax

:

Slab

Income Tax

i) Up to Rs
...
50,001 to Rs
...
50,000

iii)From Rs
...
1,50,000

Rs
...

60,000

iv) Above Rs
...
19,000 + 30% of the amount exceding Rs
...
14,000/-, if gross income
is upto Rs
...
10,500/-if gross income
is above Rs
...
500,000

143

MARKING SCHEME
SECTION A
Q
...


Q1
...
(i)

1

Subtracting we get
x - y = -1
...


Writing as

x2 + 3x +9
(x+5) (x-5)

=

=

Q3
...
angle of eyclic quad PRSQ
∠OQP = ∠RSQ -

1
P


...


R

Cash Price = Rs
...
500 + Rs
...
1535
Interest Charged = Rs
...
1000 + Rs
...
310
∴ Total Principal = Rs
...
31% approx
131
144

S

½

1
1
1

Q
...


Q5
...
1450
100
29 2
= Rs
...
1450
+ Rs
...
2700
Interest Charged

½

= Rs
...
2700

½

= Rs
...


(x - 2) is a factor of x² +ax + b
∴ 4 + 2a + b = 0 Ü 2a + b = -4
also

1+1=2

a+b=1

Solving to get a = -5

Q7
...
(i)
x² = 18y
...
NO
...


VALUE POINTS

Marks

Here a = 3 , d = 12
∴ t54 = 3 + (54 —1)
...
12 = 771
∴ n = 65

Q9
...
Where l =a +(n—1) d
∴ Sn = l + (l—d) + (l —2d) + --------+a
∴ 2 Sn = (a+l) + (a+l) + (a+l) + -------+ (a+l) = n )a+l)
n
n
Sn =
(a+l) =
[2a +(n -1) d]
2
2
Here a=1, d=2

1
1
1

Let tn = 199
∴ 1 + (n—1)
...
[2
...
2]
2
=50 [200]

1

= 10,000

½

Q10
...


3x + 4
(x+1) (x+2)

=

4
x+4

1

Ü 4 (x+1) (x+2) = ( x+4) (3x +4)

½

or 4x² + 12x + 8 = 3x² + 16x + 16

½

or x² — 4x — 8 = 0

1

Solving to get x =2 + 2œ 2—2œ
3,
3,

1

146

Q
...


VALUE POINTS

x
Q12
...
Correct Construction :

3 marks

Correct Measurement of radius :

1 mark

Q14
...

∴ 2x 22 r = 88
7
Ü r = 14 cm
Again 2πrh + 2πr² = 6512 cm²
6512
∴h=
-14 = 60 cm
88

1

1½

22
x 14 x 14 x 60
7
= 36960 cm³

Volume =

1½

Q15
...
H
...


(
=
=

sinθ + cosθ —1
sinθ

) (

sinθ + cosθ + 1
cosθ

(sinθ + cosθ)²—1
sinθ
...
cosθ

)

1

1

=2

1

L
...
S
...
H
...


1

147

Q
...


VALUE POINTS

Marks

OR
cos 35°
sin (90° - 35)°+
=

cos 35°
cos 35° +

tan 27° tan (90° - 27)°
- 3 tan²60°
sin 30°
tan 27°
...
Let A = (7, 10) ; B = (—2, 5) ; C = (3, —4)
∴ AB = œ
(—2 —7)² + (5—10)²

½
½

=œ
106
BC = œ
(3+2)² + (—4 — 5)²
=œ
106

½

CA = œ
(7—3)² + (10+4)²
= œ + 196
16
=œ
212

½

Ü AB=BC

½

and CA² = AB² + BC²

1

∴ A, B & C are vertices of an isosceles rt
...
NO
...
Let the ratio be K : 1 in which x, y divides the join of (—5, —4) and (—2, 3)
-2K -5
∴x=
K+1
y=
∴

3K -4
K+1

1

—2K —5
3K —4
= -3 (i) and
= p (ii)
K+1
K+1

1

Ü K=2 ∴ Ratio is 2:1

½

2
Putting value of K in (ii) we get p =
3
:
145
155
165 175 185 195

f
f
...
x

1

52

5

f1

725

155f1

20

f2

6

2

3300 175f2 1110 390

½

5525+155f1+175f 2

9
4325
4325
Mean = 166 26 = 26 ∴ Ûfx = 26

...
(i) Sample space = {1, 2, 3, 4, 5, 6 }

1

(ii) Numbers greater than 4 = 5, 6
∴ Probability =

2
6

=

½

1
3

1

(iii)Prime numbers = 2, 3, 5

½

3
1
=
6
2
Q20
...
108,000, Central angle = 360º
∴ Probability =

∴ Expenditure on Hockey = 108,000x
Expenditure on - cricket = 108,000 x
Expenditure on football = 108,000 x
Expenditure on Tennis = 108,000x
149

100
= Rs
...
45,000

1

= Rs
...
15000

½

Q
...


VALUE POINTS

Marks

SECTION C
Q21
...
given, To prove and Construction

½ x 4=2

Correct Proof

2

Proof : 2 ∠APB = ∠AOB

½

( ∠AOB < 180°)

Fig
...
No figure no marks
correct fig, given, to prove, construction

2 marks (½each)

correct proof

2

(ii) Proof Let side of square = a cm ∴ diagonal = œ cm
2a

½

∆ APD ∆ A QC (Equilateral)
area ∆ APD
area ∆ AQC

∴
=

fig
...
Let Tower AB = 60 m and Building be DC

Correct figure

1

In ∆ ADB -----AB
= tan 60°
BD

1

60
= 20œ m
3
œ
3
∴ CP = 20œ
3m

½

Again in ∆ ACP--------AP
CP = tan 30°

1

Ü AP = 20m

½

∴ BD =

1

Height of Building = CD = PB = AB — AP
= 60 — 20
= 40 m

1
150

Q
...


VALUE POINTS

Marks

OR
Let A and B are two positions of the aeroplane
...
5 (
) = (0
...
5) (œ km
3)
BM

∴

1½

∴ d = OM — OL = (1
...
5) œ = œ km
3
3
3
∴ speed = Distance =
time
or 415
...
Volume of toy =

[

1
3

œ
3
15
3600

1

= 240 œ km/hr
3

]

2
π(3)²
...
14 = 94
...
14) = 103
...
)

½

Let ON = x
x
3
∆ ONB ~ ∆ OMC ∴ x+8 = 9
Ü x=4
∴ height of cone = 8 + 4 = 12 cm
Volume of bucket = [π(9)²
...
4] cm³
= 312 π cm³
Slant height of cone of radius 9cm = 9² + 12² cm
∴ L = 15 cm
Slant height of cone of radius 3cm = 3² + 4² cm
l = 5 cm
Area of the copper sheet used to form bucket
= [π(9) (15) - π(3) (5) + π(3)² cm²
129π cm²
151

M 9 cm

D

C

½
1

8 cm
N 3 cm
A

1

B

½
½
½

O
1
½

Q
...


VALUE POINTS

Marks

Q25
...
[1,96,000 — 30,000] = Rs
...
[19,000 + 30% of 16,000] = Rs
...
[12 x 5,000] Rs
...
[70,000 — 60,000] = Rs
...
[70,000 x

15
] = Rs
...
[23800 — 10500] = Rs
...


13300
= Rs
...
Maths
Instalments
Income Tax

Sub-Total

-

Sub-Total

Total

G
...

Quadratic Eqns
Arith
...


Subject
Time

-

-

-

4(1)
-

-

-

3(1)

3(1)
-

-

4(1)
-

SA2

-

-

-

-

-

-

6(1)

6(1)

-

-

LA

45(11)

8(2)

8(2)

4(1)Ÿ

4(1)
4(1)Ÿ
-

4(1)
-

-

-

-

8(2)

4(1)
4(1)
-

SA1

6(2)

6(2)

-

7(1)
6(2)Ÿ

2*
3(1)
**
2
-

3(1)

3(1)
-

3(1)

3(1)

SA2

Understaning

6(1)

6(1)

-

6(1)Ÿ
-

-

-

-

-

-

-

LA

12(3)

-

-

-

-

-

-

-

-

-

-

SA1

6(2)

6(2)

-

6(2)Ÿ
-

-

-

-

-

-

-

SA2

Application

BLUE PRINT-II

6(1)

6(1)

-

6(1)

-

-

-

-

-

-

LA

12(3)

-

-

-

-

-

-

-

-

-

-

SA1

Skill

6(2)

-

-

3(1)
-

3(1)

-

-

-

3(1)

3(1)
-

SA2

8(2)

4(1)
4(1)
4(1)
-

4(1)
-

-

-

-

20(5)

4(1)
4(1)
4(1)
8(2)
-

SA1

8(2)

22(5)
10(3)
10(2)
12(3)

10(2)
3(1)

9(2)

12(3)

6(2)
6(1)

26(7)

7(2)
4(1)
4(1)
8(2)
3(1)

Total

12(4) 40(10)

-

6(2)
6(2)
6(2)

3(1)

3(1)

6(2)

6(2)
-

6(2)

3(1)
3(1)

SA2

Grand

100(25)

40(10) 30(10) 30(10)100(25)

12(2) 16(4)

-

12(2)
6(1)
6(1)

*
6(1)*
**Ÿ
6(1)
-

6(1)

6(1)

-

-

LA

Total

Class
: X
Maximum Marks : 100

Sample Question Paper-II
Class X
Subject : Mathematics

Time : 3 Hours
Max Marks : 100

General Instructions :
1
...


2
...

Section A contains 10 questions of 3 marks each, Section B is of 10 questions of 4
marks each and Sections C is of 5 questions of 6 marks each
...


There is no overall choice
...


4
...


5
...
However, you may ask for Mathematical tables
...


Sove the following system of equations graphically
5x -y = 7
x - y = -1

Q2
...


Q3
...
AC is drawn perpendicular from A to DB
...


A loan of Rs
...
If the rate of
interest is 25% per annum compounded annually, find the instalment
...


A watch is available for Rs
...
210 as cash down followed by three equal
monthly instalments
...


Q6
...
Measure the lengths of the tangents
...


A solid metallic cylinder of radius 14cm and height 21 cm is melted and recast into 72
equal small spheres
...


154

Q8
...
5m
...
)
OR
The largest sptere is carved out of a cube of side 7cm ; find the volume of the sphere
...


The following table shows the marks secured by 100 students in an examination
Marks

0-10

Number

15

10-20 20-30
20

30-40
20

35

40-50
10

Find the mean marks obtained by a student
...
A dice is thrown once
...

(i) a number greater than 3
(ii) a number less than 5
OR
A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls
...
Find the probability that it is
...
Solve for x and y
(a—b)x + (a+b)y = a² —2ab —b²
(a+b) (x+y) = a² + b²
Q12
...
C
...
of
f(x) = (x+3) (2x²—3x+a)
and g(x) = (x—2) (3x² + 10x—b)
find the value of a and b
Q13
...
Solve for x :
x—1
x—3
10
+
=
(x= 2,x=4)
x-2
x—4
3
Q15
...
Find the usual speed of the train
...
AB is a diameter of a circle with
centre O and chord CD is equal to
radius of the circle
...
Prove that ∠CPD = 60°
...
A circus tent is in the shape of a cylinder surmounted by a cone
...
If the vertex of the tent is 16m above the
ground, find the area of canvas required
to make the tent
...
Prove that :
tanθ
cotθ
+
= 1 + secθ cosecθ
1—cotθ
1—tanθ
OR
Evaluate :
sin39°
cos 51° + 2tan 11° tan 31° tan 45° tan59°
...
Find a point on the x-axis which is equidistant from the points (7, 6) and (—3, 4)
Q20
...
Find the fourth vertex D
...

SECTION C
Q21
...


156

Q22
...
At a point on the
ground, the angles of elevation of the bottom and top of the flagstaff are α and β
respectively
...
If a line is drawn parallel to one side of a triangle, prove that the other two sides are
divided in the same ratio
...

Q24
...

Using the above result, determine as under :
ABCD is a cyclic trapezium with AD || BC
...


∠B=70°, determine the other three

OR
If two circles touch each other internally or externally, prove that the point of contact
lies on the line joining their centers
...

AB is a line through P intersecting the two circles at A & B respectively
...


157

Q25
...
1,45,000/-
...
2000 per month
in his GPF and pays and annual LIC premium of Rs
...
If he pays Rs
...
Use the following for calculating income tax :
a)

Standard Deduction

(i) 40% of the total income subject to a
maximum of Rs
...
100,000/(ii) Rs
...
100,001 to Rs
...
50,000
ii) Rs
...
60,000

10% of the amount exceeding Rs
...
60,0001 to Rs
...
1000 + 20% of the amount exceeding
Rs
...


Rebate on Savings

20% of the total savings if the gross income
is upto, 150,000 subject to a maximum of
Rs
...


158

MATHEMATICS
Marking Scheme II
Q
...


Value Points

Marks

SECTION A
Q1
...


½

Let a be the first term and d, the common difference

1

∴ Third term = t3 = a + 2d = 16
...
(ii)

½+½

From (i) and (ii), getting a = 4
∴ The arithmetic progression is 4, 10, 16, 22, 28
...


½
½

Correct Figure

½

Showing ∆ DCA ~ ∆ DAB
AD
BD
=
CD
AD

1

Ü AD² = BD
...


Let the instalment be Rs x
Present values of 1st, 2nd and 3rd instalments are
4
4 2
4 3
are
x,
x,
x
5
5
5
4
4
16
∴
5 x [1 + 5 + 25 ] = 48800

( ) ( )

OR x = 25000
∴ each instalment = Rs
...
No
...


Value Points

Marks

Cash price of watch = Rs
...
210
∴ Payment to be made in instalments = Rs
...
x be each instalment
∴

½

16
16
+
[ x + x x1200x 2] +[x + x x1200x 1 ] x = Rs
...


1

Correct construction

3

2
3
Q7
...
21] cm
This has been melted to form 72 spheres
Let r be the radius of the sphere
24
72

∴

x

4
3

1

3

π r = π 196
...
5 cm
Let h cm be the rainfall on the roof

( )

Q8
...
h m³

1

Q
...


Value Points

1
22
7
x
x
m³ =
3
7
2
22
11
Ü
h=
5
3
1
11
5
Ü h=
x
=
3
22
2
=

Marks

11
m³
3

5
6

1

∴ rainfall =

5
6

cm

OR

The diameter of sphere = side of cube
7
∴ Radius of sphere =
cm
2
Volume =
=

4
3

4
x
3

Q9
...
I
0 -10
10-20
20-30
30-40
40-50

½
½

7
2

xi
5
15
25
35
45
Ífi ü

x=

1

3
πr

2211
x
7

x

7
2

x

7
2

= 179

fi
15
20
35
20
10
100

2
3

cm³

1

fixi
075
300
875
700
450
2400

Í fixi
Í fi

= 2400
100

1

Correctly finding
ö Ífixi
Í fixi
Í fi

= 24

1
½
1

½

Q10
...
No
...
of balls in the bag = 24
(i) Numbers of black balls = 7

½

7
24
(ii) Number of balls which are not green = Total - green = 24 - 4 =20
∴ Required probability =

∴ Required probability =

Q11
...
(x+3)(x-2) divides f(x)
2
∴ 2x —3x+a has a factor (x—2)
2
∴ 2(2) —3(2) + a = 0
8 — 6 + a = 0 Ü a = —2
2
Similarly, (x + 3) divides 3x +10x-b
2
∴ 3(—3) —30 — b = 0

1
1½
½

Ü b = —3
2
2
Q13
...
No
...
1 + x—2
1

+1 + x—4

=

1
+ x—4 =

10
3

Ü

x—2

Ü

x²—6x+8

2x —6

10
3
—2

1
=

4
3

4

=

½

3

2

Ü 4x —30x+50=0

1

2

Ü 2x —10x—5x+25=0, Ü 2x(x—5)—5(x—5)=0 Ü (x—5)(2x—5)=0
Ü x = 5,

Marks

5

1
½

2

Q15
...
OC = CD = OD Ü OCD is an equilateral traiangle
0
∴ ∠1 =∠2 =∠3=60

1

Again OA = OC and OB=OD
∴ ∠OAC = ∠OCA = β and ∠OBD= ∠ODB= α
0
∠5 = 180 —2 β

1

∠4 = 180 —2 α

0
0
180 — ∠1 = ∠5 + ∠4 = 120
0
0
0
120 = 360 —2(α + β) Ü α + β =120
0
∴∠6 = 60° i
...
No
...
Area of canvas required to build the tent
= curved surface area of cylindrical
part + curved surface of conical part
2 2
2
OA =5 +12 =169 Ü OA = 13 m

1

∴ Required area = 2 πrh + πrl = πr(2h+l)

½

=

1

22 x 12(22+13) m2 = 1320 m2

1½

7

Q18
...
H
...
H
...

sin θ
cos θ

½+½

OR
0
0
0
cos 51 = cos(90-39) = sin39
1
0
0
tan 79 = tan(90—11) =
tan 11°
1
0
0
tan 59 = tan (90-31) =
tan 31°
0
tan 45 = 1
0
0
0
sin 69 = sin(90-21) = cos 21

2½

∴Given expression becomes
sin + 2
...
1
39°
sin 39°

1


...
No
...
Any point P on x axis is given by (x,0)

Marks

½

(Distance) between (x, 0) and (7, 6) is given by œ (x — 7)² + 6² ……(i)

1

(Distance) between (x, 0) and (—3, 4) is given by œ (x + 3)² + 4² ……(ii)
2
2
(i) = (ii) Ü x —14x + 49 + 36 = x + 6x + 9 + 16

1
1

OR, 20x = 60
x=3
∴ The point is (3,0)

½

Q20
...
No
...
Making the table:
Correct Central angles
Activity
Duration in hours
Sleep
7
School
8
Home work
4
Play
3
Others
2

2
Central angle
0
105
0
120
0
60
0
45
0
30

Drawing correct Pie chart with markings
Q22
...
No
...
Given, to prove, construction and correct figure

½x4=2
A

Correct proof

2

B

Draw OE || AB

½

In ∆ DAB, OE || AB

AE
Ü ED

=

BO
OD

E

O

(i)

½

Similarly, In ∆ ADC, EO || AB || DC
AE
∴ ED =

AO
OC

D

(ii)

BO
From (i) and (ii), we get DO

=

AO

C

½
½

OC

Q24
...
No
...
Opp
...
Taxable income = Rs
...
1,15,000
Income tax = Rs
...
12,000

½
1

Annual savings = Rs [2000 x 12 + 15000] = Rs
...
39000

1

= Rs
...
(12000 — 7800) = Rs
...
(250 x 11) = Rs
...
(4200— 2750) = Rs

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