SOLUTIONS
QUESTION PAPER - 6
Self Assessment __________________________________
Time : 3 Hours
Maximum Marks : 90
SECTION — A
1.
Probability of an event associated with a random experiment lies between 0 and 1 (both
included). So given statement is true.
1
2.
Since, ∠A = 90° and AB = BC = CD = DA, then ABCD is a square.
1
3.
Perimeter of first square = 4x unit
Perimeter of second square = 4y unit
\
Mean =
4x + 4 y
= 2(x + y) unit.
2
4.Probability of a sure event is 1.
1
1
SECTION — B
5.
In the given figure,
∠CDB = ∠ABD = 90°
But they are alternate angles.
∴ AB || DC
Also,DC = AB = 3 cm
A quadrilateral with a pair of equal and parallel sides is a parallelogram.
Area of parallelogram
= Base × Corresponding altitude
= 3 × 4 = 12 cm2.
6.
∠BAC = 45° = ∠BDC (Angles in the same segment of a circle) ∠DBC = 55° = ∠DAC (Angles in the same segment of a circle) \
∠BAC + ∠DAC = (55° + 45°)
= 100° = ∠BAD
Q ABCD is a cyclic quadrilateral.
∴
∠BCD = 180° − ∠BAD
= 180° -100° = 80°.
½
½
½
½
½
½
½
½
MATHEMATICS Oswaal CBSE Class -9, SA-2 Examination Sample Question Paper
SAMPLE
2 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
7.
Mean =
Σfi xi
Σfi
=
4 × 5 + 6 × 10 + 9 × 10 + 10 × 7 + 15 × 8
5 + 10 + 10 + 7 + 8
=
20 + 60 + 90 + 70 + 120
40
1
=
360
= 9.
40
1
8.
Mean =
Sum of observations
No. of observations ½
145 =
Sum of all observations
5
z
Sum of all observations = 145 × 5 = 725
½
Sum of correct observations = 725 − 45 + 25 ½
= 705
∴ Correct Mean =
705
= 141.
5
½
9.(i) Probability that student was born in April
=
2
40
1
=
20
1
(ii) Total number of students = 40
Now months having 31 day are Jan, March, May, July, August, Oct, Dec.
So, total number of students born in these months is
(3 + 2 + 5 + 2 + 6 + 4 + 4) = 26
So required probability that the student was born in the month having 31 days
26
13
=
=
= 0.65
20
40
10.
n =
=
1
Vcuboid
Vcube
1
1
5×2 ×1
= 10 cubes
1×1×1
SECTION — C
11. Given, Inner dimensions of the box are 21 cm, 14 cm, 11 cm.
\ Inner volume or capacity or the box = 21 × 14 × 11 cm3
= 3234 cm3
Outer volume of the box = 25 × 18 × 15 cm3
= 6750 cm3.
½
1
1
3
Volume of wood = 6750 – 3234 = 3516 cm ½
Solutions (SA-2) | 3
12.
Let,
radius of base = r
Given,
CSA, 2prh = 94.2
Þ
1
2 × 3.14 × r × 5 = 94.2
Þ
94.2
= 3
10 × 3.14
r =
1
Volume = pr2h = 3.14 × 3 × 3 × 5
Now,
= 141.3 cm3.
1
P
13.
A
B
R
Q
In DPQR, we have :
PQ = 10 cm, PR = 26 cm
Now, by Pythagoras Theorem,
PR2 = PQ2 + QR2
Þ
(26)2 = (10)2 + QR2
Þ
QR2 = 676 – 100
Þ
QR2 = 576
1
QR = 24 cm
Þ
1
Since A and B are the mid-points of sides PQ and PR respectively in DPQR, then
AB =
1
RQ 2
Þ
AB =
24
= 12 cm.
2
y 3 = 8x +
14. ⇒8x – y 3 +
(mid-point theorem)
1
3
3 = 0
Putting x = 0, y = – 1
⇒
3 +
3 ≠ 0.
1
\ (0, – 1) is not the solution of given equation.
Putting x =
⇒
3, y=9
8 3 − 9 3 + 3 = 0, which is correct
\ ( 3 , 9) is a solution of the given equation.
1
1
15. The line passing through (2, 14) is
or,
2y = 14x
y = 7x
Infinitely many lines are there.
1
1
The equation in the form ax + by + c = 0 is
7x – y + 0 = 0.
1
4 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
Area of floor = l × b = 20 × 16
16.
= 320 m2
Area of roof = Area of floor = 320 m2
Area of floor + Area of roof = Surface area of 4 walls
lb + lb = 2 (l + b) h
h =
½
640
= 8.88 m½
72
Volume of the hall = lbh
= ( 20 × 16 × 8.88) m3
½
320 + 320 = 2 (20 +16) h
640 = 72 h
½
3
= 2841.6 m .
½
½
17. Let the radii of two spheres be r1 and r2, then according to the question
r1 + r2 = 21
r1 = 21 − r2
½
4
Volume of sphere I
= r13
3 4
= π(21 − r2 )3 3
Volume of sphere II =
\
4 3
pr2
3
4
π(21 − r2 )3
Volume of sphere I
3
=
4 3
Volume of sphere II
πr2
3
=
64
27
½
3
(21 − r2 )3 64
4 =
=  
⇒
3
r2 3
27
⇒
21 − r2
4
=
r2
3
⇒
3(21 – r2) = 4r2
½
½
63 − 3r2 = 4r2
63 = 7r2 ⇒
63
7
½
⇒ r2 = 9 cm
r1 = 21 − 9
= 12 cm.
½
Solutions (SA-2) | 5
18. Observation : 17, 23, 25, 18, 17, 23, 19, 23, 17, 26, 23
No.
Tally marks
Frequency
17
18
19
23
25
26
|||
|
|
||||
|
|
3
1
1
4
1
1
11
Mode= 23
1½
After subtracting 4 from every observation : 13, 19, 21, 14, 13, 19, 15, 19, 13, 22, 19 No.
Tally marks
Frequency
13
|||
3
14
|
1
15
|
1
19
||||
4
21
|
1
22
|
1
11
Now, New Mode =19.
1½
SECTION — D
19.
Marks
Number of Students
0 – 20
15
20 – 40
10
40 – 60
10
60 – 80
11
80 – 100
4
Frequency polygon graph is ABCDEFG.
y
N umber of st udent s
16
B
14
12
C
10
D
E
8
6
F
4
A 2
G
0
20
40
60
Marks
80
100
4
x
6 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
20. Steps of Construction :
X
M
1. Draw a line segment BC = 4 cm.
2
2. Draw a ray BX such that ∠CBX = 75°.
3. From ray BX, cut off BM = 10 cm.
A
4. Join MC.
5. Draw perpendicular bisector of MC intersecting BM at A.
2
75°
6. Join AC, then DABC is the required triangle.
B
C
21. O is centre of circle.
AB is a chord with mid-point M.
To prove : AB < CD
1
Join OM and ON ⊥ CD
DONM is right angled triangle
∴ OM > ON, (OM is hypotenuse)
1
D
O
N
Chord CD is nearer to in comparison of AB,
A
M
1
B
C
⇒ CD > AB or AB < CD
22. O is centre of the circle. Chord PQ and RQ are equi-distance from the centre O.
Then OM = ON
i.e., OM ⊥ PQ, ON ⊥ RQ
R
S
In DOMQ and DONQ
N
OM = ON (given)
O
OQ = OQ (common)
∠OMQ = ∠ONQ = 90°
P
Q
M
DOMQ ≅ ONQ, (by RHS.)
∴
∠OQM = ∠OQN
1
i.e., diameter QS is a bisector of ∠PQR.
1
1
1
1
23. In DABC,
1
AB =AC
A
∴
∠C = ∠B ...(i) (Angle opposite to equal sides of a triangle are
E
D
equal) 1
∠ADE = ∠C and ∠AED = ∠B
Again, (Exterior angle of cyclic quadrilateral BCED) 1
C
B
∴
∠ADE =∠B and ∠AED = ∠C
[by equation (i)] 1
BC || DE.
24. x + y = 10 ⇒ y = 10 – x...(i)
x
0
2
3
4
5
y
10
8
7
6
5
2x – y = 5 ⇒ y = 2x – 5
x
0
2
5
y
–5
–1
5
Plot these points on the graph paper.
....(ii)
1
Solutions (SA-2) | 7
y
(2,8)
8
(3,7)
7
(4,6)
6
5
2x – 7 = 5
(5,5)
4
3
x + y = 10
2
1
x'
–8 –7 –6 –5 –4 –3 –2 –1 0
–1 1
–2
2
x
2 3 4 5
(2, –1)
6 7
8
–3
–4
(0, –5)
–5
–6
–7
–8
y'
From graph it is clear that point of intersection is (5, 5).
25. Length of wall (l) =10 m = 1000 cm
1
Thickness of wall (b) =36 cm
Height of wall (h) = 9 m = 900 cm
1
3
× Volume of wall
4
No. of bricks (for three fourth of this wall) =
Volume of one brick
=
3 1000 × 36 × 900
×
4
36 × 15 × 9 = 5000 bricks
26.
1
1
Inner radius of hemisphere (x) = 1 m = 100 cm
Outer radius of hemisphere (R) = 100 + 1 = 101 cm
½
Volume of hemisphere (used to make the tank) =
2
π(R3 – r3)
3
1
=
2
× 3·14 × (1013 – 1003)
3
½
= 63430·09 cm3.
2
27. (i) Required number of students = 8 + 32 = 40
1
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We
notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle
(adjusted frequency) for each class.
Length of rectangle (adjusted frequency)
Frequency of Class
=
× Minimum class-size 1
Width of class
8 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
Marks
(C.I.)
Number of students (f)
Width of class
(Class-size)
0 – 10
8
10
8
×5 = 4
10
10 – 30
32
20
32
×5 = 8
20
30 – 45
18
15
45 – 50
10
5
18
×5 = 6
15
Length of rectangle
10
× 5 = 10
5
Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as
heights. 1
Histogram representing marks obtained by students in unit test of Mathematics.
Number of st udent s
y
10
9
8
7
6
5
4
3
2
1
O
10
8
6
4
10 20 30 40
Marks obtained
50
x
(iii) Hardwork and Dilligence.
28. (i)Base BC =120 m
A
Height AD =90 m
1
Area of triangular plot = × base × height
2
1
= × 120 × 90
2
= 5400 m2.
1
2
E
B
D
C
(ii) In DABC they draw median AD on base BC and divide it into two equal areas ABD and ACD. Take
any point E on AD and join BE and CE.
1
Two brothers get areas ar (DABE) and ar (DACE) and ar (DBCE) is donated to school.
(iii) Both brothers know importance of education and love their community.
1
SECTION — E
This section consists of questions asked from OTBA which are provided by schools only.
29.(3 Marks)
30.(3 Marks)
31.(4 Marks)
nnn
SOLUTIONS
QUESTION PAPER - 7
Self Assessment __________________________________
Time : 3 Hours
Maximum Marks : 90
SECTION — A
1. Given,
Area of sphere = Volume of Sphere
4pr2 = 4 pr 3
3
r = 3 cm
Diameter = 2r = 6 cm.
where r is the radius of sphere
2.
Volume of cuboid = length × bredth × height
= 3·6 × 8·2 × 11
= 324·72 cu. cm.
11 + 15 + 17 + y + 1 + 19 + y − 2 + 3
3. 14 =
7
⇒
⇒
⇒
4. Given, 98 = 64 + 2y
2y = 34
y = 17
BE = 2EC
[on solving]
1
½
½
1
EC
1
=
BC 3
⇒
and
ar (∆ABC) = 60 cm2
∴
ar (∆AEC)=
1
× 60 cm2 = 20 cm2.
3
SECTION — B
1
MATHEMATICS Oswaal CBSE Class -9, SA-2 Examination Sample Question Paper
SAMPLE
5. AD is a median of ∆ ABC
∴
area (∆ ABD) = area (∆ ACD)…(i) ½
Similarly, for ∆ EBC, ED is a Median of ∆ EBC
∴
area (∆ EBD) = area (∆ ECD) ....(ii) ½
Subtracting equ. (ii) from equ. (i)
Area (∆ ABD) − Area (∆ EBD) = Area (∆ ACD) − Area (∆ ECD)
½
Area (∆ ABE) = Area (∆ ACE).
10 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
½
6.Draw OP perpendicular to xy from the centre to a chord bisecting it.
OP ⊥ to chord BC.
⇒
BP= PC ...(i) 1
Similarly, AP= PD...(ii)
Subtracting eqn. (i) from eqn. (ii), we get
½
AP – BP= PD – PC
or AB= CD
½
7.Sum of 100 observations = 60 × 100 = 6000 ½
After replacement, sum of new observations = 6000 – 50 + 110 = 6060
½
6060
= 60.6 New mean =
1
100
8. Arranging the data in ascending order : 3, 4, 5, 5, 5, 6, 6, 7, 7, 7, x + 1, 8
Since mode is 7, So
x + 1 =7
\
x = 7 – 1 =6
9. (i) 0 boys 1
P(E)=
(ii) 2 boys ½
[means family having two girls]
275 11
=
1000 40 198
99
=
P(E)=
1000 500 ½
1
[means family having 0 girls]
1
[CBSE Marking Scheme 2015]
10. Given : r = 3·5 m, h = 12 m
1
Capacity of conical pit (V) = πr2h
3
1 22
V = ×
× 3·5 × 3·5 × 12
3 7
= 154 m3
= 154000 l.
½
½
½
½
SECTION — C
11.
r1 : r2 = 4 : 1 Þ
r1 4
=
r2 1
1
l1
=
2
l2
1
r l 
CSA1
πr l
= 1 1 =  1  1 
CSA2
πr2 l2  r2   l2 
4 1 2
= × =
1 2 1
\
\
CSA1= 2CSA2
14
12.Base radius (r) =
= 7 cm
2
l = 25 m
1
1
½
Solutions (SA-2) | 11
C.S.A.= πrl =
22
× 7 × 25
7
= 550 m2
Cost of white washing @ ` 210 per 100 m2
220 × 210
= `
100
1½
= ` 1155.
13. Given ∆ABC is an isosceles triangle with AB = AC. A circle is drawn taking AB as the diameter which
intersects the side BC at D.
To Prove that :
BD = DC
Construction : Join AD
A
B
∠ADB = 90°
Proof :
C
D
1
(angle in semi-circle is 90°)
∠ADB + ∠ADC = 180°
⇒
∠ADC = 90°
In ∆ABD and ∆ACD
AB= AC
(given)
∠ADB= ∠ADC (proved)
AD= AD ∴
∆ABD @ ∆ACD
∴
(common)
(By R.H.S. ≅)
BD= DC (c.p.c.t) 2
14. Writing in standard form
3x + 2y – 12 = 0
½
3x – 12 = 0
1
On x-axis, y = 0 ⇒
\
x = 4
Point on the x-axis = (4, 0)
½
On y-axis, x = 0 ⇒
2y – 12 = 0
y = 6
\
15. The equation is A (1, 2); B (– 1, – 16); Point on the y-axis = (0, 6).
½
y = 9x – 7
2 = 9(1) – 7
2 = 2; True
1
– 16 = 9 (– 1) – 7 = – 9 – 7
= – 16; True
C (0, – 7); ½
1
– 7 = 9 (0) – 7
= 0 – 7 = – 7; True.
1
12 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
16.
Distance (in km)
Tally Marks
Frequency
40 — 60
|
1
60 — 80
||||
5
80 — 100
|||| |||
||||
8
|||| |||
|||
8
100 — 120
120 — 140
140 — 160
5
3
17.
Inner radius (r) = 2 cm
outer radius (R) = 2.2 cm
Height (h) = 77 cm
(i) 1
C.S.A. (Inner) = 2πrh
= 2 ×
22
× 2 × 77
7
= 968 cm2
1
2 cm
77
2·20 cm
(ii) C.S.A. (Outer) = 2πRh
= 2 ×
22
× 2.2 × 77
7
= 1064.8 cm2
(iii) 1
Area of top = π(R + r)(R – r)
=
22
× 4.2 × 0.2
7
= 2.64 cm2
= Area of the bottom
∴ T.S.A. = Inner (C.S.A.) + Outer (C.S.A.)
+ Area of top + Area of bottom
= 968 + 1064.8 + 2 × 2.64
= 2038.08 cm2.
1
18.
Total no. of cars =100
5
1
(i) P (exactly 5 occupants) =
=
100 20
(ii)P (more than 2 occupants) =
23 + 17 + 5 45
9
=
=
100
100 20
(iii)P (less than 5 occupants) =
29 + 26 + 23 + 17 95 19
=
=
100
100 20 (1 + 1 + 1)
Solutions (SA-2) | 13
SECTION — D
19.
Areas of four walls =
2(l + b)h=
Total Cost
Cost/m 2
340·20
= 252 sq. m
1·35
1
2(l + b)h = 252
Area of floor = l × b
...(i)
91·80
= 108 sq. m
=
·85
12 × b = 108
b = 9 m
Eg. (i) and (ii)
2(12 + 9)h = 252
h=
[l = 12 m, Given]
...(ii)
252
= 6m .
2 ´ 21
X
V
S
P
90°
A
30°
30°
Z
Y
T
21. D and E are mid-points of AB and BC respectively.
∴
DE || AC
Similarly, DF || BC and EF ||AB
∴ ADEF, BDEF and DFCE are all parallelogram
DE is the diagonal of parallelogram BDFE.
∴
DBDE ≅ DFED
and
DEFC ≅ DFED
DADF ≅ DFED
∴ All four triangles are concurrent.
2
[CBSE Marking Scheme, 2015]
Q 2
20. Steps of Construction :
1. Draw a line segment AB = 18 cm (XY + YZ +
ZX = 18 cm)
2. Construct an angle ∠PAB = 30° at point A and
an angle ∠QBA = 90° at point B.
3. Bisect ∠PAB and ∠QBA. These bisectors inter
sect each other at point X.
4. Draw perpendicular bisector ST of AX and UV
of BX.
5. Perpendicular bisector ST intersect AB at
Y and UV intersect AB at Z. Join XY, XZ, then
DXYZ is the required triangle.
22. Let 1
B
U
1
A
1
F
D
B
E
1
C
1
Age of Amit = x years
Age of Akhil = y years
(a) According to the question the linear equation for the above situation is
x + y = 25
⇒
y = 25 – x
x
0
10
15
y
25
15
10
1
14 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
y
25 (0, 25)
20
(10, 15)
15
11
(15, 10)
10
5
0
5
10 15 20
25
x
14
2
(b) From the graph when Amit’s age = 14 years, then Akhil’s age = 11 years.
1
23. As D, E, C, B are concylic quadrilateral and ∠EDB + ∠ECB = 180° and ∠DBC + ∠DEC = 180°
1
As ∠EDB + ∠ADE = 180° and ∠AED + ∠DEC = 180°
(linear pairs)
Hence, we can say that, ∠ADE = ∠ACB and ∠AED = ∠ABC.
1
As triangle is isosceles, so we can say ∠ADE = ∠ACB = ∠AED = ∠ABC
1
So, DE is parallel to BC and AD = AE & DB = EC and DECB is an isosceles trapezium. So, DC and EB
will be equal and if they intersect at O, AO will be the median of the triangle ABC and triangle ADE
as well.
1
A
24.
Given, BQ || CR
P
B
Q
C
R
Therefore, DBCQ and DBQR are on the same base and between the same parallels BQ and CR
ar (DBCQ) =ar (DBQR)...(i) 1½
So, Also, AP || BQ (given)
Therefore, DABQ and DPBQ are on the same base BQ and between same parallels BQ and AP
∴
ar (DABQ) =ar (DPBQ)...(ii) 1½
Adding (i) and (ii), we get
(DBCQ) + ar (DABQ) =ar (DBQR) + ar (DPBQ)
1
ar (DAQC) =ar (DPBR).
25.Since, perpendicular from the centre of the circle to a chord bisects the chord.
∴ P and Q are the mid-points of AB and CD
1
1
AP= AB =
× 6 = 3 cm
2
2
1
1
CQ= CD =
× 8 = 4 cm
2
2
In right triangle OAP
A
P
B
O
1
C
Q
D
Solutions (SA-2) | 15
OA2 = OP2 + AP2
52 = OP2 + 32
OP2 = 25 – 9
1
2
OP = 16
OP = 4 cm
In right DOCQ
OC2 =OQ2 + CQ2
52 =OQ2 + 42
OQ2 =25 – 16
1
2
OQ =9
OQ =3 cm
∴
PQ =OP + OQ = 4 + 3 = 7 cm
1
26. The inner diameter of hemispherical bowl = 8 cm
Then its inner radius (r) = 4 cm
Thickness of steel = 0.2 cm
Hence, outer radius of bowl (R) = 4 + 0.2 = 4.2 cm
1
Hence, the outer curved surface area of the bowl
= 2pR2
22
= 2 × ×( 4.2 )2
7
44 42 42
=
× ×
7 10 10
= 110.88 cm2
2
Hence, the cost of polishing its outer surface area
= 2 × 110.88
= ` 221.76.
1
27. Let r and h be the radius of base and height of the cylinder, cone and hemisphere.
We know that,
\
(a)
Height = Radius of hemisphere.
h= r
Volume of cylinder = pr2h = pr2 × r = pr3
Volume of cone =
1 2
1
pr h = pr2 × r
3
3
1
= pr3
3
Volume of hemisphere =
2 3
pr
3
Volume of cone : Volume of hemisphere : Volume of cylinder
1
2
= pr3 : pr3 : pr3
3
3
1
16 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
1 2
= : :1
3 3 (Dividing by pr3)
= 1 : 2 : 3
(Multiplying by 3)
\
(b)
\
1
Required Ratio = 1 : 2 : 3.
Slant height of cone = l =
r2 + r2 = r 2
C.S.A. of cone = πrl
(
)
= πr r 2 =
C.S.A. of hemisphere = 2 πr2
C.S.A. of cylinder = 2 πrh
2 πr2
= 2 πr (r)
= 2 πr2
\ C.S.A. of cone : C.S.A. of hemisphere : C.S.A. of cylinder
=
2 πr2 : 2πr2 : 2πr2
=
2 :2:2
= 1 :
2 :
1
2
1
28. (i) For Histogram,
Y-axis =one square = one students
X-axis = one square = 10 marks
(ii) For frequency polygon, first we obtain the class marks.
Marks
Class Marks
No. of students
0 — 10
10 — 20
20 — 30
30 — 40
40 — 50
50 — 60
60 — 70
70 — 80
5
15
25
35
45
55
65
75
0
2
5
6
4
8
10
5
1
To obtain the frequency polygon, we plot the points (5, 0) (15, 2), (25, 5), (35, 6), (45, 4), (55, 8), (65, 10), (75, 5)
10
9
8
No. of St udent
½
7
6
5
4
3
2
1
10 20 30 40 50 60 70 80
Marks
Solutions (SA-2) | 17
(i) Statistics.
½
(ii) Sincerity
½
SECTION — E
This section consists of questions asked from OTBA which are provided by schools only.
29.(3 Marks)
30.(3 Marks)
31.(4 Marks)
nnn
MATHEMATICS Oswaal CBSE Class -9, SA-2 Examination Sample Question Paper
S O L U T I O N S
SAMPLE
QUESTION PAPER - 8
Self Assessment_______________________________
Time : 3 Hours
Maximum Marks : 90
SECTION — A
1.Let h and r be the height and radius of original cone and let h' and r' be the height and radius of
new cone
Given,
h' = 3h and r' = 3r
1 2
πr h
3
Volume of original cone
V =
2
V'= 3 π(r ') h '
1
Volume of new cone Volume of new cone
Volume of original cone
=
=
=
=
1
π( r ')2 h '
3
1 2
πr h
3
( r ')2 h '
r 2h
( 3r )2 3 h
r 2h
27
1 1
Hence, the ratio of new cone to the original cone is 27 : 1.
2. Mode = 27
1
3.(a + b) units
1
4. ÐADB= ÐACB = 40°
½
[Q Angles in the same segment are equal]
Now, in DDPB
Þ
ÐDPB + ÐDBP + ÐPDB = 180°
[Angle sum property]
120° + ÐDBP + 40° = 180°
Þ
ÐDBP = 180 – (120° + 40°)
Þ
ÐDBP = 20°
\
ÐCBD= ÐPBD = 20°
SECTION — B
½
Solutions (SA-2) | 19
5. Steps of Construction :
1. Draw a line segment PQ of any size (let be 7 cm)
2. Draw another line segment RS intersecting PQ at point K. (length of RS = 8 cm)
3. Measure pair of vertically opposite angles (both are equal)
4. Bisect both the angles (OX is bisector of angle QOS and OY is bisector of angle POR)
5. Yes, from the constuction it is clear that the bisecting rays froming a straight line. 1½
R
Y
P
O
Q
X
6.
S
½
[CBSE Marking Scheme, 2016]
Length OB = 4 cm ½
Radius= 5 cm
By Pythagoras theorem, O
2
2
AB = OA − OB 5 cm
2
4 cm
= ( 5 )2 − ( 4 )2 = 25 − 16 A
C
½
½
= 9
AB = 3 cm
∴
AC = 2 × AB = 2 × 3 = 6 cm. 7.
B
½
x+x+2+x+4+x+6+x+8
Mean = 13 =
5
5 x + 20
5
13 = x + 4
13=
1
½
x = 9.
½
8. Observation : 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Number of observations N = 10 (even)
th
∴
N
N

  +  + 1 
2
2
Median =
2
th
th
 10 
 10 
  +  + 1 
2
2
=
2
observation ½
observation ½
th
20 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
=
63 =
5 th + 6 th
observation
2
x+x+2
2
126 = 2x + 2 2x = 124
124
x =
= 62.
2
½
½
9. (i) At least 2 tails [Frequency of number of tails 2 + number of tails 3]
32 + 63 95 19
P(E) =
=
=
1
150
150 30 (ii) Exactly one tail
P(E) =
30 1
=
150 5 1
[CBSE Marking Scheme 2014]
10.
r =4·9 cm
4
Volume V = πr3
3
4 22
×
× 4.9 × 4.9 × 4.9
3 7
=
= 493 cm3 (approx.)
½
1
½
SECTION — C
11. Let r and h be the radius and height of the cylinder given r : h = 5 : 7
\ The radius of the cylinder (r) = 5x
and The height of the cylinder (h) = 7x
Volume of the cylinder = pr2h
\
4400=
22
× ( 5 x )2 × 7 x
7
⇒
4400=
22
× 5x × 5x × 7x
7
⇒
x3=
⇒
22 

 π = 7  1


4400 × 7
22 × 5 × 5 × 7
x3 = 8 = 23
\
x = 2
1
Hence, the radius of the cylinder = 5x
= 5(2)
= 10 cm.
1
12. Let the outer radius be R and inner radius be r.
Outer Surface Area – Inner Surface Area= 44 cm2
2πRh – 2πrh = 44
2π × 14(R – r) = 44
∴
R – r =
44 ´ 7
2 ´ 22 ´ 14
½
Solutions (SA-2) | 21
1
cm...(i) ½
2
=
Volume of metal = 99 cm2
πR2h – πr2h = 99
14π(R2 – r2) = 99
½
99 ´ 7
9
= cm2
22 ´ 14
4
R2 – r2 =
9
(R + r) (R – r) =
4
9
9
R + r = × 2 =
4
...(ii) ½
2
Adding (i) and (ii) , we get
2R =
10
2
∴
R =
5
2
r = -
9
2
5
4
=
2
2
= 2
∴ Outer radius = 2·5 cm and inner radius = 2 cm.
1
13. XY || BC and CQ || EB
A
F
E
Q
P
X
B
Y
C
∴EBCQ is a ||gm
½
Similarly PBCF is a || gm
½
These || gms are on the same base BC and between the same parallels.
1
∴
ar (EBCQ) = ar (PBCF)
½
As ar (PBCQ) is common to both parallelograms. ∴
ar (EBP) = ar (FCQ).
½
14.Given, Equation
⇒
3x – 5y – 15 = 0
5y = 3x – 15
⇒
y=
3x − 15
5
⇒
y=
3
( x − 5)
5
put x = 0, then y = –3
put x = 5, then y = 0
put x = –5, then y = –6
1
22 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
A
B
C
x
0
5
–5
y
–3
0
–6
To plot these points on graph paper, join these points. The graph of the line intersects x-axis at (5, 0)
and y-axis at (0, –3).
1
y
4
3
2
1
x'
–5 –4
–3 –2
–1
1
2
3
4
–1
5
B (5, 0)
x
–2
3x
–5
y
–1
0
5=
–3
A (0, –3)
–4
–5
1
–6
C(–5,–6)
y'
15. Equation of the sides are,
1
AB : Y = 0
BC : X = –1
CD : Y = –4
DA : X = –4
1
Area = 4 × 3
1
= 12 sq. units
16.
x
f
fx
10
3
30
12
10
120
20
15
300
25
7
175
35
5
175
Total
N = 40
Sfx = 800
2
∴
Mean =
Σfx 800
=
= 20 .
N
40
1
Solutions (SA-2) | 23
17. Total number of balls faced by Sachin =40
No. of balls on which he hit a six =12
Let E, be the event of hitting a six, ∴
No. of favourable outcomes =12
∴
½
12 3
=
40 10 P(E1) =
1
Now, Total No. of balls faced by Saurav=30.
Let E2 be the event of Saurav who did not hit the boundary.
∴
No. of favourable outcomes =30 – 9
½
= 21
∴
P(E2) =
21 7
=
30 10 1
[CBSE Marking Scheme, 2014]
18.
(a) Total number of outcomes = 250
Number of outcomes only heads = 50
∴ Probability of outcomes no tail (only head)
=
No. of outcomes no tail (only head)
Total number of outcomess
50 1
=
= 250
5
1½
(b) Number of outcomes only one head
= 125
\ Probability of outcomes only one head
125 1
=
= 250 2
1½
SECTION — D
19.(a)Given inner edge of cubical box (a) = 20 cm
Thickness of wood = 2 cm
Outer edge = 20 + 2 = 22 cm
Volume of cubical box = a3
\
Volume of the wood = (22 × 22 × 22) – (20 × 20 × 20)
= 223 – 203
= 2648 cm3
(b) Volume of air trapped in it
= 20 × 20 × 20
= 8000 cm3.
20. (i) Draw a line segment PQ = 24 cm. (ii)At P, construct an angle of 60° and at Q an angle of 45°
(iii)Draw bisectors of ∠P and ∠Q intersect at a point A.
(iv)Draw perpendicular bisectors of AP and AQ intersects PQ at B and C respectively.
2
2
½
½
½
½
24 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
M
L
M
A
1
60°
P
45°
B
C
Q
(v) Join AB and AC. (vi) ABC is the required triangle.
1
21. To prove that : Ar (gm DLOP) = Ar (gm BMOQ)
Proof : Give ABCD is a gm and AC is one if its digonal
∴ar(∆ABC) = ar(∆ADC)...(i)
[\ The diagonal of a gm divides it into two equal triangles] ½
Now, LM is parallel to AD
(∵ AD = BC and AD  BC as ABCD is a gm
∴
LM  BC ⇒ LO  CQ
PQ  DC ⇒ OQ  LC
Similarly
gm
∴ CLOQ is also a  , as its opposite sides on parallel to each other.
In gm CLOQ, CO is the the diagonal
∴ar(∆COQ) = ar(∆CLO)...(ii) ½
[ The diagonal of a gm divides it into two equal triangles]
Again,
AD  LM ⇒ AP  OM
and
AB  PQ ⇒ AM  PO
∴ APOM is also a gm, as its opposite sides one parallel to each other.
In gm AOPM, AO is the diagonal
∴ar(∆AOP) = ar(∆AOM)...(iii) ½
[ The diagonal of a gm divides it into two equal triangles]
Now using (i)
ar(∆ABC) = ar(∆ADC)
⇒ ar(∆COQ) + ar(∆AOM) + ar(BMOQ) = ar(∆CLO) + ar(AOP) + ar(DLOP)
using (ii) and (iii), we get
ar(BMOQ) = ar(DLOP)
½
Now, BMOQ is a gm as
PQ  AB ⇒ OQ  BM
BC  AD ⇒ OM  BQ
and
gm
Similarly, DLOP is also 
as
AD  LM ⇒ DP  LO
and
PQ  DC ⇒ PO  DL
∴
½
ar(gm BMOQ) = ar(gm DLOP)
½
Hence Proved. 1
Solutions (SA-2) | 25
22.Total students in the class = y
Let the boys in the class = x, then equation between the students and the boys
y =
4
x 3
1
Now
x
0
30
60
y
0
40
80
Now, draw a graph between these points.
y
80
70
60
students
(60, 80)
50
40
(30, 40)
30
20
10
0
10 20 30 40 50 60 70 80
x
boys
2
From the graph it is clear that there are 30 boys in a class of 40 students.
1
23.AD = BC (opp. sides of a ||gm)
⇒
1
1
AD = BC
3
3
1
⇒
ED = BF
Also AD || BC
⇒
ED || BF
∴ EDFB is a || gm ( a quad in which one pair of opposite sides is equal and parallel is a || gm)
2
Now ∠EDF + ∠BFD = 180° (adjacent angles of a || gm are supplementary)
1
⇒
∠BFD = 120°.
24.
y = Reflex ∠AOC = Reflex 110°
= 360° − 110° = 250° 1
1
1
∠z = ∠AOC =
× 110° = 55°
1
2
2
∠x + ∠z =180°
∠x = 180° − ∠z = 180° − 55° = 125°
1
∠w = 360° − ∠AOC − ∠OAB − ∠ABC
= 360° − 110° − 65° − 125°
25.ABCD is a trapezium.
∴
Draw a line DE || CB
∴
But = 360° − 300° = 60°.
∠1 + ∠2 =180° 1
…(i) ½
DE = CB and ∠4 = ∠3…(ii) ½
DA = CB (given)
…(iii) 26 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
D
From (ii) and (iii)
C
1
DE = DA…(iv) 1
In ∆ DAE
2
DE = DA
∴
∠2 = ∠4
∠2 = ∠3 (from ii)
4
A
3
E
B
…(v) 1
From (i) and (v)
∠1 + ∠3 =180° i.e., ABCD is cyclic.
1
26. (i) Here, radius of two cones and cylinder
=
3
= 1.5 cm
2
2 cm
2 cm
1.5 cm
8 cm
1.5 cm
2 cm
2 cm
Height of each cone, h = 2 cm
∴
Height of cylindrical portion
h´ = 12 -2 − 2
= 8 cm ½
∴ ½
Volume of the air = Volume of cylindrical part + 2 × Volume of cone
1
= πr 2 h + 2 × πr 2 h ′
3
½
= 22 × 1·5 × 1·5 × 8 + 2 × 1 × 22 × 1·5 × 1·5 × 2 3 7
7
22
4

=
× 1·5 × 1·5 8 + 
3

7
22
28
=
× 1·5 × 1·5 ×
7
3
(ii) = 66 cm3
½
Slant height of each cone = h 2 + r 2
= (2 )2 + (1·5)2 = 4 + 2·25 ½
= 6·25 = 2.5 cm
½
(iii)Mensuration
(iv)Sincerity
½
½
Solutions (SA-2) | 27
27. For frequency polygon first we obtain the class marks
1
Cost of living index
Class marks
No. of weeks
140 − 150
145
5
150 − 160
155
10
160 − 170
165
20
170 − 180
175
9
180 − 190
185
6
190 − 200
195
2
To obtain the frequency polygon we plot the point (145, 5), (155, 10), (165, 20), (175, 9), (185, 6), (195, 2) and
join the points by line segments.
1
Y
24
Number of Weeks
20
16
12
8
4
O
28. 140 150 160 170 180 190 200 210
Cost of Living Index
X
2
Cost of white washing=` 498·96
Cost of white washing
Curved Surface area of hemisphere = Rate per square meter
Volume of hemisphere
2pr2 =
2×
498·96
2 1
22 2
× r =249·48 cm2
7
249·48 × 7
= 5·67 × 7
r2 =
2 × 22
r2 = 39·69
r = 6·3 cm
2
= πr3
3
2 22
= ×
× 6·3 × 6·3 × 6·3
3 7
= 523·908 cm3.
1
½
½
1
SECTION — E
This section consists of questions asked from OTBA which are provided by schools only.
29.(3 Marks)
30.(3 Marks)
31.(4 Marks)
nnn
MATHEMATICS Oswaal CBSE Class -9, SA-2 Examination Sample Question Paper
S O L U T I O N S
SAMPLE
QUESTION PAPER - 9
Self Assessment_______________________________
Time : 3 Hours
Maximum Marks : 90
SECTION — A
1.Given, diameter of football = 5 × diameter of cricket ball
If r denotes radius of a football and r’ that of a cricket ball, then we have
2r = 5 × (2r’)
2r
= 5
2r '
or
r
= 5
r'
½
2
Now, 4 πr 2
25
r
=  =
ratio of surface areas =
4 π(r ')2  r ' 
1
= 25 : 1
½
2. Let the angle of a quadrilateral = 2x, 3x, 6x, 7x then, we know that
2x + 3x + 6x + 7x = 360°
18x = 360°
x = 20°
So, largest angle of a quadrilateral
=7x = 7 × 20°
= 140°
3.
(
)
3
Volume of cube =(Edge)3 = 3 3a = 81 3a 3 4.
140
A
E
B
1
1
SECTION — B
5.
1
D
C
Solutions (SA-2) | 29
In ∆ABC, AD is a median.
ar (∆ABD) = ar (∆ACD)
...(i) ½
In ∆BEC, ED is a median.
ar (∆BDE) = ar (∆CDE)
...(ii) ½
Subtracting the equation (ii) from equation (i), we get
ar (∆ABD) – ar (∆BDE) = ar (∆ACD) – ar (∆CDE)
½
∴
ar (∆ABE) =ar (∆ACE).
½
6.
∠ADB = ∠ACB = 70° (Angles in same segment of a circle) ½
In ∆ ADB,
∠DAB + ∠ADB + ∠DBA =180° ½
60° + 70° + ∠DBA = 180°
∠DBA = 180° − 130°
½
∠DBA = 50°.
1
7. 2, 2, 2, 2, 4, 4, 4, 5, 6, 6, 8, 10
1
Mode = 2
1
Sum of all observations
½
8.
Mean =
Total number of observations
23 =
x+x+2+x+4+x+6+x+8
5
5x + 20 = 23 × 5 = 115
5x = 115 − 20 = 95
95
x =
= 19 ½
5
Last four observations
= 19 + 2, 19 + 4, 19 + 6, 19 + 8
= 21, 23, 25, 27
21 + 23 + 25 + 27
∴
Mean of last four observations =
4
=
96
= 24.
4
9.Total No. of families = 184 + 714 + 425
= 1323
(i) Probability that chosen family has exactly one girl
714 34
=
=
1323
63 184
(ii) Probability that chosen family has 2 boys =
1323 10.
Total Surface area of cube = 6 (side)2
= 6 × 10·5 × 10·5
= 661·5 mm2
= 6·615 cm2
½
½
1
1
1
1
SECTION — C
11.
Total surface area =2πR2 + 2πr2 + (πR2 – πr2)
=3πR2 + πr2
=π(3R2 + r2) = [3(8)2 + 62]
1
30 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
22
× [228] cm2
7
=
=716·57 cm2
∴ Cost of painting the vessel all over =` (716·57 × 2)
=` 1433·14. 12. Let r be the base radius and h be the height of the cylinder
Given, h = 12
For base radius
2pr = 66
22
2 × × r = 66
7
\
r=
66 × 7
22 × 2
\
r=
21
cm.
2
1
1
1½
Volume of cylinder = pr2h
2
22  21 
×   × 12
=
7  2 
22 21 21
=
× × × 12
7 2 2
= 4158 cm3.
1½
13. Given : OA =OB = AB
OB =OA (radius)
∴ OAB is an equilateral triangle.
∴
∠AOB = 60° (angle of equilateral triangle are 60° each)
1
D
C
∴
a + ∠AOB=180° (linear pair)
b
∴
a + 60°=180°
a
⇒
a = 120°
½
O
B
Reflex angle BOD =2∠BCD (angle subtended by an arc at the centre is twice
at the circumference) 1
A
∴
360° – a =2b
⇒
2b =240°
½
∴
b =120°
Hence,
a =120°, b = 120°
[CBSE Marking Scheme, 2014]
14. 3x = y + 3; three solutions are x = 1, y = 0; x = 2, y = 3 and x = 0, y = – 3.
1½
y
4
3
(2, 3)
2
1
(1, 0)
x
–1
–2
–3
1 2 3
4
(0, – 3)
From graph it is clear that line meets x-axis at (1, 0) and y-axis at (0, – 3).
1½
Solutions (SA-2) | 31
15. Total cost = value of pen × number of pens
\
y = 16x...(i)
Table of value of (i) is
x
0
1
2
3
4
5
6
y
0
16
32
48
64
80
96
From the graph, cost of 6 pens = ` 96.
y
y = 16x
128
112
96
80
64
48
32
16
0
1
2
3
4
5
6
7
8
x
16. Mean monthly salary of 12 employees = ` 14,500
Sum of monthly salary of 12 employees
= 14,500 × 12
= ` 1,74,000
\Sum of monthly salary of 13 employees
= 1,74,000 + 18,400
= ` 1,92,400
\Mean of monthly salary of 13 employees
1, 92 , 400
=
= ` 14,800
13
17. (i) (ii) 8 + 12 20 2
Probability of less than 41 =
=
= 90
90 9
Probability of more than 50 =
1+½+½+1
1
1
1
1
20 + 13 + 17 + 05
55 11
=
=
1
90
90 18
(iii) Probability of marks between 41 and 80
=
15 + 20 + 13 + 17
90
=
65 13
½
=
90 18
18.
½
Total Number of tosses = 100
Number of outcomes of an odd number= 20 + 20 + 20 = 60
½
Number of outcomes of a prime number = 15 + 20 + 20 = 55½
No. of outcomes of an even prime number = 15
½
32 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
(a) Probability of getting an odd number =
60 3
= .½
100 5
(b) Probability of getting a prime number=
55 11
=
.
100 20
½
(c)Probability of getting an even prime number =
½
15
= 0·15
100
SECTION — D
19. Circumference of the base of conical heap = 132 cm
2pr =132
r =21 cm
Amount of Tarpaulin used =prl(l = 35 cm, Given)
22
´ 21 ´ 35
7
=
=2310 cm2
1
h2= l 2 - r 2 = (35)2 - (21)2
= 28 2
Þ
1
h =28 cm
1
3
Volume of grains = pr 2 h
1
3
= ´
22
´ 21 ´ 21 ´ 28
7
=12936 cm3
1
Volume of grains distibuted = 1 ´ 12936
2
=6468 cm3
Values : Sensitivity, Generosity,
Care for protection of grains
1
[CBSE Marking Scheme 2015]
X
20. Steps of Construction :
D
1.Draw BC = 6 cm.
2.
Draw ∠CBX = 90° and cut off BD = 10 cm.
Y
X
meeting BD at A.
4.
Join AC, then ABC is the required triangle.
⇒ ½
3. Join CD and draw its perpendicular bisector
21. ½
A
½
B
C
½ 2
2x – y = 4
y = 2x – 4
1
Solutions (SA-2) | 33
x
0
2
1
y
–4
0
–2
x + y = 2
⇒
1
y = 2 – x
x
0
2
1
y
2
0
1
y
4
2x–
y=
4
3
(0, 2)
A
2
(1, 1)
1
x'
–4
–3
–2
–1
1
O
2
C
–1
(2,0)
–2
(1, –2)
–3
3
x
4
x+y=2
(0, –4)
B
–4
y'
1
1
From the graph, ∆ABC is the required triangle and its vertices are A(0, 2), B(0, – 4) and C(2, 0).
22. We have to prove ∠ECB + ∠EDB =180°
AB = AC ⇒ ∠1 = ∠2
AD = AE ⇒ ∠3 = ∠4
A
D
∠A + ∠1 + ∠2 = 180° = ∠A + ∠3 + ∠4
3
1
4 E
2∠1 =2∠3
∠1 =∠3 = ∠2 = ∠4
⇒
DE || BC
∴
∠1 + ∠BDE = 180° ∠1 + ∠CED = 180°,
B
2
1
C
1
(Corresponding angle)
(∠BDE = ∠CED, as ∠3 = ∠4) 1
∴ B, C, E, D are concyclic.
1
23. Given : C is the mid-point of AB, AD and BE are perpendiculars to the line m meeting at D and E.
To prove : CD = CE,
Construction : Draw CM ⊥ m
Proof : Since AD ⊥ m and BE ⊥ m,
∴
AD || BE (transversal AB and DE intersect the parallels AD and BE)
and
AC = CB (∵ C is mid-point of AB)
1
34 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
∴
DM =ME
C
In DCMD and DCME,
1
A
B
DM =EM
∠CMD=∠CME (each 90°)
1
CM =CM (common)
DCDM ≅ DCEM (SAS congruency)
Then,
D
M
E
m
Hence Proved.
CD =CE (by C.P.C.T.)
1
(CBSE Marking Shceme, 2014)
24.Join A to C.
Draw AM ⊥ DC and CN ⊥ AX
∴
∴
AB || DC
AM = CN
1
ar (ADC) = × base × height
2
A
B
N
X
1
=
1
× DC × AM...(i) 1
2
Again, 1
× AB × CN
2
ar (ABC) = D
M
C
...(ii)
Add (i) and (ii), we get
ar (ADC) + ar (ABC) =
1
[DC × AM + AB × CN]
2
ar (ABCD) =
1
[DC × AM + AB × AM]
2
Area of trapezium =
=
25. Volume of liquid =
Given,
and r=
1
× AM (DC + AB)
2
1
× height × sum of parallel side
2
18
= 9 cm
2
h = 15 cm
1
Volume of liquid = p(9)2 15
3
and height of small bottle
\
1
1 2
pr h
3
\
= 405 p cm3
Also, radius of small bottle
3
(r’)=
= 1.5 cm
2
1
(h’) = 4 cm
1
Volume of small bottle = p(r’)2h’
3
1
= p(1.5)24
3
1
Solutions (SA-2) | 35
= 3p
405 p
Number of bottles =
= 135
3p
1
\
1
1
Amount earned = 135 × 5 = ` 675
26. Area of three adjacent face of cuboid are lb, bh and hl, where l, b and h are length, breadth and height
of cuboid respectively, then
1
2
2
2
lb = 15 cm , bh = 20 cm , hl = 12 cm 1
lb × bh × hl =15 × 20 × 12
(lbh)2 = 3 × 5 × 4 × 5 × 3 × 4
1
Volume of cuboid =lbh = 3 × 4 × 5 = 60 cm2
1
27. Pocket Money
No. of Children
0–10
12
10–20
23
20–30
35
30–40
20
40–50
10
1
Y - axis
No. of children
50
40
30
20
10
0
10
20
30
40
50
Pocket money
X - axis
3
28. To prove : ∠APB = 45°
Proof : AN = NB = 1 cm
and OB =
2 cm
B
In DONB,
⇒
⇒
OB2 =ON2 + NB2 (by pythagoras theorem)
( 2)
2
=ON2 + 12
ON =1
P
O
N
M
A
1
∴ DONB is an isosceles triangle.
∠ONB =90°. (∵ ON is the perpendicular bisector of chord AB)
∴
∠NOB =∠NBO = 45°
Similarly,
∠AON =45°
∠AOB=∠AON + ∠NOB
1
36 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
=45° + 45°
=90°
1
∠APB= ∠AOB
2
(chord subtends an angle is twice angle subtends by an arc.) 1
=
1
× 90 = 45°
2
Hence Proved. 1
[CBSE Marking Scheme 2014]
SECTION — E
This section consists of questions asked from OTBA which are provided by schools only.
29.(3 Marks)
30.(3 Marks)
31.(4 Marks)
nnn
SOLUTIONS
QUESTION PAPER - 10
Self Assessment __________________________________
Time : 3 Hours
Maximum Marks : 90
SECTION — A
1.
Volume of right circular cone =
1 2
pr h
3
1 22
= ×
× ( 6 )2 × 7
3 7
1 22
= ×
× 36 × 7
3 7
= 264 cm3
1
2.Let radii of cylinders be 2x and 3x and heights be 5y and 3y respectively.
∴
Ratio of volumes =
π(2 x )2 × 5 y
π(3x )2 × 3 y
4x2 × 5
=
2
9 x × 3
= 20 : 27
3. Given
1
2
½
ar (BCP) = 15 cm ar (ABCD) = 2 × ar (BCP)
= 2 × 15
= 30 cm2.
4.
½
Range =Highest value – lowest value = 32 – 6 = 26.
SECTION — B
5.
∠ACB = 70°
∠ADB= ∠ACB
⇒
∠ADB = 70° (Angles in the same segment of a circle) ½
1
[CBSE Marking Scheme 2014]
MATHEMATICS Oswaal CBSE Class -9, SA-2 Examination Sample Question Paper
SAMPLE
In ∆DAB,
∠DAB + ∠ADB + ∠DBA = 180° ⇒
60° + 70° + ∠DBA = 180°
⇒
∠DBA = 50°.
(Angle sum property of triangles) 1
½
6. Given,
Radius of cone (r) = 2.1 cm
Height of cone (h) = 8.4 cm
Let the radius of the sphere be R.
\
Volume of cone = Volume of sphere
⇒
1 2
4
pr h = pR 3
3
3
⇒
r2h = 4R3
⇒R3=
r2 h
4
⇒R3=
(2.1)2 (8.4 )
4
1
R3 = (2.1)2 × (2.1)
⇒R3 = (2.1)3
⇒
R = 2.1 cm.
1
7.Arranging the data in ascending order
0.02, 0.03, 0.03, 0.04, 0.05, 0.05, 0.05, 0.07, 0.08, 0.08, 1.00, 1.03, 1.04
Mode = 0.05
1
Range = 1.04 – 0.02 = 1.02.
1
8. Observation : 3, 5, 7, 4, 7, 8, 3, 6, 7, 4, 7, 3
1
Mode of observation =7
After adding,
New observation : 8, 10, 12, 9, 12, 13, 8, 11, 12, 9, 12, 8
1
New mode =12
9. Let the probability of winnig a game = p
and probability of lossing a game = q
We know that
According to question,
p + q = 1
p= 2 q −
...(i)
1
3
⇒6q – 3p = 1
1
...(ii)
On solving (i) and (ii), we get
∴
q=
Probability of winning a game =
4
5
and p =
9
9
5
9
1
Solutions (SA-2) | 39
10.
C
D
B
A
ABCD is a parallelogram and opposite angles of it are equal.
∴
1
∠B= ∠D
∠B + ∠D = 180°
But
(Opp. angles of a cyclic quadrilateral)
⇒
2∠D = 180°
⇒
∠D = 90°
⇒ ABCD is a rectangle.
1
SECTION — C
11.
Vol. of cuboid = lbh
= 24 × 18 × 4
= 1728 cu.cm.
Edge of a cube =
3
1728 = 12 cm
LSA = 4a2
= 4 × 12 × 22 = 576 sq. cm.
[CBSE Marking Scheme, 2013]
Detailed Solution :
Let x be the edge of cube.
According to the question, we have
Þ
Volume of cube = Volume of cuboid
Þ
x3 = 24 × 18 × 4
Þ
x3 = 1728
Þ
x = 12
\
\
Edge of the cube = 12 cm
2
2
Lateral surface area of cube = 4x
= 4(12)2 = 576 cm3.
12. I cone
1
II cone
r = radius
r’ = radius
½
l = slant height
l’ = slant height
½
h = height
h’ = height
C.S.A. = πrl
C.S.A. = πr’l’
½
Given : As per Question :
C.S.A of first cone = 2 × C.S.A. of second one
πrl = 2πr’l’
πrl =2πr’2l[l’ = 2l, Given] ½
r
2 p ´ 2l 4
=
=
r ' pl
1
r : r’ = 4 : 1.
½
1
40 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
13.Given
7x – 3y = 15
⇒
15 + 3 y
7
x=
½
x = 0;
½
\
7(0) – 3y = 15
½
⇒
0 – 3y = 15
½
At y-axis \
14.
15
= – 5
−3
y=
½
Construction : Draw OA and O'B perpendicular to CD from O and O' respectively.
O'
O
C
A
P
B
D
OA ⊥ CD
Proof : ∴ OA bisects the chord CP (perpendicular from the centre to the chord bisects the chord)
∴ AP=
or CP = 2AP
Similarly O'B ⊥ PD
∴ BP=
1
CP
2
...(i)
1
PD
2
or
PD= 2BP...(ii)
CD = CP + DP
= 2AP + 2BP
= 2(AB)
...(iii)
1
OA = O'B
(two lines ⊥ to same line I)
AB = OO'
∴ ABO'O is  gm (Given)
AB = OO'
(from (i) and (ii))
In quadrilateral ABO'O
1
(opp. sides of parallelogram are equal)
∴ CD = 2AB = 2OO'
1
[CBSE Marking Scheme, 2015]
15. 3x + 4y = 7
x
1
–3
y
1
4
3x – 2y = 1
x
1
–1
y
1
–2
Solutions (SA-2) | 41
y
3x – 2y = 1
(1, 1)
3x
x
+
4y
=
7
½+½+1+1
From the graph, point of intersection is (1, 1).
16. Arranging data in ascending order :
8 , 13, 17, 20, 24, 24, 24, 26, 26, 30, 41
Here, \
n =11 (odd)
 n+1
Median=

 2 
th
 11 + 1 
=

 2 
term
1
th
term
=6th term
\
Median = 24
Mode = 24
(maximum frequency) 1
If we replace one 24 by 26, the given data will be (in ascending order) :
8, 13, 17, 20, 24, 24, 26, 26, 26, 26, 30, 41
Again
\
n = 11 (odd)
Median = 6th term
= 24
and
Mode = 26
(maximum frequency) 1
17. (a) Probability (non occurrence of exactly 2 heads)
=
216 + 270 + 130 616
=
= 0·616
1000
100
1
(b) Probability (3 heads) =
216
= 0·216
1000
1
(c) Probability (no head) =
130
= 0·13
1000
1
18. (i) P (weight less than 65 kg) =
(ii) P (weight between 61 and 64) =
(iii) P (weight equal to or more than 64) =
5 + 18 + 4 + 16 + 5 48 4
=
=
60
60 5
4 + 16 20 1
=
=
60
60 3
5 + 12 17
=
60
60
(1 + 1 + 1)
42 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
SECTION — D
19.
l = 140 cm
b = 110 cm
h = 80 cm
Surface area of open box = lb + 2(bh + hl)
1
1
75
[154 + 2(88 + 112)] × 100
100 × 100
Cost of painting box = `
3
= [ 554] = 3(138·5)
4
= ` 415·5
2
[CBSE Marking Scheme, 2014]
20. Constructions : (i) Draw a line segment AB = 13 cm
P
2
60°
30°
30°
60°
Q
A
R
B
(ii)At A construct an angle at 30° and at B an angle 60°.
(iii)Draw bisectors of ∠A and ∠B.
(iv)Bisectors meet at P.
(v) Draw perpendicular bisector of PA and PB which intersect AB at Q and R respectively.
(vi)Join P to Q and R.
(vii)PQR is the required triangle.
2
A
21.
F
B
G
D
E
C
AD is the median of ∆ABC
∴
½
ar (∆ABD) = ar (∆ACD)...(i)
In ∆GBC, GD is the median
∴
ar (∆GBD) = ar (∆GCD)...(ii) ½
From (i) and (ii),
ar (∆ABD) – ar (∆GBD) = ar (∆ACD) – ar (∆GCD)
⇒
ar (∆AGB) = ar (∆AGC)...(iii) ½
Similarly we can prove that
ar (∆AGB) = ar (∆BGC)...(iv) ½
Form (iii) and (iv),
ar (∆AGB) = ar (∆BGC) = ar (∆AGC) ...(v) ½
Solutions (SA-2) | 43
½
Now ar (∆ABC)= ar (∆AGB + ar (∆BGC) + ar (∆AGC)
= ar (∆AGB) + ar (∆AGD) + ar (∆AGD)
= 3ar (∆AGB),
1
ar (∆ABC),
3
Hence
ar (∆AGB)=
Hence ar (∆AGB) = ar (∆AGC) = ar (∆BGC)
1
= ar (∆ABC)
3
1
[CBSE Marking Scheme, 2012]
22. Through O, draw AB  PS.
Also 1
PA  BS
PABS is a parallelogram.
1
ar(PABS)
2
ar (POS)=
A
P
Q
O
S
R
B
(Triangle and a parallelogram are on the same base and between the same parallels) Similarly,
ar (QOR)=
1
2
23.
=
1
ar (PQRS).
2
1
3x + 4 = 5x + 8
2x = – 4
⇒ x = – 2
⇒ 1
ar (POS) + ar (QOR) = [ar (PABS) + ar (QABR)]
∴ 1
ar (QABR)
2
1
1
(i)On the number line the point P(– 2, 0) represent the solution
p
–3 –2 –1
0
1
2
3
1
x=–2
(ii) On the cartesian plane x = – 2 is a line x = –2 is represented as
It has infinite solutions.
y
x'
–3 –2
(–2, 2)
1
2
1
(–2, 0)
–1
0 1
–1
2
3
x
–2
y'
1
44 | OSWAAL CBSE (CCE) , Mathematics, Class – 9
24. Given : AB is diameter of the circle with centre O.
CD is equal to radius.
To prove : ∠APB = 60°,
Construction : Join OC, OD and BC.
Proof : D ODC is equilateral
∴
∠COD = 60°
1
∠CBD = ∠COD
2
½
1
(angle subtended by an arc at the centre is double the angle subtended by it at any point on the
remaining part of the circle)
P
D
C
A
⇒
Now In DCPB, ⇒
25.
⇒
B
O
∠CBD =30°
∠ACB = 90° (angle in the semi-circle)
∠BCP =180° − ∠ACB = 90°
∠CPB =90° − 30° = 60°
∠APB = 60°. ∠AEB =90° = ∠AED (semi-circle)
½
½
½
½
½
C
P
D
Q
∠EAC + ∠ACD + ∠CDE + ∠AED = 360° (sum of angles in Qua.)
∠EAC = 360° – 90° – 90° – 90°
1
1
A
E
= 90°
Each angle = 90°
O
1
EACD is rectangle.
26. (i) B
AC = ED.
1
Volume of soft drink in tin can = Volume of cuboid
= lbh = 5 × 4 × 15 1
3
= 300 cm
Volume of soft drink in cylinderical can
=πr2h =
22 7 7
× × × 10 1
7 2 2
= 385 cm3
Volume of soft drink is greater in cylinderical can than tin can by
= (385 − 300) = 85 cm3.
1
(ii) Mensuration ½
(iii) Honesty ½
27.
Volume of bigger cube =(side)3
3
½
3
3
= (12) cm = 12 × 12 × 12 = 1728 cm ½
Solutions (SA-2) | 45
∴
∴
Volume of one small cube =
1728
= 216 cm3½
8
Side of one small cube = 3 Volume ½
= 3 216 = 6 cm
Surface area of one small cube = 6 × side2
= 6 × 6 × 6
½
Sum of surface area of 8 smaller cubes = 8 × 6 × 6 × 6
= 1728 cm2
Surface area of bigger cube = 6 × 12 × 12
= 864 cm2
∴
½
½
Required ratio = 1728 : 864
= 2 : 1.
½
Probability of less than 30 marks=
7 + 10
17
=
90
90 1
(ii) Probability of marks 60 or more marks =
15 + 8
23
=
90
90 1
(iii)Probability of marks between 40 and 70 =
20 + 20 + 15
55
=
90
90 1
8
4
=
90
45 1
28. (i) (iv)
Probability of marks 70 or more =
SECTION — E
This section consists of questions asked from OTBA which are provided by schools only.
29.(3 Marks)
30.(3 Marks)
31.(4 Marks)
nnn