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PIONEER GUESS PAPER
th
9 CBSE (SA-II)
MATHEMATICS
“Solutions”
TIME: 2:30 HOURS
MAX. MARKS: 80
GENERAL INSTRUCTIONS & MARKING SCHEME
All questions are compulsory.
The questions paper consists of 34 questions divided into four sections A, B, C and D.
Section A contains 10 questions of 1 mark each, which are multiple choice type questions,
Section B contains 8 questions of 2 mark each, Section C contains 10 questions of 3 marks each, Section D contains
6 questions of 4 marks each.
•
There is no overall choice in the paper. However, internal choice is provided in one question of 2 marks, 3
questions of 3 marks and two questions of 4 marks.
•
Use of calculators is not permitted.
•
•
•
NAME OF THE CANDIDATE
PHONE NUMBER
L.K. Gupta (Mathematics Classes)
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Section−
−A
1.
x = 5, y = 2 is a solution of the linear equation
(A) x + 2 y = 7
(B)
5x + 2y = 7
(C)
x+y=7
(D)
5x+y=7
Sol: (C)
2.
The equation x = 7, in two variables, can be written as
(A)1 . x + 1. y = 7
(B) 1. x + 0. y = 7
(C) 0. x + 1. y = 7 (D) 0. x + 0. y = 7
Sol: (B)
3.
Three angles of a quadrilateral are 750, 900 and 750. The fourth angle is
(A) 900
(B) 950
(C) 1050
(D) 1200
Sol: (D)
4.
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order,
is
(A) a rhombus
(B) a rectangle
(C) a square
(D) any parallelogram
Sol: (B)
5.
ABCD is a trapezium with parallel with sides AB = a cm & DC = b cm. E & F are the mid-
points of the non-parallel sides. The ratio of ar (ABFE) & ar (EFCD) is
(A) a : b
(B) (a+3b) : (3a+b)
(C) (3a+b) : (a+3b)
(D) (2a+b) : (3a+b)
Sol: (C)
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1
h
×
+
+
×
a
b
2l
(
)
ar ( ABFE ) 2
2 = a + b + 2l ….. (1)
=
ar ( EFCD) 1 × b + b + 2l × h
2b + 2l
(
)
2
2
As ∆ AOD and ∆ EMD are similar
l h/2
=
⇒ x = 2l
x
h
⇒
Put in (1)
a−b
a+b+
ar ( ABFE ) a + b + x
2  3a + b

=
=
=
ar ( EFCD)
2b + x
 a − b  a + 3b
2b + 

 2 
6.
[ ∵ a = b + 2x ⇒ x =
a−b
]
2
Diagonals of a parallelogram ABCD intersect at O. If
∠ BOC = 900 and ∠ BDC = 500 , then ∠ OAB is
(A)
900
(B)
500
(C)
400
(D)100
Sol: (C)
7.
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of
AB from the centre of the circle is :
(A) 17cm
(B) 15cm
(C) 4cm
(D) 8cm
Sol: (D)
8.
With the help of a ruler and a compass, it is possible to construct an angle of :
(A)350
(B)400
(C)37.50
(D)47.50
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Sol: (C)
9.
The radius of a sphere is 2r, then its volume will be
(A) 4 /3 πr3
(B) 4πr3
(C) 8πr3 / 3
(D) 32/ 3 πr3
(C)140
(D)145
Sol: (D)
10. The class-mark of the class 130−150 is :
(A)130
(B)135
Sol: (C)
SECTION B
Question numbers 11 to 18 carry 2 marks each.
11. Determine the point on the graph of the equation 2x + 5y = 20 whose x-coordinate is
5
2
times its ordinate.
Sol:
As the x-coordinate of the point is
5
5
times its ordinate, therefore, x = y .
2
2
5
Now, putting x = y in 2x + 5y = 20, we get, y = 2. Therefore, x = 5. Thus, the required point is
2
(5, 2).
12. Three angles of a quadrilateral ABCD are equal. Is it a parallelogram?
Sol:
It need not be a parallelogram, because we may have ∠ A = ∠ B = ∠ C = 800 and
∠ D = 1200 . Here, ∠ B ≠ ∠ D .
13. ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (Fig.). If AQ
intersects DC at P, show that ar (BPC) = ar (DPQ)
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Sol:
ar ( ACP ) = ar ( BCP )
(1)
[Triangles on the same base and between same parallels]
ar ( ADQ ) = ar ( ADC )
(2)
ar ( ADC ) − ar ( ADP ) = ar ( ADQ ) − ar ( ADP )
ar ( APC ) = ar ( DPQ )
(3)
From (1) and (3), we get
ar ( BCP ) = ar ( DPQ )
14. In the Fig., OD is perpendicular to the chord AB of a circle whose centre is O.
If BC is a diameter, show that CA = 2OD.
Sol:
Since OD ⊥ AB and the perpendicular drawn from the centre to a chord bisects the chord.
∴ D is the mid-point of AB
Also, O being the centre, is the mid-point of BC.
Thus, in ∆ ABC,D and O are mid-points of AB and BC respectively. Therefore,
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OD AC
and OD =
1
CA
2
∴ Segment joining the mid − point s of two 
sides of a triangle is half of the third side 


⇒ CA = 2OD
15. Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line
segment.
Sol:
Steps of Construction:
(i) Draw a line segment AB = 5.8 cm.
(ii) With A is centre and radius more than half of AB, draw arcs, one on each side of AB.
(iii) From B (with the same radius) cut both arcs at P and Q.
(iv) Join PQ to meet AB at C.
(v) AC= BC = 2.9cm.
⇒ C bisects AB.
16. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the
curved surface of the pillar at the rate of 12.50 per m2.
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Sol:
Radius = 25 cm =
25
m
100
Height = 3.5m
C.S.A. = 2π rh = 2 ×
22 25 35 11
×
×
=
m
7 100 10 2
Cost = 12.50 per m2
11
11 1250 275
m=
×
=
= Rs.68.75
2
2 100
4
17. A soft drink is available in two packs –
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container
has greater capacity and by how much?
Sol:
Volume of rectangular tin can = l × b × h
= 5 × 4 × 15 = 300 cm2
Volume of plastic cylinder π r2h
=
22 7 7
× × × 10 = 385cm3
7 2 2
The plastic cylinder has greater capacity by 85 cm3 [=385–300]
18. Calculate the mean for the following distribution:
x:
5
6
7
8
9
f:
4
8
14
11
3
Sol:
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xf
x:
f:
5
4
20
6
8
48
7
14
98
8
11
88
9
3
27
Total
40
281
∑ f = 40
∑ x f = 281
∑ x f = 281 = 7.025
x=
∑ f 40
i
i i
i i
i
SECTION−
−C
Question numbers 19 to 28 carry 3 marks each.
19. Draw a graph of each of the following equations :
(i) x + 2 = 0
(ii) 2y + 3 = 0
(iii)4x − 6 = 0
Sol:
(i) x + 2 = 0 ⇒x = −2
Clearly, it does not contain y. So, its graph is a line parallel to y-axis passing through the point
(−2, 0). In fact it passes through every point whose x-coordinate is −2.
Plotting any two points, say (−2, 0) and (2, 1) given by above table on the graph paper and
joining them, we obtain the straight line as shown.
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(ii) 2y + 3 = 0⇒ 2y=−3
3
⇒y = − =−1.5
2
The equation y=−1.5 does not contain x. So, its graph is a line parallel to x-axis passing through
the point (0, −1.5). Clearly, y=−1.5 means that for all values of the abscissa x, the ordinate y is
−1.5
Thus, we have the following table exhibiting the abscissa and ordinate of the points lying on
the line represented by given equations.
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Plotting the points (0, −1.5), (1, −1.5) and (2, −1.5) on the graph paper and joining them by a
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line, we obtain the graph of the line represented by the given equation as shown in graph.
(iii) 4x – 6 = 0
⇒4x = 6
⇒x =
6
= 1.5
4
20. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.
Sol:
If point (3, 4) lies on the graph of 3y = ax + 7, then it must satisfy it.
⇒ 3 ( 4 ) = a (3) + 7
⇒ 12 = 3a + 7 ⇒ 3a = 5 ⇒ a =
5
3
21. In a quadrilateral ABCD, CO and DO are the bisectors of ∠ C and ∠ D respectively.
Prove that ∠ COD =
1
(∠ A + ∠ B) .
2
Sol:
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Given: Quadrilateral ABCD , CO and DO are the bisectors ∠C and ∠D.
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Proof: In ∆ COD
o
∠COD = 180 – ( ∠ OCD + ∠ ODC)
1
o
∠COD = 180 − (∠ C + ∠ D)
2
1
o
∠COD = 180 − [360o − [(∠ A + ∠ B )]
2
1
o
o
∠COD = 180 – 180 + (∠ A + ∠ B )
2
1
∠ COD = (∠ A + ∠ B ) Hence proved.
2
22. If ABCD is a parallelogram, the prove that
ar ( ∆ ABD ) = ar ( ∆ BCD ) = ar ( ∆ ABC ) = ar ( ∆ ACD ) =
1
ar gm ABCD
2
(
)
Sol:
1
To prove: ar ΔABD = ar(ΔBCD) = ar(ΔABC) = ar(ΔACD) = ar(gm ABCD)
2
Constructions: Draw DP ⊥ AB, BQ ⊥ CD.
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Proof:
1

In ∆ABD, ar ΔABD =  .AB.DP  …. (1)
2

1

In ∆BCD, ar ΔBCD =  DC.BQ 
2

….. (2)
As AB = CD, PD = BQ = distance between parallel line.
So (1) and (2) are equal.
∴ ar ΔABD = ar ΔBCD …. (3)
Similarly ar ∆ABC = ar ∆CDA
ar || ABCD = ar ΔABD + ar ΔBCD (but they are equal)
ar || ABCD = 2 ar ΔABD
1
ar || ABCD = ar ΔABD …. (4)
2
From (3) and (4)
1
ar ΔABD = ar(ΔBCD) = ar(ΔABC) = ar(ΔACD) = ar(gm ABCD)
2
Hence proved.
23. Find the length of a chord which is at a distance of 5 cm from the centre of a circle of
radius 10 cm.
Sol:
Let AB be the chord and O is the centre of the circle
from O draw OL ⊥ AB
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Join OA. Since , perpendicular from the centre of the circle bisects the chord.
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1
∴ AL = LB = AB
2
In ΔOAL ,
OA2 = OL2 + AL2
⇒ (10)2 = (5)2 + AL2
⇒ AL2 = 100 − 25 = 75
⇒ AL = 75 = 3 × 5 × 5 = 5 3 cm
So AB = 2 × AL = 2 × 5 3 = 10 3 cm
24. How many liters of water flow out of a pipe having an area of cross-section of 5 cm2 in
one minute, if the speed of water in the pipe is 30 cm/sec?
Sol:
Speed of water in the pipe = 30 cm / sec
volume of flowing water in 1 second = π r2h = 5 × 30 = 150cm3
volume of water flowing in 1 minute
= 150 × 60 = 9000 cm3 =9L
[ 1cm3 = 1000L ]
25. A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all
around it to a width of 7.5 m to form an embankment. Find the height of the embankment.
Sol:
Volume of earth dug out = π r2h =
22
84
× 5 × 5×
= 660 m3
7
10
Area of embankment= π(R 2 − r2 )
=
22
2
2
× (12.5) − (5)
7
(
)
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=
22
× (156.25 − 25)
7
=
22 13125
×
7
100
=
5775
= 412.5 m2
14
Height =
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volume
660 660 × 10
=
=
= 1.6 m
Area
412.5
4125
26. The ages of ten students of a group are given below. The age have been recorded in
years and months:
8 – 6, 9 – 0, 8 – 4, 9 – 3, 7 – 8, 8 – 11, 8 – 7, 9 – 2, 7 – 10, 8 – 8
(i) What is the lowest age?
(ii) What is the highest age?
(iii) Determine the range?
Sol:
(i) 7 years 8 months
(ii) 9 years 3 months
(iii) Range =Max − Min= 9 years 3 months –7 years 8 months = 1 year 7 months
27. Following are the ages of 360 patients getting medical treatment in a hospital on a day:
Construct a Cumulative frequency distribution
Age (in years):
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
No. of Patients :
90
50
60
80
50
30
Sol:
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Marks
No of students
Cumulative Frequency
10–20
90
90
20– 30
50
140
30– 40
60
200
40–50
80
280
50–60
50
330
60–70
30
360
28. The percentage of marks obtained by a student in monthly unit tests are given below :
Unit test :
I
II
III
IV
V
Percentage of marks obtained:
69
71
73
68
76
Find the probability that the student gets:
(i) more than 70% marks
(ii) less than 70% marks
(iii) a distinction.
Sol:
(i) P(more than 70% marks)
= P(A) =
Total no.of percentagesabove70% 3
= = 0.6
Total numberof pecentages
5
(ii) P(less than 70% marks) = P(A) =
Total no.of percentagesbelow70% 2
= = 0.4
Total numberof pecentages
5
1
(iii) A distinction = = 0.2
5
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Section−
−D
Question numbers 29 to 34 carry 4 marks each
29. Draw the graphs of the lines represented by the equations x + y = 4 and 2x −y = 2 in the
same graph. Also, find the coordinates of the point where the two lines intersect.
So, the common point is (2, 2) which is the point of intersection of both lines.
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30. ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that
AE = BF = CG = DH. Prove that EFGH is a square.
Sol:
Given ABCD is a square ,E,F,G,H are the mid points of AB , BC ,CD and DA
AE = BF = C G = DH
To prove : - EFGH is a square .
Proof : AE = BF = CG = DH = x
∴ BE = CF = DG = AH = y
In ΔAEHandΔBFE
A E = BF (given) , ∠A = ∠B = [ each 90 o , AH = BF [ given]
By SAS ≅ , triangles are ≅
ΔAEH ≅ ∆ BFE
∠1 = ∠2 and ∠3 = ∠4
But, ∠1 + ∠3 = 90 o
and ∠2 + ∠ 4 = 90 o
∠1 + ∠3 + ∠2 + ∠4 = 90 o + 90 o
∠1 + ∠4 +∠1 + ∠4 = 180 o
2 ( ∠1 + ∠4) = 180 o
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∠1 + ∠4 = 90
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o
∠HEF = 90 o
Similarly , ∠F = ∠G = ∠H = 90 o
Hence, EFGH is a square
31. In a ∆ ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that:
(i) ar ( ∆ PBQ ) = ar ( ∆ ARC )
(ii) ar ( ∆ PRQ ) =
1
ar ( ∆ ARC )
2
(iii) ar ( ∆ RQC ) =
3
ar ( ∆ ABC ) .
8
Sol:
Given : P and Q are the mid points of AB and BC R is the midpoint of AP
To prove: (i) ar (PBQ ) = ar ( ∆ ARC ) (ii) ar ( ∆ PRQ ) =
(iii) ar ( ∆ RQC ) =
1
ar(ARC)
2
3
ar( ∆ ABC)
8
Proof : We know that each Median of a triangle divides it into 2
triangles of equal area
(i) Since CR is the median of ∆ CAP
1
∴ar( ∆ CRA) = ar( ∆ CAP) ..(i)
2
also , CP is a median of of ∆ CAB
∴ ar( ∆ CAP) = ar(CPB)..(ii)
from (i) and (ii) we get
1
∴ar( ∆ ARC) = ar( ∆ CPB)..(iii)
2
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PQ is the median of ∆ PBC
∴ar( ∆ CPB) = 2ar( ∆ PBQ)..(iv)
from (iii) and (iv) we get
∴ar( ∆ ARC) = ar( ∆ PBQ)..(v)
(ii) Since QP and QR medians of triangles QAB and QAP
∴ar( ∆ QAP) = ar( ∆ QBP)..(vi)
and ar ∆(QAP) = 2ar( ∆ QRP) ..(vii)
from (vi) and (viii) we get
1
ar ( ∆ PRQ = ar( ∆ PBQ) ...(viii)
2
1
ar ( ∆ PRQ = ar( ∆ ARC) …(viii)
2
(iii) since CR is a median of ∆ CAP
1
∴ar( ∆ ARC) = ar( ∆ CAP)
2
=
1 1

 ar(ABC)[∵CPis themedianof ∆ ABC]
2 2

=
1
ar( ∆ ABC) ...(ix)
4
Since RQ is median of ∆ RBC
1
∴ar( ∆ RQC) = ar( ∆ RBC)
2
=
1
{ar(ABC) − ar( ∆ ARC)}
2
=
1
1
{ar( ∆ ABC) − ar[∆ ABC}[from(ix)]
2
4
=
3
ar( ∆ ABC)
8
Hence proved.
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32. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F
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respectively. Prove that the angles of ∆ DEF are 900 −
A
B
C
, 900 − and 900 − .
2
2
2
Sol:
We have,
∠ BED =∠ BAD
⇒ ∠ BED =
∠A
2
Also, ∠ BEF = ∠ BCF
1
⇒ ∠BEF = ∠ C
2
1
1
∴ ∠ BED + ∠ BEF = ∠A + ∠ C
2
2
⇒ ∠ DEF =
1
( ∠ A + ∠C)
2
⇒ ∠ DEF =
1
1800 − ∠ B
2
(
)
∴ ∠ A + ∠ B + ∠ C = 1800 
1
⇒ ∠ DEF = 900 − ∠B
2
We observe that AF subtends angles ∠ ADF and ∠ ACF at points D and C in the same segment.
∴ ∠ ADF = ∠ ACF
⇒ ∠ ADF =
1
∠C
2
Also,
∠ ADE = ∠ ABE
⇒ ∠ ADE =
1
∠B
2
1
1
∴ ∠ADE + ∠ ADF = ∠ B + ∠ C
2
2
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⇒ ∠EDF =
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1
( ∠ B + ∠C)
2
⇒ ∠ EDF =
1
1
(180 − ∠ A ) = 900 − ∠ A
2
2
Similarly, we have
1
∠ DEF = 900 − ∠C
2
33. A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is
105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m
wide to make the required tent.
Sol:
Slant height of cone = 53m
Height of cylinder = 3m
Diameter = 105 m
Radius =
105
= 52.5
2
C.S.A. of conical portion = πrl
C.S.A. of cylindrical portion = 2πrh
Total C.S.A of circus tent = πrl + 2πrh
= πr(l + 2h)
22
= × 52.5(53 + 2(3))
7
22
= × 52.5(59)
7
= 9735m2
Area of canvas = l × b
So, 9735 = l × 5
⇒
9735
=l
5
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So, length of the canvas =1947m .
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34. The time taken, in seconds, to solve a problem by each of 25 pupils is as follows:
16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20
(A) Construct a frequency distribution for these data using a class interval of 10 seconds.
(B) Draw a histogram to represent the frequency distribution.
Sol:
(a) Frequency distribution
(b)
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We represent the class limit along X-axis on a suitable scale and the frequencies along y-axis
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on a suitable scale.
Since the scale on X-axis starts at 15, a kink (break) is indicated near the origin to signify that
the graph is drawn to scale beginning at 15, and not at the origin.
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