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MODEL QUESTION PAPER
SA-II
CLASS-X
SUBJECT- MATHS
TIME:3 to 3.1/2 hr
MM: 80
General Instructions:
iv.
v.
All questions are compulsory.
The question paper consists of 34 questions divided into four section- A,B,C and D.
Section A contains 10 question of 1 mark each which are multiple choice type questions. Section B
contains 8 question of 2 marks each, Section C contains 10 question of 3 marks each and Section D
contains 6 question of 4 marks each.
There is no overall choice in the paper. However, internal choice is provided in one question of 2 marks ,
three questions of 3 marks and two questions of 4 marks.
Use of calculator is not permitted.
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i.
ii.
iii.
SECTION A
Question Numbers1 to 10 carries 1 mark each. For each of the question number 1to 10, four alternative choice have
been provided, of which only one is correct. Select the correct choice.
The perimeter (in cm) of square circumscribing a circle of radius a cm, is
A.
B.
C.
D.
3.
4.
5.
13
26
169
238
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If A and B are the points (-6,7) and (-1,-5) respectively, then the distance 2 AB is equal to
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2.
8a
4a
2a
16 a
.s
t
A.
B.
C.
D.
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1.
If the common difference of an A.P is 3, then
is.
A. 5
B. 3
C. 15
D. 20
If P ( a/2,4) is the midpoint of the line segment joining the points A(-6,5) and B(-2,3), then the value of
a is .
A. -8
B. 3
C. -4
D. 4
In Figure 1 .○ O is the centre of a circle, PQ is a chord and PT is the tangent at P. If <PQR=700, then <TPQ
is equal to
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68
Figure 1
550
700
450
350
A.
B.
C.
D.
In Figure 2, AB and AC are tangent to the circle with centre O such that <BAC =400 Then <BOC is equal to
B
A
400
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C
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6.
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The roots of the equation x2-3x-m(m+3)=0, where m is a constant are
w
7.
400
500
1400
1500
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Figure- 2
A.
B.
C.
D.
A. m, m+3
B. –m, m+3
C. m, -(m+3)
D. –m , -(m+3)
8.
9.
A tower stands vertically on the ground. From a point on the ground which is 25m away from the foot of
the tower, the angle of elevation of the top of the tower is found to be 450. Then the height (in meters)
of the tower is
A. 25√2
B. 25√3
C. 25
D. 12.5
The surface area of solid hemisphere of radius r cm (in cm2) is
A.
B. 3
C. 4
D.
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69
10.
A card is drawn from a well-shuffled deck of 52 playing cards. The probability that the card is not a red
king is
A.
B.
C.
D.
SECTION – B
Question Numbers 11 to 18 carry 2 marks each.
Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting
cuboid.
OR
A cone of height 20 cm and radius of base 5 cm is made up of modeling clay. A child reshapes it in the form of a
sphere. Find the diameter of the sphere.
12.
Find the value of y for which the distance between the points A(3,-1) and B(11,y) is 10 units.
13.
A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the
selected ticket has a number which is a multiple of 5.
14.
Find the value of m so that quadratic equation mx(x-7) + 49 =0 has two equal roots.
15.
In figure 3 , a circle touches all the four side of a quadrilateral ABCD whose sides are AB=6 cm, BC=9cm and
CD=8 cm. Find the length of side AD.
D
C
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11.
A
B
Figure-3
16.
Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that
17.
Which term of the A.P 3,14,25,36,… will be 99 more that its 25th term?
18.
=
.
In Figure 4, a semi circle is drawn with O as center and AB as diameter. Semi-circle are drawn with AO and OB as
diameters. If AB=28 m, Find the perimeter of the shaded region. [Use
Figure- 4
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70
SECTION-C
Question Numbers 19 to 28 carry 3 marks each.
19.
20.
21.
Draw a right triangle ABC in which the sides (other than hypotenuse) are of length 4 cm and 3 cm. Then
construct another triangle whose sides are 3/5 times the corresponding sides of the given triangle.
Solve the following quadratic equation for x. p2x2-(p2- q 2) x –q2 =0
OR,
Find the value of ‘k’ so that the quadratic equation 2kx 2- 40x +25 = 0 has real and equal roots.
Find the sum of first 15 terms of an AP whose nth term is given by 9 -5n..
OR
Find the sum of all 3digit numbers which leaves the remainder 3 when divided by 5.
A bag contains 4 red balls, 5 black ball and 6 white balls. A ball is drawn from the bag at random. Find the
probability that the ball drawn is:
a. red
b. white
c. not black
d. red or black
23.
If the area of ∆ ABC whose vertices are (2,4),(5,k) and (3,10) is 15 sq units, find the value of ‘k’
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22.
OR
Find relation between x and y such that the point(x,y) is equidistant from the points(3,6) and (-3,4).
A kite is flying at height of 60m above the ground. The string attached to the kite is temporarily tied to a point
on the ground. The inclination of the string to the ground is 600. Find the length of the string assuming that
there is no slack in the string.
25.
Three spherical metal ball of radii 6cm,8 cm and x cm are melted and recasted into single sphere of radius 12
cm. Find the radius of 3rd sphere.
26.
XY and X’Y’ are two parallel tangents to a circle with center O and another tangent AB with point of contact C,
intersecting XY at A and X’Y’ at B, is drawn. Prove that AOB=900
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24.
27. ABCD is Square Park of side 10 m. It is divided into 4 equal squares parks. Semi circular flower beds are made as
shown by the shaded region. Find the area of shaded region.
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71
28.
Two vertices of a triangle ABC are A(-7,6) and B(8,5). If the midpoint of the side AC is at 5/2,2 , Find the coordinate of vertex C.
SECTION-D
29.
30.
.
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QUESTION NUMBERS 29 TO 34 CARRY 4 MARKS EACH
Prove that the length of tangents drawn from an external point to a circle are equal.
A shopkeeper buys a number of books for RS 1200. If he had bought 10 more books for the same amount, each
book would have cost him Rs.20 less. How many books did he buy?
OR,
A passenger train take one hour less for a journey of 360 km if its speed is increased by 5 km/ hr. from its usual
speed. Calculate the usual speed of the train.
A man standing on the deck of a ship which is 10m above the water level, observes the angle of elevation of the
top of a hill as 600 and angle of depression of the base of the hill as 300. Calculate the distance of the hill from
ship and the height of the hill.
32.
A chord of a circle of diameter 24 cm subtends an angle 600 at the center of the circle. Find the area of the
corresponding.
(a)
Minor sector
(b)
Minor segment
33.
How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid of dimensions 11 cm x 10
cm x 7 cm.
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31.
OR
If the radii of the circular ends of a bucket in the shape of frustum of a cone which is45 cm high are 28 cm and
7cm , find the capacity of the bucket.
34.
The sum of first 15 terms of an AP is 105 and the sum of the next 15 terms is 780. Find the first 3 terms of the
arithmetic progression.
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72
SA-II
MARKING SCHEME
CLASS-X (MATHS)
EXPECTED ANSWERS/VALUE POINTS
SECTION-A
Q . No
Marks
2. (B):26
3.(C):15
4. (A) : -8
5. (D):350
7. (B): -m, m+3
8. (C):25
1X10 =10
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1.(A):8a
6. (C):1400
9. (B): 3
10.(D)
Getting side of cube=3 cm
1/2m
1/2m
.s
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11.
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SECTION-B
w
= 2(6x3+3x3+3x6) cm2
=90 cm2
w
 Surface area
w
Dimensions of the resulting cuboids are 6cm x 3cm x 3cm
Volume of cone =1/3
 4/3
r2=1/3
1m
OR
(5)2.20cm2
(5)2.20
½m
r=5cm
1m
Diameter=10 cm
12.
13.
AB=
1/2m
=
1/2m
=> 65+y2+2y=100 or y2+2y-35=0
 y=-7 or y=5
1/2m
1m
Total number of tickets =40
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73
Number of ticket with number divisible by 5=8
 Required probability =8/40 or 1/5
1m
1m
mx(x-7)+49=0 => mx2-7mx+49=0
14.
For equal roots B2-4AC=0
½m
½m
(-7m) -4(m)(49)=0
2
=>m=0 or m=4
½m
But m≠0 m=4
½m
Let P,Q ,R,S be the points of contact
D
 AP=AS
PB=BQ
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15.
R
C
DR=DS
S
Q
CR=CQ
A
p
B
½m
ie
 (AP+PB)+(DR+CR)=(AS+DS)+(BQ+CQ)
½m
ud
AB+CD=AD+BC
w
For writing (or using AP:PB=3:2
17.
For correct construction
1½m
w
16.
1m
w
.s
t
6+8=AD+9=>AD=5CM
a25=3+24x11=3+264=267
1/2m
267+99=366=3+(n-1)11
1m
n=34
18.
1/2m
Perimeter( of shaded region)=circumference of semicircle of radius 14 cm
+2(circumference of semicircle of radius 7 cm)
=
(14)+2
x7=28
=28 x 22/7 =88cm
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1m
1/2m
1m
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74
SECTION-C
19.
Construction ∆ABC
Equal angles, construction ║lines
1m
1m
Construction of similar triangle
20.
1m
P2x2 – P2x + q2x- q2 =0
½m
P2x(x-1)+ q2(x-1)=0
1m
(p2x+q2) (x-1)=0
1m
½m
x=1, -q2/P2
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OR
For real and equal roots
b2-4ac=0
a=2k,b=-40,c=25
(-40)2-4(2K)(25)=0
16000-200K=0
½m
1m
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200K=1600
1m
½m
=4 ;
= -1,
w
=9-5n
= -6
w
21.
w
.s
t
K=8
(A.P)
½m
a=4;d=-1-4=-5
½m
Sn=n/2[ 2a+(n-1)d]
½m
=15/2[2(4)+(15-1)(-5)]
½m
=15/2 [8-70]=15/2 x -62=15x-31 =-465
1m
OR
3 digits numbers ae
100,101___________________ 999
Nos. that leave remainder 3 on dividing
by 5 are: 103:108__________998
a= 103 d=5
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½m
½m
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Sn=n/2[ 2a+(n-1)d]
n=?
a=a + (n-1) d
998=103 + (n-1)5
N=180
1m
sn=180/2 [2(103)+(180-1)5]
=90[206+895]
1m
22.
Total no. of balls = R
B
W
4 + 5 + 6 =15
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=99090
= 4
P(Red ball)=
½m
½m
P(White ball)=
ud
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15
½ x 4=2 m
w
=
w
w
P(Not black)=
.s
t
=
P(red or black)=
23.
=
(
,
)
(2,4)
(
,
)
(5,K)
Area of ∆ =1/2 [
(
(
)+
,
) = (3,10)
(
)+
=>½ [ s(k-10)+5(10-4) +3(4-k)] =15
½m
(
)]
1m
½m
=>2k -20 +50 -20 +12 -3k =30
1m
=> -k+22= + 30
K=-8 or 52
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76
OR
Distance formulae
½m
√ [(x-3)2 +(y-6)2 ] = √ [(x+3)2+(y-4)2 ]
½m
2m
Solving and getting 3x+y=5
A
Correct fig. :
½m
24.
string
60 m
600
C
666
0
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B
AB=ht of the kite above the
ground level =60
AC=String
<ACB=angle of elevation=600
ie
=cosec
1m
=
AC=
25.
=40
+
+
+
=
1m
=
63 + 83 +x3= 123
X 3 =123 -63 - 83
= 1728 – 216 – 512 = 1000.

1m
m
w
=
w
w
= =cosec
.s
t
ud
=
½m
X3 =1000
X= 10 cm
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1m
1m
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77
26.Join OC (length of tangents from Ext point)
∆AOP ∆ AOC and ∆ BOQ
∆BOC (RHS )
1m
POA =COA and COB=BOQ (cpct)
1m
POA +COA+COB+BOQ= 1800 AOB=900
1m
27. Side of the square =10m
Diameter of semi circle = 5m
Radius = 5/2 m
2 semi circles of equal radii =1 circle
4 semi circles of equal radii =2 circles
1m
Hence Area of shaded region. = 2
= 2x22/7x(5/2)2
= 275/7
= 39.3 m2
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1m
1m
ie
(-7 , 6)
28. ( , )
( , )
(8,5)
?
( , )
Let the coordinates of C be(x,y)&
A( -7,6)
.s
t
= -5/2,
1m
w

1m
ud
Mid point of AC is
w
 X=2,
= 2 => y = -2 .
Hence coordinate of C are (2,-2)
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1m
SECTION-D
29. (Correct figure)
(Given to prove)
(Correct proof)
1m
1m
2m
30.Total amount spent= Rs 1200
Let the number of books originally bought =x
Cost of 1 book = Rs
1m
If the no. of books =x+10
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78
Cost of each books = Rs
-
m
=20
+10x-600=0
(x+30)(x-20)=0
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X=20, -30
Rejecting x=-30
We get number of books=20
OR
-
=1
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ie
Let the usual speed =x km/hr
½m
1½ m
.s
t
X2 +5x -1800=0
2m
w
w
D=b2-4ac=7225
w
X=-45(rejected 2m
Or,X=40 km/hr
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79
A
31.
hill
P
D
10m
Correct figure
B
AB=hill
1m
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P=position of the man on the deck
PC=10m =DB
APD=600; BPD=300
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C
1m
PD=10
w
w
∆PDB, tan 300 =
w
.s
t
To find A B and BC
∆PDB, tan 600 =
AD=PD
= 10
x
=30m
 Distance of the hill from the ship
=BC=PD=10
=17.3m Approx
Ht of the hill=AB=AD+DB
1m
1m
=30+10=40m
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80
32.
o
A
B
p
x
=
x
x12 x 12
=75.4 cm2
Area of ∆ OAB =1/2 x r2 sin
2m
ie
/2
ud
= 1/2x 12x12x
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Diameter=24 cm
Radius=12cm.
Area of the minor sector
=62.35 cm 2
.s
t
1m
 area of minor segment =area of minor sector- Area of ∆
w
=75.4 – 62.35 =13.05 cm2
w
w
1m
33.Let the number of coins =n
D= 1.75 cm ; r=
= 0.875 cm
H=2mm =0.2 cm
Vol of each coin =
=
vol of n coins =
x 0.2 cm3
2m
x 0.875 x 0.875 x0.2 x n
Vol of ‘n’ coins =volume of cuboid
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1m
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81
x 0.875 x 0.875 x0.2 x n =11x10 x 7
1m
n= 1600
OR,
h(
Volume (frustum) =
+
)
1m
x 45[(28)2 +(7)2 +(28)(7)]
= x
= 48510cm3
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34.Sum of first 15 terms =105
Sum of next 15 terms =780
Sum of first 30 terms =780+105
=885
2m
1m
1m
1m
[ 2a+(15-1)d]=105 and
ie

2a+14d=14 and
2a+29d=59
a=-14 and d=3
1m
w
w
w

.s
t
ud
[ 2a+(30-1)d]=885
The first three terms are
1m
a,a+d,a+2d……….
i.e. -14,-11,-8
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