Solution Of Class 10 CBSE SA-II Board (Set-1)Mathematics th

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Solution Of Class 10th CBSE SA-II
Board (Set-1)Mathematics
1.
(2k – 1) – k = (2k + 1) – (2k – 1)
2k – 1 – k = 2k + 1 – 2k + 1
k–1=2
k 3
Ans. (B)
2.
APB 900
Ans. (D)
3.
AB = 5 cm
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BC = 12 cm
AC = ?
Pythagoras theorem
= AC2 = AB2 + BC2
= 52 + 122
= 25 + 144
AC2 = 169
AC = 13 cm
BC = x + 12 – x
AB = x + 5 – x
BP = BQ
..(1) (Tangents drawn from an external point to a circle are equal)
AP = AR
..(2)
CO = CR
..(3)
AC = 13
5 – x + (12 – x) = 13
5 – x + (12 – x) = 13
–2x + 17 = 13
– 2x = – 4
x=
14
=x=2
2
Radius = 2 cm
Ans. (C)
4.
No of children = 3
Outcomes = BBB
GGG
BBG
BGB
GBB
BGG
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GBG
GGB
Probability =
No. of favourablecases
Total no. of cases
Total cases = 8
=
5.
7
8
Ans. (A)
In ABC
150
BC
AB
tan30o
BC
1
3
BC 150 3 m
Ans.(B)
6.
Total numbers= 15
No. divisible by 4 are
4, 8, 12
Outcomes =
3
15
1
5
Ans. (C)
7.
Distance of BD
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BD= (0 4)2 (0 3)2
= 16 9
25 5 cm
Ans. (A)
8.
AB = ?
Pythagoras theorem
AB2 = AO2 + BO2
= 102 + 102
= 100 + 100
AB2 = 200
AB = 10 2 cm
Ans.(b)
9.
As given equal roots
D=0
b2 –4ac =0
a = 4, b = p, c =3
p2 – 48 =0
p2 = 48
p
48
p= 4 3
10. The numbers divisible by both 2 and 5 are
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110, 120 ------------------------990
d = 120–110 =10
a 110
d 10
an=110
an = a + (n –1)d.
990 = 110 + (n –1)10
990 –110 = (n –1)10
880 + 10n –10.
890
n
10
n 89
11. As tangents drawn from an external point to the circle are equal in length
So
EA = EC …………..(1)
EB = ED ……….(2)
Adding (1) and (2)
EA + EB =EC + ED
AB = CD.
Hence. Proved.
12.
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AB = AC (Given)
BD = BF (length of the tangent drawn from an external point to a circle are equal)…..(1)
AF = AE
…………….(2)
DC = CE
………….(3)
Now, AB = AC
AF + BF = AE + CE
AE + BD = AE + DC
(From equation (1) , (2) & (3))
BD DC
13. (1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
(6, 1)
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
(6, 4)
(1, 5)
(2, 5)
(3, 5)
(4, 5)
(5, 5)
(6, 5)
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 6)
(i) p( No. of each dice is even)=
favourable cases
9
36
1
4
(2,2),(2,4),(2,6),(4,2),(4,4)(4,6),(6,2),(6,4),(6,6)
(ii) P( sum appearing on both dice is 5)=
Favourable
4
36
1
9
(1,4),(2,3),(3,2),(4,1)
14. 3 r2 462
3
r2
22 2
r 462
7
462 7
22 3
21 7
r2
3
r2 49
r 7
Vol. of hemisphere=
2 3
r
3
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15.
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=
2 22
7 7 7
3 7
=
44
2156
49
718.6 cm3 .
3
3
16
15
1
x
x 1
16 x
x
15
x 1
(16 – x) (x +1) = 15x
16x + 16 –x2 –x = 15x
–x2 +15x + 16 = 15x
–x2 + 15x + 16 -15x=0
x2 = 16
x
4
16. a5 + a9 = 30
(a + 4d) + (a + 8d) = 30
an a (n 1)d
2a + 12d = 30
a + 6d = 15 ……………(1)
a25 = 3a8.
a + 24d = 3(a + 7d)
a + 24d = 3a + 21d
24d – 21d = 3a –a.
3d = 2a.
d=
2a
…………….(2)
3
Put d in the equation (1)
a+ 6
2a
3
15
a + 4a = 15
5a =15
a 3
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d=
2(3)
3
6
2
3
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d=2
A.P. = 3, 5, 7,-----------17. Step of construction
(i) Draw a ABC with the sides given such that AC = 5.5 cm, AB = 5 cm, BC=6.5cm
(ii) Then draw acute BAX
(iii) Make 5 arcs on line AX such that AA1=A1A2=……….=A4A5
(iv) Join BA5
(v) Draw B’ A3 parallel to BA5
(vi) Similarly drawn C’ B’ parallel to CB.
Hence
C'AB' is the required triangle
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18. In
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PAB
tan60o
3
3000 3
y
3000 3
y
y 3000m
In
A' B'P
tan30o
1
3
3000 3
x y
3000 3
x y
x y 3000 3
3
x+3000=9000
x 6000m
From B to B’
time taken by aeroplane is 30 second
S
dis tance
time
6000
200 m / sec
30
19. As given PA = PB
(K 1 3)2 (2 K)2
(K 1 K)2 (2 5)2
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Squaring both sides
(K –4)2 + (2 – K)2 = 1 + 9
K2 + 16 – 8K + 4 + K2 – 4K =10
2K2 –12K + 10=0
K2 –6K + 5 =0
K2– 5K –K+ 5=0
K(K–5) –1(K–5)=0
K= 1, 5.
20. Let the ratio be K:1
Let P divides AB in the ratio K : 1
2K 3 7K 3
,
K 1 K 1
P
7K 3
0
K 1
But
7K = 3.
K=
3
7
Ratio 3 : 7.
Co-ordinates of pt. P
3
3 7
3
7
,
3
3
1
1
7
7
2
3
7
3
,0
2
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21.
Area of major sector OAYB =
=
5
6
300
360
π (42)2
300
360
π (21)2
π (42)2
= 4620 cm2
Area of major sector OCZD =
=
5
6
π 441
= 1155 cm2
Area of shaded region
= Area of major section OAYB -Area of major sector OCZD
= 4620 – 1155
= 3465 cm2
22.
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a = 7 cm
Let R = radius of the sphere
2R = a
2R = 7
7
cm
2
R=
Volume of the wood left
= Volume of cube – Volume of sphere
4
π R3
3
= a3
4
π
3
3
= (7)
7
2
3
4
73
π
3
8
= 73
3
7
π 73
6
73 1
π
6
=163.33cm3
23. Speed = 4 km/hr
=
4 1000 m
60
min
=
2
100 m/min
3
=
200
m/min
3
S
D
T
200
3
D
D
10
200
10
3
D
2000
m
3
Volume of the water flowing in a canal
= Volume of water used for irrigation
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2000
3
6 1.5
6 15 200
3
A
A
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8
100
100
8
A
30 200 100
8
A
15
× 200 × 100
4
A
15 50 100
A
75000 m2
24.
Area of trapezium
=
1
(sum of parallel sides) × Height
2
24.5
1
(AD BC) AB
2
24.5
1
(10 4) AB
2
49 = 14 AB
AB
49
14
AB
7
cm
2
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Area of quadrant =
=
1
4
22
7
=
22
7
4 4
=
11
7
8
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1
π (AB)2
4
49
4
77
cm2
8
Area of shaded region
= Area of trapezium – Area of quadrant
= 24.5 –
77
8
= 24.5 – 9.625
= 14.875 cm2
25.
x 2
x 3
x 4
x 5
10
x
3
3,5
(x 2)(x 5) (x 4)(x 3)
(x 3)(x 5)
10
3
x2 5x 2x 10 x2 3x 4x 12 10
(x 3)(x 5)
3
2x2 5x 2x 3x 4x 10 12
x2 5x 3x 15
10
3
2x2 14x 22 10
x2 8x 15
3
3(2x2 – 14x + 22) = 10(x2 – 8x + 15)
6x2 – 42x + 66 = 10x2 – 80x + 150
10x2 – 6x2 – 80x + 42x + 150 – 66 = 0
4x2 – 38x + 84 = 0
2x2 – 19x + 42 = 0
Quadratic formula
x=
b
b2 4ac
2a
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x
x
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19
(19)2 4 2 42
4
19
361 336
4
x
19 5
4
x
24 14
,
4 4
x 6,
7
2
26. Total no. of trees planted
= 2[2 + 4 + 6 + …….24]
=2
12
[2 2
2
12 1 2]
= 12[4 + 22]
[Sum of n terms of on A.P. =
n
[2A + (n – 1) D]
2
= 12 × 26
= 312
27. Let BD be the flagstaff
BC be the tower
In
ABC
BC
AC
tan 450
BC
120
1
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BC = 120 m
In
ADC
DC
AC
tan 600
DC
120
3
DC
120 3 m
DB = DC – BC
= 120 3
120
= 120( 3 1)
= 120(1.73 1)
= 120(0.73)
= 87.60 m
28. Total cards = 52
Cards removed = 4
52 – 4 = 48
Cards removed
2 red queens
2 blocks Jacks
(1) Probability of a king =
No. of favourable cases
Total no.of cases
Outcomes = 4
=
4
48
1
12
(2) Probability of a red colour card
Outcomes = 24
=
24
48
1
2
(3) Probability of a face card
Outcomes = 8
=
8
48
1
6
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(4) Probability of a queen
Outcomes = 2
=
2
48
1
24
29. A(–3, 5), B(–2, –7), C(1, –8) and D(6, 3)
Area =
ACD
1
x1 (y 2 y 3 ) x2(y 3 y 1 ) x3(y 1 y 2 )
2
1
( 3)(( 8 3)) 1(3 5) 6( 5( 8))
2
1
( 3)( 11) 1( 2) 6(13)
2
=
1
109
2
Area of
109
cm2
2
ABC
1
( 3)(( 7) ( 8)) ( 2)(( 8) 5) 1(5 ( 7))
2
1
( 3)(1) ( 2)( 13) 1(12)
2
=
1
3 26 12
2
=
35
cm2
2
1
(35)
2
ar of quad. ABCD
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= area of
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ACD + area of
109
2
35
2
144
2
ABC
72 cm2
30. Let the speed of the stream = x km/hr
Upstream speed = (18 – x) km/hr
Downstream speed = (18 + x) km/hr
Dis tance
time
Dis tance
time
Speed
Speed
Upstream time = 1 + Downstream time
24
24
1
18 x
18 x
24
18 x
24
24
18 x
1
18 x 18 x
18 x 18 x
1
24 × 2x = 324 – x2
x2 + 48x – 324 = 0
x
x
48
2304 4 324
2
48
3600
2
x
48 60
2
x
48 60 48 60
,
2
2
x = 6 km/hr
Speed of the stream = 6 km/hr
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31.
To is the angle bisector of
OT
PTQ
PQ
OT bisects PQ
PR = RQ = 8 cm
Also
TPO = 900
In PRO
OP2 = OR2 + PR2
(10)2 = OR2 + 82
OR = 6 cm
Let TP = x, TR = y
In
PRT
PT2 = PR2 + TR2
x2 = 82 + y2
In
..(1)
PTO
OT2 = TP2 + OP2
(6 + y)2 = x2 + 102
36 + y2 + 12y = x2 + 100
x2 = y2 + 12y – 64
..(2)
Solving (1) & (2)
y2 + 12y – 64 = 64 + y2
12y = 128
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y=
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128
12
y
32
3
x2 = 64 + y2
x2
32
= 64 +
3
x2 = 64 +
x2
x2
2
1024
9
64 9 1024
9
576 1024
9
x2
1600
9
x
40
cm
3
TP
40
cm
3
32.
To Prove: OP
AB
Construction : Take any point Q, other than P on the tangent AB .Join OQ.
Suppose OQ meets the circle at R.
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Proof : We know that among all line segments joining the point O to a point on AB, the
shortest one is perpendicular to AB. So to prove OP
AB, it is sufficient to prove that
OP is shorter than any other segment joining O to any point of AB
OP = OR (Radii of the same circle)
OQ = OR + RQ
OQ > OR
OQ > OP
OR < OQ
Thus OP is shorter than any other segment joining O to any point of AB
OP
AB
33. Radius of spherical marbles = 0.7 cm
Radius of cylinder = 3.5 cm
Volume of one spherical marble
4
= π
3
7
10
4 3
πr
3
3
Volume of 150 spherical marbles = Volume of water risen in the cylinder
4
7
150
π
3
10
3
π
150
4
3
7
103
1
h
4
150
4
3
7
103
4
5 16 7
100
h
h
7
2
2
h
h
h
560
100
5.6 cm
34. r1 = 8 cm
r2 = 20 cm
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h = 24 cm
Volume of frustum =
=
1
3
22
24 8
7
2
1
πh r12 r22 r1r2
3
20
8 20
22
8 64 400 160
7
=
176
624 = 15689.142 cm3
7
= 15.689 litres
1 litre of milk costs
15.689 litres …….
Rs 21
21
15.689
1
= Rs 329.47
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