OBILE Solution Of Class 10 CBSE SA-II Board (Set...
Transcript of OBILE Solution Of Class 10 CBSE SA-II Board (Set...
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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.
0APB 90
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
2 = x = 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
= 7
8 Ans. (A)
5. In ABC
oABtan30
BC
150 1
BC 3BC 150 3 m
Ans.(B)
6. Total numbers= 15
No. divisible by 4 are
4, 8, 12
Outcomes = 3 1
15 5
Ans. (C)
7. Distance of BD
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BD= 2 2(0 4) (0 3)
= 16 9 25 5cm
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.
890n
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)=9 1
36 4
favourable cases (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)=4 1
36 9
Favourable (1,4),(2,3),(3,2),(4,1)
14. 23 r 462
2223 r 462
7
2
2
2
462 7r
22 3
21 7r
3
r 49 r 7
Vol. of hemisphere= 32r
3
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=2 22
7 7 73 7
= 344 215649 718.6 cm
3 3.
15. 16 15
1x x 1
16 x 15
x 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 na 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
3…………….(2)
Put d in the equation (1)
a + 2a
6 153
a + 4a = 15
5a =15
a 3
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d = 2(3) 6
23 3
d = 2
A.P. = 3, 5, 7,------------
17. Step of construction
(i) Draw a ABCwith 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 PAB
o 3000 3tan60
y
3000 33
y
y 3000m
In A' B'P
o 3000 3tan30
x y
1 3000 3
x y3
x y 3000 3 3
x+3000=9000
x 6000m
From B to B’
time taken by aeroplane is 30 second
distance 6000S 200 m / sec
time 30
19. As given PA = PB
2 2 2 2(K 1 3) (2 K) (K 1 K) (2 5)
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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
P ,K 1 K 1
But 7K 3
0K 1
7K = 3.
K = 3
7
Ratio 3 : 7.
Co-ordinates of pt. P
3 32 3 7 3
37 7, , 0
3 3 21 17 7
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21.
Area of major sector OAYB = 2300π (42)
360
= 25π (42)
6
= 4620 cm2
Area of major sector OCZD = 2300π (21)
360
= 5
π 4416
= 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 R = 7
2 cm
Volume of the wood left
= Volume of cube – Volume of sphere
= 3 34a π R
3
= 3
3 4 7(7) π
3 2
= 3 3
3 3 34 7 π 7 π7 π 7 7 1
3 8 6 6
=163.33cm3
23. Speed = 4 km/hr
= 4 1000 m
60 min
= 2
1003
m/min
= 200
3m/min
DS
T
200 D
3 10
200D 10
3
2000D m
3
Volume of the water flowing in a canal
= Volume of water used for irrigation
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2000 8
6 1.5 A3 100
6 15 200 100
A3 8
30 200 100
A8
15
A4
× 200 × 100
A 15 50 100
2A 75000 m
24.
Area of trapezium
= 1
2 (sum of parallel sides) × Height
1
24.5 (AD BC) AB2
1
24.5 (10 4) AB2
49 = 14 AB
49
AB14
7
AB cm2
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Area of quadrant = 21π (AB)
4
= 1 22 49
4 7 4
= 22
74 4
= 211 777 cm
8 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 4 10
x 3 x 5 3 x 3,5
(x 2)(x 5) (x 4)(x 3) 10
(x 3)(x 5) 3
2 2x 5x 2x 10 x 3x 4x 12 10
(x 3)(x 5) 3
2
2
2x 5x 2x 3x 4x 10 12 10
x 5x 3x 15 3
2
2
2x 14x 22 10
x 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=2b b 4ac
2a
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219 (19) 4 2 42
x4
19 361 336
x4
19 5
x4
24 14
x ,4 4
7
x 6,2
26. Total no. of trees planted
= 2[2 + 4 + 6 + …….24]
= 12
2 [2 2 12 1 2]2
= 12[4 + 22] [Sum of n terms of on A.P. = n
2[2A + (n – 1) D]
= 12 × 26
= 312
27. Let BD be the flagstaff
BC be the tower
In ABC
0BCtan 45
AC
BC1
120
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BC = 120 m
In ADC
0DCtan 60
AC
DC3 DC 120 3 m
120
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 1
48 12
(2) Probability of a red colour card
Outcomes = 24
=24 1
48 2
(3) Probability of a face card
Outcomes = 8
=8 1
48 6
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(4) Probability of a queen
Outcomes = 2
=2 1
48 24
29. A(–3, 5), B(–2, –7), C(1, –8) and D(6, 3)
Area = ACD
1 2 3 2 3 1 3 1 2
1x (y y ) x (y y ) x (y y )
2
1
( 3)(( 8 3)) 1(3 5) 6( 5( 8))2
1
( 3)( 11) 1( 2) 6(13)2
= 1
1092
2109cm
2
Area of ABC
1
( 3)(( 7) ( 8)) ( 2)(( 8) 5) 1(5 ( 7))2
1( 3)(1) ( 2)( 13) 1(12)
2
= 1 1
3 26 12 (35)2 2
= 235cm
2
ar of quad. ABCD
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= area of ACD + area of ABC
2109 35 14472 cm
2 2 2
30. Let the speed of the stream = x km/hr
Upstream speed = (18 – x) km/hr
Downstream speed = (18 + x) km/hr
DistanceSpeed
time
Distancetime
Speed
Upstream time = 1 + Downstream time
24 24
118 x 18 x
24 24
118 x 18 x
18 x 18 x
24 118 x 18 x
24 × 2x = 324 – x2
x2 + 48x – 324 = 0
48 2304 4 324
x2
48 3600
x2
48 60
x2
48 60 48 60
x ,2 2
x = 6 km/hr
Speed of the stream = 6 km/hr
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31.
To is the angle bisector of PTQ
OT 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 ..(1)
In 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 = 128
12
32
y3
x2 = 64 + y2
x2 = 64 + 2
32
3
x2 = 64 + 1024
9
2 64 9 1024x
9
2 576 1024x
9
2 1600x
9
40
x cm3
40
TP cm3
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 34πr
3
= 3
4 7π
3 10
Volume of 150 spherical marbles = Volume of water risen in the cylinder
3 2
4 7 7150 π π h
3 10 2
3
4 7 1150 h
3 10 4
3
4 7150 4 h
3 10
5 16 7
h100
560
h100
h 5.6 cm
34. r1 = 8 cm
r2 = 20 cm
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h = 24 cm
Volume of frustum = 2 21 2 1 2
1πh r r r r
3
= 21 22
24 8 20 8 203 7
228 64 400 160
7
= 176
6247
= 15689.142 cm3
= 15.689 litres
1 litre of milk costs Rs 21
15.689 litres ……. 21
15.6891
= Rs 329.47