2 Answer key & Solution

Velocity

Timet

v

07. According to newton's third law every action has an

opposite and equal reaction. So A gun recoils when

a bullet is fired.

08. Unit of force, weight and change in momentum per

second is Newton. But the unit of work done per

second is watt.

09. The G = Universal Gravitational constant never

changes.

But the acceleration due to gravity varies according

to the planet

10. The pressure due to a liquid does not depends on

the area of surface.

It only depends on volume imeresed, density of liq-

uid, 'g' at that point and depth of that point.

11. Time of flight is 8 sec.

Then time to reach maximum height is 4 sec

So at time of 6 sec it has fallen for 2 sec from the

maximum height so the total distance will be

��Maximum height – height dropped in 2 sec

22u 1

gt2g 2

here given time of height

2u8

g and t = 2 sec.

u = 40 m/s (after maximum height)

240 40 1d 10 2

2 10 2

= 80 – 20 = 60m.

12. According to achemedez principle.

Apparent weight = Actual weight - up thurst

(decrease in wieght)

So.

WApp.

=WAct.

–ViP

Lg here v

i = volume imersed

app s s i LW V P G V P g PL = density of liquid

g = acc. due to gravity

Vs = volume of ball

Ps = density of ball

If a ball floats in a liquid the its density should be

less or equal to the density of that liquid so the

density Ball densityA if the density of ball is

greater than the density of liquid then it sinks.

So, Densityball

> DensityB

then, DensityA > Density

Ball > Density

B

13. Relative density of a an object is given by

Relative density (R.S.) = absolutedensityof object

densityof water at 4º C

So. (R.D.)1 = 7 and (R.S.)

2 = 3.5

the 1 1

2 2

R.D.Density 7 2

Density (R.D.) 3.5 1

14. The unit of power is watt

11Joule1watt J sec

1sec

also P VI volt Ampere

so

watt hour (wh) will be unit of energy.

15. Distance travelled = 90 m

acceleration = 6m/sec2

initial velocity = 3m/s.

then.

final velocity = ?

from 3rd equation of motion,

v2 = u2 + 2as

16. Train is slowing down because due to inertia the

coin will have a horizontal velocity when it was

tossed. Then the relative acceleration of coin is

positive with respect to train acceleration.

17. As we know

FF ma a

m

here m is mass of block the mass per unit lenght

M

L

mass of lenght M

x xL

Let, T is the tension of P

FT

x L–x

T

Applying newton's second law

3 Answer key & Solution

MF T x.a

L

M FF T x.

L m

FxT F

L

L xT F

L

18. Electronic configuration

K L M

2 8 3

= 2+8+3 = 13 e– = Aluminium

19. No. of electrons in K+ = 18e- = 19–1 (Positive charge)

20.

Solid State

Inter particle force of attraction is maximum

.

21. Mercury -Hg

22. Hydrogen atom Proton = 1Electron = 1Neutron = 0

23. Value of n = 4

Maximum Number of e- = 2n2

= 2(4)2

= 2×16

= 32 electrons

24. Sublimation2 2Gaseous CO Solid CO Dry ice

25. E.Rutherford

26. 24SO

27. d-block element = Fe (II)

Fe (III)

28. Centrifugation

29. 10–15

30. NaHCO3

31. Maximum Number of electrons = 6

32.Charge

Fixed for electronMass

33. E. Goldstein

34. Colloidal solution

35. Bilateral symmetry - Body can be divided along a

median longitudinal plane into two mirrored portion

right and left halves.

Ex. - Platyhelminthes to chordata

36. Spongilla is a fresh water sponge.

37. Flame cells (Protonephridia) are the part of excre-

tory system of platyhelminthes and it helps in ex-

cretion and osmoregulation.

38. All are examples of phylum arthopoda

39. Lateral meristem is responsible for secondary growth

40. Parenchyma, Collenchyma and Sclerenchyma are

supporting tissue of plants.

Function -

(i) Parenchyma - Storage of food and provide sup-

port

(ii) Collenchyma - It provides mechanical support

and elasticity.

(iii) Sclerenchyma - Main mechanical tissue which

provides mechanical support.

41. Cytokinesis is not a stage of Meiosis I

42. Amitosis is derived from greek words which means

without threads. It is mainly found in prokaryotes,

protozoans, lower algei etc.

43. Statements (iii) (iv) and (v) are correct

44. Round worm also known as aschehelminthes are

pseudocoelomate.

45. Memory based.

46. Fishes are of two types based on the nature of their

endoskeleton.

(a) Cartilagenous fish

(b) Bony fishes

(a) Cartilagenous fish also known as -

Chondrichthyes which possess cartilagenous endos

keleton.

47. Mammals - Latinword - Mamma - Breast -

Mammary glangs are found in females for body

feeding. Body is covered by hairs.

48. Kangaroo, Hedgehog. Dolphin and loris are mam-

mal.

49. Loligo, sepia and octopus are examples of Mallusca

50. Ribosome is the non-membranous organelle. It is

also a organelle within organelle.

4 Answer key & Solution

51. Let angles are 2x, 4x and 9x

��2x + 4x + 9x = 180°

= 15x = 180°

x = 12°

��angles are 24°, 48°, 180°

��exterior angles are 156°, 132°, 72°

��difference = 132° - 72° = 60°

52.

P 5 Q 10 U

S

T

5

3R

PQ = 5, QR = 5, RS = 10, ST = 3In �PUT :-By pythogoros theoremPT2 = PU2 + UT2

PT2 = 152 + 82

PT = 17

53. � x + y + z = 0

(x y)(y z)(z x) ( z)( x)( y)1

xyz xyz

54. M

x

B

h

C d A

h + d - x

Given :- BM + AM = BC + AC .......(1)AM = h + d - xIn �ACM :-by pythogoros :-AM2 = AC2 + MC2

= (h + d - x)2 = (h + x)2 + d2

= d2 = (h + d - x)2 - (h + x)2

= d2 = (d - 2x) (2h + d)

hdx

2h d

55.1 1 1 1

8 4 3 21 6 10 7 4 3 5 2 6

22 2 2

1 1 1 1

( 2 3)( 6 2) ( 15 6) (2 3)

1 1 1 1

6 2 15 6 2 3 3 2

6 2 15 6 2 3 3 2

4 9 1 1

6 2 15 62 2

4 9

56. b + c - a = 7, c + a - b = 10, a + b - c = 3 .....(1)

= (b + c - a) + (c + a - b) + (a + b - c) = 20

= a + b + c = 20 .....(2)

from equation (1) and (2) :-

13 17a , b 5,c

2 22 2 2 2 2 2 4 4 42b c 2c a 2a b a b c

2 2 2 2 4 42 2 417 17 13 13 13 17

2(5) 2 2 (5) 52 2 2 2 2 2

= 4200

57.60° 60°

60°

60° 60°

60°

E

B C

AD F

��ABC is an equilateral

let AB = BC = AC = a units

so �DEF will be equilateral

In �ABD :-

B

a

A D60°

ABtan 60

AD

aAD

3

In �AFC :- A

a

C F60°

a 2asin 60 AF

AF 3

so DF = AD + AF = a 2a

3a3 3

2

2

3( 3a)area of DEF 34 3 :1

area of ABC 13(a)

4

58.

A B

CD 20

10

5

In a quadrilateral :-sum of all three sides > largest fourth sideso :-

5 Answer key & Solution

AD + 5 + 10 > 20 5 + 10 + 20 > ADAD > 5 AD < 35so :-5 < AD < 35so AD will have 29 possible values.

59. a 2 3 5 , b 3 3 5

a 2 3 5 ...(1) b 3 3 5 ...(2)2 2(a 2) ( 3 5) 2 2(b 3) ( 3 5)

2a 4a 4 8 2 15 ...(3)2b 6a 9 8 2 15 ...(4)

equation (3) + (4) : -

a2 + b2 - 4a - 6a = 3

60. ax2 + bx + c

���������b/a��������c/a

1 1( )

bab

ac

a

b b a c( ) b

a c ac

1 1 11 1

c a2

a c

2 2 2c a 2ac (a c)

ac ac

so polynomial 2

2 a c (a c)k x b x

ac ac

61. a2(b + c) +b2(a + c) + c2(a + b) + 3abc

a2b + a2c + b2a + b2c + c2a + c2b + 3abc

(a2b + a2c + abc)+(b2a + b2c + abc)+(c2a + c2b +abc)

a(ab + ac + bc) +(ab + bc + ac) +c(ac + bc + ab)(ab + bc + ca) (a + b + c)

62. let polynomial is f(x)

f(x) = Q(x) (x2 - x - 6) + (ax + b) {by division algoritham}

f(x) = Q(x) (x + 2) (x - 3) + (ax + b) ....(1)put x = -2 f(-2) = -2a + b

-2a + b = 2 ....(2)put x = 3f(3) = 3a + b

3a + b = 7 ....(3)from equation (2) and (3) :- a = 1, b = 4

63. Unit digit in = 2123 × 61234 × 812345

= 23 × 6 × 8

unit digit in = 8 × 6 × 8 = 384

so unit digit = 4

64. 6 4 2x x x 5 0 ve 0

ve ve ve

It has no real values.

65. |x + y| = 2

x + y = 2 x + y = -2

so both lines are parallel.

66.2 1 5

0x 2y 2x y 9

let 1 1

A, Bx 2y 2x y

52A B 0

9.....(1)

9 64 0

x 2y 2x y

9A + 6B + 4 = 0 .....(2)

solve following equations we get 1

x ,y 22

67. y = -3x2 + 2x + 7

so maximum value = D

4a

2(b 4ac) (4 84) 22

4a 4 3 3

68.

A

CD

B

O12 16

AB2 = AO2 + OB2

AB2 = 122 + 162

AB 144 2 6 20 cm

so perimeter of ABCD = 4 × 20 = 80 cm

69. C

B A

6 cm

4 cm

D

BD2 = AD × CD26 AD 4

AD = 9 cm

6 Answer key & Solution

70. A

B CD

E F

� AD is bisector of ��

so BD AB

DC AC

BD 12 BD 2

DC 18 DC 3so ar (ADB) = 2x

� ar (ADC) = 3x

� ar (�DCF) = 3

x2

� ar (�AED) = x

3xar( DCF) 32

ar( AED) x 2

71. a = b2 = c2 = d4

a = b2, a = c3, a = d4

abcd = a.(a)1/2 . (a)1/3 . (a)1/4

1 1 1 12 6 4 3 251

2 3 4 12 12a a a

2512a a

25log abcd log a

12

72.

r8 cm

10 cm

A B

CD

In right angled �BDC :-BD2 = 102 + 82

BD = 2 41 cm

1r (sum of leg sides hypotenuse)

2�

1r (10 8 2 41)

2�

9 41 40r 9 41

9 41 9 41

73. ar (ABCD) = ar (FECG)

ar (AQED) + ar (QBCE) = ar (FGBQ) + ar (QBCE)ar (AQED) = ar (FGBQ) ....(1)

� AQED will be parallelogram�ar (AQED) = 2 × ar (�AQE) = 2 × 48 = 96 cm2

so ar (FGBQ) = 96 cm2

74. x° = 360° - 70 = 290°

75. A

B CD

In �ADB :- AB + BD > AD ....(1)in �ADC :- AC + CD > AD ....(2)equation (1) and (2) :-

AB + AC + BC > 2AD

76.

30° 40°

40° 35°

A

B

C D

E

F G

F' G'

�AF'D = 30° + 40° = 70

�AG'C = 35° + 40° = 75

��BAE = 180° - 70 - 75 = 35°

77.

p'Q R

P

HI

In �PQP' :-�QPP' = 180° - 90° - 80°�QPP' = 10°PI is angle bisector of �P

so1

QPI P2

1(180 80 40 )

2

1QPI 60 30

2so �HPI = 30° - 10° = 20°

78. A

B CD

1

2G13

cm

13 cm

10 cm

By appoloniory theorem :-

132 + 132 = 2(AD2 + 52)169 + 169 = 2(AD2 + 25)AD2 = 169 - 25

AD = 12

2AG 12 8 cm

3

7 Answer key & Solution

79. A

C B(0, 0)

(4, 3)

(5, 1)

ar 4 3 1

1( ABC) 0 0 1

25 1 1

1 1 1| (4 15) | |11 | 11 5.5

2 2 2 units

80. Area of shaded region = area of ||gm ABCD - ar (�AED)

117 8 2 8

2 = 136 - 8

= 128 cm2

81. Area of x region = 2

21 5 25cm

2 2 8

Area of y region =2 21 36

6 18 cm2 2

Area of z region =2

21 13 1 169 169cm

2 2 2 4 8

z = x + y

82. ab = 1, a + b = 2

so a = 1 and b = 12 2

2 21 1a b (1 1) (1 1) 4 4 8

a b

83. Unit digit = (316)2000 × (125)40000 × (514)2555

= 6 × 5 × 4

unit digit = 120

so unit digit = 0

84. 2005 + p = q2 ....(1)

2005 + q = p2 ....(2)

equation (1) - (2) :-

p - q = q2 - p2

(p - q) = (q - p) (q + p)p + q = -1 ....(3)

square of equation (3) :- (p + q)2 = (-1)2

p2 + q2 + 2pq = 1(2005 + q) + (2005 + p) + 2pq = 12(2005) + (p + q) + 2pq = 12(2005) - 1 + 2pq = 12005 + pq = 1

2014 + pq = 10

85. (100a + 10b + c)2 = (a + b + c)5

2 5(abc) (a b c) abc 3 digit number

52abc (a b c)

122abc (a b c) (a b c)

let a + b + c = 9

abc 81 3

abc 243so a = 2, b = 4, c = 3a × b - c = 2 × 4 - 3 = 8 - 3 = 5

86.a(a 1) (p 1) a

(a 1)(p 1) p

2a a p 1 a

ap a p 1 p

2 2 2 2a p ap p p a p a ap a

2 2a a (p p) 0

both roots are equal so D = 0b2 - 4ac = 0

2

2

1 4(p p) 0

4p 4p 1 0

2 12(2p 1) 0 p

given m

pn

so m = 1, n = 211m + 13m = 11(1) + 13(2)

= 11 + 26 = 37

87. a × b × c × d = 40320 .....(1)

ab + a + b = 322

ab + a + b + 1 = 323(a + 1) (b + 1) = 323(a + 1) (b + 1) = 17 × 19a + 1 = 17 or b + 1 = 19

a = 16 b = 18

bc + b + c = 398

bc + b + c + 1 = 399(b + 1) (c + 1) = 19 × 21b + 1 = 19 or c + 1 = 21b = 18 c = 20

from equation (1)

40320d

16 18 20

d = 7

88. x + y + z = 22 ....(1)

xy + yz + zx = 91 xyz

xy yz zx91

xyz

8 Answer key & Solution

1 1 191

x y z .....(2)

equation (1) × (2) :-

1 1 1(x y z) 22 91

x y z

x x y y z z1 1 1 2002

y z x z x y

x y x z y z1999

y x z x z y ....(3)

2 2 2 2 2 2x(y z ) y(z x ) z(z y )

xyz

y z z x x y

z y x z y x

= 1999so a + b + c + d = 1 + 9 + 9 + 9 = 28

89. f(x) = ax2 + bx + c

x = 0

f(0) = c

c = 7

x = 2

f(2) = 4a + 2b + c

= 4a + 2b + c = 9

4a + 2b = 2

2a + b = 1 .....(1)

x = -3

f(-3) = 9a - 3b + c

9a - 3b + c = 499a - 3b = 423a - b = 14 .....(2)

from equation (1) and (2) :-a = 3, b = -5, c = 7(3a + 5b + 2c)2 = (9 - 25 + 14)2 = (-2)2 = 4

90. 2x + 3y = 763

763 3yx

2.....(1)

from equation (1) :-y = 1, 3, 5, 7, .......... 253253 = 1 + (n - 1)2n = 127

91.

P Q

RS

5

4

m

PQM ~ RSM(by AAA rule)

so 2

ar( PQM) QM

ar( RSM) SM

2QM 32

SM 50

QM 32 4

SM 50 5......(1)

so 4

ar( PQS) 329

2ar( PQS) 72 cm

5ar( RSQ) 50

92ar( RSQ) 90 cm

area of trap PQRS = ar (�PQS) + ar (�RSQ)= 72 + 90 = 162 cm2

92.

12 cm6 cm QP

RS

T

N M

L

In �PLM :-

TQ || ML

so �PQT ~ �PLM

TQ PQ

ML PL

TQ 6 TQ 1

ML 18 ML 3

TQ 1TQ 4 cm

12 3

ar 21( PQT) 6 4 12 cm

2

93. Let side of �1 = a1, side of �6 = a6

2 21 6

3 3a a 968 3

4 4

2 21 6

3a a 968 3

4

1 6 1 6(a a )(a a ) 968 4 ....(1)

6 6 6 6(9 3a a )(9 3a a ) 968 4

26a (9 3 1)(9 3 1) 968 4

26a 242 968 4

6a 4

9 Answer key & Solution

h1 = altitude of �1Let

h1 = x, h2 = 3x , h3 = 3x ,h4= 3 3x , h5= 9x, h6 =

9 3x

1

6

a 9 3x

a x

1 6a 9 3a

perimeter of triangle = 4 × 3 = 12 units94. Let x = 1540

x 1 (mod 7) ....(1)

x 1 (mod 11) ....(2)

x 1 (mod 17) ....(3)

from equation (3) :- x = 17a + 1 ....(4)

from eeuation (2) :- 17a + 1 1 (mod 11)a = 11b

so x = 17 × 11 b + 1from equation (1) :-

17 × 11 × b + 1 = 1 (mod 7)b = 7cx = 17 × 11 × 7c + 1x = 1309c + 1

so remainder = 1

95. Let x 31997

x 3 (mod 4) .....(1)x 13 (mod 25) .....(2)from equation (2) :-

x = 25a + 13from equation (1) :-

25a + 13 = 3 (mod 4)a = 4b + 2x = 25 (4b + 2) + 13x = 100b + 63

so last two digits = 63

96. (2x - 4)3 + (4x - 2)3 = [(2x - 4) + (4x - 2)]3

x 3 x 3 x 3 x 3 x x x x(2 4) (4 2) (2 4) (4 2) 3(2 4)(4 2)(2 4 6)

x x x x(2 4)(4 2)(2 4 6) 0

2x - 4 = 0 x = 24x - 2 = 0 x = 1/22x + 4x - 6 = 0 x = 1

sum = 72

13

2

97. (x2 + ax + 20) (x2 + 17x + b) = 0; a, b are integer

according to question :-a > 0 and b > 0 (� sum of roots <0 and product >0)since ���������a

����������so 20 (1 × 20), (2 × 10) or (4, 5)

min. a = 9 ......(1)

( , ) ( 1, 16),( 2, 15) ( 8, 9)

min. b = 16(a + b)min. = amin + bmin = 9 + 16 = 25

98. abccba is divisible by 7

if abc - cba is divisible by 7

abc - cba = 99(a - c) = 7M 7/(a - c)so (a, c) = {(9, 2), (8, 1), (7, 0), (2, 9), (1, 8), (9, 9), (8,8), (7, 7) (6, 6), (5, 5), (4, 4), (3, 3), (2, 2), (1, 1)}no of pair of (a, b) = 14also no of b's can be = 5

total number of 6 digit number of 14 × 5 = 70

99. from given equations by eliminating 'c' we get

a2 + b2 - (a + b - 1)2 = -1

-2ab + 2(a + b) - 1 = -1ab - a - b = 0 (a - 1) (b - 1) = 1a - 1 = 1 and b - 1 = 1a = b = 2c = 3

or a - 1 = -1 and b - 1 = -1a = b = 0c = - 1a2 + b2 + c2 = 17 or 1

100. As a + b is a root of x2 + ax + b = 0, (a + b)2 + a(a + b) + b = 0

2a2 + 3ab + b2 + b = 0

23b b 8ba

4

so b2 - 8b must be a perfect square for some wholenumber = k2, K No

(b - 4)2 - 16 = K2

(b - 4)2 - K2 = 16(b - 4 - K) ( b - 4 + K) = 16

Now we have four possibilities :-(i) b - 4 + K = 8, b - 4 - K = 2 (b, K) = (9, 3)(ii) b - 4 + K = 4, b - 4 - K = 4 (b, K) = (8, 0)(iii) b - 4 + K = -2, b - 4 -K -8 (b, K) = (-1, 3)(iv) b - 4 + K = -4, b - 4 - K = -4 (b, K) = (0, 0)Now maximum possible value of b = 9

b2 = 81