test no. 3 · 2019-11-25 · all india aakash test series-2020 (class ix) test-3 (answers &...

Test No. 3

Olympiads & Class IX-2020for

24-11-2019

A, B & C

Test-3 (Answers) All India Aakash Test Series-2020 (Class IX)

1/10

All India Aakash Test Series (Junior) - 2020 (Class IX) TEST - 3

Test Date : 24-11-2019 ANSWERS SECTION-I (Code-A)

1. (4) 2. (4) 3. (3) 4. (2) 5. (3) 6. (2) 7. (1) 8. (4) 9. (2) 10. (3) 11. (1) 12. (3) 13. (1) 14. (3) 15. (3) 16. (2) 17. (4) 18. (2) 19. (2) 20. (2)

21. (3) 22. (2) 23. (3) 24. (4) 25. (2) 26. (1) 27. (4) 28. (3) 29. (2) 30. (1) 31. (4) 32. (1) 33. (3) 34. (3) 35. (2) 36. (3) 37. (4) 38. (2) 39. (1) 40. (4)

41. (2) 42. (3) 43. (3) 44. (4) 45. (1) 46. (3) 47. (2) 48. (3) 49. (3) 50. (3) 51. (2) 52. (2) 53. (4) 54. (4) 55. (3) 56. (4) 57. (3) 58. (2) 59. (4) 60. (3)

61. (4) 62. (3) 63. (3) 64. (2) 65. (1) 66. (3) 67. (3) 68. (4) 69. (1) 70. (4) 71. (4) 72. (3) 73. (4) 74. (2) 75. (3) 76. (2) 77. (1) 78. (3) 79. (1) 80. (4)

81. (3) 82. (2) 83. (4) 84. (1) 85. (2) 86. (3) 87. (2) 88. (4) 89. (3) 90. (1) 91. (2) 92. (3) 93. (2) 94. (1) 95. (3) 96. (4) 97. (2) 98. (3) 99. (4) 100. (1)

SECTION-II (Code-B)

1. (2) 2. (4) 3. (3) 4. (4) 5. (3) 6. (4)

7. (2) 8. (3) 9. (3) 10. (4) 11. (3) 12. (3)

13. (3) 14. (4) 15. (2) 16. (3) 17. (4) 18. (3)

19. (4) 20. (2) 21. (3) 22. (2) 23. (3) 24. (2)

25. (2) 26. (4) 27. (4) 28. (3) 29. (2) 30. (3)

SECTION-III (Code-C)

1. (3) 2. (4) 3. (4)

4. (4) 5. (2) 6. (2)

7. (4) 8. (2) 9. (1)

10. (3) 11. (4) 12. (2)

13. (2) 14. (1) 15. (4)

All India Aakash Test Series-2020 (Class IX) Test-3 (Answers & Hints)

2/10

Answers & Hints SECTION-I (Code-A)

1. Answer (4) PEi = 12 × 0.75 PEi = 9 J KEf = PEi

21 92

mv =

181

v =

= 18 = 3 2 m/s 2. Answer (4) 3. Answer (3) P = Fv

5 × 103 = F × 20 F = 0.25 × 103 N = 250 N 4. Answer (2) By conservation of momentum mv = (m + M)v ′

mvvm M

′ =+

21

1KE2

mv=

22

1KE ( )2

m M v ′= +

2 2

21 ( )2 ( )

m vm Mm M

= ++

= 2 2

2( )m vM m+

∆KE = KE2 – KE1

= 2 2

21 12 ( ) 2

m v mvm M

−+

= 21 12

mmvm M

− +

= 21

2 ( )m Mvm M

−+

5. Answer (3) Wnet = ∆K

WF = mgh – σVgh

= Vgh (ρ – σ) 6. Answer (2)

21

12

W Kd=

2

22

1 3 12 2 2

dW K Kd = −

⇒ 2 22

9 1 14 2 2

W Kd Kd= −

⇒ 22

1 9 12 4

W Kd = −

⇒ 2

254 2

KdW =

2 154

W W=

⇒ 2 14 5=W W

7. Answer (1) 8. Answer (4) F = Mg + 2200 = 1000 × 9.8 + 2200 = 12000 N P = Fv P = 12000 × 2 P = 24000 W

= 24000746

= 32.17 hp 9. Answer (2) Work done by heart in 1 min = P × t = 2 × 60 = 120 J Number of beat

= Work done by heart in one minuteWork done by heart in each beat

= 120 80 times1.5

=

3/10

Test-3 (Answers & Hints) All India Aakash Test Series-2020 (Class IX)

10. Answer (3) 11. Answer (1) 12. Answer (3) 13. Answer (1) 14. Answer (3) 15. Answer (3) 16. Answer (2) W = ∆K

21 112 102 2

× × = mv

60 25

v ×=

= 24 = 2 6 m/s 17. Answer (4)

(2 1)2aS u n= + −

1 2aS =

232aS =

352aS =

W1 : W2 : W3 = FS1 : FS2 : FS3 = S1 : S2 : S3 = 1 : 3 : 5 18. Answer (2)

140 10 4 m

100h = × =

240 4 1.6 m

100h = × =

19. Answer (2) Pi = P Pf = 2P

2

2iPK Km

= =

2

2f

fP

Km

=

= 2

42Pm

= 4K

4100 100K K KK K

∆ −× = ×

= 300%

20. Answer (2)

Wnet = ∆K

(6gh – 5h) = 2 21 16 8 3 82 2

× × + × ×

3(60 5) 64 32

h − = +

⇒ 9 642 55

h = ×

= 576110

Work done by friction = 5 576110×

= 26.18 J

21. Answer (3)

22. Answer (2)

Heat3 2

Calcium CarbonLimestoneoxide dioxide

CaCO CaO CO→ +

23. Answer (3)

24. Answer (4)

25. Answer (2)

26. Answer (1)

Hydrogen carbonate is 3HCO − and it has

5 atoms.

27. Answer (4)

28. Answer (3)

Potassium ion has unipositive charge.

29. Answer (2)

30. Answer (1)

Baking powder is NaHCO3


4/10

31. Answer (4)

The valency of Z in ZPO4 will be 3.

The formula of sulphate of Z will be

The formula of chloride of Z will be

32. Answer (1)

We know that, 3 g of carbon reacts with

= 8 g of oxygen.

1 g of carbon will react with = 83

g of oxygen

∴ 12 g of carbon will react with = 8 123

×

= 32 g of oxygen

33. Answer (3)

34. Answer (3)

Sulphite ion = 23SO − , Sulphate ion = 2

4SO − ,

Sulphide ion = S2– , Phosphate ion = 34PO −

35. Answer (2)

36. Answer (3)

The compounds possible by combining the given ions are – CaO, Al2O3, Na2O, CaCl2, AlCl3, NaCl, NaOH, Al(OH)3, Ca(OH)2

37. Answer (4)

38. Answer (2)

Formula of magnesium nitride

Formula of magnesium nitrate

39. Answer (1)

According to law of conservation of mass –

Mass of reactants = Mass of products

Mass of NaCl + Mass of AgNO3 = Mass of NaNO3

+ Mass of AgCl

117 g + 340 g = 170 g + x g

x 287 g=

40. Answer (4)

Quicklime is CaO and the ratio by mass of Ca and

O is 40 : 16 or 5 : 2

∴ x 2=

The ratio by mass of hydrogen and oxygen in

water is 2 : 16 or 1 : 8

∴ z 8=

The ratio by mass of magnesium and sulphur in

magnesium sulphide is 24 : 32 or 3 : 4

∴ y 4=

On putting the values of x, y and z in option (4)

(y)x = 2z

(4)2 = 2 × 8

16 = 16

41. Answer (2)

42. Answer (3)

43. Answer (3)

44. Answer (4)

45. Answer (1)

46. Answer (3)

47. Answer (2)

48. Answer (3)

49. Answer (3)

50. Answer (3)

51. Answer (2)

52. Answer (2)

Marasmus is caused by deficiency of proteins, fats

and carbohydrates in diet.

5/10


53. Answer (4)

54. Answer (4)

55. Answer (3)

56. Answer (4)

57. Answer (3)

58. Answer (2)

59. Answer (4)

Leishmania causes kala-azar and it is transmitted by sand fly.

60. Answer (3)

61. Answer (4)

Supplementary pair of 63° = 180° – 63° = 117°

∴ rd2

3of 117° = 78°

⇒ Complementary pair of 78° = 90° – 78° = 12°

62. Answer (3)

63. Answer (3)

64. Answer (2)

∆ABC and ∆BDE are equilateral triangles and D is

the mid-point of BC. So, DE || AC (i.e., E is the mid-point of AB)

∴ ar(∆BDE) = 14

ar (∆ABC)

= 1 60 154

× = cm2

65. Answer (1)

66. Answer (3)

In ∆ADB,

AB + BD > AD …(i)

In ∆BEC,

BC + CE > BE …(ii)

In ∆ACF,

AC + AF > CF …(iii)

Adding (i), (ii) and (iii), we get

(AB + BC + AC + BD + CE + AF) > AD + BE + CF

…(iv)

Since, AD, BE and CF are medians, then

1 ,2

BD BC= 12

CE AC= and 12

AF AB=

Put these values in (iv), we get

⇒ 32

(AB + BC + AC) > (AD + BE + CF)

⇒ (AB + BC + AC) > 23

(AD + BE + CF)

67. Answer (3)

68. Answer (4)

∠EPA = ∠DPB [Given]

Adding ∠EPD both sides

∴ ∠EPA + ∠EPD = ∠DPB + ∠EPD

⇒ ∠APD = ∠BPE …(i)

Now, In ∆APD and ∆BPE

∠APD = ∠BPE [From equation (i)]

AP = BP [P is the mid-point of AB]

∠BAD = ∠ PBE (Given)

∆APD ≅ ∆BPE [By ASA congruency rule]

∴ AD = BE [By CPCT]

∴ 18 cmBE =

69. Answer (1)


6/10

70. Answer (4)

ar(∆DPC) = 12

ar(ABCD) …(i)

[Triangle on same and between same parallels] ar(ABCD) = 2 ar(GFED) …(ii) From equation (i) and (ii), we get

ar(∆BPC) = 12

ar(ABCD) = ar(GFED)

⇒ ar(∆DPC) = 96 cm2 71. Answer (4) 72. Answer (3) 73. Answer (4) 74. Answer (2) 75. Answer (3) 76. Answer (2) 77. Answer (1) In ∆PQR, ∠P + ∠Q + ∠R = 180° 2x° + 3x° + 4x° = 180° 9x° = 180

20x° = °

∴ ∠P = 40° …(i) TS || PR, then ∠QUS = ∠QPR = 40° [From equation (i)] ∴ Complement of ∠QUS = 90° – 40° = 50° 78. Answer (3)

ar ( ) 12

ABD∆ = ar(∆ABC)

[AD is the median of ∆ABC] …(i)

ar ( ) 12

BDO∆ = ar(∆ABD)

[BO is the median of ∆ABD] …(ii) From equation (i) and (ii), we get

ar ( ) ( )1 1ar 724 4

BDO ABC∆ = ∆ = ×

= 18 cm2

79. Answer (1)

80. Answer (4)

81. Answer (3)

(31)2, (41)2, (51)2, (61)2, (71)2, (81)2

82. Answer (2)

(1)2 × 1 + 1 = 2 (5)2 × 5 + 5 = 130

(2)2 × 2 + 2 = 10 (6)2 × 6 + 6 = 222

(3)2 × 3 + 3 = 30 (7)2 × 7 + 7 = 350

(4)2 × 4 + 4 = 68 (8)2 × 8 + 8 = 520

83. Answer (4)

AIATS | AIATS | AIATS

84. Answer (1)

85. Answer (2)

abc : (a × c) + b

86. Answer (3)

87. Answer (2)

88. Answer (4)

89. Answer (3)

90. Answer (1)

60 m

A

BC

D

45 m 36 m

91. Answer (2)

ROSE → 18 + 15 + 19 + 5 + 4 = 61

92. Answer (3)

RIGHT → 18 + 9 + 7 + 8 + 20 + 5 = 67

93. Answer (2)

94. Answer (1)

95. Answer (3)

Total number of odd days = 31 + 28 + 31 + 30 + 31 + 30 + 20 = 201

That implies net 5 odd days

⇒ Saturday + 5 day → Thursday

7/10


96. Answer (4)

( )1130 H M H 62

= =

360 8M 32 minute11 11

= =

97. Answer (2) 98. Answer (3) 99. Answer (4) 100. Answer (1)

SECTION-II (Code-B) 1. Answer (2)

2. Answer (4)

3. Answer (3)

4. Answer (4)

Hib vaccine and TAB vaccine are given to prevent from influenza type B and typhoid respectively.

5. Answer (3)

6. Answer (4)

7. Answer (2)

8. Answer (3)

9. Answer (3)

10. Answer (4)

11. Answer (3)

12. Answer (3)

13. Answer (3)

14. Answer (4)

15. Answer (2)

16. Answer (3)

17. Answer (4)

Monocots have fibrous root system whereas dicots have tap root system.

18. Answer (3)

19. Answer (4)

20. Answer (2)

21. Answer (3)

22. Answer (2)

23. Answer (3)

24. Answer (2)

25. Answer (2)

26. Answer (4)

27. Answer (4)

28. Answer (3)

29. Answer (2)

Thallophytes contain pyrenoids.

30. Answer (3)

SECTION-III (Code-C) 1. Answer (3)

ar(∆PQR) = ar(∆PQS) + ar(∆PRS)

⇒ 1 1 12 2 2

PQ RX PQ SY PR SZ× × = × × + × ×

⇒ ( )1 12 2

PQ RX PQ SY SZ× × = × × +

[PQ = PR]

⇒ RX = SY + SZ

Squaring both sides,

RX2 = SY2 + SZ2 + 2SY × SZ 2. Answer (4)

Here,

ar(∆LPZ) = ar(∆PQZ) = ar(∆QZN) = x (let)

[ LP = PQ = QN]

∴ ar(∆LZQ) = 2x …(i)


8/10

Now,

( )( )

ar 1 1ar 2 2

MQZ MZ LPQZN NZ PN

∆ = = = ∆

⇒ ( ) ( )arar2 2QZN xMQZ ∆

∆ = =

⇒ 2ar(∆PZN) = x …(ii)

and ar(∆PZN) = 2x …(iii)

and ar(MZQP) = ar(∆QPZ) + ar(∆PZM)

= x + ar(∆LPZ)

[ ar(∆PZM) = ar(∆LPZ) as LM || PZ]

= x + x = 2x …(iv)

and ( )( )

ar 1 1ar 2 2

LMZ MZLNZ NZ

∆ = = ∆

⇒ ( ) ( )ar 3ar2 2LNZ xLMZ ∆

∆ = =

∴ ar(∆LMN) = ar(∆LNZ) + ar(∆LMZ)

= 3 932 2x xx + =

⇒ ( )1 3ar3 2

xLMN∆ = …(iv)

∴ ar(∆LZQ) = ar(∆PZN) = ar(MZQP)

[Using (i), (iii) and (iv)]

3. Answer (4)

4. Answer (4)

BC = CE (Given)

C is the mid-point of BE …(i)

and AB || CD

∴ AB || CF …(ii)

From equation (i) and (ii),

F is the mid-point of AE [Converse of MPT]

In ∆ADF and ∆CFE, [ AD || CE]

AF = EF [F is the mid-point of AE]

∠DAF = ∠CEF [ AD || CE]

AD = CE

∆ADF ≅ ∠CEF [By SAS congruency rule]

∴ DF = CF [By CPCT]

Now,

1ar ( ) 2ar ( )

CF hBCFABCD AB h

× ×∆=

×

=

112

2 4

CF h

CF h

× ×=

×

5. Answer (2) 6. Answer (2) 7. Answer (4)

Draw SV || QU. In ∆PSV, T is the mid-point of PS

and TU || SV. ∴ U is the mid-point of PV

⇒ PU = UV …(i) In ∆QUR, S is the mid-point of QR and SV || QU ∴ V is the mid-point of UR ∴ UV = VR …(ii) From equation (i) and (ii), we have

PU = UV = VR = 13

PR …(iii)

34

PR PQ= [Given] …(iv)

From equation (iii) and (iv), we have

1 33 4

PU UV VR PQ = = =

∴ 14

=PU PQ

8. Answer (2) Join XC

∴ 21ar( ) ar ( ) 24 cm2

XBC XQCB∆ = =

9/10


∆XBC and ∆XBA are drawn on same base XB and between same parallels XB and AC. Then,

ar(∆XBC) = ar(∆XBA) = 24 cm2 …(i) Since ∆XBC and ∆YBC are on same base BC and

between the same parallel BC and XY

∴ ar(∆XBC) = ar(∆BCY) = 24 cm2 …(ii) Clearly, ∆BCY and ∆ACY are on the same base

CY and between the same parallels AB and CY. ∴ ar(∆BCY) = as (∆ACY) …(iii) From equation (ii) and (iii), we get

ar(∆ACY) = ar(∆XBC) = 24 cm2 Since QY is the median of ∆ACY

∴ 21 1ar( ) ar ( ) (24) 12 cm2 2

AQY ACY∆ = ∆ = × =

⇒ ar(∆APQ) = ar(∆AQY) = 12 cm2

(AQ is the median of ∆APY)

∴ ar(∆APQ) + ar(∆CQY) = 12 + 12

= 24 cm2 9. Answer (1)

ABCD is an isosceles trapezium (i.e., AD = BC) AX = BY = 3 cm

Now, 2 2AC AY CY= +

= 2 210 4 116 2 29 cm+ = =

In isosceles trapezium length of both the diagonals are equal

∴ AC + BD = 2 29 2 29+

= 4 29 cm

10. Answer (3) RE || BPQ E is the mid-point of DB.

In ∆DQB, DR = QR

Since RE ⊥ CF

∆RGC ≅ ∆EGC

∠CRG = ∠CEG [By CPCT]

∴ RQPE is an isosceles trapezoid

PE = QR

2RQ = DQ

∴ DQ = 2PE

11. Answer (4)

Let ∠ACB = x,

then, ∠ABC = 2x and ∠BAC = 3x

∴ x + 2x + 3x = 180°

x = 30° i.e. ∠ACB = 30°, ∠ABC = 60° and ∠BAC = 90°

Now, ∠BAN = ∠CAN = 12

∠BAC = 45°

[AN is the bisector ∠BAC]

In ∆AMB,

∠B + ∠M + ∠A = 180° ⇒ ∠A = 30°

∴ AB > AM > BM …(i)

In ∆ANC,

∠C + ∠N + ∠A = 180° ⇒ ∠N = 105°

∴ AC > NC > AN …(ii) In ∆AMN,

∠M + ∠A + ∠N = 180° ⇒ ∠N = 75° and

∠A = 15°

∴ AN > AM > MN …(iii)

In ∆ABC,

BC > AC > AB …(iv)

From equation (i), (ii), (iii) and (iv), we get

NC AM>

12. Answer (2)


10/10

13. Answer (2)

14. Answer (1)

Perimeter of the rhombus = 4 × side = 50 cm

∴ 50 25Side cm4 2

= =

Now, ( )2 2

21 2 side2 2d d

+ =

[diagonals bisect each other at right angle]

⇒ 22 2

215 252 2 2

d + =

⇒ 22225 625

4 4 4d

+ =

⇒ 22 400d =

⇒ d2 = 20 cm

15. Answer (4)

1800-102-2727TOLL FREE NUMBERRegistered Office / DLP Division: Aakash Tower, 8, Pusa Road, New Delhi-110005.

E-mail: [email protected]

www.aakash.ac.in

Our Result in Medical & Engineering Entrance Exams-2019

Our Result in Olympiads / Scholarship Exams 2018-2019

481

in IMO

466 Classroom 15 Distance & Digital

Level-I 2018-2019

585

in KVPY Aptitude Test 2018-2019


12

in INO 2019

11 Classroom 1 Distance

791

in NSO


Level-I 2018-2019

in JEE (Adv.)

69826 Classroom10255 Distance & Digital

80081

in NEET-UG

688

for AIIMS

576 Classroom112 Distance & Digital

1421

in PRMO


2019

751

in NTSE


Stage - I 2018-2019

366

in NTSE


Stage - II 2019

576

in NSEs


2018

90

in RMO

89 Classroom 1 Distance

2018

445

in KVPY


Fellowship Award 2018-2019

7879

in JEE ( )Main



1633

test no. 3 · 2019-11-25 · all india aakash test series-2020 (class ix) test-3 (answers &...

Documents