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