test - 1 (code-c) (answers) all india aakash test …...sol : no. of eq. of acid = 1.22 n 36.5 eq....

Test - 1 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/13

1. (2)

2. (3)

3. (3)

4. (3)

5. (1)

6. (3)

7. (1)

8. (4)

9. (3)

10. (1)

11. (4)

12. (3)

13. (1)

14. (4)

15. (4)

16. (4)

17. (3)

18. (1)

19. (4)

20. (2)

21. (3)

22. (1)

23. (4)

24. (3)

25. (3)

26. (2)

27. (4)

28. (3)

29. (1)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (3)

33. (3)

34. (3)

35. (2)

36. (3)

37. (3)

38. (4)

39. (2)

40. (2)

41. (2)

42. (3)

43. (1)

44. (2)

45. (2)

46. (1)

47. (4)

48. (1)

49. (3)

50. (2)

51. (2)

52. (4)

53. (4)

54. (4)

55. (1)

56. (4)

57. (4)

58. (1)

59. (4)

60. (4)

61. (3)

62. (1)

63. (2)

64. (1)

65. (1)

66. (1)

67. (3)

68. (3)

69. (4)

70. (3)

71. (1)

72. (2)

73. (1)

74. (2)

75. (3)

76. (2)

77. (4)

78. (2)

79. (3)

80. (4)

81. (3)

82. (2)

83. (3)

84. (4)

85. (1)

86. (1)

87. (2)

88. (3)

89. (3)

90. (2)

Test Date : 07/10/2018

ANSWERS

TEST - 1 - Code-C

All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-C) (Hints & Solutions)

2/13

1. Answer (2)

Hint : True value = Mean value

Mean absolute error = 1 2

...

na a a

n

Sol : Mean value, a = 46 s

Mean absolute error = 1.0 s

2. Answer (3)

Hint : If Z = Ap B

q, then Z p A q B

Z A B

Sol :

1 1

3 91

Q P RK

1 1

3 9

Q P R

Q P R

3. Answer (3)

Hint : Use dimensional analysis.

Sol : Let L = FaVbMc

L = (MLT–2)a(LT–1)bMc

Solving a = –1, b = +2, c = +1

4. Answer (3)

Hint :

1 2

1 1 1

R R R

Sol :1 2

1 2

6 124

18

R RR

R R

1 2

2 2 2

1 2

R RR

R R R

0.3 0.34 100%

36 144

R

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

30 30 5 54% 4%

36 144 6 24

R

R

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10 54.1%

3 6

5. Answer (1)

Hint : Take log on both sides and differentiate.

Sol :3

ln ln2

Z A kt

3

2

Z A tkt

Z A t

32 1.25 4.25%

2

PART - A (PHYSICS)

6. Answer (3)

Hint : a = mt + c [c < 0, m > 0]

Sol : ∫v adt

7. Answer (1)

Hint : Rules of significant figures for addition.

Sol : 0.000026

0.00032+

0.000346

There can be only five digits after decimal.

Result is 3.5 × 10–5

8. Answer (4)

Hint : Speed × time = distance

Sol : 50 × 20 = 40 × t2

2

50 2025 s

40t

9. Answer (3)

Hint : All terms of the equation must have same

dimensions. Exponents are dimensionless.

Sol : [a] = LT–1, [b] = T–1

[a2

b3] = L2T–2T–3 = L2T–5

10. Answer (1)


Sol : v × 30 = xx

v × 18 = 2x

2 18 3

30 5

x

x

2 3

10

x

x

11. Answer (4)

Hint : th

(2 1)

2n

a ns u

Sol : th

(2 1)

2n

a ns u

12. Answer (3)

Hint :21

2h gt

Sol : h1

= 5 × 62 = 180 m

h2

= 5 × 32 = 45 m

h1

– h2

= 135 m

Test - 1 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/13

13. Answer (1)

Hint :D

tv

Sol :10

4 5 60

x x

10km

3x

14. Answer (4)

Hint : Look at signs of x and v.

Sol : If x < 0, v > 0

If x > 0, v < 0

15. Answer (4)

Hint :21

2s ut at

s = vt

v = u + at, if a = 0

Sol : 1 22

fft t t

2

= 2t1

2 2 2

1 1 1 1 1

1 1 310 2

2 2 2s ft ft t f ft ft t

1

27 .

ss ft t ft

f

2

1

1

2s ft

1

2st

f

2 2 2 249

ss f t

f

249

2

sft

2

2

49

fts

16. Answer (4)

Hint : For retardation, av < 0

Sol : 2 2dv

a tdt

v = t2 – 2t

17. Answer (3)

Hint :5

3 2

dva v

ds v

Sol :5

3 2

dvvds v

10 5

2

0 0

(3 2 ) 5v v dv ds ∫ ∫

103 2

0

( ) 5 sv v

1100 = 5s s = 220 m

18. Answer (1)

Hint :dv

a vdx

Sol :dv

a vdx

2

02

x

vadx ∫

2

202

v = 40 m/sv

19. Answer (4)

Hint :dv

a vdx

Sol :dv

a vdx

a = (–2.5x + 10 )(–2.5)

ax=1

= 6.25 – 25 = –18.75 m/s2

20. Answer (2)

Hint : vAR

= vA – v

R

Distance = Speed × Time

Sol :9 60

(60 )60 4

v = 15

15 6060 100

9v

v = 40 km/h

21. Answer (3)

Hint :

12 feet

4 feet

x1

x2

1= 2 m/s

dx

dt

2

I

dxv

dt

Sol :2 12

3 m/ s12 4

sv


4/13

22. Answer (1)

Hint : Consider velocity profile of the two balls.

Sol :

1 2 3 4

t

5t1

2 = 5

t1

= 1 s

t2

= 0.5 s

23. Answer (4)

Hint :

0.5 m

B

AO2 m

0.5

2

y

x

Sol :0.5

2

y

x

2

1.

dy dx

dt dtx

h = 0.5 m

y

O

2 m

x

= 2

110

(0.5)

= – 40 cm/s

24. Answer (3)

Hint : vt = distance

Sol : 160 90

x x

2 1802 72 km

5x

1

360.6 h

60t , t

2

= 0.4 h

211 72

2a

a = 144 km/h2

144t = 90

90 5

h144 8

t

25. Answer (3)

Hint :dv

adt

Sol :0

0 0

v t

btdv a e dt

∫ ∫

10 0

0

1bt bta a

v e eb b

⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦

26. Answer (2)

Hint : arel

= a1

– a2

Sol : arel

= 0

vrel

= constant

At t = 2 s, vA = 60 – 10 × 2 = 40 m/s,

vB = 25 m/s, v

AB = 15 m/s

27. Answer (4)

Hint :21

52

H a

Sol :25

2H a , u = a5

2255 12.5 5 (12.5)

2a a

2

25 (12.5) 5 12.510.41m/s

6 12.5 6a

28. Answer (3)

Hint :Displacement

Average velocityTime

DistanceAverage speed

Time

Sol : x

t (s)3/8 1.5 2.5

4

8Average velocity

2.25

32m/ s

9

32Average speed m/ s

3


5/13

PART - B (CHEMISTRY)

31. Answer (4)

Hint : mol of H2

O2

= mol of I2

= 2 2 3

mol of Na S O

2

Sol : From reaction (I),

H2

O2

+ 2KI I2

+ 2KOH

mol of H2

O2

= mol of I2

mol of H2

O2

= 100 × 0.1 = 10 mmol

From reaction (II),

2(mol of I2

) = mol of Na2

S2

O3

2 × 10 = 0.2 × x

x = 100 mL

mol of KOH = 2(mol of H2

O2

)

mol of KOH = 20 × 10–3 mol

Mass of KOH = 56 × 20 × 10–3 = 1.12 g

mol of I2

= 10 × 10–3 mol

Mass of I2

= 254 × 10 × 10–3 = 2.54 g

mol of NaI = 2(mol of I2

)

= 2 × 10 × 10–3 = 0.02 mol

32. Answer (3)

Hint : Due to mixing, volume become 600 mL.

Mass of solution A = 600 g

Mass of H2

SO4

= 49 g

and Mass of solution B = 140 g

Mass of H2

SO4

= 19.6 g

Sol : Molarity = 2 4

Moles of H SO

Volume of solution

Total mol of H2

SO4

= 0.5 + 0.2 = 0.7

Total volume = 600 = 0.6

Molarity = 0.7

0.6 = 1.167 M.

Normality = 1.167 × 2 = 2.33 N

Mass of solution A = 500 × 1.2 = 600 g

29. Answer (1)

Hint : dy = sec2d

Sol : dy = sec2d

25 3.6

16 180y

25 2

3.1416 100

= 0.0981

30. Answer (4)

Hint : (i) v = a3

(ii) Rule of significant figures.

Sol : v = (4.40)3

v = 85.2 cm3

Mass of H2

SO4

= 49 g

Mass of solution B = 140 g

Mass of H2

SO4

in B = 19.6 g

(w/w)% of H2

SO4

= 68.6

100 9.27740

(w/w)% of H2

SO4

= 68.6

100 11.43600

33. Answer (3)

Hint :2 3 2 2 3 2

Na CO ·xH O Na CO xH O

Sol : mol of Na2

CO3

·

xH2

O = 2.32

106 18x

2 3 2 2 3 2Na CO ·xH O Na CO xH O

1 mol of Na2

CO3

·

xH2

O

= Produce x mol of H2

O

mol of Na2

CO3

·

xH2

O = 2.32

106 18xMass of water

18x 2.321.26

106 18x

18x × 2.32 = 133.56 + 22.68x

19.08x = 133.56

x = 7

34. Answer (3)

Hint : Eq. of acid = Eq. of base

Sol : No. of eq. of acid = 1.22 n

36.5

Eq. of NaOH = 4

0.140

Eq. of acid = Eq. of base

1.22 n 4 1

36.5 40

n = 3

Formula of acid is H3

A.

Molar mass of H3

A = 36.5 g

So, atomic mass of A = 33.5 g


6/13

35. Answer (2)

Hint : Average relative atomic mass =

n

i i

i 1

MX

100

∑

Sol : Relative mass approach to 40 so most of the

Ar is present as 40Ar.

36. Answer (3)

Hint : M + 2HCl MCl2

+ H2

Metal belongs to 2nd group.

n-factor = 2

Sol : Equivent mass of metal = 0.338 × 35.5

= 11.99 g � 12

Metal is M,

Eq. of M in 2.4 g = 2.4

0.212

mol of H2

= 2.4

0.122.4

mol of H2

= 0.1

M + x HCl MClx

+ 2

xH

2

x1

2

x = 2

37. Answer (3)

Hint : A = Li, x = +2, n = 4

Sol : Separation energy = 7.65 eV

–En

= 7.65

2

2

13.6 Z

n= 7.65 …(i)

E(n+2)

– En

= –19

h

1.6 10

2 2 34 15

2 2 19

13.6 Z 13.6 Z 6.63 10 1.025 10

n (n 2) 1.6 10

2

2

7.65 n7.65

(n 2)

= 4.25

22

2

n 1

1.5(n 2)

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ n = 4

From equation (i),

Z = 3

A+x = Li+2

38. Answer (4)

Hint : Spin quantum number was not given by

Schrodinger.

Sol : Quantum numbers n, �, m were derived from

solution of Schrodinger wave equation.

39. Answer (2)

Hint : mol of CaCO3

= mol of CO2

= mol of CO

2

Sol : Let mol of I2

= x

mol of CO = 5x

mol of CO2

= 5x

2

mol of CaCO3

= 2.5x

Mass of CaCO3

= 2.5 × 100 × x

mol of I2

= 12.7

0.05154

mol of CO = 0.25

mol of CO2

= 0.25

2

mol of CaCO3

= 0.25

2

Mass of CaCO3

sample = 0.25 100

2 0.8

40. Answer (2)

Hint : Fe + O2

FeO + Fe2

O3

Use POAC

Sol : Fe(s) + O2

FeO + Fe2

O3

mol of Fe = 1.68

56 = 0.03

mol of Fe in FeO and Fe2

O3

= 0.03

mol of oxygen (O) atom = 0.64

16 = 0.04

mol of 'O' in FeO and Fe2

O3

= 0.04

mol of FeO = x

mol of Fe2

O3

= y

So, mol of Fe = x + 2y = 0.03

mol of 'O' = x + 3y = 0.04

x 0.01

y 0.01

⎡ ⎤⎢ ⎥⎣ ⎦


7/13

41. Answer (2)

Hint :2

2 2

1 2

1 1R Z

n n

⎛ ⎞ ⎜ ⎟⎝ ⎠

n1

for Balmer = 2

Sol : x = 1 1

R4 9

⎛ ⎞⎜ ⎟⎝ ⎠

x = 5

R36

⎛ ⎞⎜ ⎟⎝ ⎠

y = 1 1

R 4 3R1 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 1 8RR 1

z 9 9

⎛ ⎞ ⎜ ⎟⎝ ⎠

z = 9

8R

R = 36x

5

xyz = 5 9R 3R

36 8R

15R 15 36x 108x

32 32 5 32

42. Answer (3)

Hint : n = 5 to 1

Sol : Possible line = 5 4

102

43. Answer (1)

Hint : Possible value of n

2 and 3

Possible value of � = 0 to 2, m = –� to +�

Sol : n = 3, � = 2, m = –2

n = 3, � = 1, m = –1

n = 3, � = 0, m = 0

n = 2, � = 1, m = 0

So, Zn has maximum electron.

44. Answer (2)

Hint : = 2 × 0.53 × n

Z

⎛ ⎞⎜ ⎟⎝ ⎠

Sol : For He+,

P = 3

2 0.532

⎛ ⎞ ⎜ ⎟⎝ ⎠

Q = 1

2 0.531

⎛ ⎞ ⎜ ⎟⎝ ⎠

R = 2

2 0.533

⎛ ⎞ ⎜ ⎟⎝ ⎠

So, order is P > Q > R.

Hence, the correct answer is (2).

45. Answer (2)

Hint : Spin only magnetic moment

Number of unpaired electron

Sol : For maximum , n must be maximum and

in d-block that is possible for Mn.

For n(n 2) BM

For maximum , n = 5

For M2+, n = 5 only possible with Mn.

46. Answer (1)

Hint : Metal and solid.

Sol : Most of the elements are metal and solid.

47. Answer (4)

Hint : Velocity of electron in H-atom is minimum for

the 1st orbit.

P = 1

n = 2, 3, …

Sol : Series belongs to Lyman series.

1 1 1 4R R 1216 Å

1 4 3

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

1 1 1R 1 912 Å

R

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

So, emitted photon have wavelength

between 912 Å to 1216 Å.

48. Answer (1)

Hint : Velocity of electron more than ‘c' is not

possible.

Sol : If P = x,

P·P = h

4

mV = h

4

V = 1 h

m 4V = 7.98 × 1012 m/s

Not possible.


8/13

For excitation from transition,

� must be = ±1

S = 1

2 for any electron.

49. Answer (3)

Hint : Closest distance of approach

KE of alpha particle = Potential energy

Sol : R =

2

0

2Ze

4 KE

Z = 79 for gold

0

1

4 = 9.1 × 109

KE = 7.7 MeV = 7.7 × 106 × 1.6 × 10–19 J

50. Answer (2)

Hint : n = 2(n – 1)

Sol : n = 2(n – 1)

n = 2n – 2

n = 2

Possible shell is 2p with Z = 7

51. Answer (2)

Hint : Mass of H2

SO4

= 98 mol of NaOH

2

Mass of SO3

in 1 g oleum

= 2 4

(Total mass of H SO after dilution 1g) 80

18

Sol : mmol of NaOH = 222.449 × 0.1 = 22.2449

mmol of H2

SO4

= 22.2449

2

Mass of H2

SO4

= 22.2449 98

2 1000

Mass of H2

SO4

= 1.09 g

Mass of water added = 0.09 g

mol of water = mol of SO3

= 0.09

18

Mass of SO3

= 0.09 80

18

= 0.4 g

% of SO3

= 40%

52. Answer (4)

Hint : If gap between IEn

and IEn+1

is maximum,

then n is valence electron.

Sol : Gap in IE4

and IE5

is maximum so valence

electron is 4, hence element is Si.

53. Answer (4)

Hint : Isoelectronic ions have same number of

electrons.

Sol : Isoelectronic ions have different size.

54. Answer (4)

Hint & Sol :

Mass of NH4

Cl = 280.1 – 224.3 g = 55.8 g

Mass of water = 1239.5 – 280.1 = 959.4 g

Mass of solution = 1015.2 g

4

w 55.8 100% of NH Cl 5.496%

w 1015.2

⎛ ⎞ ⎜ ⎟⎝ ⎠

55. Answer (1)

Hint : Size E.N.

Sol : E.N. : F > Cl > C > Si

56. Answer (4)

Hint : Group number = 3

Sol : Group number, n = 3

Period number = 4

Element X is scandium (Sc).

57. Answer (4)

Hint : Size of cation is always smaller than its

parent atom.

Sol : Size K+ < K.

58. Answer (1)

Hint : f = V

2 r = 8.23 × 1014

Sol : Frequency of revolution =

18

34 3

2.18 10 2

6.63 10 n

⎛ ⎞

⎜ ⎟ ⎝ ⎠

n = 2

2s has one radial node and zero angular node.

x = 1, y = 0

59. Answer (4)

Hint : IE depends on size and electronic

configuration for period element.

Sol : I.E.: F > N > O > C > B


9/13

PART - C (MATHEMATICS)

61. Answer (3)

Hint : A B = {x : x A or x B} and total

number of subsets of set A with n elements

= 2n

Sol : Here, A = {2, 3} and B = {– 3, – 2, – 1, 0,

1, 2, 3}

A B = {– 3, – 2, – 1, 0, 1, 2, 3}

Number of subsets of A B = 27 = 128

62. Answer (1)

Hint : Total number of subsets of A × B = 2n(A) × n(B)

and use number of subsets of set A with

n elements taking r elements = nCr

Sol : A × B has 2 × 3 = 6 elements

Total number of subsets of A × B = 26 = 64

Number of subsets with zero elements

= 6

C0

= 1

Number of subsets with one element = 6C1

= 6

Number of subsets with two element = 6

C2

6! 6 5 4!15

2! 4! 2 1 4!

Required number of subsets of A × B

= 64 – (1 + 6 + 15)

= 64 – 22 = 42

60. Answer (4)

Hint : Total mass of Na+ is calculated from Na2

CO3

and NaCl.

% of Na+

= 2 3

Mass of Na from (Na CO NaCl) 100

total mass

% of NaCl = y

100x y

Sol : mol of Na2

CO3

= x

106

mol of Na+ = 2x

106

mol of NaCl = y

58.5

mol of Na+ = y

58.5

Total mol of Na+ = 2x y

106 58.5

Mass of total Na+ = 2x y

23106 58.5

⎛ ⎞⎜ ⎟⎝ ⎠

% of Na+ =

2x y23 100

106 58.5

x y

⎛ ⎞ ⎜ ⎟⎝ ⎠

% of NaCl = y 100

x y

By solving, we get

x = 106 g

y = 58.5 g

mol of Na2

CO3

= 1 mol

mol of NaCl = 1 mol

63. Answer (2)

Hint : n(A B) = n(A) + n(B) – n(A B) which is

maximum if n(A B) is minimum and is

minimum if n(A B) is maximum.

Sol : n(A B) = n(A) + n(B) – n(A B) and

n(A B) is maximum if n(A B) is

minimum and n(A B) is minimum if

n(A B) is maximum.

Maximum n(A B) = 15 + 10 – 0 = 25

Mininum n(A B) = 15 + 10 – 10 = 15

Their sum = 25 + 15 = 40

64. Answer (1)

Hint : Required number of elements = n(A B) +

n(B C) + n(C A) – 3n(A B C)

Sol A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

B = {– 2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}

C = {2, 3, 5, 7, 11, ...}

A B = {1, 2, 3, 4, 5, 6, 7},

B C = {2, 3, 5, 7}, C A = {2, 3, 5, 7}

and A B C = {2, 3, 5, 7}

The required number

= n(A B) + n(B C) + n(C A)

– 3n(A B C)

= 7 + 4 + 4 – 3 × 4 = 3


10/13

65. Answer (1)

Hint : Using properties of log, find a relation in

p and q squaring both sides, find a relation

in p and q, then solving them find values of

p and q.

Sol : Given, log10

(p – 2) + log10

q = 0

q(p – 2) = 100 = 1

p – 2 = 1

q p =

12

q ...(i)

Given, 2p q p q

Squaring, we get

p + q – 2 + 2 ( 2)p q = p + q

p(q – 2) = 1

1

2qp

...(ii)

From (i) and (ii), p = q

From (i),

p2 – 2p – 1 = 0

(p – 1)2 = 2

1 2p

But q 2

1 2p q

[p + q] = 4

66. Answer (1)

Hint : {x + n} = {x}, if n I

Sol : Here, f(x) = {x} + {x} + ... to 10 terms = 10{x},

since {x + [x]} = {x}

( 2) 10{ 2} 10{1.414}f

= 10 × 0.414 10 414 414

4.141000 100

[ ( 2)] 4f

67. Answer (3)

Hint : Use properties of |x|.

Sol : Given, x2 – |x| – 6 0

(|x| – 3)(|x| + 2) 0

x x+ +–

– 2 3

| | [ 2, 3]x

But |x| 0 |x| [0, 3] 0 |x| 3

0 x 3 and 0 – x – 3

0 x – 3 and 0 x 3

[ 3, 3]x

Number of integral values of x = 7.

68. Answer (3)

Hint : Solving equations, [x] = 4 and y = 11

Sol : Given, y = 2 [x] + 3 ...(i)

and y = 3[x] – 1 ...(ii) [From (ii) equation]

By (ii) and (i),

[x] = 4 and so from (i) y = 11 when [x] = 4

4 x < 5

[x + y] = 15

69. Answer (4)

Hint : From given equation x2 + 3x – 4 0 and for

log function to be defined x(x + 3) > 0.

Then solving them find values of x.

Sol : For log function to be defined,

x2 + 3x > 0

x(x + 3) > 0

– – 3 0 +

+ ve + ve

– ve

( , 3) (0, )x ...(i)

and from given inequation log2

(x2 + 3x) 2.

– 4 1

+ ve + ve– ve

x2 + 3x – 4 0

x [–4, 1] ...(ii)

From (i) and (ii), x [– 4, – 3) (0, 1],

taking intersection of (i) and (ii).

70. Answer (3)

Hint : Use property of |x|.

Sol : From given equation,

||2 – |x – 2|| = 1

|x – 2| – 2 = 1, – 1

|x – 2| = 3, 1

x – 2 = – 1, 1, 3, – 3

x = 1, 3, 5, –1


11/13

71. Answer (1)

Hint : Take pair (a, b), a A, b B, such that

|a – b| is even number.

Sol : Here, R = {(1, 5), (1, 7), (1, 9), (2, 8), (3, 5),

(3, 7), (3, 9), (4, 8)}

Number of elements in relation R = 8

72. Answer (2)

Hint : logg(x)

f(x) is defined if f(x) > 0, g(x) > 0,

g(x) 1.

Sol : Here, for log function to be defined 5

02

x

and 5

12

x and

25

02 3

x

x

⎛ ⎞ ⎜ ⎟⎝ ⎠.

– –5 3

2

even even

+ve +ve +ve

3 3

( , 5) 5, ,2 2

x⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

...(i)

Also, 5 3

and2 2

x x ...(ii)

From (i) and (ii),

5 3 3, ,

2 2 2x

⎛ ⎞ ⎧ ⎫ ⎨ ⎬⎜ ⎟⎝ ⎠ ⎩ ⎭

73. Answer (1)

Hint : Draw the graph of y = x2 and y = 1 – |x – 5|

and see the point of intersection of these two

graphs.

Sol : From given equation, x2 = 1 – |x – 5|

Now, draw the graph of y = x2 and

y = 1 – |x – 5| as shown below.

x x

y

O

1 (5, 1)

y x = 2

It is evident from the graph that two curves

do not intersect at any point.

Number of real solutions = 0

74. Answer (2)

Hint : 12 × 10 = 15 × 8

Sol : f(120) = 12f(10) + 10f(12) = 15f(8) + 8f(15)

12 19 + 10 52 = 15f(8) + 8 26

f(8) = 36

75. Answer (3)

Hint : Use replacement property.

Sol : ∵ 3f(x) – 22

5 3f x xx

⎛ ⎞ ⎜ ⎟⎝ ⎠...(i)

Replacing x by 2

x

, we get

2

2 2 45 ( ) 3 3f x f

x x x

⎛ ⎞ ⎜ ⎟⎝ ⎠...(ii)

Solving (i) and (ii) for f(x), we get

f(x) = 2

2

1 10 2024 3 3

16x x

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

f(1) = 34 17

16 8

76. Answer (2)

Hint : Use replacement property and {n + x} = {x},

if n N and logam

n = nlogam and log

aa = 1

Sol : Here, 410

( ) (10 )3

f f f⎛ ⎞ ⎜ ⎟⎝ ⎠

4

10

10[ ] 3 2 log 10

3

⎛ ⎞⎧ ⎫ ⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

14 3 2 4 1

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

77. Answer (4)

Hint : n[(A × B) B × A)] = [n(A B]2

Sol : A = {2, 3, 5, 7}, B = {1, 2, 3, 4, 5}

A B = {2, 3, 5}

n[(A × B) (B × A)] = [n(A B)]2 =

32 = 9 ordered pairs

78. Answer (2)

Hint : Put y = p(x)

2

2 2 2 2

14

4 ( 1)

( 1) 11

x

x x x

x x

x

x

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

4 4 1use

41

tt x

xtt

t


12/13

Sol : Put

2

2 2 2 2

14

4 ( 1)( )

( 1) 11

xx x x

y p xx x

xx

⎛ ⎞⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟

⎝ ⎠

Put 1

x t

x

, for x > 0, t 2

2

4 4

1 /1

ty

t tt

But for t 2, 1 1 5

22 2

tt

1 2

1 / 5t t

4 2 4

1/ 5t t

80

5y

and y = 0 for x = 0 also

8 8

0 0,5 5

y y⎡ ⎤ ⇒ ⎢ ⎥⎣ ⎦

79. Answer (3)

Hint : First find range of

4

2

1x

x

using AM GM.

Sol : Here,

4

2

2 2

1 12

x

x

x x

14

22 22

1log log 2

x

x

f(x) 1

2

1

( ) ,2

f x⎡ ⎞ ⎟⎢⎣ ⎠

80. Answer (4)

Hint :( )

( ) , ( ) 0( )

f g

f f xx g x x D D

g g x

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol : Here, 7,

2f

D⎡ ⎞ ⎟⎢ ⎠⎣

and Dg = (–, 1]

f gD D ( )

fx

g

⎛ ⎞⎜ ⎟⎝ ⎠

is undefined,

since ( )

( )( )

f f xx

g g x

⎛ ⎞⎜ ⎟⎝ ⎠

is possible only if

( ) 0g x and f gx D D

But, here x Df D

g

81. Answer (3)

Hint : Use transformations of graphs.

Sol : T1

: Graph of y = log|x| = log , 0

log ( ) , 0

e

e

x x

x x

⎧⎨ ⎩

Graph of y = log|x| is as follows

y

y

xx(1, 0)( 1, 0)–

y x = log(– ) y x = ln

T2 : Graph of y = ln|x| – 1

When y = 0, ln|x| = 1 |x| = e x = ± e

when x = 0, y –

and1 | | 1

0| |

dy x

dx x x x for x > 0 < 0 for

x < 0

y

y

xx ( , 0)e( , 0)–e

T3 : Finally graph of f(x) = |ln|x| – 1|

y

y

xx ( , 0)e( , 0)–e

82. Answer (2)

Hint : f(x) is defined if [x – 5] 0 and x – [x] 0.

Sol : Here, f(x) is defined if [x – 5] 0

( 5) [0,1)x

[5, 6)x

and x – [x] 0 x I

Df = (–, 5) (6, ) – I


13/13

83. Answer (3)

Hint : If any vertical line parallel to y-axis cuts the

graph at only one point, then that graph

represents a function.

Sol : If the vertical lines parallel to y-axis

intersects the graph of curve at only one

point, then that is the graph of function. If we

draw any line parallel to y-axis in the graph

of option (3) cuts at only one point.

Graph in option (3) represents a function.

84. Answer (4)

Hint :25 24 23 22018

( ) ( 1)( ... )1

f x x x x x x xx

Sol : f(x) = x25 – 2018{(x24 – x23 + x22 + ... + x2 – x)}

= x25 –2018

1 x{(x24 – x23 + x22 + ... + x2 – x)

+ (x25 – x24 + x23 + ... + x3 – x2)}

= x25 – 2018

1 x(x25 – x)

= x25 –

252018( )

1

x x

x

f(2017) = 201725 –

252018(2017 2017)

2018

= 2017

85. Answer (1)

Hint : Add all three equations.

Sol : Adding, (log2

(ABC))2 = 144

log2

(ABC) = ±12

So, log2

(ABC) = 12 ABC = 212

log2

A = 4, log2

B = 1, log2

C = 7

A = 24, B = 2, C = 27

86. Answer (1)

Hint : Vertex is (1, –1) and coefficient of x2 = 1 > 0

parabolic curve is vertically upward.

Sol : Here, f(x) = x2 – 2x + 1 – 1 = x(x – 2)

curve cuts x-axis at (0, 0) and (2, 0), vertex

is (1, –1) and graph is vertically upward as

coefficient of x2 = 1 > 0.

87. Answer (2)

Hint : Draw graphs of y = 4–|x| and y = 3 – |x|

Sol : From given equation, 3 – |x| = 4–|x|

Here, draw the graphs of

y = 4–|x| ...(i)

and y = 3 – |x| ...(ii)

y

y

x x

(0, 1)

O (0, 3)(–3, 0)

(0, 3)

y x = 3 – | |

y = 4–| |x

From graph, we observe that curves (i) and (ii)

intersects at two points only.

Two real solutions (–3, 3).

88. Answer (3)

Hint : x – [x] = {x} and 0 {x} < 1

Sol : Range of sgn(x – [x]) is {0, 1}, since

0 x – [x] < 1.

Range of (sgn(x – [x]) + 1) is {1, 2}.

89. Answer (3)

Hint : n(A × B) = n(A) × n(B)

Sol : n(A × B × A) = n(A) × n(B) × n(A)

= 2 × 3 × 2 = 12

Total number of proper subsets of

A × B × A = 212 – 1.

90. Answer (2)

Hint : Divide both sides by log2

x log4

x log

6

x and

use properties of log.


6 2 4

1 1 11 , 1

log log logx

x x x

1 log 6 log 2 log 4x x x

1 log 48x

x = 48

Also, x = 1 satisfies given equation.

Sum of solutions = 48 + 1 = 49

� � �

Test - 1 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/13

1. (4)

2. (1)

3. (3)

4. (4)

5. (2)

6. (3)

7. (3)

8. (4)

9. (1)

10. (3)

11. (2)

12. (4)

13. (1)

14. (3)

15. (4)

16. (4)

17. (4)

18. (1)

19. (3)

20. (4)

21. (1)

22. (3)

23. (4)

24. (1)

25. (3)

26. (1)

27. (3)

28. (3)

29. (3)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (4)

33. (1)

34. (4)

35. (4)

36. (1)

37. (4)

38. (4)

39. (4)

40. (2)

41. (2)

42. (3)

43. (1)

44. (4)

45. (1)

46. (2)

47. (2)

48. (1)

49. (3)

50. (2)

51. (2)

52. (2)

53. (4)

54. (3)

55. (3)

56. (2)

57. (3)

58. (3)

59. (3)

60. (4)

61. (2)

62. (3)

63. (3)

64. (2)

65. (1)

66. (1)

67. (4)

68. (3)

69. (2)

70. (3)

71. (4)

72. (3)

73. (2)

74. (4)

75. (2)

76. (3)

77. (2)

78. (1)

79. (2)

80. (1)

81. (3)

82. (4)

83. (3)

84. (3)

85. (1)

86. (1)

87. (1)

88. (2)

89. (1)

90. (3)

Test Date : 07/10/2018

ANSWERS

TEST - 1 - Code-D

All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020 Test - 1 (Code-D) (Hints & Solutions)

2/13

1. Answer (4)

Hint : (i) v = a3

(ii) Rule of significant figures.

Sol : v = (4.40)3

v = 85.2 cm3

2. Answer (1)

Hint : dy = sec2d

Sol : dy = sec2d

25 3.6

16 180y

25 2

3.1416 100

= 0.0981

3. Answer (3)

Hint :Displacement

Average velocityTime

DistanceAverage speed

Time

Sol : x

t (s)3/8 1.5 2.5

4

8Average velocity

2.25

32m/ s

9

32Average speed m/ s

3

4. Answer (4)

Hint :21

52

H a

Sol :25

2H a , u = a5

2255 12.5 5 (12.5)

2a a

2

25 (12.5) 5 12.510.41m/s

6 12.5 6a

5. Answer (2)

Hint : arel

= a1 – a

2

PART - A (PHYSICS)

Sol : arel

= 0

vrel

= constant

At t = 2 s, vA = 60 – 10 × 2 = 40 m/s,

vB = 25 m/s, v

AB = 15 m/s

6. Answer (3)

Hint :dv

adt

Sol :0

0 0

v t

btdv a e dt

∫ ∫

10 0

0

1bt bta a

v e eb b

⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦7. Answer (3)

Hint : vt = distance

Sol : 160 90

x x

2 1802 72 km

5x

1

360.6 h

60t , t

2 = 0.4 h

211 72

2a

a = 144 km/h2

144t = 90

90 5

h144 8

t

8. Answer (4)

Hint :

0.5 m

B

AO2 m

0.5

2

y

x

Sol :0.5

2

y

x

2

1.

dy dx

dt dtx

h = 0.5 m

y

O

2 m

x

= 2

110

(0.5)

= – 40 cm/s

Test - 1 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/13

9. Answer (1)

Hint : Consider velocity profile of the two balls.

Sol :

1 2 3 4

t

5t1

2 = 5

t1 = 1 s

t2 = 0.5 s

10. Answer (3)

Hint :

12 feet

4 feet

x1

x2

1= 2 m/s

dx

dt

2

I

dxv

dt

Sol :2 12

3 m/ s12 4

sv

11. Answer (2)

Hint : vAR

= vA – v

R

Distance = Speed × Time

Sol :9 60

(60 )60 4

v = 15

15 6060 100

9v

v = 40 km/h

12. Answer (4)

Hint :dv

a vdx

Sol :dv

a vdx

a = (–2.5x + 10 )(–2.5)

ax=1

= 6.25 – 25 = –18.75 m/s2

13. Answer (1)

Hint :dv

a vdx

Sol :dv

a vdx

2

02

x

vadx ∫

2

202

v = 40 m/sv

14. Answer (3)

Hint :5

3 2

dva v

ds v

Sol :5

3 2

dvvds v

10 5

2

0 0

(3 2 ) 5v v dv ds ∫ ∫

103 2

0

( ) 5 sv v

1100 = 5s s = 220 m

15. Answer (4)

Hint : For retardation, av < 0

Sol : 2 2dv

a tdt

v = t2 – 2t

16. Answer (4)

Hint :21

2s ut at

s = vt

v = u + at, if a = 0

Sol : 1 22

fft t t

2 = 2t

1

2 2 2

1 1 1 1 1

1 1 310 2

2 2 2s ft ft t f ft ft t

1

27 .

ss ft t ft

f

2

1

1

2s ft

1

2st

f

2 2 2 249

ss f t

f

249

2

sft

2

2

49

fts


4/13

17. Answer (4)

Hint : Look at signs of x and v.

Sol : If x < 0, v > 0

If x > 0, v < 0

18. Answer (1)

Hint :D

tv

Sol :10

4 5 60

x x

10km

3x

19. Answer (3)

Hint :21

2h gt

Sol : h1 = 5 × 62 = 180 m

h2 = 5 × 32 = 45 m

h1 – h

2 = 135 m

20. Answer (4)

Hint : th

(2 1)

2n

a ns u

Sol : th

(2 1)

2n

a ns u

21. Answer (1)


Sol : v × 30 = xx

v × 18 = 2x

2 18 3

30 5

x

x

2 3

10

x

x

22. Answer (3)

Hint : All terms of the equation must have same

dimensions. Exponents are dimensionless.

Sol : [a] = LT–1, [b] = T–1

[a2b

3] = L2T–2T–3 = L2T–5

23. Answer (4)


Sol : 50 × 20 = 40 × t2

2

50 2025 s

40t

24. Answer (1)

Hint : Rules of significant figures for addition.

Sol : 0.000026

0.00032+

0.000346

There can be only five digits after decimal.

Result is 3.5 × 10–5

25. Answer (3)

Hint : a = mt + c [c < 0, m > 0]

Sol : ∫v adt

26. Answer (1)

Hint : Take log on both sides and differentiate.

Sol :3

ln ln2

Z A kt

3

2

Z A tkt

Z A t

32 1.25 4.25%

2

27. Answer (3)

Hint :

1 2

1 1 1

R R R

Sol :1 2

1 2

6 124

18

R RR

R R

1 2

2 2 2

1 2

R RR

R R R

0.3 0.34 100%

36 144

R

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

30 30 5 54% 4%

36 144 6 24

R

R

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10 54.1%

3 6

28. Answer (3)

Hint : Use dimensional analysis.

Sol : Let L = FaVbMc

L = (MLT–2)a(LT–1)bMc

Solving a = –1, b = +2, c = +1


5/13

PART - B (CHEMISTRY)

31. Answer (4)

Hint : Total mass of Na+ is calculated from Na2CO

3

and NaCl.

% of Na+

= 2 3

Mass of Na from (Na CO NaCl) 100

total mass

% of NaCl = y

100x y

Sol : mol of Na2CO

3 =

x

106

mol of Na+ = 2x

106

mol of NaCl = y

58.5

mol of Na+ = y

58.5

Total mol of Na+ = 2x y

106 58.5

Mass of total Na+ = 2x y

23106 58.5

⎛ ⎞⎜ ⎟⎝ ⎠

% of Na+ =

2x y23 100

106 58.5

x y

⎛ ⎞ ⎜ ⎟⎝ ⎠

% of NaCl = y 100

x y

By solving, we get

x = 106 g

y = 58.5 g

mol of Na2CO

3 = 1 mol

mol of NaCl = 1 mol

32. Answer (4)

Hint : IE depends on size and electronic

configuration for period element.

Sol : I.E.: F > N > O > C > B

33. Answer (1)

Hint : f = V

2 r = 8.23 × 1014

Sol : Frequency of revolution =

18

34 3

2.18 10 2

6.63 10 n

⎛ ⎞

⎜ ⎟ ⎝ ⎠

n = 2

2s has one radial node and zero angular node.

x = 1, y = 0

34. Answer (4)

Hint : Size of cation is always smaller than its

parent atom.

Sol : Size K+ < K.

35. Answer (4)

Hint : Group number = 3

Sol : Group number, n = 3

Period number = 4

Element X is scandium (Sc).

36. Answer (1)

Hint : Size E.N.

Sol : E.N. : F > Cl > C > Si

37. Answer (4)

Hint & Sol :

Mass of NH4Cl = 280.1 – 224.3 g = 55.8 g

Mass of water = 1239.5 – 280.1 = 959.4 g

Mass of solution = 1015.2 g

4

w 55.8 100% of NH Cl 5.496%

w 1015.2

⎛ ⎞ ⎜ ⎟⎝ ⎠

38. Answer (4)

Hint : Isoelectronic ions have same number of

electrons.

Sol : Isoelectronic ions have different size.

29. Answer (3)

Hint : If Z = Ap B

q, then Z p A q B

Z A B

Sol :

1 1

3 91

Q P RK

1 1

3 9

Q P R

Q P R

30. Answer (2)

Hint : True value = Mean value

Mean absolute error = 1 2

...

na a a

n

Sol : Mean value, a = 46 s

Mean absolute error = 1.0 s


6/13

39. Answer (4)

Hint : If gap between IEn and IE

n+1 is maximum,

then n is valence electron.

Sol : Gap in IE4 and IE

5 is maximum so valence

electron is 4, hence element is Si.

40. Answer (2)

Hint : Mass of H2SO

4 =

98 mol of NaOH

2

Mass of SO3 in 1 g oleum

= 2 4

(Total mass of H SO after dilution 1g) 80

18

Sol : mmol of NaOH = 222.449 × 0.1 = 22.2449

mmol of H2SO

4 =

22.2449

2

Mass of H2SO

4 =

22.2449 98

2 1000

Mass of H2SO

4 = 1.09 g

Mass of water added = 0.09 g

mol of water = mol of SO3 =

0.09

18

Mass of SO3 =

0.09 80

18

= 0.4 g

% of SO3 = 40%

41. Answer (2)

Hint : n = 2(n – 1)

Sol : n = 2(n – 1)

n = 2n – 2

n = 2

Possible shell is 2p with Z = 7

42. Answer (3)

Hint : Closest distance of approach

KE of alpha particle = Potential energy

Sol : R =

2

0

2Ze

4 KE

Z = 79 for gold

0

1

4 = 9.1 × 109

KE = 7.7 MeV = 7.7 × 106 × 1.6 × 10–19 J

43. Answer (1)

Hint : Velocity of electron more than ‘c' is not

possible.

Sol : If P = x,

P·P = h

4

mV = h

4

V = 1 h

m 4V = 7.98 × 1012 m/s

Not possible.

For excitation from transition,

� must be = ±1

S = 1

2 for any electron.

44. Answer (4)

Hint : Velocity of electron in H-atom is minimum for

the 1st orbit.

P = 1

n = 2, 3, …

Sol : Series belongs to Lyman series.

1 1 1 4R R 1216 Å

1 4 3

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

1 1 1R 1 912 Å

R

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠

So, emitted photon have wavelength

between 912 Å to 1216 Å.

45. Answer (1)

Hint : Metal and solid.

Sol : Most of the elements are metal and solid.

46. Answer (2)

Hint : Spin only magnetic moment

Number of unpaired electron

Sol : For maximum , n must be maximum and

in d-block that is possible for Mn.

For n(n 2) BM

For maximum , n = 5

For M2+, n = 5 only possible with Mn.

47. Answer (2)

Hint : = 2 × 0.53 × n

Z

⎛ ⎞⎜ ⎟⎝ ⎠


7/13

Sol : For He+,

P = 3

2 0.532

⎛ ⎞ ⎜ ⎟⎝ ⎠

Q = 1

2 0.531

⎛ ⎞ ⎜ ⎟⎝ ⎠

R = 2

2 0.533

⎛ ⎞ ⎜ ⎟⎝ ⎠

So, order is P > Q > R.

Hence, the correct answer is (2).

48. Answer (1)

Hint : Possible value of n

2 and 3

Possible value of � = 0 to 2, m = –� to +�

Sol : n = 3, � = 2, m = –2

n = 3, � = 1, m = –1

n = 3, � = 0, m = 0

n = 2, � = 1, m = 0

So, Zn has maximum electron.

49. Answer (3)

Hint : n = 5 to 1

Sol : Possible line = 5 4

102

50. Answer (2)

Hint :2

2 2

1 2

1 1R Z

n n

⎛ ⎞ ⎜ ⎟⎝ ⎠

n1 for Balmer = 2

Sol : x = 1 1

R4 9

⎛ ⎞⎜ ⎟⎝ ⎠

x = 5

R36

⎛ ⎞⎜ ⎟⎝ ⎠

y = 1 1

R 4 3R1 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

1 1 8RR 1

z 9 9

⎛ ⎞ ⎜ ⎟⎝ ⎠

z = 9

8R

R = 36x

5

xyz = 5 9R 3R

36 8R

15R 15 36x 108x

32 32 5 32

51. Answer (2)

Hint : Fe + O2 FeO + Fe

2O

3

Use POAC

Sol : Fe(s) + O2 FeO + Fe

2O

3

mol of Fe = 1.68

56 = 0.03

mol of Fe in FeO and Fe2O

3 = 0.03

mol of oxygen (O) atom = 0.64

16 = 0.04

mol of 'O' in FeO and Fe2O

3 = 0.04

mol of FeO = x

mol of Fe2O

3 = y

So, mol of Fe = x + 2y = 0.03

mol of 'O' = x + 3y = 0.04

x 0.01

y 0.01

⎡ ⎤⎢ ⎥⎣ ⎦

52. Answer (2)

Hint : mol of CaCO3 = mol of CO

2 =

mol of CO

2

Sol : Let mol of I2 = x

mol of CO = 5x

mol of CO2 =

5x

2

mol of CaCO3 = 2.5x

Mass of CaCO3 = 2.5 × 100 × x

mol of I2 =

12.70.05

154

mol of CO = 0.25

mol of CO2 =

0.25

2

mol of CaCO3 =

0.25

2

Mass of CaCO3 sample =

0.25 100

2 0.8

53. Answer (4)

Hint : Spin quantum number was not given by

Schrodinger.

Sol : Quantum numbers n, �, m were derived from

solution of Schrodinger wave equation.


8/13

54. Answer (3)

Hint : A = Li, x = +2, n = 4

Sol : Separation energy = 7.65 eV

–En = 7.65

2

2

13.6 Z

n= 7.65 …(i)

E(n+2)

– En =

–19

h

1.6 10

2 2 34 15

2 2 19

13.6 Z 13.6 Z 6.63 10 1.025 10

n (n 2) 1.6 10

2

2

7.65 n7.65

(n 2)

= 4.25

22

2

n 1

1.5(n 2)

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ n = 4

From equation (i),

Z = 3

A+x = Li+2

55. Answer (3)

Hint : M + 2HCl MCl2 + H

2

Metal belongs to 2nd group.

n-factor = 2

Sol : Equivent mass of metal = 0.338 × 35.5

= 11.99 g � 12

Metal is M,

Eq. of M in 2.4 g = 2.4

0.212

mol of H2 =

2.40.1

22.4

mol of H2 = 0.1

M + x HCl MClx + 2

xH

2

x1

2

x = 2

56. Answer (2)

Hint : Average relative atomic mass =

n

i i

i 1

MX

100

∑

Sol : Relative mass approach to 40 so most of the

Ar is present as 40Ar.

57. Answer (3)

Hint : Eq. of acid = Eq. of base

Sol : No. of eq. of acid = 1.22 n

36.5

Eq. of NaOH = 4

0.140

Eq. of acid = Eq. of base

1.22 n 4 1

36.5 40

n = 3

Formula of acid is H3A.

Molar mass of H3A = 36.5 g

So, atomic mass of A = 33.5 g

58. Answer (3)

Hint :2 3 2 2 3 2

Na CO ·xH O Na CO xH O

Sol : mol of Na2CO

3

· xH

2O =

2.32

106 18x

2 3 2 2 3 2Na CO ·xH O Na CO xH O

1 mol of Na2CO

3

· xH

2O

= Produce x mol of H2O

mol of Na2CO

3

· xH

2O =

2.32

106 18x

Mass of water

18x 2.321.26

106 18x

18x × 2.32 = 133.56 + 22.68x

19.08x = 133.56

x = 7

59. Answer (3)

Hint : Due to mixing, volume become 600 mL.

Mass of solution A = 600 g

Mass of H2SO

4 = 49 g

and Mass of solution B = 140 g

Mass of H2SO

4 = 19.6 g

Sol : Molarity = 2 4

Moles of H SO

Volume of solution

Total mol of H2SO

4 = 0.5 + 0.2 = 0.7

Total volume = 600 = 0.6

Molarity = 0.7

0.6 = 1.167 M.


9/13

PART - C (MATHEMATICS)

61. Answer (2)

Hint : Divide both sides by log2 x log

4

x log

6 x and

use properties of log.


6 2 4

1 1 11 , 1

log log logx

x x x

1 log 6 log 2 log 4x x x

1 log 48x

x = 48

Also, x = 1 satisfies given equation.

Sum of solutions = 48 + 1 = 49

62. Answer (3)

Hint : n(A × B) = n(A) × n(B)

Sol : n(A × B × A) = n(A) × n(B) × n(A)

= 2 × 3 × 2 = 12

Total number of proper subsets of

A × B × A = 212 – 1.

63. Answer (3)

Hint : x – [x] = {x} and 0 {x} < 1

Sol : Range of sgn(x – [x]) is {0, 1}, since

0 x – [x] < 1.

Range of (sgn(x – [x]) + 1) is {1, 2}.

64. Answer (2)

Hint : Draw graphs of y = 4–|x| and y = 3 – |x|

Sol : From given equation, 3 – |x| = 4–|x|

Here, draw the graphs of

y = 4–|x| ...(i)

and y = 3 – |x| ...(ii)

y

y

x x

(0, 1)

O (0, 3)(–3, 0)

(0, 3)

y x = 3 – | |

y = 4–| |x

From graph, we observe that curves (i) and (ii)

intersects at two points only.

Two real solutions (–3, 3).

65. Answer (1)

Hint : Vertex is (1, –1) and coefficient of x2 = 1 > 0

parabolic curve is vertically upward.

Sol : Here, f(x) = x2 – 2x + 1 – 1 = x(x – 2)

curve cuts x-axis at (0, 0) and (2, 0), vertex

is (1, –1) and graph is vertically upward as

coefficient of x2 = 1 > 0.

Normality = 1.167 × 2 = 2.33 N

Mass of solution A = 500 × 1.2 = 600 g

Mass of H2SO

4 = 49 g

Mass of solution B = 140 g

Mass of H2SO

4 in B = 19.6 g

(w/w)% of H2SO

4 =

68.6100 9.27

740

(w/w)% of H2SO

4 =

68.6100 11.43

600

60. Answer (4)

Hint : mol of H2O

2 = mol of I

2 =

2 2 3mol of Na S O

2

Sol : From reaction (I),

H2O

2 + 2KI I

2 + 2KOH

mol of H2O

2 = mol of I

2

mol of H2O

2 = 100 × 0.1 = 10 mmol

From reaction (II),

2(mol of I2) = mol of Na

2S

2O

3

2 × 10 = 0.2 × x

x = 100 mL

mol of KOH = 2(mol of H2O

2)

mol of KOH = 20 × 10–3 mol

Mass of KOH = 56 × 20 × 10–3 = 1.12 g

mol of I2 = 10 × 10–3 mol

Mass of I2 = 254 × 10 × 10–3 = 2.54 g

mol of NaI = 2(mol of I2)

= 2 × 10 × 10–3 = 0.02 mol


10/13

66. Answer (1)

Hint : Add all three equations.

Sol : Adding, (log2(ABC))2 = 144

log2(ABC) = ±12

So, log2(ABC) = 12 ABC = 212

log2A = 4, log

2B = 1, log

2C = 7

A = 24, B = 2, C = 27

67. Answer (4)

Hint :25 24 23 22018

( ) ( 1)( ... )1

f x x x x x x xx

Sol : f(x) = x25 – 2018{(x24 – x23 + x22 + ... + x2 – x)}

= x25 –2018

1 x{(x24 – x23 + x22 + ... + x2 – x)

+ (x25 – x24 + x23 + ... + x3 – x2)}

= x25 – 2018

1 x(x25 – x)

= x25 –

252018( )

1

x x

x

f(2017) = 201725 –

252018(2017 2017)

2018

= 2017

68. Answer (3)

Hint : If any vertical line parallel to y-axis cuts the

graph at only one point, then that graph

represents a function.

Sol : If the vertical lines parallel to y-axis

intersects the graph of curve at only one

point, then that is the graph of function. If we

draw any line parallel to y-axis in the graph

of option (3) cuts at only one point.

Graph in option (3) represents a function.

69. Answer (2)

Hint : f(x) is defined if [x – 5] 0 and x – [x] 0.

Sol : Here, f(x) is defined if [x – 5] 0

( 5) [0,1)x

[5, 6)x

and x – [x] 0 x I

Df = (–, 5) (6, ) – I

70. Answer (3)

Hint : Use transformations of graphs.

Sol : T1 : Graph of y = log|x| =

log , 0

log ( ) , 0

e

e

x x

x x

⎧⎨ ⎩

Graph of y = log|x| is as follows

y

y

xx(1, 0)( 1, 0)–

y x = log(– ) y x = ln

T2 : Graph of y = ln|x| – 1

When y = 0, ln|x| = 1 |x| = e x = ± e

when x = 0, y –

and1 | | 1

0| |

dy x

dx x x x for x > 0 < 0 for

x < 0

y

y

xx ( , 0)e( , 0)–e

T3 : Finally graph of f(x) = |ln|x| – 1|

y

y

xx ( , 0)e( , 0)–e

71. Answer (4)

Hint :( )

( ) , ( ) 0( )

f g

f f xx g x x D D

g g x

⎛ ⎞ ⎜ ⎟⎝ ⎠

Sol : Here, 7,

2f

D⎡ ⎞ ⎟⎢ ⎠⎣

and Dg = (–, 1]

f gD D ( )

fx

g

⎛ ⎞⎜ ⎟⎝ ⎠

is undefined,

since ( )

( )( )

f f xx

g g x

⎛ ⎞⎜ ⎟⎝ ⎠

is possible only if

( ) 0g x and f gx D D

But, here x Df D

g


11/13

72. Answer (3)

Hint : First find range of

4

2

1x

x

using AM GM.

Sol : Here,

4

2

2 2

1 12

x

x

x x

14

22 22

1log log 2

x

x

f(x) 1

2

1

( ) ,2

f x⎡ ⎞ ⎟⎢⎣ ⎠

73. Answer (2)

Hint : Put y = p(x)

2

2 2 2 2

14

4 ( 1)

( 1) 11

x

x x x

x x

x

x

⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞ ⎜ ⎟⎝ ⎠

2

4 4 1use

41

tt x

xtt

t

Sol : Put

2

2 2 2 2

14

4 ( 1)( )

( 1) 11

xx x x

y p xx x

xx

⎛ ⎞⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟

⎝ ⎠

Put 1

x t

x

, for x > 0, t 2

2

4 4

1 /1

ty

t tt

But for t 2, 1 1 5

22 2

tt

1 2

1 / 5t t

4 2 4

1/ 5t t

80

5y

and y = 0 for x = 0 also

8 8

0 0,5 5

y y⎡ ⎤ ⇒ ⎢ ⎥⎣ ⎦

74. Answer (4)

Hint : n[(A × B) B × A)] = [n(A B]2

Sol : A = {2, 3, 5, 7}, B = {1, 2, 3, 4, 5}

A B = {2, 3, 5}

n[(A × B) (B × A)] = [n(A B)]2 =

32 = 9 ordered pairs

75. Answer (2)

Hint : Use replacement property and {n + x} = {x},

if n N and logam

n = nlogam and log

aa = 1

Sol : Here, 410

( ) (10 )3

f f f⎛ ⎞ ⎜ ⎟⎝ ⎠

4

10

10[ ] 3 2 log 10

3

⎛ ⎞⎧ ⎫ ⎨ ⎬⎜ ⎟⎝ ⎠⎩ ⎭

14 3 2 4 1

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

76. Answer (3)

Hint : Use replacement property.

Sol : ∵ 3f(x) – 22

5 3f x xx

⎛ ⎞ ⎜ ⎟⎝ ⎠...(i)

Replacing x by 2

x

, we get

2

2 2 45 ( ) 3 3f x f

x x x

⎛ ⎞ ⎜ ⎟⎝ ⎠...(ii)

Solving (i) and (ii) for f(x), we get

f(x) = 2

2

1 10 2024 3 3

16x x

x x

⎛ ⎞ ⎜ ⎟⎝ ⎠

f(1) = 34 17

16 8

77. Answer (2)

Hint : 12 × 10 = 15 × 8

Sol : f(120) = 12f(10) + 10f(12) = 15f(8) + 8f(15)

12 19 + 10 52 = 15f(8) + 8 26

f(8) = 36

78. Answer (1)

Hint : Draw the graph of y = x2 and y = 1 – |x – 5|

and see the point of intersection of these two

graphs.

Sol : From given equation, x2 = 1 – |x – 5|

Now, draw the graph of y = x2 and

y = 1 – |x – 5| as shown below.

x x

y

O

1 (5, 1)

y x = 2

It is evident from the graph that two curves

do not intersect at any point.

Number of real solutions = 0


12/13

79. Answer (2)

Hint : logg(x)

f(x) is defined if f(x) > 0, g(x) > 0,

g(x) 1.

Sol : Here, for log function to be defined 5

02

x

and 5

12

x and

25

02 3

x

x

⎛ ⎞ ⎜ ⎟⎝ ⎠.

– –5 3

2

even even

+ve +ve +ve

3 3

( , 5) 5, ,2 2

x⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

...(i)

Also, 5 3

and2 2

x x ...(ii)

From (i) and (ii),

5 3 3, ,

2 2 2x

⎛ ⎞ ⎧ ⎫ ⎨ ⎬⎜ ⎟⎝ ⎠ ⎩ ⎭

80. Answer (1)

Hint : Take pair (a, b), a A, b B, such that

|a – b| is even number.

Sol : Here, R = {(1, 5), (1, 7), (1, 9), (2, 8), (3, 5),

(3, 7), (3, 9), (4, 8)}

Number of elements in relation R = 8

81. Answer (3)

Hint : Use property of |x|.


||2 – |x – 2|| = 1

|x – 2| – 2 = 1, – 1

|x – 2| = 3, 1

x – 2 = – 1, 1, 3, – 3

x = 1, 3, 5, –1

82. Answer (4)

Hint : From given equation x2 + 3x – 4 0 and for

log function to be defined x(x + 3) > 0.

Then solving them find values of x.

Sol : For log function to be defined,

x2 + 3x > 0

x(x + 3) > 0

– – 3 0 +

+ ve + ve

– ve

( , 3) (0, )x ...(i)

and from given inequation log2 (x2 + 3x) 2.

– 4 1

+ ve + ve– ve

x2 + 3x – 4 0

x [–4, 1] ...(ii)

From (i) and (ii), x [– 4, – 3) (0, 1],

taking intersection of (i) and (ii).

83. Answer (3)

Hint : Solving equations, [x] = 4 and y = 11

Sol : Given, y = 2 [x] + 3 ...(i)

and y = 3[x] – 1 ...(ii) [From (ii) equation]

By (ii) and (i),

[x] = 4 and so from (i) y = 11 when [x] = 4

4 x < 5

[x + y] = 15

84. Answer (3)

Hint : Use properties of |x|.

Sol : Given, x2 – |x| – 6 0

(|x| – 3)(|x| + 2) 0

x x+ +–

– 2 3

| | [ 2, 3]x

But |x| 0 |x| [0, 3] 0 |x| 3

0 x 3 and 0 – x – 3

0 x – 3 and 0 x 3

[ 3, 3]x

Number of integral values of x = 7.

85. Answer (1)

Hint : {x + n} = {x}, if n I

Sol : Here, f(x) = {x} + {x} + ... to 10 terms = 10{x},

since {x + [x]} = {x}

( 2) 10{ 2} 10{1.414}f

= 10 × 0.414 10 414 414

4.141000 100

[ ( 2)] 4f


13/13

86. Answer (1)

Hint : Using properties of log, find a relation in

p and q squaring both sides, find a relation

in p and q, then solving them find values of

p and q.

Sol : Given, log10

(p – 2) + log10

q = 0

q(p – 2) = 100 = 1

p – 2 = 1

q p =

12

q ...(i)

Given, 2p q p q

Squaring, we get

p + q – 2 + 2 ( 2)p q = p + q

p(q – 2) = 1

1

2qp

...(ii)

From (i) and (ii), p = q

From (i),

p2 – 2p – 1 = 0

(p – 1)2 = 2

1 2p

But q 2

1 2p q

[p + q] = 4

87. Answer (1)

Hint : Required number of elements = n(A B) +

n(B C) + n(C A) – 3n(A B C)

Sol A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

B = {– 2, – 1, 0, 1, 2, 3, 4, 5, 6, 7}

C = {2, 3, 5, 7, 11, ...}

A B = {1, 2, 3, 4, 5, 6, 7},

B C = {2, 3, 5, 7}, C A = {2, 3, 5, 7}

and A B C = {2, 3, 5, 7}

The required number

= n(A B) + n(B C) + n(C A)

– 3n(A B C)

= 7 + 4 + 4 – 3 × 4 = 3

� � �

88. Answer (2)

Hint : n(A B) = n(A) + n(B) – n(A B) which is

maximum if n(A B) is minimum and is

minimum if n(A B) is maximum.

Sol : n(A B) = n(A) + n(B) – n(A B) and

n(A B) is maximum if n(A B) is

minimum and n(A B) is minimum if

n(A B) is maximum.

Maximum n(A B) = 15 + 10 – 0 = 25

Mininum n(A B) = 15 + 10 – 10 = 15

Their sum = 25 + 15 = 40

89. Answer (1)

Hint : Total number of subsets of A × B = 2n(A) × n(B)

and use number of subsets of set A with

n elements taking r elements = nCr

Sol : A × B has 2 × 3 = 6 elements

Total number of subsets of A × B = 26 = 64

Number of subsets with zero elements

= 6C

0 = 1

Number of subsets with one element = 6C1 = 6

Number of subsets with two element = 6C

2

6! 6 5 4!15

2! 4! 2 1 4!

Required number of subsets of A × B

= 64 – (1 + 6 + 15)

= 64 – 22 = 42

90. Answer (3)

Hint : A B = {x : x A or x B} and total

number of subsets of set A with n elements

= 2n

Sol : Here, A = {2, 3} and B = {– 3, – 2, – 1, 0,

1, 2, 3}

A B = {– 3, – 2, – 1, 0, 1, 2, 3}

Number of subsets of A B = 27 = 128

test - 1 (code-c) (answers) all india aakash test …...sol : no. of eq. of acid = 1.22 n 36.5 eq....

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