16-12-2024

1001CJM202023240013 JM

PART-1 : PHYSICS

SECTION-I

1) Four particles of mass 5, 3, 2, 4 kg are at the points (1, 6), (–1, 5), (2, –3), (–1, –4). Find the
coordinates of their centre of mass.

(A)

(B)

(C)

(D) None

2) A force of 0.5 N is applied on the upper block as shown in figure. The coefficient of static friction
between the two blocks is 0.1 and that between the lower block and the surface is zero. The work
done by the lower block on the upper block for a displacement of 3 m of the upper block is :-

(A) 1 J
(B) –1 J
(C) 2 J
(D) –2 J

3) A uniform chain is held on a frictionless table with L/4 hanging over. Knowing total mass of the
chain is M and total length is L, the work required to pull hanging part back to the table is :

(A)

(B)

(C)

(D)

4) A ball of mass m moving horizontally at a speed u collides with the bob of a simple pendulum at

rest. The mass of the bob is . If the collision is perfectly inelastic the height to which the composite
ball rise after the collision is :-
(A)

(B)

(C)

(D)

5) Three identical blocks A, B and C are placed on a fixed horizontal surface which is perfectly
smooth. The block B and C are at rest while the block A is approaching towards B with a uniform
velocity of 10 m/s. The coefficient of restitution for all the collisions is 0.5. The block C just after the

collision moves with the speed of :

(A) 3.75 m/s

(B) 5.6 m/s
(C) 7.5 m/s
(D) zero

6) A thin and circular disc of mass M and radius R is rotating in a horizontal plane about an axis
passing through its centre and perpendicular to its plane with an angular velocity ω. If another disc
of same dimensions but of mass M/4 is placed gently on the first co-axially, then the new angular
velocity of the system is :-

(A)
ω

(B)
ω

(C)
ω

(D)
ω

7) A disc has mass 9m. A hole of radius R/3 is cut from it as shown in the figure. The moment of
inertia of remaining part about an axis passing through the centre 'O' of the disc and perpendicular

to the plane of the disc is:

(A) 8 mR2
(B) 4 mR2

(C)
mR2
(D)
mR2

8) A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point

of contact, B is the centre of the sphere and C is its topmost point. Then
(A)
(B)
(C)
(D)

(A) A,B
(B) B,C
(C) C,D
(D) B,C,D

9) A rod of length L is hinged at one end. It is brought to a horizontal position and released. Then
angular velocity of the rod when it is in vertical position is

(A)

(B)

(C)

(D)

10) A solid sphere of uniform density and radius R applies a gravitational force of attraction equals
to F1 on a particle placed at P, distance 2R from the centre. O of the sphere. A spherical cavity of
radius R/2 is now made in a sphere as shown in figure. The sphere with cavity now applies a

gravitational force F2 on the same particle placed at P. The ratio F2/F1 will be -

(A) 1/2
(B) 7/9
(C) 3
(D) 9/7

11) An asteroid of mass m is approaching earth initially at a distance of 10 Re from surface of earth
and speed Vi. It hits the earth with a speed Vƒ (Re and Me are radius and mass of earth), then-
(A)

(B)

(C)

(D)

12) The orbital velocity of an artificial satellite in a circular very close to earth is V. The velocity of a
geo-sationary satellite orbiting in circular orbit at an altitude of 3R from earth's surface will be :-

(A)

(B)

(C)

(D)

13) The equation of motion of a particle executing SHM is + 25 x = 0. The time period of the
particle will be :-

(A)

(B)

(C)

(D)

14) The time period of a simple pendulum inside a stationary lift is . What will be the time
period when the lift moves upwards with an acceleration g/4 ?

(A)
(B) 2

(C)

(D) 2 s

15) The phase difference between the two simple harmonic oscillations, y1 = sinωt + cosωt
and y2 = sinωt + cosωt is :-

(A)

(B)

(C)

(D)

16) The ratio of the lengths of two wires A and B of same material is 1 : 2 and the ratio of their
diameter is 2 : 1. They are stretched by the same force, then the ratio of increase in length will be :-

(A) 2 : 1
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1

17) To what depth below the surface of sea should a rubber ball be taken as to decease its volume by
0.1% ?
[Take : density of sea water = 1000 kgm–3,
Bulk modulus of rubber = 9 × 108 Nm–2 ;
acceleration due to gravity = 10 ms–2]

(A) 9 m
(B) 18 m
(C) 180 m
(D) 90 m

18) A wire of length 2L and radius r is stretched between A and B without the application of any
tension. If Y is the Young's modulus of the wire and it is stretched like ACB, then the tension in the

wire will be :

(A)

(B)

(C)

(D)

19) If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble
of volume 2V from the same soap solution will be :-
(A) W/2
(B)
(C)
(D)

20) Consider a water jar of radius R that has water filled up to height H and is kept on a
stand of height h (see figure). Through a hole of radius r(r < < R) at its bottom, the water leaks out
and the stream of water coming down towards the ground has a shape like a funnel as shown in the
figure. If the radius of the cross-section of water stream when it hits the ground is x. Then :

(A)

(B)

(C)

(D)

SECTION-II

1) When a bomb of mass 5kg initially at rest, explodes in to two fragments of mass 2kg and 3kg. If
KE of 2kg mass is 600 J then KE of 3kg will be

2) A bullet weighing 10 gm and moving at 300 m/s strikes a 5 kg block of ice and drops dead. The ice
block is sitting on smooth surface. The speed (in cm/s) of the block, after the collision is :-

3) The ratio of the K.E. required to be given to the satellite to escape earth's gravitational field to the
K.E. required to be given so that the satellite moves in a circular orbit just above earth atmosphere
is:-

4) The maximum velocity of simple harmonic motion represented by y = 2 sin (100t + π/6) is given
by :

5) A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a

height R above the earth's surface where R is the radius of the earth. The value of is :-
 PART-2 : CHEMISTRY

SECTION-I

1) KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI,
iodide ion is oxidised to :

(A) I2
(B) IO–
–
(C) IO3
–
(D) IO4

2) On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is obtained
which is highly explosive in nature. Identify the compound from the following.

(A) Mn2O7
(B) MnO2
(C) MnSO4
(D) Mn2O3

3) Which of the following pairs of compound have nearly equal size :-

(a) Y and La (b) Zr and Hf
(c) Ti and Zr (d) Nb and Ta

(A) a and b
(B) b and c
(C) b and d
(D) a, b, c

4) The metallic bond strength in first transition series increases from -

(A) Sc to Mn
(B) Sc to Cr
(C) Cr to Zn
(D) Sc to Cu

5) Transition elements act as catalyst because -

(A) Their melting points are higher

(B) Their I.P. values are higher
(C) They have high density
(D) They can show variable oxidation states

6) The equilibrium
Cr2O72– ⇌ 2CrO42– is shifted to right in

(A) An acidic medium

(B) A basic medium
(C) A neutral medium
(D) It does not exist

7) The equilibrium constant for the reaction

N2(g) + O2(g) 2NO(g) is 4 × 10–4 at 200 K. In presence of a catalyst, equilibrium is attained ten
times faster. Therefore, the equilibrium constant in presence of the catalyst at 200 K is :-

(A) 40 × 10–4
(B) 4 × 10–4
(C) 4 × 10–3
(D) difficult to compute without more data

8) For which of the following reaction is product formation favoured by low pressure and low
temperature?

(A) CO2(g)+C(s) 2CO(g) ; ΔH° = 172.5 kJ

(B) CO(g)+2H2(g) CH3OH(l) ; ΔH° = –21.7 kJ
(C) 2O3(g) 3O2(g) ; ΔH° = –285 kJ
(D) H2(g) + F2(g) 2HF(g) ; ΔH° = 541 kJ

9) Two solid compounds X and Y dissociates at a certain temperature as follows

X(s) ⇌ A(g) + 2B(g) ; = 9 × 10–3 atm3
Y(s) ⇌ 2B(g) + C(g) ; = 4.5 × 10–3 atm3
The total pressure of gases over a mixture of X and Y is :-

(A) 4.5 atm

(B) 0.45 atm
(C) 0.6 atm
(D) None of these

10) On heating a mixture of SO2Cl2 and CO, two equilibrium are simultaneously established :

On adding more SO2 at equilibrium what will happen ?

(A) Amount of CO will decrease

(B) Amount of SO2Cl2 and COCl2 will increase
(C) Amount of CO will reamin unaffected
(D) Amount of SO2Cl2 and CO will increase

11) A3B2 is a sparingly soluble salt of molar mass M (g mol–1) and solubility x g litre–1. The ratio of the
molar concentration of B3– to the solubility product of the salt is :-
(A)

(B)

(C)

(D) None of these

12) the pKaof acetic acid and pKbof ammonium hydroxide are 4.76 and 4.75 respectively. Calculate
the pH of ammonium acetate solution at 25°C :-

(A) 6.955
(B) 7.005
(C) 6.055
(D) 8.01

13) The highest pH value is of :

(A) 0.1 M NaCl

(B) 0.1 M NH4Cl
(C) 0.1 M CH3COONa
(D) 0.1 M CH3COONH4

14) Identify the product (P) in the following reaction:

(A)

(B)

(C)

(D)

15) Match List-I with List-II.

List-I List-II
(Reactions) (Products)

(a) (I)
 (b) (II)

(c) (III)

(d) (IV)

Choose the correct answer from the options given below :

(A) (a)-(III), (b)-(II), (c)-(I), (d)-(IV)
(B) (a)-(IV), (b)-(II), (c)-(III), (d)-(I)
(C) (a)-(I), (b)-(IV), (c)-(II), (d)-(III)
(D) (a)-(II), (b)-(IV), (c)-(I), (d)-(III)

16) Identify the reagents used for the following conversion

A = LiAlH4, B = NaOH(aq),
(A)
C = NH2–NH2/KOH, ethylene glycol
(B) A = LiAlH4, B = NaOH(alc), C =Zn/HCl
A = DIBAL-H, B= NaOH(aq),
(C)
C = NH2–NH2/KOH, ethylene glycol
(D) A = DIBAL-H, B = NaOH(alc), C = Zn/HCl

17) Given below are two statements :

Statement I : Acidity of α-hydrogens of aldehydes and ketones is responsible for Aldol reaction.
Statement II : Reaction between benzaldehyde and ethanal will NOT give Cross – Aldol product.
In the light of above statements, choose the most appropriate answer from the options given below

(A) Both Statement I and Statement II are correct.

(B) Both Statement I and Statement II are incorrect.
(C) Statement I is incorrect but Statement II is correct.
(D) Statement I is correct but Statement II is incorrect.
18) Identify the product in the following reaction :

(A)

(B)

(C)

(D)

19)
This reduction reaction is known as:

(A) Rosenmund reduction

(B) Wolff-Kishner reduction
(C) Stephen reduction
(D) Etard reduction

20) The final product A, formed in the following multistep reaction sequence is:

(A)

(B)

(C)
(D)

SECTION-II

1) For Mn+x the magnetic moment (spin only) is 3.92 B.M then find x ?

2) When mixture of NaCl and K2Cr2O7 is gently warmed with concentrated H2SO4 then compound X is
formed. What is the oxidation state of central atom of X ?

3) The dissociation constant of weak acid is 1 × 10–4. In order to prepare a buffer solution with pH =

5 the ratio is then what will be (x + y) :-

–1
4) A compound (x) with molar mass 108 g mol undergoes acetylation to give product with molar
mass 192 g mol–1. The number of amino groups in the compound (x) is ______.

5) In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the

amount of benzaldehyde required is ______ g. (Nearest integer)

PART-3 : MATHEMATICS

SECTION-I

1) Let
.
Then the number of elements in S is :

(A) 4
(B) 0
(C) 2
(D) 1

2) Let a1, a2, ….. a10 be 10 observations such that

and . Then the standard deviation of a1, a2, .., a10 is equal to :

(A) 5
(B)
(C) 10
(D)

3) If and
then the S.D. of observations x1, x2 ........ x10 is :-

(A)

(B)

(C)

(D) None of these

4) The H.M. of following frequency distribution is

(A) 5
(B) 10
(C) 15
(D) None

5) Mean and median of four numbers a, b, c and d(b < a < d < c) is 35 and 25 respectively then the
value of b + c – a – d will be :-

(A) 90
(B) 115
(C) 40
(D) 10

6) = x ; then the value of x will be:

(A)

(B)

(C)

(D)

7) Maximum value of is :-

(A)
(B)

(C) 3
(D) 4

8)

If x ∈ (π, 2π) &

then 'a' equals :-

(A)

(B)

(C)

(D) None of these

9) The number of solutions of equation

6cos2θ + 2cos2θ/2 + 2sin2θ = 0, –π < θ < π is :

(A) 3
(B) 4
(C) 5
(D) 6

10) In any ΔABC, is equal to :-

(A)

(B)

(C)

(D) None of these

11) If tn = (n + 2) (n + 3) for n = 1,2,3 ..., then

=?

(A)

(B)

(C)
(D) None

12) Concentric circles of radii 1, 2, 3, …..100 cm are drawn. The interior of smallest circle is
coloured red and the annulas regions are coloured alternately green and red, so that no two
adjacent regions are of the same colour. Then the total area of green regions in sq. cm is equal to :-

(A) 100 π
(B) 5050 π
(C) 4950 π
(D) 5151 π

13) The arithmetic mean of the nine numbers in the given set {9, 99, 999, ........, 999999999} is a 9
digit number N, then number N doesn't contain the digit ?

(A) 0
(B) 2
(C) 5
(D) 9

14) A man saves Rs. 100 in the first month of his service. In each of the subsequent months his
saving increases by twice of the saving of immediately previous month. His total saving from the
start of service will be Rs. 409500 after

(A) 10 months
(B) 14 months
(C) 12 months
(D) 19 months

15) Find the number of number lying between 100 and 500 that are divisible by 7 but not by 21 :

(A) 57
(B) 38
(C) 39
(D) 26

16) Let x2 –(m – 3) x + m = 0 (m ∈ R) be a quadratic equation. Find the values of m for which one
root is smaller than '2' & other root is greater than '2' :-

(A) m < 10
(B) m > 10
(C) [9, 10)
(D) (0, 1]

17) The number of real solutions of the equation

are where t = x2 – 2 | x |

(A) 0
(B) 1
(C) 2
(D) 4

18) If α, β, γ are the roots of the equation

8x3 + 1001x + 2008 = 0 then the value of
(α + β)3 + (β + γ)3 + (γ + α)3 is :

(A) 251
(B) 751
(C) 735
(D) 753

19) If exp. is always

negative for every real 'x' then the interval in which 'a' lies :-

(A)

(B)

(C)

(D) None of these

20) If graph of expression y = x2 + bx + d is following Now ΔOBC is right

2
angled isosceles Δ and A is mid point of OB then roots of equation x + bx + d = 0 is :

(A) 2, 4
(B) 0, 0
(C) 1, 2
(D) 4, 8

SECTION-II

1) The variance σ2 of the data

xi 0 1 5 6 10 12 17

fi 3 2 3 2 6 3 3
is _______.

2) A person standing on the bank of a river observes that the angle of elevation of the top of a tree
on the opposite bank of the river is 60º and when he retires 40 m away from the tree the angle of
elevation becomes 30º. The breadth of the river is :

3) If a1, a2, a3, .... a21 are in A.P. and

a3 + a5 + a11 + a17 + a19 = 10 then the value of is :-

4)

If the equation

(a2 – 3a + 2)x2 +(a2 – 5a + 6)x + (a2 – 4) = 0

satisfies by more than two value of 'x' then a =

5) If α,β are the roots of the equation

x2 – 3x – 15 = 0, and ƒ(n) = αn + βn, then

is equal to :-
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. A B C C B C B B A B C C B D C C D B D D

SECTION-II

Q. 21 22 23 24 25
A. 400 60 2 200 2

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. C A C B D B B C B D C B C A D D D D A B

SECTION-II

Q. 46 47 48 49 50
A. 4 6 11 2 318

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. C B B C C D D A B B A B A C B B D D A C

SECTION-II

Q. 71 72 73 74 75
A. 29 20 42 2 8
 SOLUTIONS

PART-1 : PHYSICS

1)

2)
0.5 = 3a ⇒ a = 5/30
f = 2a = 2(5/30)= 1/3N
fmax = 0.1 × 10 = 10N
∴ f = 1/3 N will act
work done by lower block = –f(s)
= –1/3(3) = –1 Joule

3) Mass of hanging part =

Depth of centre of mass from level of table =

Hence, work required to pull hanging part to table = change in gravitational

P.E. = =
Alternate Solution :
Consider an element dx at a distance x from table level.

work required to pull element to table

dw = (dm)g(x)

dw = gx
Hence, work required to pull element to table

w=

w= ⇒w=
4)

mu = ⇒v=

h=

5)
A&B :

10 = v1 + v2 ....(1)

e= = 0.5 ⇒ v2–v1 = 5 ....(2)

∴ v2 = 7.5 m/s
B&C :

7.5 = v3 + v4 ...(1)

e= = 0.5
v4 – v3 = 3.75 ...(2)
2v4 = 11.25 ⇒ v4 = 5.6 m/s

6) According to conservation of angular momentum I1ω1 = I2ω2

⇒ MR2ω = ω2

∴ ω2 = ω

10) F1 =
Force applied by remains sphere = force applied by complete sphere – force applied by remain
sphere.

F2 =
= =

11) From COME

0 0
12) V = ⇒ V ∝

= , V2 =

13)

14)
when lift move upwards then

geff = g +

so T' = 2π

15)

y1 = sin ωt + cos ωt

A1 =
tan α1 = ⇒ α1 = π/3
y1 = sin (ωt + π/3) ...(i)
y2 = sin ωt + sin (ωt + π/2)
∴ A2 = =
tan α2 = 1 ⇒ α2 = π/4
...(ii)
phase diff. (from (i) & (ii))
= π/3 – π/4 = π/12.

16) ⇒ ⇒

17) Bulk modulous

K=

Here × 100 = 0.1

ρ = 1000 kg/m3
⇒ h = 90 m g = 10 m/s2

18) T =
Increase in length of one segment of wire

l=

So, T =

21)

23) K.E. required for satellite to escape from earth's gravitational field

K.E. required for satellite to move in circular orbit

The ratio between these two energies = 2

24) Maximum velocity of a S.H.M.

 Vmax = Aω
= 2 × 100 = 200

25)

then

PART-2 : CHEMISTRY

26)

KMnO4 + KI

27)

KMnO4 Mn2O7
explosive

28) Size of 4 d series 5 d series in group no 4 to 11

29) Number of unpaired e– µ metallic bond strength.

30) Transition elements act as catalyst because they can show variable oxidation states.

31) ​

34) Let x is partial pressure of A and y is partial pressure of C when both equilibrium
simultaneously established in a vessel
 ⇒ x = 0.1 atm;
y = 0.05 atm
Total pressure of gases = PA + PB + PC
= 3(x + y)
= 0.45 atm.

35)

On adding SO2 at new equilibrium

SO2Cl2(g) SO2(g) + Cl2(g)
↑ ↑ ↓
CO(g) + Cl2(g) COCl2(g)
↑ ↓ ↓

36) Solubility S=
3A (aq.) + 2B3–(aq.)
2–
A3B2(s)
3s 2s
2 2 5
Ksp = (3S) (2S) = 108S

37) For salt of weak acid and weak base

pH = 7 + [pKa – pKb]

= 7 + (4.76 – 4.75)

= 7 + (0.01)
= 7 + 0.005
= 7.005

39) HVZ Reaction

40)
41)

42) Aldehyde and ketones having acidic α-hydrogen show aldol reaction

43)

44)
It is known as rosenmund reduction that is the partial reduction of acid chloride to aldehyde.
45)

46)

Mn+x, μ = 3.92 BM so
No. of unpaired e– = 3
Mn = 4s2 3d5
Sp. Mn+4 = 3d3
SO, x = + 4

47)

48) HPO42– and PO4–3 are conjugate acid and base so

Ka × Kb = 10–14

Ka(HPO42–) = = 4.17 × 10–13

pKa = – log Ka = 12.38

or pH = 7 + pKa + log C
pH = 13.04

49)
Gain in molecular weight after acylation with one –NH2 group is 42.
Total increase in molecular weight = 84

∴ Number of amino group in

50) Claisen Schmidt reaction

 mw of benzaldehyde = 106
106 × 3 = 318 gm. Benzaldehyde is required to give 1.5 mole (or 351 gm) product

PART-3 : MATHEMATICS

51)

Let

t2 – 10t + 1 = 0

x = 2 or x = –2
Number of solutions = 2

52)

a1 + a2 + … + a10 = 50 ….(i)

....(ii)
If a1 + a2 + … + a10 = 50.
(a1 + a2 + … + a10)2 = 2500

⇒
⇒

, Standard deviation ‘σ’

54) H.M. =

55) ∵ mean = 35

a + b + c + d = 140 ...(1)
and medium of b, a, d, c is 25

∴
a + d = 50 ...(2)
From (1) and (2)
b + c = 90 ...(3)
∵ b + c – a – d = 90 – 50 = 40

56)

57) sin6x + cos6x = 1 – 3 sin2xcos2x

=1–

58)

∴
= =

59) 6cos 2θ + 2cos2 + 2sin2θ = 0

⇒ 12cos2θ – 6 + 1 + cosθ + 2 – 2cos2θ = 0
⇒ 10 cos2θ + cosθ – 3 = 0

⇒ cosθ = ,

⇒ θ = , π – cos–1 ,
cos5θ is – ve in I & II quadrant so total 4 solutions.

60) We have,

61)

62) Required Area

= 5050 π sq. cm

63)
⇒ A = 1 + 11 + 111 + …… + 111111111
⇒ A = 123456789 [doesn't Contain zero]

64) Total savings :

⇒ 100 + 200 + 400 + .... n monts (let) = 409500
⇒ 100(2n – 1) = 409500
⇒ 2n = 4096 = 212
⇒ n = 12

65) n(7) = 105, 112, 119, ........ 497

n(21) = 105, 126, 147, ....... 483
n(7) = 497 = 105 + (n – 1)7 or n = 57
n(21) = 483 = 105 + (m – 1)21 or m = 19
n(7) – n(21) ⇒ 57 – 19 = 38

66)
f (2) < 0
4 – 2 (m – 3) + m < 0
⇒

67)

Let =y

So, y +
So, t = + 1 or – 1
x2 – 2| x | = 1 or x2 – 2| x | = – 1
has four Real roots.

68) 8x3 + 1001x + 2008 = 0 → α, β, γ

S.O.R ⇒ α + β + γ = 0 ⇒ α3 + β3 + γ3 = 3αβγ … (1)

Now, (α + β)3 + (β + γ)3 + (γ + α)3 = – γ3 – α3 – β3

⇒
⇒ 753

69) <0∀x∈R
2 2
⇒ [x – 8x + 32 = (x – 4) + 16 > 0 ∀ x ∈ R]
⇒ ax2 + 2(a + 1)x + (9a + 4) < 0 ∀ x ∈ R
⇒a<0 & D<0
4(a + 1)2 – 4.9(9a + 4) < 0
a2 + 2a + 1 – 9a2 – 4a < 0
–8a2 – 2a + 1 < 0
8a2 + 2a – 1 > 0
8a2 + 4a – 2a – 1 > 0
4a(2a + 1) – 1(2a + 1) > 0
(4a – 1) (2a + 1) > 0
 a∈

70) Let f(x) = x2 + bx + d

f(0) = d
point C (0, d)

now B(d, 0), A

roots of f(x) = 0 are & d.

product of roots ×d=d ⇒

Roots are 1, 2

71)

xi fi fixi fixi2

0 3 0 0

1 2 2 2

5 3 15 75

6 2 12 72

10 6 60 600

12 3 36 432

17 3 51 867

Σfi =
Σfixi2 = 2048
22
∴ Σfixi = 176

So

for

= 93.090964 – 64 = 29.0909

72) = tan30º =
 and, = tan60º =
⇒ = 40 + x ⇒ x = 20 ml

73)

Let a3 + a19 = a5 + a17 = 2a11 = k

given a3 + a5 + a11 + a17 + a19 = 10

⇒ k = 10 ⇒ k = 4
so, a1 + a21 = 4 ...(1)

= 42

74)

It is possible if eq. becomes identity

2
⇒ a –3a+2 = 0 ⇒ a = 2, 1
and a2–5a + 6 = 0 ⇒ a = 2, 3
and a2 – 4 = 0 ⇒ a = 2, –2

common value of a = 2

75)

x2 – 3x – 15 = 0

f(n) = αn + βn