PHYSICS - gradeup.co April Shift _ 1.pdf-74.pdf · Formula of solid is A 4 B 4 O 8 = ABO 2 12. What...

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PHYSICS

1. Breakdown voltage of Zener diode = 5.6V. Load Resistance = 800 Ω, unregulated

voltage = 9V, dropping resistance is 200 Ω.

Find the current passing through the Zener diode.

A. 1 Amp

B. 2 Amp

C. 0.1 Amp

D. 0.01 Amp

Ans. D.

Sol:

3 4

200

5 6

800

=

= =

.

.

I =

L

V

RV

IR

Current passing through the Zener diode

3 4 5 6

200 800

13 6 5 6 8 10 01

800 800 100

= −

−= = = =

. .

. .. A

2. In YDSE ratio of amplitude of wave is 3 : 1. Find the ratio of Imax : Imin is

A. 1: 4

B. 1 : 2

C. 4 : 1

.

D. 2 : 1

Ans. C.

Sol:

( )

( )

( )

( )

2

2

2

2

3 1 164

43 1

+

−

+= = =

−

max

min

I=

a b

I a b

3. Find the dimension of

0

0 .

A. M–1 L–2 T3 A2

B. M–2 L–1 T2 A3

C. M–1 L–1 T2 A3

D. M–1 L–1 T2 A2

Ans. A.

Sol:

μo = [M L T–2 A–2]

ϵ0 = [M–1 L–3 T4 A2]

0

0

-1 –3 4 2

–2 –2

–1–1 –3–1 4+2 2+2

–2 –4 6 4

ε=

μ

M L T A=

ML T A

= M L T A

= M L T A

= M–1 L–2 T3 A2

4. VAC = 220 sin 100 πt, A resistor of 50 ohm is connected across it. What is the time

taken to reach peak value of current from half of the peak value.

Sol: Given that, VAC = 220 sin 100 πt and as we know that the AC voltage is V = Vosinωt

.

From the standard equation, Vo = 220 V, R = 50 Ohm, and

2204.4

50

oo

VI Amp

R= = =

4

4 44

⎯⎯→

⎯⎯→

TMaximum

T.

2

6 100 6

=

π

π

T

T =

1sec

300

5. What is the wavelength of carrier wave in optical fibre.

A. 900 nm

B. 1500 nm

C. 2700 nm

D. 3200 nm

Ans. B.

6. Four identical particles each of mass m are moving in a circle under mutual

gravitational attraction and the side of the square inscribed in the circle is a. Find the

velocity of the particle

A. ( )GM

V = 1 + 2 2R

B. ( )1 GM

V = 1 + 2 22 R

C. ( )1 GM

V = 1 + 1 22 R

.

D. ( )GM

V = 2 + 2 2R

Ans. B.

Sol:

Net gravitation force on each particle = Centripetal force

Fcos45° + Fcos45° + F’ =

2MV

R

( ) ( )

( )

2 2

2 2

22

1

2 22

1 1

4 2

2 4

4 2

11 2 2

2

+ + =

+

+

GM GM MV

RRR

GM

R

GM

R

GM

R

= 2 ×

= + = MV

= V =

= V =

7. Find the potential difference between the points A and B.

A. 14/5

B. 12/5

C. 13/5

D. 11/5

Ans. A.

.

8. If the voltage across parallel plate capacitor is 500 V having a dielectric which can

handle maximum electric field of 6V/m and cross-sectional area of 10–4 m. Find the

dielectric constant if the capacitance of capacitor is 15 pf.

A. 14 × 10–12

B. 15 × 10–14

C. 12 × 10–12

D. 15 × 10–12

Ans. D.

Sol:

( )

12

4120

500

500

6

1015 10

2503

=

=

dEd

d

k

–0

––

ε kA= 15×10

=

ε

9. If 1022 no. of atoms per second having mass 10–26 kg are following on an area of 1m2

with the velocity of the atoms 104 m per second then find the pressure applied over

the surface. (assume perfectly reflecting surface)

Solution-

22 –26 4

2

2 mvP =

E

A

2 10 10 10P

1

P 2 N / m

=

=

10. Energy required for transition in hydrogen atom from 1st energy level to the 2nd

energy level is E. Then which of the following transitions is possible for some energy in

helium ion?

A. 2 → 3

.

B. 2 → 4

C. 1 → 2

D. 1 → 3

Ans. B.

Solution. Use the formula

2

2 2

1 2

1 1E cZ

n n

= −

11. Starting from rest a particle is moving along x-axis. The variation of force with is

shown in the figure. Find the kinetic energy of particle when it is at

x = 3m.

A. 3.3 J

B. 6.5 J

C. 5 J

D. 4.2 J

Ans. B.

Sol: Area = 4 + 2.5 = 6.5

k = 6.5 J

12. A ring has N turns carrying current I in xz plane direction of magnetic field is ˆBi.

Find the torque on the ring.

A. zero

B. B nI πr2

.

C. BIr2

D. NB πr2I

Ans. D.

Sol: Torque on the ring is

2

| |=|M × B|

= NI r B

13. A bob is suspended vertically having charge 2 Coulomb, mass 200 kg and a

horizontal electric field of 2000 V/m is applied then find angle of pendulum with

vertical.

A. tan–1 (0.5)

B. tan–1 (5)

C. tan–1 (2)

D. tan–1 (4)

Ans. C.

Sol:

T sinθ = qE

T cosθ = mg

1

2 200

200 10

2

=

–

tanθ

tanθ =

θ = tan

qE

mg

14. Which of the following is incorrect regarding first law of thermodynamics in a given

process.

.

A. Isothermal, Q = –W

B. Adiabatic, Q = – ΔU

C. Isothermal, Q = ΔU

D. Cyclic, Q = –W

Ans. B.

Sol: For adiabatic process Q = 0

15. In an electromagnetic wave electric field is along y-axis and at a point its

component along x-axis is 6V/m. find component of magnetic field at that point.

A. 3 × 10–8

B. 9 × 10–6

C. 2 × 10–8

D. 3 × 10–6

Ans. C.

Sol:

0

8

8

6

3 10

2 10

=

=

E

c0

–

B =

16. Two particles are travelling at right angles having de-broglie wavelength λ1 and λ2.

They are collide inelastically. Find the final λ of the particles after collision.

Sol:

.

2 2

2 21 2

2 22

h

h

h h

11

22

2 21 2

eq

2

P =λ

P =λ

P = P + P

h= +

λ

1 1 1+ +

λ λ λ

17. A ray is incident on an optical fibre at an angle of 40°. It is 2m in length and 40 m in

diameter. Refractive index of the optical fibre is 1.31 find the number of reflections it

makes before exiting.

Ans. 57803

18. The Electric field of plane polarized E.M. Wave in free space at t = 0 is

( ) ( )+ˆE x, z = 10 j cos 6x 8z . The magnetic field ( )B x, z

A. ( )

++

ˆˆ–8i 6kcos 6x 8z

C

B. ( )

++

ˆ ˆ– 8i 6 jsin 6x 8z

C

C. ( )

++

ˆˆ8i 6kcos 6x 8z

C

D. ( )+

10j cos 6x 8z

C

Ans. A.

Sol: As we know that the magnetic field is

.

( )0

00

6x

EB

C

n

ˆB = B n cos + 8z

=

ˆ ˆˆ = C × E

CHEMISTRY

1. which amine is prepared by Gabriel phthalimide

A. t- butyl amine

B. neo butyl amine

C. Tertiary butyl amine

D. butyl amine

Ans. D

(8th April 2019 (First Shift), based on memory, as shared by students)

Sol. Gabriel synthetic is used to prepare primary amines from primary alkyl halides.

C

C

O

N-H

O

KOH⎯⎯→

C

C

O

N

O

Br

COO

COO

NH2 3H O+

⎯⎯

C

C

O

N

O

CH2

CH2-CH-CH3

2. What is wrong regarding to human about responsibility to protect our environment.

A. Utilising plastic bags

B. Using dust bin

C. No cleanliness

.


3.

C

C

O

O

O Cl

3

2

AlCl

H O⎯⎯→


Sol.

C

C

O

O

O

3AlCl⎯⎯→

C

C

O

O

O

AlCl3

C

C

O

O

Cl

O

AlCl3

C O

O

C

O

Cl

Cl

4. Increasing order of basic strength

A. ( )2 5 2C H NH

B. 2 5 2C H NH

C. 3NH

D. ( )2 5 3C H N

(A) A > B > C > D (B) A > D > B > C

(C) B > A > D > C (D) B > D > C > A

Ans. (B)


.

Sol. (C2H5)2 NH > (C2H5)3 N > (C2H5)1 NH2 > NH3

A > D > B > C

5.

N

NH2

OO

A. Tridentate

B. Tetradentate

C. Hexadentate

D. Bidentate

Ans. B


6. A & B two liquid mole fraction B = 0.5, 0BP 600= , 0

AP 400= b find total pressure and

respective mole fraction of A & B in air

A. 200atm, 0.4, 0.6

B. 400atm, 0.5, 0.5

C. 300atm, 0.5, 0.5

D. 500atm, 0.4, 0.6

Ans. D


Sol. PT = ( ) ( )0.5 600 0.5 400 +

= 300 + 200 = 500 atm

For Air

BB

T

P 300 3x

P 500 5= = =

.

0 = 0.6

AA

T

P 200x 0.4

P 500= = =

7. Benzene diazonium chloride reacts with -Napththol to give


Sol.

N2Cl

OH

N N

OH

benzene diazonium chloride

2-Naphthol anisole dye(orange-red dye)

8. Which of the following amino acids is most basic

A. Arg

B. Lys

C. Gly

D. Asp

Ans. A


9. What is the solubility product of

Zr3(PO4)4?


.

Sol. ( ) 4 33 4 44 3S 4S

Zr PO 3Zr 4PO+ −+

( ) ( )3 4

spK 3s 4s=

= 3 424s 256s

= 6912 S7

10. What is the IUPAC name of the following compound

CH3 CH CH CH2 COOH

CH3 HO


Sol.

CH3 CH CH CH2 COOH

CH3 HO

5 4 3 2 1

3-hydroxy-4-methyl pentatonic acid

11. If B belong to ccp and A belong octahedral void and oxygen belong to tetrahedral

void then the formula of solid is

A. A2B2O4

B. A2BO

C. AB2O4

D. ABO2

Ans. D


Sol. Effective no. of B atom = 4

Effective no. of A atom = 4

.

Effective no. of O atom = 8

Formula of solid is A4B4O8

= ABO2

12. What is the compound A which gives the following tests positive

A in H2O is Insoluble

A in 5% HCL is Insoluble

A in 10% HCL is soluble

A in NaHCO3 is Insoluble

A. Oxalic Acid

B. Benzyl amide

C. m-Cresol

D. O-toluidine

Ans. D


13. Maltose on treatment with HCl gives

A. D-Fructose

B. D-Glucose

C. D-Fructose, D-Glucose

D. Galactose

Ans. B

14.

OCH3

CH=CH2

o

conc.HBr

25 C⎯⎯⎯→

.

A.

OH

CH2-CH2Br B.

OH

Br-CH2-CH3 C.

Br

Br-CH-CH3 D.

Br

CH2-CH2-Br

15. Arrange in order of increasing energy

A. n = 4, l = 2, m = , s = -1/2

B. n = 4, l = 1, m = , s = +1/2

C. n = 3, l = 2, m , s = -1/2

D. n = , l = 1, m = , s = -1/2

Sol. A > B > C > D

16. Which of the following forms coloured ions?

A. 3Gd +

B. 3La +

C. 3Nd +

D. 3Lu +

Ans. C


Sol. Nd3+ = 3xe 4f

Note: 3 7Gd xe 4f+ = , the wavelength required is grater than visible region so

coloureless.

17. Hydration enthalpy of Na+ , Li+ , K+, Rb+, Cs+ is in order


Sol. q

r

= charge density

| hydration enthalpy|

q is same for all

lesser r, higher the |Hhyd|

Li Na k Rb Cs+ + + + +

.

18. What make effect on size of iso-electronic species Cl–, Ar, Ca+2?

A. Nuclear charge

B. Principle quantum number

C.

D.


Sol. Zeff = z -

is constant for isoelectronic species (because no of e–(s) is same)

Zeff Z (no of proton)

eff

1r

Z

19. Carbon dioxide on heating with lime stone ore, in metallurgy this is called

A. Smelting

B. Ore dressing

C. Calcination

D. Roasting

Ans. C


Sol. Calcination → ore is heated strongly in absence or limited supply of air at a

temperature that ore dose not melt, for removal of volatile impurity such as CO2

moisture. Etc

3 2CaCO CaO CO⎯⎯→ +

2 3 2 2 3 2Al O .2H O Al O 2H O⎯⎯→ +

20. What is the use of Ellingham diagram

A. Electrolysis

.

B. Thermal Reduction

C. Zone refining

Ans. B


Sol. Plot of G & T

→ Determine the reactive case of reducing a given metallic oxide to metal

→ This diagram show the stability of the oxides as function of T.

21. When Diborane (B2H6) reacts with O2 and H2O independently, then the products

formed are.


Sol. 2 6 2 2 3 2B H 3O B O H O+ → +

( )2 6 2 23B H 6H O 2B OH 6H+ → +

22. White phosphorus possesses which composition in vapours state

A. P6

B. P8

C. P2

D. P4

Ans. D


.

Sol. Vapour density study confirm its formula P4 upto 800oC and above 800oC P2

molecules exists.

23. On hydrolysis, PCl5 would produce

A. H3PO3

B. H3PO4

C. HPO3

D. H4P2O7

Ans. B


Sol.

5 2 3

3 2 3 4

PCl H O POCl 2HCl

POCl 3H O H PO 3HCl

+ → +

+ → +

PCl5 → it hydrolyses to POCl3 & finally converted to H3PO4

MATHEMATICS

1. If and are roots of the quadratic equation − + =2x 2x 2 0 and

=

n

1 . Then

find the least possible integral value of n.


Ans. 4

Sol.

.

( )

( )

( )

2

2

2

n

n

x 2x 2 0

x 1 1 0

x 1 1

x 1 i

x 1 i

1 i1

1 i

i 1

n 4

− + =

− + =

− = −

− =

=

+ =

−

=

=

2. The minimum distance between the curve = −2y x 2 and line =x y is?


Ans.

Sol.

Shortest distance b/w = −2y x 2 and y = x will be along common normal. A line which

is normal to = −2y x 2 and perpendicular to y = x

Slope of normal = –1

Equation of Normal to ( )

= −

2 1y 4 x 2

4 is

( ) 3y m x 2 2am am= − − − at ( )22 am , 2am+ −

9 1,

4 2

Distance b/w y = x and 9 1

,4 2

.

−

=

9 174 2

2 4 2

3. The local maxima and minima of function ( ) = + − +4 3 2f x 9x 12x 36x 25 is (Local

maxima is represented by set S1 and local minima is represented by set S2)

S1 = ? and S2 = ?


Ans.

Sol. ( ) = + −3 2f ' x 36x 36x 72x

( ) ( )= + −2f ' x 36x x x 2

( )( )36x x 2 x 1= + −

=1S 0

= −2S 2,1

4. Integrate ( )( )

−

/4

/4

g f x where

.

( )−

=+

2 x cosxf x

2 xcosx , ( ) = eg x log x


Ans.

Sol. ( )( )−

= +

2 xcoxg f x ln

2 xcosx

( )( )+

− = −

2 xcosxg f x ln

2 xcosx

( )g of x is an odd function

( )

−

=4

4

gof x dx 0

5. Integrate

5xsin

2

xsin

2


Ans.

Sol.

5x 3x 3x x xsin sin dx sin sin dx sin dx

2 2 2 2 2x x x

sin sin sin2 2 2

− −

+ +

= + + 2cos2x dx 2cosx dx dx

= + + +sin2x 2sinx x c

6. The sum of the roots of the equation

.

( )− + − + =x 2 x x 4 2 0


Ans.

Sol. Let,

x t 0=

− + − + =2t 2 t 4t 2 0

− =

=

=

− + =

=

=

2

2

t 3t 0 if t 2

t 0,3

t 0;t 3

t 5t 4 0 if t 2

t 1,4

t 4;t 1

=

= =

= =

t 3,1

x 3; x 1

x 9;x 1

Sum of roots = 10

7. The differential equation

( ) ( )+ + + =22 2dy

1 x 2y x 1 x 1dx

with conditions ( ) =f 0 0 and ( )

=af 132

. Then =a ?


.

Ans.

Sol.

( )

( )++

+ =+ +

= = = +

22

22 2

2xdx ln 1 x 21 x

dy 2yx 1

dx 1 y 1 x

I.F e e 1 x

( )( )

( ) + = ++

2 2

22

1y. 1 x 1 x dx

1 x

( )

( ) ( )

( )

2 1

1

2

y 1 x tan x c

tan xf 0 0 f x

1 x

a.f 1 a.32 8 32

1a

16

−

−

+ = +

= =+

= =

=

8. Let O(0, 0), A(0,1) and P be point such that the perimeter of triangle =OAP 4 , then

the locus of point P = ?


Ans.

Sol.

.

( )

( )

+ + + − =

+ − = − +

22 2 2

22 2 2

2

x y x y 1 3

x y 1 3 x y

x + 2y − + = + 22y 1 9 x + 2y − +2 26 x y

+ = +2 26 x y 8 2y

+ = +2 23 x y 4 y

+ = + +2 2 29x 9y 16 y 8y

+ − − =2 29x 8y 8y 16 0

( )2 29x 8 y y 16+ − =

+ − + =

2 2 19x 8 y y 18

4

+ − =

2

2 19x 8 y 18

2

−

+ =

2

2

1y

x 21

924

9. Arrange these numbers 1, 1, 2, 2, 2, 2, 3, 4, 4 by using all numbers at a time such

that the odd numbers occupy even places

.


Ans.

Sol.

1, 1, 3 can be arranged in 2nd, 4th, 6th, 8th position in 4

3P

2l ways = 12 ways.

Remaining 2, 2, 2, 2, 4, 4 can be arranged in =6l

154l 2l

ways

Total ways = 12 × 15 = 180 ways

10. → − +

2

x 0

sin xlim

2 1 cosx


Ans.

Sol. By rationalize the denominator

( )→

→

+ +

−

=

2

x O

2

x O

sin x 2 1 cosxIt

1 cosx

sin x 2It

( )2

2 2 xsin

2

( )( )→

= =

x 0

sinx sinx 2It

x xsin sin

2 2

2 2 2 4 2

.

11. If ( )+ =3

cos5

and ( )− =5

sin13

, then ( ) =tan 2 ?


Ans.

Sol.

( ) ( )

( ) ( )

( ) ( ) ( )( )( ) ( )( ) ( )

3 4cos tan

5 35 5

sin tan15 12

tan 2 tan

tan tan

1 tan tan

4 5633 12

20 16136

+ = + =

− = − =

= + + −

+ + −=

− + −

+

= =

−

12. If − =

1 3cos

5 and −

=

1 5tan

12 then − = ?


Ans. 1 33tan

56−

− =

Sol.

= =3 4

cos tan5 3

.

=5

tan12

( )−

− − = = =

+ + +

4 5tan tan 113 12tan

201 tan tan 56136

1 33tan

56−

− =

13. Find the sum of series –

( ) ( ) ( ) ( )+ + + +20 20 20 200 1 2 202 C 5 C 8 C ........... 62 C


Ans. 192 64

Sol.

( )20

20r

r 0

3r 2 . C=

+

20 2020 20

r rr 1 r 0

3 r. C 2. C= =

= +

2019 20

r 1r 1

3 20. C 2.2−

=

= +

= + 19 21 1960.2 2 2 64

14. If AB (A is a subset of B), then

A. P(A/B) = 1

B. P(A/B) = P(B) – P(A)

C. P(A/B) P(A)

D. P(A/B) P(A)


.

Ans. C

Sol.

A B

( )( )

( )( )

( )

= =

P A B P AAP P A

B P B P B

( )

AP P A

B

As ( ) P B 1

15. The contrapositive statement of

“If you are born in India, then you are citizen of India.”


Sol.

If you are not a citizen of India then you are not born in India.

16. If − −

= +

2

1 3 cosx sinx2y cot ,

cosx 3 sinx then find

dy

dx.


Ans.

Sol.

− −

= +

2

1 3 tanx2y cot

1 3 tanx

− = −

2

12y cot tan x3

.

2 2

12y cot cot x 2y x6 6

−

= + = +

2 =dy

2dx

+

x

6

= +

dyx

dx 6

17. if −

=

cos sinA

sin cos and

− =

320 1

A1 0

, then what is the value of ?


Ans. 64

=

Sol.

− − = =

32cos32 sin32 0 1

Asin32 cos32 1 0

=cos32 0 and =sin32 1

( )

= +2n 164

( )

= +4n 164

64

=

18. If f(x) is twice differentiable in [0, 2] and f"(x)>0 and ( ) ( ) ( ) = + −x f x f 2 x , then

( ) x is

.

A. Increasing in [0, 2]

B. Decreasing in [0, 2]

C. Increasing in [0, 1] and decreasing in [1, 2]

D. Decreasing in [0, 1] and increasing in [1, 2]


Ans. D

Sol.

( ) ( ) ( ) = − −' x f ' x f ' 2 x

As ( ) ( ) f" x 0 f' x is an increasing function

If ( ) x is increasing then ( )' x 0

( ) ( ) −f ' x f' 2 x

−

x 2 x

x 1

( ) x is increasing in [1, 2]

and interval in which ( ) x is decreasing

( ) ( ) ( )' x 0 f' x f' 2 x x 2 x

x 1

− −

( )' x is decreasing in [0, 1]

( )' x is decreasing in [0,1] and increasing in [1, 2]

option (d) is correct.

19. Area of region described by

0 x 3, 0 y 4 for 2y x 3x +


.

Ans. 59/6

Sol.

( )= + + 1 3

2

0 1x 3x dx 4dx

= + +1 3

83 2

=59

6

20. In an ellipse + =2 24x y 8 , a tangent is drawn at (1, 2) which is perpendicular to

tangent at point (a, b). Then value of a2 is.


Ans. =2 2a

17

Sol.

Tangent at (1, 2) is ( ) ( )+ =4x 1 y 2 8

+ =

= −

2x y 4

slope 2

Slope of Tangent perpendicular to tangent at (a, b) is 1

2.

Tangent at (a, b) is + =4ax by 8

Slope =−

= = −4a 1

b 8ab 2

Also, + =2 24a b 8 + =2 24a 64a 8

.

=

=

2

2

68a 8

2a

17

21. If ( )−

= +

1 xf x n

1 x; ( ) =

+ 2

2xg x

1 x find ( )( )f g x in terms of ( )f x .


Ans. 2f(x)

( )( )( )

− − += =

+ ++

2

2

2x11 g x 1 xfog x n n

2x1 g x 11 x

( )

( )

−=

+

2

2

1 xn

1 x

− =

+

1 x2 n

1 x

( )= 2f x

22. Sum of all the coefficients of even degree terms in expansion of

+ − + − −

6 63 3x x 1 x x 1


Ans. 24

Sol.

( ) ( )

( )

+ − + − + −

26 6 6 4 3 6 2 30 2 4

36 0 36

c x c x x 1 c x x 12

c x x 1

.

( )

( )

+ − + + − + − − +

6 7 4 2 6 3

9 6 3

x 15x 15x 15x x 1 2x2

x 1 3x 3x

+ − + + − + − − +

6 7 4 8 2 5

9 6 3

x 15x 15x 15x 15x 30x2

x 1 3x 3x

+ + − − − + + −

9 8 7 6 5 4

3 2

x 15x 15x 2x 30x 15x2

3x 15x 1

=sum 2 15 − −2 15 + −

=

15 1

24

23. If the mean and variance of seven numbers are 8 and 16 respectively. If the five

observations are 2, 4, 10, 12, 14. Then, the product of other two numbers is ?


Ans. =1 2x x 48

Sol.

Mean = + + + − − − −

= =1 2 3 7x x x xx 8

7

+ + − − − =1 2 7x x x 56

+ =1 2x x 14

Variance = ( )2

22 ix

x 16n

= − =

22i

i

x80 x 560

7

= =

.

+ + + + + + =2 21 2x x 4 16 100 144 196 560

2 21 2x x 100+ =

And + + =2 21 2 1 2x x 2x x 196

=1 2x x 48

24. What is the sum of squares of intercepts made by all chords of the line x + y = n

(where n is a natural number) on the circle + =2 2x y 16


Ans. 210

Sol.

Let =PQ

In OAQ

=n

OA2

=OQ 4

= −2n

AQ 162

.

2nPQ 2 16

2= = −

= − = −

22 2n

4 16 64 2n2

If n N n = 1, 2, 3, 4, 5

Sum of squares of all lengths

=

= − = − =5

2

n 1

64 2n 320 110 210

25. Find sum of all ‘n’ natural numbers where 100<n<200 and HCF (91, n) > 1


Ans. 3121

Sol.

Sum = Sum of all multiple of 13 between 100 and 200 + Sum of all multiple of 7

between 100 and 200 – sum of all multiples of 91 between 100 & 200

( ) ( )+ − − − + − − −= + −

104 195 105 196182

8terms 14terms

+ − =1196 2107 182 3121

PHYSICS - gradeup.co April Shift _ 1.pdf-74.pdf · Formula of solid is A 4 B 4 O 8 = ABO 2 12. What...

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