CG Pre MCA 2021 Question Paper

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Question Booklet No.
SET – A


A Subject Code : 32102/UE – CA/ENT – E

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Number of Pages in this Question Booklet : 64 Number of Questions in this Question Booklet : 200

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11. Bg àíZ nwpñVH$m _| Mma ^mJ hmo§Jo :- 11. This Question booklet contains four Parts :
(i) ^mJ I :- J{UV - 1-100 100 A§H$ (i) Part I : – Mathematics – 1-100 100 Marks
(ii) ^mJ II :- H$åß`yQ>a OmJê$H$Vm - 101-140 40 A§H$ (ii) Part II : – Computer Awareness – 101-140 40 Marks
(iii) ^mJ III :- {díbofU Ed§ VH©$ epŠV - 141-180 40 A§H$ (iii) Part III : – Analytical ability and – 141-180 40 Marks
(iv) ^mJ IV :- gm_mÝ` AÜ``Z - 181-200 20 A§H$ Logical Reasoning
àË`oH$ àíZ 1 A§H$ H$m h¡ & g^r àíZ hb H$aZm A{Zdm`© h¡ & (iv) Part IV : – General Awareness – 181-200 20 Marks
Each question contains 1 mark. All questions are compulsory.
12. `{X A§JO
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be considered authentic.
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-2- Set-A

PART – I
Mathematics J{UV
1.
In solving the following LPP by simplex 1. {gåßboŠg {d{Y Ûmam {ZåZ LPP H$mo hb H$aZo _|
method the first outgoing vector is àW_ OmdH$ g{Xe h¡
max. z = 2x1 + 8x2 + 6x3 max. z = 2x1 + 8x2 + 6x3
s.t. 3x1 + 2x2 + 4x3 ≤100, s.t. 3x1 + 2x2 + 4x3 ≤100,
x1 + 4x2 + 2x3 ≤100, x1 + 4x2 + 2x3 ≤100,
x1 + 3x2 + x3 ≤100, x1 + 3x2 + x3 ≤100,
x1, x2, x3 ≥0 x1, x2, x3 ≥0
(A) (0, 1, 0) (B) (0, 0, 1) (A) (0, 1, 0) (B) (0, 0, 1)
(C) (1, 0, 0) (D) None of these (C) (1, 0, 0) (D) BZ_| go H$moB© Zht

2.
Using simplex method the minimum 2. {gåßboŠg {d{Y H$m Cn`moJ H$aHo$ {ZåZ LPP _|
value of Z in the following LPP is Z H$m Ý`yZV_ _mZ h¡
min. Z = 4x + 8y + 3z min. Z = 4x + 8y + 3z
s.t. x + y ≥ 2, s.t. x + y ≥ 2,
2x + z ≥ 5, 2x + z ≥ 5,
x, y, z ≥0 x, y, z ≥0
(A) –10 (B) 10 (A) –10 (B) 10

(C) 5 (D) None of these (C) 5 (D) BZ_| go H$moB© Zht
2 2

3. The number of seven digit integers, 3. A§H$ 1, 2 VWm 3 go ~ZZo dmbr gmV A§H$m| dmbo
with sum of digits equal to 10 and nyUmªH$m| H$s g§»`m {Og_| A§H$mo§ H$m `moJ 10 Ho$
formed by using the digits 1, 2 and 3 ~am~a h¡, hmoJr
only, is
(A) 60 (B) 81 (A) 60 (B) 81
(C) 56 (D) 77 (C) 56 (D) 77

4. The number of total three digit natural 4. VrZ A§H$mo§ dmbr Hw$b àmH¥$V g§»`mAm| H$s g§»`m
numbers having only two consecutive {Og_| Ho$db Xmo bJmVma A§H$ g_mZ h¡, hmoJr
digits identical is
(A) 153 (B) 161 (A) 153 (B) 161
(C) 162 (D) None of these (C) 162 (D) BZ_| go H$moB© Zht

-3- Set-A

5. The value of λ and µ for which the 5. λ VWm µ Ho$ _mZ {OZHo$ {bE g_rH$aUm|
equations x + y + z = 3, x + 3y + 2z = 6 x + y + z = 3, x + 3y + 2z = 6 VWm
and x + λy + 3z = µ have a unique x + λy + 3z = µ H$m A{ÛVr` hb hmo, h¡
solution
(A) λ = 5, µ ≠ 9 (A) λ = 5, µ ≠ 9
(B) λ ≠ 5, µ ∈  (B) λ ≠ 5, µ ∈ 
(C) λ = 5, µ = 9 (C) λ = 5, µ = 9
(D) None of these (D) BZ_| go H$moB© Zht

6. If α, β, γ are the roots of equation 6. `{X α, β, γ g_rH$aU x3 + ax2 – b = 0 Ho$ _yb
x3 + ax2 – b = 0, then the determinant α β γ
α β γ h¢, Vmo gma{UH$ ∆ = β γ α ~am~a h¡
∆ = β γ α equals γ α β
γ α β
(A) – a3 (B) a3 – 3b (A) – a3 (B) a3 – 3b
(C) a2 + 3b (D) a3 (C) a2 + 3b (D) a3

7. The variables X and Y are connected 7. Ma X VWm Y g_rH$aU 2X + 3Y + 5 = 0 go
by the equation 2X + 3Y + 5 = 0. Then gå~Õ h¢& V~ BZHo$ ~rM ghgå~ÝY r h¡
the correlation r between them is
(A) –1 (B) +1 (A) –1 (B) +1

(C) 0 (D) + 1 (C) 0 (D) + 1
2 2

8. The angle θ between two lines of 8. Xmo g_ml`U aoImAm|, {OgHo$ {bE r = ± 1, Ho$
regression for which r = ± 1 is ~rM H$m H$moU θ h¡
(A) π (B) 0 (A) π (B) 0
2 2
π π π π
(C) 4 (D) 6 (C) 4 (D) 6

-4- Set-A

2 + x, x ≥ 0 2 + x, x ≥ 0
9. If f( x ) =  , then choose 9. `{X f(x) =  , Vmo gË` H$WZ
2 − x, x < 0 2 − x, x < 0
the correct statement. Mw{ZE &
(A) lim f( x ) exists (A) lim f( x ) H$m ApñVËd h¡
x →0 x →0
(B) f(x) is continuous at x = 0 (B) f(x), x = 0 na gVV h¡
(C) f(x) is not differentiable at x = 0 (C) f(x), x = 0 na AdH$bZr` Zht h¡
(D) All of these (D) CnamoŠV g^r

 1 + e x , x ≤ 0  1 + e x , x ≤ 0
10. If f( x ) =  , then 10. `{X f(x) =  , Vmo
 2 − x , x > 0  2 − x , x > 0
(A) f(x) is differentiable at x = 0 (A) f(x) , x = 0 na AdH$bZr` h¡
(B) f(x) is differentiable at x = 2 (B) f(x) , x = 2 na AdH$bZr` h¡
(C) f(x) is dis-continuous at x = 0 (C) f(x) , x = 0 na AgVV h¡
(D) f(x) is continuous at x = 2 (D) f(x) , x = 2 na gVV h¡
2
11. The maximum value of the function 11. \$bZ f(x) = x4e–x H$m CpÀMîQ>> _mZ h¡
2
f(x) = x4e–x is
(A) 4e2 (B) 4e–2 (A) 4e2 (B) 4e–2
(C) e2 (D) None of these (C) e2 (D) BZ_| go H$moB© Zht

dx dx
12. ∫ x ( x + 1) =
4
12. ∫ x ( x + 1) =
4

1  x 4 + 1 1  x 4 + 1
(A)
4
log  x 4  + c (A)
4
log  x 4  + c

1  x4  1  x4 
(B) log  4 +c (B) log  4 +c
4  x + 1 4  x + 1
1
(C)
4
( )
log x 4 + 1 + c (C)
1
4
(
log x 4 + 1 + c )
(D) None of these (D) BZ_| go H$moB© Zht

-5- Set-A

13. If cos (cot–1(x+1)) = sin(tan–1x), then x = 13. `{X cos (cot–1(x+1)) = sin(tan–1x),V~ x =
1 1
(A) 0 (B) (A) 0 (B)
2 2
1 1 1 1
(C) – (D) (C) – (D)
2 2 2 2

14. The angle of depression of a ship 14. EH$ 60 _rQ>a D±$Mo _rZma H$s MmoQ>r go _rZma H$s
proceeding towards tower from the top Amoa ~‹T> aho OhmO H$m AdZ_Z H$moU 3 AM H$mo
of a tower of height 60 meter is 60° at 60° VWm 4 AM H$mo 30° h¡ & V~ 3:30 AM H$mo
3 AM and 30° at 4 AM. Then what was
the angle of elevation of the top of the
OhmO go _rZma H$s MmoQ>r H$m CÞ`Z H$moU Š`m Wm ?
tower from ship at 3:30 AM ?
−1 3 π −1 3 π
(A) tan (B) (A) tan (B)
2 2 2 2

−1 3 −1 3
(C) cot (D) None of these (C) cot (D) BZ_| go H$moB© Zht
2 2

15. A person finds angle of elevation of the 15. EH$ ì`pŠV EH$ Ka >Ho$ MmoQ>r H$m CÞ`Z H$moU 30°
top of a house 30° and when he moves nmVm h¡ Am¡a O~ dh Cg Ka H$s Amoa 60 _rQ>a
60 meter towards that house then he ~‹T>Vm h¡, V~ dh CÞ`Z H$moU 60° nmVm h¡ & V~
gets the angle of elevation 60°. Then
Cg Ka H$s D±$MmB© Wr
height of that house was
(A) 30 meter (B) 30 3 meter (A) 30 _rQ>a (B) 30 3 _rQ>a

(C) 60 3 meter (D) None of these (C) 60 3 _rQ>a (D) BZ_| go H$moB© Zht

 2  2   2  2 
16. cot  cos −1   + sin−1    = 16. cot  cos −1   + sin−1    =
 3  3   3  3 
1 2 1 2
(A) (B) (A) (B)
3 3 3 3
1 1
(C) (D) 0 (C) (D) 0
3 3

-6- Set-A

( 2 + 1) + ( 2 − 1) ~am~a h¡
4 4
( ) ( )
4 4
17. 2 +1 + 2 −1 is equal to 17.
(A) An irrational number (A) EH$ An[a_o` g§»`m
(B) A rational number (B) EH$ n[a_o` g§»`m
(C) A negative integer (C) EH$ F$UmË_H$ nyUmªH$
(D) None of these (D) BZ_| go H$moB© Zht

`{X ( x − 21x ) Ho$ {dñVma _| V¥Vr` Ed§ MVwW© nXm|
n
18. If the ratio of coefficients of third 18.
and fourth term in the expansion of
Ho$ JwUm§H$m| H$m AZwnmV 1 : 2 h¡, Vmo n H$m _mZ
( x − 21x ) is 1 : 2, then the value of n
n
hmoJm
will be
(A) 18 (B) – 16 (A) 18 (B) – 16
(C) 12 (D) – 10 (C) 12 (D) – 10

19. The sum of the coefficients of all the 19. (1 + 2 x )40 Ho$ {dñVma _| x Ho$ nyUmªH$s` KmV
integral powers of x in the expansion dmbo nXm| Ho$ JwUm§H$m| H$m `moJ h¡
of (1 + 2 x )40 is

(A) 340 + 1 1 (340 + 1)
(B) 2 (A) 340 + 1 1 (340 + 1)
(B) 2
1 (340 – 1)
(C) 2 (D) 340 – 1 1 (340 – 1)
(C) 2 (D) 340 – 1

20. The last two digits of the number 20. g§»`m 9200 Ho$ A§{V_ Xmo A§H$ h¡
9200 is
(A) 01 (B) 10 (A) 01 (B) 10
(C) 31 (D) None of these (C) 31 (D) BZ_| go H$moB© Zht

21. If 2i − j + 2k , i + j + k and 3i + j − 2k are 21. `{X 2i − j + 2k, i + j + k VWm 3i + j − 2k
coterminous edges of a parallelepiped, EH$ g_mÝVa fQ²>\$bH$ H$s EH$ {~ÝXþJm_r EOog hmo,
then volume of this parallelepiped is V~ Bg g_mÝVa fQ²>\$bH$ H$m Am`VZ h¡
(A) 15 (B) 13 (A) 15 (B) 13
(C) 15 (D) 13 (C) 15 (D) 13

-7- Set-A

 
Forces F1 = 2i − j + 3k and
22. 22. `{X EH$ H$U na bJZo dmbo ~b F1 = 2i − j + 3k
  
F 2 = 3i + j − 2k acting on a particle VWm F 2 = 3i + j − 2k Bgo {~ÝXþ r1 = i − j + k
 
displace it from a point r1 = i − j + k go {~ÝXþ r 2 = 2i + 3j + 4k na {dñWm{nV H$aVo

to the point r = 2i + 3j + 4k , then the
2 hmo, V~ BZ ~bm| Ûmam {H$`m J`m H$m`© h¡
work done by these forces is
(A) 13 (A) 13

(B) 8 (B) 8

(C) 5 3 (C) 5 3
(D) 7 (D) 7

23. If 2i − j + k , i + j − k and 3i + 2j + 3k 23. `{X 2i − j + k, i + j − k VWm 3i + 2j + 3k
are respectively position vectors of H«$_e: EH$ ∆ABC Ho$ erfm] A, B VWm C Ho$
vertices A, B and C of a ∆ABC, then
pñW{V g{Xe hmo, V~ ∆ABC h¡
∆ABC is
(A) Isosceles (A) g_{Û~mhþ
(B) Equilateral (B) g_~mhþ
(C) Right angled (C) g_H$moU
(D) None of these (D) BZ_| go H$moB© Zht

24. If i + j, 2j + 3k and 2i − k are position 24. `{X i + j, 2j + 3k VWm 2i − k H«$_e: EH$
vectors of vertices A, B and C ∆ABC Ho$ erfm] A, B VWm C Ho$ pñW{V g{Xe
respectively of a ∆ABC, then the hmo, V~ A go _mpÜ`H$m H$s bå~mB© hmoJr
length of median through A will be
(A) 1 (A) 1
(B) 2 (B) 2
(C) 3 (C) 3
(D) None of these (D) BZ_| go H$moB© Zht

-8- Set-A

25. The LPP represented by the following 25. {ZåZ AmaoI Ûmam {Zê${nV LPP H$m hb h¡
graph has solution

(A) x = 0, y = 7 (A) x = 0, y = 7
(B) x = 5, y = 0 (B) x = 5, y = 0
(C) x = 1.2, y = 2.3 (C) x = 1.2, y = 2.3
(D) x = 1.6, y = 2.4 (D) x = 1.6, y = 2.4

26. An analysis of result of a subject of 26. EH$ hr narjm Ho$ {bE {Z`{_V Ed§ ñdmÜ`m`r
regular and private students for same N>mÌm| Ho$ EH$ {df` Ho$ n[aUm_ H$m {díbofU
exam was as follows :
{ZåZdV² Wm :
Regular Private {Z`{_V ñdmÜ`m`r
No. of students : 18 12
N>mÌm| H$s g§»`m : 18 12
Average marks : 30 25
Am¡gV A§H$ : 30 25
Variance of distribution
A§H$m§o Ho$ {dVaU H$m àgaU : 16 49
of marks : 16 49
Then the variance of the distribution V~ XmoZm| àH$mam| Ho$ g^r N>mÌm| Ho$ A§H$m| Ho$ {dVaU
of marks of all students of both types H$m EH$ gmW àgaU h¡
together is
(A) 32.5 (B) 35.2 (A) 32.5 (B) 35.2
(C) 30.5 (D) None of these (C) 30.5 (D) BZ_| go H$moB© Zht

-9- Set-A

27. The following graph represent the LPP 27. {ZåZ{b{IV AmaoI {H$gH$m LPP {Zê${nV H$aVm
of h¡ ?

(A) max. z = 2x + 3y (A) max. z = 2x + 3y
s.t. –x + 2y ≥ 4, s.t. –x + 2y ≥ 4,
x + y ≥ 6, x + y ≥ 6,
  x + 3y ≥ 9,   x + 3y ≥ 9,
     x, y ≥ 0      x, y ≥ 0
(B) max. z = 2x + 3y (B) max. z = 2x + 3y
s.t. –x + 2y ≤ 4, s.t. –x + 2y ≤ 4,
x + y ≥ 6, x + y ≥ 6,
  x + 3y ≤ 9,   x + 3y ≤ 9,
     x, y ≥ 0      x, y ≥ 0
(C) max. z = 2x + 3y (C) max. z = 2x + 3y
s.t. –x + 2y ≤ 4, s.t. –x + 2y ≤ 4,
x + y ≤ 6, x + y ≤ 6,
  x + 3y ≤ 9,   x + 3y ≤ 9,
     x, y ≥ 0      x, y ≥ 0
(D) max. z = 2x + 3y (D) max. z = 2x + 3y
s.t. –x + 2y ≤ 4, s.t. –x + 2y ≤ 4,
x + y ≤ 6, x + y ≤ 6,
  x + 3y≥ 9,   x + 3y≥ 9,
     x, y ≥ 0      x, y ≥ 0

-10- Set-A

28. If median value is 48, then values of 28. `{X _mpÜ`H$m _mZ 48 h¡, V~ {ZåZ ~maå~maVm
f1, f2 for following frequency distribution ~§Q>Z Ho$ {bE f1, f2 Ho$ _mZ h¢ H«$_e:
are respectively :
Class : 10-20 20-30 30-40 dJ© : 10-20 20-30 30-40
Frequency : 15 25 f1 ~maå~maVm : 15 25 f1

40-50 50-60 60-70 70-80 40-50 50-60 60-70 70-80
45 f2 20 25 45 f2 20 25
Total frequency : 190 Hw$b ~maå~maVm : 190
(A) 20, 40 (A) 20, 40
(B) 19, 41 (B) 19, 41
(C) 18, 42 (C) 18, 42
(D) None of these (D) BZ_| go H$moB© Zht

29. For a group of 30 students, mean 29. 30 N>mÌm| Ho$ EH$ g_yh Ho$ {bE àmßVm§H$m| Ho$ _mÜ`
and variance of scores were 8 and 16 Ed§ àgaU H«$_e: 8 VWm16 Wo& Om±M Ho$ Xm¡amZ
respectively. During checking it was `h nm`m J`m {H$ Xmo àmßVm§H$m| 41 VWm 12
found that two scores 41 and 12 were
H$mo H«$_e: 14 VWm 21n‹T >{b`m J`m Wm& V~
misread as 14 and 21 respectively.
Then the corrected variance is g§emo{YV àgaU h¡
(A) 17.3 (A) 17.3
(B) 6.18 (B) 6.18
(C) 45.64 (C) 45.64
(D) None of these (D) BZ_| go H$moB© Zht

30. If mean and variance of binomial 30. `{X {ÛnX ~§Q>Z Ho$ {bE _mÜ` Ed§ àgaU H«$_e:
distribution are 2 and 2 respectively, 2 VWm 2 h¢, V~ P(X ≥ 1) h¡
3 3
then P(X ≥ 1) is
(A) 26 (B) 27 (A) 26 (B) 27
27 26 27 26

(C) 23 (D) 24 (C) 23 (D) 24
23 24 23
24

-11- Set-A

31.
In solving the following LPP by simplex 31. {gåßboŠg {d{Y Ûmam {ZåZ LPP H$mo hb H$aZo _|
method the first incoming vector is àW_ AmdH$ g{Xe h¡
max. z = 3x1 + 5x2 + 4x3
max. z = 3x1 + 5x2 + 4x3
s.t. 2x1 + 3x2 ≤8,
s.t. 2x1 + 3x2 ≤8,
2x2 + 5x3 ≤20, 2x2 + 5x3 ≤20,
3x1 + 4x2 + 5x3 ≤60, 3x1 + 4x2 + 5x3 ≤60,
x1, x2, x3 ≥0 x1, x2, x3 ≥0

(A) (2, 0, 3) (A) (2, 0, 3)

(B) (3, 2, 4) (B) (3, 2, 4)

(C) (0, 3, 5) (C) (0, 3, 5)

(D) (1, 1, 1) (D) (1, 1, 1)

32. In solving the following LPP by simplex 32. {gåßboŠg {d{Y Ûmam {ZåZ LPP H$mo hb H$aZo _|
method the first incoming vector is àW_ AmdH$ g{Xe h¡
min. z = x1 – 4x2 + 3x3 min. z = x1 – 4x2 + 3x3

s.t. 3x1 – x2 + 4x3 ≤7, s.t. 3x1 – x2 + 4x3 ≤7,

–2x1 + 4x2 ≤8, –2x1 + 4x2 ≤8,

– 4x1 + 3x2 + 8x3 ≤10, – 4x1 + 3x2 + 8x3 ≤10,

x1, x2, x3 ≥0 x1, x2, x3 ≥0
(A) (4, 0, 8) (B) (3, –2, –4) (A) (4, 0, 8) (B) (3, –2, –4)
(C) (–1, 4, 3) (D) None of these (C) (–1, 4, 3) (D) BZ_| go H$moB© Zht

33. If in a ∆ABC, ∆ = b2 – (c – a)2, then 33. `{X EH$ ∆ABC _|, ∆ = b2 – (c – a)2, V~
cot B = cot B =
8 15 8 15
(A) (B) (A) (B)
15 8 15 8
8 17 8 17
(C) (D) (C) (D)
17 8 17 8

-12- Set-A

34. If in a ∆ ABC, b2 = c2 + a2, 2s = a + b + c, 34. `{X EH$ ∆ ABC _|, b2 = c2 + a2, 2s = a + b + c,
then s(s – a) (s – b) (s – c) = V~ s(s – a) (s – b) (s – c) =
1 2 2 1 2 2
(A) ab (A) ab
4 4
(B) a2b2 (B) a2b2
1 2 2 1 2 2
(C) ca (C) ca
4 4
(D) c2a2 (D) c2a2

35. If (cotα – 1) (cotβ – 1) = 2 cotα⋅cotβ, 35. `{X (cotα – 1) (cotβ – 1) = 2 cotα⋅cotβ,
then the general value of α + β = V~ α + β H$m ì`mnH$ _mZ h¡
π π
(A) nπ – (A) nπ –
4 4
π π
(B) nπ + (B) nπ +
4 4
π π
(C) nπ – (C) nπ –
2 2
π π
(D) nπ + (D) nπ +
2 2

1 1
36. If cosθ = – , tanθ = 1, then the `{X cosθ = –
36. , tanθ = 1, V~ θ H$m
2 2
most general value of θ is gdm©{YH$ ì`mnH$ _mZ h¡
π π
(A) 2nπ ± (A) 2nπ ±
4 4
5π 5π
(B) 2nπ ± (B) 2nπ ±
4 4
3π
3π (C) (2n + 1)π ±
(C) (2n + 1)π ± 4
4
5π 5π
(D) (2n + 1)π ± (D) (2n + 1)π ±
4 4

-13- Set-A

37. On the basis of following table, maximum 37. {ZåZ Vm{bH$m Ho$ AmYma na Xmo _erZm| M1 VWm
profit z for a manufacturer for producing M2 Ûmam CZHo$ A{YH$V_ CnbãYVm Ho$ gmW A
A and B through two machines M1 and VWm B Ho$ CËnmXZ Ho$ {bE {H$gr CËnmXH$ Ho$ {bE
M2 with their maximum availability can
be formulated as A{YH$V_ bm^ z àmßV H$aZo H$m gyÌ h¡

Maximum A{YH$V_ CnbãY
A B
A B available time g_` ({_ZQ>m| _|)
(in minutes)
M1 1 1 400
M1 1 1 400
M2 2 1 600
M2 2 1 600
bm^ z é. 2 é. 3 –
Profit z Rs. 2 Rs. 3 –

(A) max. z = 2A + 3B (A) max. z = 2A + 3B
s.t. A + 2B ≤ 400, s.t. A + 2B ≤ 400,
A + B ≤ 600, A + B ≤ 600,
A, B ≥ 0 A, B ≥ 0

(B) max. z = 2A + 3B (B) max. z = 2A + 3B
s.t. A + 2B ≥ 400, s.t. A + 2B ≥ 400,
A + B ≥ 600, A + B ≥ 600,
A, B ≥ 0 A, B ≥ 0

(C) max. z = 2A + 3B (C) max. z = 2A + 3B
s.t. A + B ≤ 400, s.t. A + B ≤ 400,
2A + B ≤ 600, 2A + B ≤ 600,
       A, B ≥ 0        A, B ≥ 0

(D) max. z = 2A + 3B (D) max. z = 2A + 3B
s.t. A + B ≥ 400, s.t. A + B ≥ 400,
2A + B ≥ 600, 2A + B ≥ 600,
    A, B ≥ 0    A, B ≥ 0

-14- Set-A

38. The LPP represented by the following 38. {ZåZ AmaoI Ûmam {Zê${nV LPP h¡
graph has

(A) an unbounded solution (A) EH$ An[a~Õ hb
(B) no solution (B) H$moB© hb Zht
(C) solution x = 2, y = 2 (C) hb x = 2, y = 2
(D) solution x = 3, y = 4 (D) hb x = 3, y = 4
dy dy
39. The curve satisfying y = 2x is a 39 . y = 2x H$mo g§VwîQ> H$aZo dmbr dH«$ h¡
dx dx
(A) Family of parabola (A) nadb` H$m Hw$b
(B) Family of circles (B) d¥Îm H$m Hw$b
(C) Family of straight lines (C) gab aoImAm| H$m Hw$b
(D) None of these (D) BZ_| go H$moB© Zht

40. Solution of differential equation 40. AdH$b g_rH$aU
(2xy + 3y2)dx – (2xy + x2)dy = 0 is (2xy + 3y2)dx – (2xy + x2)dy = 0 H$m
hb h¡
(A) x2 – xy = cy2 (A) x2 – xy = cy2
(B) x2 + xy = cy3 (B) x2 + xy = cy3
(C) y2 + xy = cx3 (C) y2 + xy = cx3
(D) None of these (D) BZ_| go H$moB© Zht

-15- Set-A

41. Solution of differential equation 41. AdH$b g_rH$aU
(cosx –xcosy) dy – (siny +ysinx)dx = 0 is (cosx –xcosy) dy – (siny +ysinx)dx = 0
H$m h>b h¡
(A) xcosy – ysinx = c (A) xcosy – ysinx = c
(B) ycosx – xsiny = c (B) ycosx – xsiny = c
(C) ycosy – xsinx = c (C) ycosy – xsinx = c
(D) None of these (D) BZ_| go H$moB© Zht
42.
Solution of differential equation dy y
AdH$b g_rH$aU dx + x = x H$m h>b h¡
2
dy y 42.
+ = x 2 is
dx x
1 4 1 4
(A) xy = y + c (A) xy = y +c
4 4
1 4 1 4
(B) x + y = x + c (B) x + y = x + c
4 4
1 4 1 4
(C) x − y = x + c (C) x − y = x + c
4 4
1 4 1 4
(D) xy = x + c (D) xy = x + c
4 4

43. The three lines x – 2y + 1 = 0, 43. VrZ aoImE± x – 2y + 1 = 0, 2x – 5y + 3 = 0
2x – 5y + 3 = 0 and 5x – 9y + k = 0 are VWm 5x – 9y + k = 0 g§nmVr h¡, ¶{X k
concurrent, if k equals to ~am~a h¡
(A) 3 (B) 4 (A) 3 (B) 4
(C) 2 (D) 1 (C) 2 (D) 1

44. The equations of the tangents to the 44. d¥Îm x2 + y2 – 6x – 4y + 5 = 0 Ho$ ñne© aoIm
circle x2 + y2 – 6x – 4y + 5 = 0 which H$m g‘rH$aU Omo X-Aj Ho$ gmW 45° H$m H$moU
makes an angle of 45° with the X-axis is {Z‘m©U H$aVm h¡, hmoJm
(A) y=x+5 (A) y = x + 5
(B) y=x–5 (B) y = x – 5
(C) x=y+8 (C) x = y + 8
(D) x=y–8 (D) x = y – 8

-16- Set-A

45. Which of the following equations 45. {ZåZ ‘| go H$m¡Z-gm g‘rH$aU g‘H$moUr¶
does not represent a rectangular A{Vnadb¶ H$mo {Zê${nV Zhr§ H$aVm ?
hyperbola ?
(A) xy = c2 (B) x2 – y2 = a2 (A) xy = c2 (B) x2 – y2 = a2
2
(C) y 2 − x 2 = 0 (D) x = ct, y = c
2
(C) y 2 − x 2 = 0 (D) x = ct, y = c
2 2

b a t b a t
46. L is the normal to the parabola y2 = 4x 46. L nadb¶ y2 = 4x na A{^b§~ h¡ VWm {~ÝXþ
and passes through the point (1, 2). If (1, 2) go hmoH$a JwOaVm h¡ & ¶{X A{^b§~ H$s
the slope of the normal is positive, then àdUVm YZmË‘H$ h¡, Vmo CgH$m g‘rH$aU hmoJm
its equation is
(A) x + y = 3 (A) x + y = 3
(B) x + y + 3 = 0 (B) x + y + 3 = 0
(C) x – y = 3 (C) x – y = 3
(D) y – x – 3 = 0 (D) y – x – 3 = 0

47. The minimum value of sin6θ + cos6θ is 47. sin6θ + cos6θ H$m Ý`yZV_ _mZ h¡
1 1
(A) 0 (B) (A) 0 (B)
4 4
(C) 1 (D) None of these (C) 1 (D) BZ_| go H$moB© Zht

48. (cot2θ – tan2θ) (1 – sec2θ cosec2θ) 48. (cot2θ – tan2θ) (1 – sec2θ cosec2θ)
– sec2θ tan2θ + cosec2θ cot2θ = – sec2θ tan2θ + cosec2θ cot2θ =
(A) –1 (B) 0 (A) –1 (B) 0
(C) 1 (D) 3 (C) 1 (D) 3

49. If 3cos2α + 2cos2β = 4 and 49. `{X 3cos2α + 2cos2β = 4 VWm
3 sin α 2 cos β 3 sin α 2 cos β
= ,where α and β are = , Ohm± α VWm β
sin β cos α sin β cos α
positive acute angles, then α + 2β = YZmË_H$ Ý`yZ H$moU h¢, V~ α + 2β =
π π
(A) 0 (B) (A) 0 (B)
4 4
π π
(C) (D) π (C) (D) π
2 2

-17- Set-A

50. If the two acute angles α and β are 50. `{X Xmo Ý`yZ H$moU α VWm β g_rH$aU
solutions of the equation 2cos2θ + 3sin2θ = 4 Ho$ hb hmo,
2cos2θ + 3sin2θ = 4, V~ sin2α + sin2β =
then sin2α + sin2β =
21 21
(A) (A)
13 13
13 13
(B) (B)
21 21
(C) 2 (C) 2

(D) None of these (D) BZ_| go H$moB© Zht

51. The most plausible values of X and Y 51. {ZåZ{b{IV g_rH$aUm| go X VWmY Ho$ gdm©{YH$
from the following equations are : Cn`wŠV _mZ h¢ : X + Y = 3.01, 2X – Y = 0.03,
X + Y = 3.01, 2X – Y = 0.03, X + 3Y = 7.03, 3X + Y = 4.97.
X + 3Y = 7.03, 3X + Y = 4.97.
(A) X = 1, Y = 2.01 (A) X = 1, Y = 2.01
(B) X = 1.0003, Y = 2.0007 (B) X = 1.0003, Y = 2.0007
(C) X = 1.03, Y = 2 (C) X = 1.03, Y = 2
(D) None of these (D) BZ_| go H$moB© Zht

52. Fit straight line on following data : 52. {ZåZ Am±H$‹S>m| na Amg§{OV gab aoIm h¡ :
x 1 2 3 4 5 x 1 2 3 4 5
y 2 7 9 10 11 y 2 7 9 10 11

(A) y = 5.9 + 1.5 x (A) y = 5.9 + 1.5 x

(B) y = 5x + 9 (B) y = 5x + 9

(C) y = 2x + 3 (C) y = 2x + 3

(D) None of these (D) BZ_| go H$moB© Zht

-18- Set-A

lim 2 − cosec x ~am~a h¡
2

lim 2 − cosec x equals
2
53. 53.
x→ π 1 − cot x x→ π
4
1 − cot x
4

(A) 0 (B) 2 (A) 0 (B) 2
(C) 3 (D) None of these (C) 3 (D) BZ_| go H$moB© Zht

3x + 4 tan x 3x + 4 tan x
54. lim = 54. lim =
x→0 x x→0 x
(A) 7 (B) 0 (A) 7 (B) 0
(C) ∞ (D) None of these (C) ∞ (D) BZ_| go H$moB© Zht

55. For the function 55. \$bZ
e − 1
1
e x − 1
x 1

 , x≠0  , x≠0
f(x ) =  e 1x + 1 f(x ) =  e 1x + 1
 0 , x=0  0 ,
  x=0

Choose the correct statement. Ho$ {bE gË` H$WZ Mw{ZE &
(A) lim f( x ) does not exist (A) lim f( x ) H$m ApñVËd Zht h¡
x →0 x →0

(B) f(x) is continuous at x = 0 (B) f(x), x = 0 na gVV h¡

(C) lim f( x ) = 1 (C) lim f( x ) = 1
x→0 x→0

(D) lim f( x ) exists, but f(x) not (D) lim f( x ) H$m ApñVËd h¡, {H$ÝVw f(x), x = 0
x →0 x →0

continuous at x = 0 na gVV Zht h¡

3x − 4, 0 ≤ x ≤ 2 3x − 4, 0 ≤ x ≤ 2
56. Let f( x ) =  56. _mZm f(x) = 
 x + l, 2 ≤ x ≤ 9  x + l, 2 ≤ x ≤ 9
If f(x) is continuous at x = 2, then value `{X f(x), x = 2 na gVV h¡, Vmo l H$m _mZ h¡
of l is
(A) 2 (B) – 2 (A) 2 (B) – 2
(C) 0 (D) None of these (C) 0 (D) BZ_| go H$moB© Zht

-19- Set-A

57. If cos–1x + cos–1y + cos–1z = 3π, then 57. `{X cos–1x + cos–1y + cos–1z = 3π, V~
xy – yz + zx = xy – yz + zx =

(A) 0 (B) 1 (A) 0 (B) 1
(C) 2 (D) 3 (C) 2 (D) 3

58. A 60 cm long rod is broken randomly 58. EH$ 60 go_r bå~o N>‹S> H$mo `mÑpÀN>>H$ ê$n go VrZ
into three parts. Then the probability ^mJm| _| Vmo‹S>m OmVm h¡& V~ BZ VrZ ^mJm| go EH$
that a triangle can be formed from {Ì^wO ~Zm`o Om gH$Zo H$s àm{`H$Vm hmoJr
these three parts will be
(A) 1 (B) 2 (A) 1 (B) 2
3 3 3 3
(C) 1 (D) 3 (C) 1 (D) 3
4 4 4 4

59.
If E1, E2, ..., E50 are independent events 59. `{X E1, E2, ..., E50 ñdV§Ì KQ>ZmE± Bg àH$ma h¢
1 1
such that P(Ei ) = , 1 ≤ i ≤ 50 , {H$ P(Ei ) = , 1 ≤ i ≤ 50 ,
i+1 i+1
then the probability that none of these V~ BZ 50 KQ>ZmAm| _| go {H$gr Ho$ ^r Z
50 events occurs is KQ>Zo H$s àm{`H$Vm h¡
(A) 50 (B) 1 (A) 50 (B) 1
51 4 51 4
2 2
(C) 3 (D) None of these (C) 3 (D) BZ_| go H$moB© Zht

60. Average of two non-negative numbers 60. Xmo F$UoÎma g§»`mAm| H$m Am¡gV n h¡& V~ BZHo$
is n. Then the chance that their product
JwUZ\$b H$m BZHo$ _hÎm_ JwUZ\$b Ho$ 34 JwUm
is not less than 3 times their maximum go H$_ Zht hmoZo H$s g§^mdZm h¡
4
product is
1 1 (B) 3
(A) (B) 3 (A)
2
2 4 4
(C) 1 (D) 2 (C) 1 (D) 2
4 3 4 3

-20- Set-A

61. If following table shows cost for 61. `{X {ZåZ Vm{bH$m H«$_e… x VWm y H$s Amdí`H$VmAm|
production of A and B with maximum
output subject to requirements x and y
Ho$ eVm] Ho$ AYrZ A{YH$V_ CËnmXZ Ho$ gmW A
respectively, then the formulation for VWm B Ho$ CËnmXZ bmJV H$mo {Zê${nV H$aVm h¡,
maximum profit z is V~ bm^ z Ho$ A{YH$V_ hmoZo Ho ${bE gyÌ h¡

Requirement Requirement Maximum Amdí`H$Vm Amdí`H$Vm A{YH$V_
x y Output
x y CËnmXZ
Prod. A 20 13 1000
CËnmXZ A 20 13 1000
Prod. B 17 19 2000
CËnmXZ B 17 19 2000
Cost in 5 17 –
Rs. bmJV é._| 5 17 –

(A) max. z = 5x + 17y (A) max. z = 5x + 17y
s.t. 20x + 17y ≤ 1000, s.t. 20x + 17y ≤ 1000,
13x + 19y ≤ 2000, 13x + 19y ≤ 2000,
x, y ≥ 0 x, y ≥ 0

(B) max. z = 5x + 17y (B) max. z = 5x + 17y
s.t. 20x + 13y ≤ 1000, s.t. 20x + 13y ≤ 1000,
17x + 19y ≤ 2000, 17x + 19y ≤ 2000,
x, y ≥ 0 x, y ≥ 0

(C) max. z = 5x + 17y (C) max. z = 5x + 17y
s.t. 20x + 13y ≥ 1000, s.t. 20x + 13y ≥ 1000,
17x + 19y ≥ 2000, 17x + 19y ≥ 2000,
x, y ≥ 0 x, y ≥ 0

(D) max. z = 5x + 17y (D) max. z = 5x + 17y
s.t. 20x + 17y ≥ 1000, s.t. 20x + 17y ≥ 1000,
13x + 19y ≥ 2000, 13x + 19y ≥ 2000,
x, y ≥ 0 x, y ≥ 0

-21- Set-A

62. If x and y denote regular and private 62. `{X x VWm y {df`m| M, H VWm PS _| gpå_{bV
students appearing in subjects M, H hmoZo dmbo {Z`{_V Ed§ ñdmÜ`m`r N>mÌm| H$s g§»`m H$mo
and PS and their maximum result R is
to be obtained subject to the maximum
gy{MV H$aVm h¡ VWm A{YH$V_ N>mÌm| Ho$ gpå_{bV
students appearing, then on the basis hmoZo Ho$ eVm] Ho$ AYrZ R CZH$m A{YH$V_ [aOëQ>
of following table the formulation for h¡, V~ {ZåZ Vm{bH$m Ho$ AmYma na A{YH$V_
maximum result R is [aOëQ> R Ho$ {bE gyÌ h¡

Maximum A{YH$V_ gpå_{bV N>mÌ
x y students appeared x y
M 500 800 1000 M 500 800 1000
H 600 900 1200 H 600 900 1200
PS 700 900 1300 PS 700 900 1300
R 900 1700 – R 900 1700 –

(A) max. R = 900x + 1700y (A) max. R = 900x + 1700y
s.t. 5x + 8y ≤ 10, s.t. 5x + 8y ≤ 10,
2x + 3y ≤ 4, 2x + 3y ≤ 4,
7x + 9y ≤ 13, 7x + 9y ≤ 13,
x, y ≥ 0 x, y ≥ 0
(B) max. R = 900x + 1700y (B) max. R = 900x + 1700y
s.t. 5x + 8y ≥ 10, s.t. 5x + 8y ≥ 10,
2x + 3y ≥ 4, 2x + 3y ≥ 4,
7x + 9y ≥ 13, 7x + 9y ≥ 13,
x, y ≥ 0 x, y ≥ 0
(C) max. R = 900x + 1700y (C) max. R = 900x + 1700y
s.t. 5x + 8y ≥ 10, s.t. 5x + 8y ≥ 10,
2x + 3y ≤ 4, 2x + 3y ≤ 4,
7x + 9y ≤ 13, 7x + 9y ≤ 13,
  x, y ≥ 0   x, y ≥ 0
(D) max. R = 900x + 1700y (D) max. R = 900x + 1700y
s.t. 5x + 8y ≤ 10, s.t. 5x + 8y ≤ 10,
2x + 3y ≥ 4, 2x + 3y ≥ 4,
7x + 9y ≥ 13, 7x + 9y ≥ 13,
  x, y ≥ 0 x, y ≥ 0

-22- Set-A

1 1
e− x e− x
63. ∫0 1 + e− x dx = 63. ∫0 1 + e− x dx =
 1+ e  1  1+ e  1
(A) log   − +1 (A) log   − +1
 2e  e  2e  e
 1+ e  1  1+ e  1
(B) log   − +1 (B) log   − +1
 e  e  e  e
 1+ e  1  1+ e  1
(C) log   + +1 (C) log   + +1
 e  e  e  e
 1+ e  1  1+ e  1
(D) log   + +1 (D) log  + +1
 2e  e  2e  e

64. The area enclosed between the curves 64. dH«$m| y2 = x VWm y = |x| Ho$ _Ü` {Kam hþAm joÌ
y2 = x and y = |x| is H$m joÌ\$b h¡
1 1 1 1
(A) (B) (A) (B)
3 6 3 6
1 1
(C) 2 (D) 3 (C) 2 (D) 3
2π 2π

65. ∫ (sin x + sin x ) dx = 65. ∫ (sin x + sin x ) dx =
0
0

(A) 4 (B) 0 (A) 4 (B) 0
(C) 2 (D) None of these (C) 2 (D) BZ_| go H$moB© Zht

66. Area of the region bounded by 66. nadb`m| x = – 2y2 VWm x = 1 – 3y2 Ho$ _Ü`
parabolas x = – 2y2 and x = 1 – 3y2 is n[a~Õ joÌ H$m joÌ\$b h¡

(A) 2 (B) 4 (A) 2 (B) 4
3 3 3 3
1 5 1 5
(C) (D) 3 (C) (D) 3
3 3

-23- Set-A

67. A manufacturer plans to produce 67. EH$ {Z_m©Vm Xmo àH$ma Ho$ {Ibm¡Zm| M1 VWm M2
two types of toys M1 and M2. He has ~ZmZo H$s `moOZm ~ZmVm h¡ & CgHo$ nmg BVZm
ingredients sufficient to produce 2000
units of M1 and 4000 units of M2 but he
n`m©ßV gm_J«r h¡ {H$ M1 H$m 2000 BH$mB© VWm
is allowed to produce only 4500 units M2 H$m 4000 BH$mB© H$m CËnmXZ H$a gHo$ naÝVw
of either of M1 or M2. He can prepare Cgo M1 `m M2 Ho$ A{YH$V_ 4500 BH$mB©`m| Ho$
100 units of M1 in 3 hours, 100 units of CËnmXZ H$s BOmOV h¡ & dh M1 H$m 100 BH$mB© 3
M2 in one hour and only 60 hours are K§Q>m| _|, M2 H$m 100 BH$mB© EH$ K§Q>m _| V¡`ma H$a
available for this operation. The profit is
gH$Vm h¡ Am¡a Cgo Bg H$m_ Ho$ {bE Ho$db 60 K§Q>o
Rs. 5 per unit for M1 and Rs. 4 per unit
for M2. If the manufacturer produces x CnbãY h¢ & M1 Ho$ {bE à{V BH$mB© 5 é. VWm M2
units of M1 and y units of M2, then the Ho$ {bE à{V BH$mB© 4 é. H$m bm^ àmßV hmoVm h¡ &
formulation for maximization of total `{X {Z_m©Vm M1 H$m x BH$mB© VWm M2 H$m y BH$mB©
profit z for this LPP is H$m CËnmXZ H$aVm h¡, V~ Bg LPP Ho$ {bE Hw$b
bm^ z Ho$ A{YH$V_ _mZ Ho$ {bE gyÌ h¡
(A) max. z = 5x + 4y (A) max. z = 5x + 4y
s.t. x + 3y ≤ 6000, s.t. x + 3y ≤ 6000,
x + y ≤ 4500, x + y ≤ 4500,
x ≤ 2000, y ≤ 4000, x ≤ 2000, y ≤ 4000,
x, y ≥ 0 x, y ≥ 0
(B) max. z = 4x + 5y (B) max. z = 4x + 5y
s.t. 3x + y ≤ 6000, s.t. 3x + y ≤ 6000,
x + y ≤ 4500, x + y ≤ 4500,
x ≤ 2000, y ≤ 4000, x ≤ 2000, y ≤ 4000,
x, y ≥ 0 x, y ≥ 0
(C) max. z = 5x + 4y (C) max. z = 5x + 4y
s.t. 3x + y ≤ 6000, s.t. 3x + y ≤ 6000,
x + y ≤ 4500, x + y ≤ 4500,
x ≤ 2000, y ≤ 4000, x ≤ 2000, y ≤ 4000,
    x, y ≥ 0     x, y ≥ 0
(D) max. z = 5x + 4y (D) max. z = 5x + 4y
s.t. 3x + y ≥ 6000, s.t. 3x + y ≥ 6000,
x + y ≥ 4500, x + y ≥ 4500,
x ≤ 2000, y ≤ 4000, x ≤ 2000, y ≤ 4000,
x, y ≥ 0 x, y ≥ 0

-24- Set-A

68. If x, y, z are daily diet requirements for 68. `{X x, y, z H«$_e… Amhma 1, Amhma 2, Amhma 3
diet 1, diet 2, diet 3 respectively and Ho$ {bE X¡{ZH$ Amhma H$s Oê$aV| h¢ VWm BZ àH$ma Ho$
minimum daily requirements for these
type of diet for Vitamin A, B, C and cost
Amhmam| Ho$ {bE {dQ>m{_Z A, B, C Ho$ {bE Ý`yZV_
in Rs. are shown in the following table, X¡{ZH$ Oê$aVo§ VWm bmJV é. _| {ZåZ Vm{bH$m go
then the minimum cost Z for this diet àX{e©V {H$`o J`o h¢, V~ Bg Amhma g_ñ`m Ho$ {bE
problem can be formulated as Ý`yZV_ bmJV z àmßV H$aZo H$m gyÌ h¡
Minimum Amhma 1 Amhma 2 Amhma 3 Ý`yZV_
diet 1 diet 2 diet 3 daily
requirement
X¡{ZH$
Oê$aVo§
Vitamin A 1 2 4 2 mg 1 2 4 2 mg
{dQ>m{_Z A
Vitamin B 10 5 1 3 mg
{dQ>m{_Z B 10 5 1 3 mg
Vitamin C 20 50 10 5 mg
{dQ>m{_Z C 20 50 10 5 mg
Cost in Rs. 2 3 4 –
bmJV é. _| 2 3 4 –

(A) min. z = 2x + 3y + 4z (A) min. z = 2x + 3y + 4z
s.t. x + 2y + 4z ≤ 2, s.t. x + 2y + 4z ≤ 2,
10x + 5y + z ≤ 3, 10x + 5y + z ≤ 3,
20x + 50y + 10z ≤ 5, 20x + 50y + 10z ≤ 5,
x, y, z ≥ 0 x, y, z ≥ 0
(B) min. z = 2x + 3y + 4z (B) min. z = 2x + 3y + 4z
s.t. 4x + 2y + z ≥ 2, s.t. 4x + 2y + z ≥ 2,
x + 5y + 10z ≥ 3, x + 5y + 10z ≥ 3,
10x + 50y + 20z ≥ 5, 10x + 50y + 20z ≥ 5,
x, y, z ≥ 0 x, y, z ≥ 0
(C) min. z = 2x + 3y + 4z (C) min. z = 2x + 3y + 4z
s.t. x + 10y + 20z ≤ 2, s.t. x + 10y + 20z ≤ 2,
2x + 5y + 50z ≤ 3, 2x + 5y + 50z ≤ 3,
4x + y + 10z ≤ 4, 4x + y + 10z ≤ 4,
x, y, z ≥ 0 x, y, z ≥ 0
(D) min. z = 2x + 3y + 4z (D) min. z = 2x + 3y + 4z
s.t. x + 2y + 4z ≥ 2, s.t. x + 2y + 4z ≥ 2,
10x + 5y + z ≥ 3, 10x + 5y + z ≥ 3,
20x + 50y + 10z ≥ 5, 20x + 50y + 10z ≥ 5,
x, y, z ≥ 0 x, y, z ≥ 0

-25- Set-A

69. If O is the origin and Q (–2, –4) is a 69. ¶{X O ‘yb q~Xþ h¡ VWm Q (–2, –4), OP na
point on OP such that OQ = 1 OP , pñWV {~ÝXþ Bg àH$ma h¡ {H$ OQ = 1 OP , Vmo
3 3
then co-ordinates of P is P Ho$ {ZX}em§H$ h¡
(A) (6, –12) (B) (–6, –12) (A) (6, –12) (B) (–6, –12)
(C) (–6, 12) (D) (6, 12) (C) (–6, 12) (D) (6, 12)

70. Area of the parallelogram formed by the 70. aoImAm| 4y – 3x – 2 = 0, 3y – 4x + 2 = 0,
lines 4y – 3x – 2 = 0, 3y – 4x + 2 = 0, 4y – 3x – 6 = 0 Ed§ 3y – 4x + 4 = 0 go
4y – 3x – 6 = 0 and 3y – 4x + 4 = 0 is {Z{‘©V g‘m§Va MVw^w©O H$m joÌ’$b h¡
(A) 8 (B) 7 (A) 8 (B) 7
7 8 7 8

(C) 4 (D) None of these (C) 4 (D) BZ‘| go H$moB© Zht
7 7

71. Straight lines 3x + 4y = 5 and 4x – 3y = 15 71. gab aoImE± 3x + 4y = 5 VWm 4x – 3y = 15
intersect at the point A. Points B and q~ÝXþ A na à{VÀN>oX H$aVo h¢ & BZ aoImAm| na
C are chosen on those lines such that {~ÝXþAm| B VWm C H$m M¶Z Bg àH$ma {H$¶m
AB = AC. The equation of line BC OmVm h¡ {H$ AB = AC & {~ÝXþAm| B VWm C go
passing through the points B and C. JwOaZo dmbr aoIm BC H$m g‘rH$aU h¡
(A) x – 7y + 13 = 0 (A) x – 7y + 13 = 0
(B) 7x + y – 9 = 0 (B) 7x + y – 9 = 0
(C) Both (A) and (B) (C) (A) VWm (B) XmoZm|
(D) None of these (D) BZ‘| go H$moB© Zht

72. The point of intersection of the lines 72. S ≡ 3x2 + xy – 4y2 + 10x + 4y + 8 = 0
represented by go {Zê${nV aoImAm| H$m à{VÀN>oX {~ÝXþ h¡
S ≡ 3x2 + xy – 4y2 + 10x + 4y + 8 = 0
is

( )
(A) −2 , 12 (
(B) 2 , −12 ) ( −72 , 127 ) 7 )
( 27 , −12
(A) (B)
7 7 7 7

(C) ( 7 , 7 ) (D) ( −12 , 2 ) (C) ( 7 , 7 ) (D) ( −12 , 2 )
12 −2 12 −2
7 7 7 7

-26- Set-A

73. If X is a Poisson variate such that 73. `{X X EH$ ßdm°gmo {dMa Bg àH$ma h¡ {H$
4P(X = 2) = 2P(X = 1) + 3P(X = 3), 4P(X = 2) = 2P(X = 1) + 3P(X = 3),
then the coefficient of skewness is V~ d¡få` JwUm§H$ h¡
(A) 2 (B) 1 (A) 2 (B)1
2 2
(C) 3 (D) 1 (C) 3 (D) 1
9 9

74. For a normal distribution, which of the 74. EH$ àgm_mÝ` ~§Q>Z Ho$ {bE, {ZåZ _| go H$m¡Z-gm
following is not always correct ? gX¡d ghr Zht h¡ ?
(A) Mean = Median = Mode (A) _mÜ` = _mpÜ`H$m = ~hþbH$
(B) Q.D. : M.D. : S.D. :: 10 : 12 : 15 (B) Q.D. : M.D. : S.D. :: 10 : 12 : 15
(C) 7 > 0 (C) 7 > 0
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht
75. The differential equation dy 1− y2
dy 1− y2 75. AdH$b g_rH$aU = EH$ d¥Îm Hw$b
= determined a family of dx y
dx y
circles with H$mo {Zê${nV H$aVm h¡, Ohm±

(A) Variable radii and a fixed center (A) Ma {ÌÁ`m VWm (0, 1) Ho$ÝÐ {~ÝXþ na pñWa h¡
at (0, 1)
(B) Variable radii and a fixed center (B) Ma {ÌÁ`m VWm (0, –1) Ho$ÝÐ {~ÝXþ na pñWa h¡
at (0, –1)
(C) A fixed radius 1 and variable (C) AMa {ÌÁ`m 1 h¡ VWm Ma Ho$ÝÐ x-Aj Ho$
center along the x-axis AZw{Xe h¡
(D) None of these (D) BZ_| go H$moB© Zht

76. The degree of the differential equation 76. 1 − x 2 + 1 − y 2 = a(x − y ) H$mo g§Vîw Q> H$aZo
satisfying 1 − x 2 + 1 − y 2 = a(x − y ) is dmbo AdH$b g_rH$aU H$m KmV h¡
(A) 1 (B) 2 (A) 1 (B) 2
(C) 3 (D) 4 (C) 3 (D) 4

-27- Set-A

77. A curve that passes through (2, 4) and 77. (2, 4) go JwOaZo dmbo VWm Ma AYmob§~ 8 BH$mB©
having subnormal of constant length dmbo dH«$ h¡
8 units is
(A) y2 = 16x + 8 (A) y2 = 16x + 8
(B) y2 = 16x – 24 (B) y2 = 16x – 24
(C) x2 = 16y – 8 (C) x2 = 16y – 8
(D) None of these (D) BZ_| go H$moB© Zht

d2 y
78.
The solution of the differential equation 78. AdH$b g_rH$aU x = 1 H$m hb O~{H$
d2 y dy dx 2
x 2 = 1 , given that y = 1, = 0, dy
dx dx {X`m h¡, y = 1, = 0 , V~ x = 1, hmoJm
when x = 1, is dx
(A) y = xlogx + x + 2 (A) y = xlogx + x + 2
(B) y = xlogx – x + 2 (B) y = xlogx – x + 2
(C) y = xlogx + x (C) y = xlogx + x
(D) y = xlogx – x (D) y = xlogx – x

79. If x = 2, y = 3, z = 1 is a feasible solution 79. `{X x = 2, y = 3, z = 1 LPP
of the LPP
max. z = x + 2y + 4z max. z = x + 2y + 4z

s.t. 2x + y + 4z = 11, s.t. 2x + y + 4z = 11,
3x + y + 5z = 14, 3x + y + 5z = 14,
x, y, z ≥ 0,
x, y, z ≥ 0,
then which of the following is its one H$m g§^mì` hb h¡, V~ {ZåZ _| go H$m¡Z-gm BgH$m
BFS ? EH$ BFS h¡ ?
1 5 1 5
(A) x = 2 , y = 0, z = 2 (A) x =
2
, y = 0, z =
2
(B) x = 0, y = –1, z = 3 (B) x = 0, y = –1, z = 3

11 11
(C) x = 3, y = 0, z = (C) x = 3, y = 0, z =
4 4
(D) no BFS possible (D) H$moB© BFS g§^mì` Zht

-28- Set-A

80. The following graph represent the LPP 80. {ZåZ{b{IV AmaoI {H$gH$m LPP {Zê${nV
of H$aVm h¡ ?

(A) min. z = 2x + 3y (A) min. z = 2x + 3y
s.t. x + y ≤ 4, s.t. x + y ≤ 4,
3x + y ≥ 4, 3x + y ≥ 4,
x + 5y≥4, x + 5y≥4,
     x ≤ 3, y ≤ 3, x ≥ 0, y ≥ 0      x ≤ 3, y ≤ 3, x ≥ 0, y ≥ 0
(B) min. z = 2x + 3y (B) min. z = 2x + 3y
s.t. x + y ≤ 4, s.t. x + y ≤ 4,
3x + y ≥ 4, 3x + y ≥ 4,
x + 5y≥4, x + 5y≥4,
     x≥3, y≥3, x ≥ 0, y ≥ 0      x≥3, y≥3, x ≥ 0, y ≥ 0
(C) min. z = 2x + 3y (C) min. z = 2x + 3y
s.t. x + y ≥ 4, s.t. x + y ≥ 4,
3x + y ≤ 4, 3x + y ≤ 4,
x + 5y≥4, x + 5y≥4,
     x ≤ 3, y ≤ 3, x ≥ 0, y ≥ 0      x ≤ 3, y ≤ 3, x ≥ 0, y ≥ 0
(D) min. z = 2x + 3y (D) min. z = 2x + 3y
s.t. x + y ≤ 4, s.t. x + y ≤ 4,
3x + y ≤ 4, 3x + y ≤ 4,
x + 5y ≤ 4, x + 5y ≤ 4,
     x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0      x ≥ 3, y ≥ 3, x ≥ 0, y ≥ 0

-29- Set-A

81. If the sum of the roots of the equation 81. `{X g_rH$aU ax2 + bx + c = 0 Ho$ _ybm| H$m
ax2 + bx + c = 0 be equal to the sum `moJ, _ybm| Ho$ dJm] Ho$ `moJ Ho$ ~am~a hmo, Vmo
of their squares, then
(A) b(a + b) = 2ac (A) b(a + b) = 2ac
(B) a(a + b) = 2bc (B) a(a + b) = 2bc
(C) b(a + b) = ac (C) b(a + b) = ac
(D) a(a + b) = bc (D) a(a + b) = bc

82. The number of real solutions of the 82. g_rH$aU log(–x) = 2log(x + 1) Ho$ dmñV{dH$
equation log(–x) = 2log(x + 1) is hbm| H$s g§»`m h¡
(A) 1 (B) 2 (A) 1 (B) 2
(C) 3 (D) None of these (C) 3 (D) BZ_| go H$moB© Zht

83. If x, 2y, 3z are in A.P., where the 83. `{X x, 2y, 3z g.lo. _| h¡, Ohm± x, y, z {^Þ-{^Þ
distinct numbers x, y, z are in G.P., g§»`mE± Jw.lo. _| h¡, Vmo Jw.lo. H$m gmdm©ZwnmV h¡
then the common ratio of G.P. is
(A) 1 (B) 1 (A) 1 (B) 1
2 3 2 3
(C) 2 (D) None of these (C) 2 (D) BZ_| go H$moB© Zht

84. If a, b, c are in G.P., then a + b, 2b, 84. `{X a, b, c Jw.lo. _| h¡, Vmo a + b, 2b, b + c
b + c are in hmoJm
(A) A.P. (B) G.P. (A) g.lo. (B) Jw.lo.
(C) H.P. (D) None of these (C) h.lo. (D) BZ_| go H$moB© Zht

(i + j) . ( 2j × k ) + (j + k ) . ( 3i × 4k ) +
85. (i + j) . ( 2j × k ) + (j + k ) . ( 3i × 4k ) +
85.


(k + i) . ( 2j × 3i) =
(k + i) . ( 2j × 3i) =
(A) 16 (B) –16 (A) 16 (B) –16
(C) 14 (D) –14 (C) 14 (D) –14

-30- Set-A

     
86. If a, b,
c are non-coplanar vectors and 86. `{X a, b, c Ag_Vbr` g{Xe hmo VWm
  
   
   
if a ′, b ′, c ′ are vectors reciprocal to a ′, b ′, c ′ BZHo$ ì`wËH«$_ g{Xe hmo, V~
       

( )(
them, then a + b + c . a′ + b′ + c′ = )  

(a + b + c ) . (a′ + b′ + c′ ) =
(A) 0 (B) 1 (A) 0 (B) 1
(C) 2 (D) 3 (C) 2 (D) 3

   
87. If a = 2i + j − k , b = 5i + 6j + 7k 87. `{X a = 2i + j − k, b = 5i + 6j + 7k VWm
     
and c = i − j + k , then c = i − j + k , V~ a + b, b + c, c − a  =
     
a + b, b + c, c − a  =
 
(A) –32 (B) 0 (A) –32 (B) 0
(C) 32 (D) 84 (C) 32 (D) 84

88. If 2i + j − k and i − 2j + 5k are two 88. `{X 2i + j − k VWm i − 2j + 5k EH$ g_mZmÝVa
adjacent sides of a parallelogram MVw^w©O ABCD H$s Xmo AmgÝZ ^wOmE± hmo, V~ Bg
ABCD, then area of this parallelogram
g_mZmÝVa MVw^w©O H$m joÌ\$b h¡
is
(A) 7 3 (B) 91 (A) 7 3 (B) 91
(C) 139 (D) 35 (C) 139 (D) 35

89. Solution of differential equation 89. AdH$b g_rH$aU dx + xdy = e–y sec2ydy
dx + xdy = e–y sec2ydy is H$m >hb h¡ &
(A) yex = tanx + c (A) yex = tanx + c

(B) ye–x = tanx + c (B) ye–x = tanx + c

(C) xey = tany + c (C) xey = tany + c

(D) xe–y = tany + c (D) xe–y = tany + c

-31- Set-A

90. The differential equation of the family 90. dH«$VmAm| y = Ae3x + Be5x Ho$ n[adma H$s AdH$b
of curves y = Ae3x + Be5x , where A and B g_rH$aU, Ohm± A Am¡a B _mZH$ h¢
are parameters, is

(A) d2 y dy (A) d2 y dy
2
+8 + 15y = 0 2
+8 + 15y = 0
dx dx dx dx
d2 y dy d2 y dy
(B) 3 2 − 8 + 5y = 0 (B) 3 2 − 8 + 5y = 0
dx dx dx dx
d2 y dy d2 y dy
(C) +5 + 6y = 0 (C) 2
+5 + 6y = 0
dx 2
dx dx dx
d2 y dy d2 y dy
(D) 2
−8 + 15y = 0 (D) 2
−8 + 15y = 0
dx dx dx dx

91. The solution of the differential equation 91. AdH$b g_rH$aU x2dy + y(x + y) dx = 0 H$m
x2dy + y(x + y) dx = 0 is hb h¡
(A) x2y = c2(2x + y) (A) x2y = c2(2x + y)
(B) xy2 = c2(2x + y) (B) xy2 = c2(2x + y)
(C) x2y = c2(x + 2y) (C) x2y = c2(x + 2y)
(D) xy2 = c2 (x + 2y) (D) xy2 = c2 (x + 2y)

92. The order of the differential equation 92. nadb` y = x2 Ho$ g^r ñne© aoImAm| H$m AdH$b
of all the tangent lines to the parabola g_rH$aU H$s H$mo{Q> h¡
y = x2 is
(A) 1 (B) 2 (A) 1 (B) 2
(C) 3 (D) 4 (C) 3 (D) 4

93. IfABCDEF
   isa regular
hexagon and
93. `{X ABCDEF EH$ g_ fQ²>^wO h¡, VWm
    
AB = a, bC = b , then CD = AB = a, bC = b , V~ CD =
   
  (A) 2 b (B) a + b
(A) 2 b (B) a + b
   
(C) b − a (D) None of these (C) b − a (D) BZ_| go H$moB© Zht

-32- Set-A

   
94. If a = 3i − j + 4k , b = 2i + 3j − 2k , then 94. `{X a = 3i − j + 4k, b = 2i + 3j − 2k , V~
  
a−b = a−b =

(A) 53 (A) 53

(B) 37 (B) 37

(C) 17 (C) 17

(D) None of these (D) BZ_| go H$moB© Zht

95. If 2i + 3j − 4k and 3i − j + 2k are 95. `{X 2i + 3j − 4k VWm 3i − j + 2k H«$_e:
position vectors of two points A and Xmo {~ÝXþAm| A VWm B Ho$ pñW{V g{Xe hmo, V~ {~ÝXþ C
B respectively, then what will be
the position vector of point C, where
H$m pñW{V g{Xe Š`m hmoJm, Ohm± C, AB H$mo
C divides AB internally in the ratio 1 : 3 Ho$ AZwnmV _| A§V: {d^m{OV H$aVm h¡ ?
1:3?
9 5 9 5
(A) i + 2j + k (A) i + 2j + k
4 2 4 2

5   5  
(B) i+ j−k (B) i+ j−k
2 2

(C) 2 i − j + 3k (C) 2 i − j + 3k

(D) 4 i + j + k (D) 4 i + j + k
   
If α = 2i + j − k , β = 3i + j + 2k and
96. 96. `{X α = 2i + j − k, β = 3i + j +2k VWm
      
( )(
γ = i + j − 2k , then α × β . β × γ = ) 
γ = i + j − 2k , V~ (α × β ) . (β × γ ) =

     
(
(A) –14 α . β ) (B) γ . α (A) –14 α . β ( )  
(B) γ . α
   
(C) β . γ (D) 0 (C) β . γ (D) 0

-33- Set-A

97. The value of λ, for which the line 97. λ H$m dh ‘mZ {OgHo$ {bE aoIm 2
y
2x − 8 λy = −3 is a normal to the 2x − 8 λy = −3 , em§H$d x 2 + = 1 na
3 3 4
y2
conic x +
2
= 1 is A{^b§~ h¡, hmoJm
4

(A) 2 (B) −2 (A) 2 (B) −2
3 3 3 3

(C) 1 (D) 3 (C) 1 (D) 3
4 8 4 8

If the eccentricity of an ellipse be 5
98. 98. ¶{X EH$ XrK©d¥Îm H$s CËHo$ÝÐVm 5 h¡ VWm BZHo$
8 8
and the distance between its foci be Zm{^¶m| Ho$ ‘ܶ Xÿar 10 h¡, Vmo BgH$m Zm{^b§~
10, then its latus rectum is h¡
(A) 39 (B) 12 (A) 39 (B) 12
4 4
(C) 15 (D) 37 (C) 15 (D) 37
2 2

99. On the ellipse 4x2 + 9y2 = 1, the points 99. XrK©d¥Îm 4x2 + 9y2 = 1 na pñWV dh {~ÝXþ {Og
at which the tangents are parallel to the na ñne©Á¶m aoIm 8x = 9y Ho$ g‘m§Va h¡, hmoJm
line 8x = 9y are

( )
(A) 2 , 1
5 5
( )
(B) −2 , 1
5 5
(A) ( 25 , 51 ) (B) ( −52 , 51 )
(C) ( 5 , −15 ) (D) ( 2 , 3 ) (C) ( 5 , −15 ) (D) ( 2 , 3 )
−2 −2
5 5 5 5

100. The equation 2x2 – 3y2 – 12x + 12y = 0 100. g‘rH$aU 2x2 – 3y2 – 12x + 12y = 0 {Zê${nV
represents H$aVm h¡
(A) A parabola (A) EH$ nadb¶
(B) An ellipse (B) EH$ XrK©d¥Îm
(C) A hyperbola (C) EH$ A{Vnadb¶
(D) A rectangular hyperbola (D) EH$ g‘H$moUr¶ A{Vnadb¶

-34- Set-A

PART – II
Computer Awareness H$åß`yQ>a OmJê$H$Vm
101. Interrupts form an important part of 101. B§Q>aßQ>g __________ {gñQ>_ H$m EH$ _hËdnyU©
__________ systems. {hñgm h¡ &
(A) Batch processing (A) ~¡M àmogoqgJ
(B) Multitasking (B) _ëQ>rQ>mpñH¨$J
(C) Real-time Processing (C) ar`b Q>mB©_ àmogoqgJ
(D) Multi-user (D) _ëQ>r`yga

102. The main job of the interrupt system 102. B§Q>aßQ> {gñQ>_ H$m _w»` H$m_ B§Q>aßQ> Ho$ _________
is to identify the __________ of H$s nhMmZ H$aZm h¡ &
interrupt.
(A) Signal (B) Device (A) {g½Zb (B) {S>dmBg
(C) Source (D) Peripheral (C) òmoV (D) ~mø CnH$aU

103. Interrupts initiated by an instruction is 103. {ZX}e Ûmam ewê$ {H$E JE ì`dYmZ H$mo __________
called as H$hm OmVm h¡ &
(A) Internal (A) Am§V[aH$
(B) External (B) ~mhar
(C) Hardware (C) hmS>©do`a
(D) Software (D) gmâQ>do`a

104. In general the two interrupt request 104. gm_mÝ` Vm¡a na Xmo B§Q>aßQ> AZwamoY bmBZ|
lines are __________ h¡ &
(A) Maskable and non-maskable (A) _mñH$ H$aZo `mo½` Am¡a J¡a-_mñH$ H$aZo `mo½`
interrupts B§Q>aßQ>g
(B) Blocked and non-maskable (B) AdéÕ Am¡a ZmZ-_mñHo$~b B§Q>aßQ>g
interrupts
(C) Maskable and Blocked interrupts (C) _mñH$ H$aZo `mo½` Ed§ AdéÕ B§Q>aßQ>g
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-35- Set-A

105. What is the name of component of a 105. H$åß`yQ>a Ho$ Cg KQ>H$ H$m Zm_ Š`m h¡, Omo S>oQ>m
computer that is used to both read and
H$mo n‹T>Zo Am¡a {bIZo XmoZm| Ho$ {b`o à`moJ {H$`m
write data ?
OmVm h¡ ?
(A) RAM (B) ROM (A) a¡_ (B) amo_
(C) Hard Drive (D) Flash Memory (C) hmS>© S´>mBd (D) âb¡e _o_moar

106. The PROM is more effective than ROM 106. PROM, ROM {Mn H$s VwbZm _| _____ Ho$
chips in regard to H$maU A{YH$ à^mdr h¡ &
(A) Memory Management (A) _o_moar _¡ZoO_|Q>
(B) Cost (B) bmJV
(C) Speed of operation (C) Am°naoeZ H$s J{V
(D) Both (B) and (C) (D) (B) Am¡a (C) XmoZm|

107. 1 peta byte is equal to 107. 1 noQ>m ~mB©Q> _____ Ho$ ~am~a hmoVm h¡ &
(A) 103 TB (B) 106 TB (A) 103 TB (B) 106 TB
(C) 103 GB (D) 109 GB (C) 103 GB (D) 109 GB

108. Antivirus is a/an 108. EÝQ>rdm`ag EH$ _____ h¡ &
(A) System software (A) {gñQ>_ gm°âQ>do`a
(B) Utility software (B) `y{Q>{bQ>r> gm°âQ>do`a
(C) Application software (C) EßbrHo$eZ gm°âQ>do`a
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

109. A ‘C’ variable name can start with a 109. ‘C’ do[a`o~b Zm_ EH$ ________ go ewé hmo
gH$Vm h¡ &
(A) Number (A) Z§~a
(B) Plus sign (+) (B) ßbg gmB©Z (+)
(C) Underscore (C) A§S>añH$moa
(D) Asterisk (∗) (D) EoñQ>o[añH$ (∗)

-36- Set-A

110. Name the loop that executes at least 110. Cg byn H$m Zm_ Xo Omo H$_ go H$_ EH$ ~ma {Zînm{XV
once
hmoVm h¡
(A) For (B) If (A) For (B) If
(C) Do . . . While (D) While (C) Do . . . While (D) While

111. The prototype of a function can be 111. \§$ŠeZ Ho$ àmoQ>moQ>mBn H$m Cn`moJ {H$`m Om
used for gH$Vm h¡
(A) Define a function (A) EH$ \§$ŠeZ H$s n[a^mfm _|
(B) Declare a function (B) EH$ \§$ŠeZ H$s KmofUm _|
(C) Erase a function (C) EH$ \§$ŠeZ {_Q>mZo _|
(D) Call a function (D) EH$ \§$ŠeZ ~wbmZo _|

112. In ‘C’, what is the correct hierarchy of 112. ‘C’ _| A§H$J{UVr` g§{H«$`mAm| H$m ghr nXmZwH«$_
arithmetic operation ? Š`m h¡ ?
(A) ∗ / + – (B) ∗ + – / (A) ∗ / + – (B) ∗ + – /
(C) / ∗ + – (D) + – / ∗ (C) / ∗ + – (D) + – / ∗

113. (48)10 = (?)8 the value is 113. (48)10 = (?)8 BgH$m _mZ h¡
(A) 58 (B) 22 (A) 58 (B) 22
(C) 26 (D) 60 (C) 26 (D) 60

114. A ____ enables you to view data from 114. EH$ _________ {deof _mZX§S> na Q>o~b go S>mQ>m
a table based on a specific criterion. {XImVm h¡ &
(A) Form (B) Query (A) µ\$m_© (B) Šdoar
(C) Macro (D) Report (C) _¡H«$mo (D) [anmoQ>©

115. MS-Excel is an example of 115. E_.Eg.-EŠgb_____ H$m EH$ CXmhaU h¡ &
(A) Web browser (A) do~ ~«mCOa
(B) Application software (B) EpßbHo$eZ gm°âQ>do`a
(C) An operating system (C) Am°naoqQ>J {gñQ>_
(D) An input device (D) BZnwQ> {S>dmB©g

-37- Set-A

116. Choose the incorrect option related to 116. H$åß`yQ>a Ho$ ~w{Z`mXr g§MmbZ Ho$ g§~§Y _| JbV
basic operations of a computer. {dH$ën MwZ| &
(A) Processing (B) Storing (A) àmogoqgJ (B) ñQ>mo[a¨J
(C) Analyzing (D) Input (C) EZmbmBqOJ (D) BZnwQ>

117. Logical extension of multiprogramming 117. _ëQ>ràmoJ«mq_J Am°naoqQ>J {gñQ>_ H$m bm°{OH$b
operating system is EŠñQ>|eZ h¡
(A) Time sharing (A) Q>mB©_ eo`[a¨J
(B) Multi tasking (B) _ëQ>rQ>mpñH¨$J
(C) Single Programming (C) EH$b àmoJ«mq_J
(D) Both (A) and (B) (D) (A) Am¡a (B) XmoZm|

118. Multiprogramming of the computer 118. H$åß`yQ>a {gñQ>_ H$s _ëQ>ràmoJ«mq_J ~‹T> OmVr h¡
system increases
(A) Memory (A) _o_moar
(B) CPU utilization (B) gr. nr. `y. `y{Q>bmBOoeZ
(C) Storage (C) ñQ>moaoO
(D) Cost (D) H$s_V

119. Multiprogramming system 119. _ëQ>ràmoJ«mq_J {gñQ>_
(A) Are easier to develop than single (A) EH$b àmoJ«mq_J {gñQ>_ H$s VwbZm _| {dH${gV
programming system H$aZm AmgmZ h¡
(B) Execute each job faster (B) àË`oH$ H$m`© H$mo VoOr go {Zînm{XV H$a|
(C) Execute more jobs in the same (C) EH$ hr g_` _| A{YH$ H$m`© {Zînm{XV H$a|
time
(D) Are used only on large main frame (D) Ho$db ~‹S>o _w»` \«o$_ H$åß`yQ>am| na Cn`moJ
computers {H$`m OmVm h¡

-38- Set-A

120. In a time-sharing operating system, 120. Q>mB©_ eo`[a§J Am°naoqQ>J {gñQ>_ _|, O~ {H$gr àmoggo
when the time slot given to a process H$mo {X`m J`m g_` ñbm°Q> nyam hmo OmVm h¡, Vmo àmogog
is completed, the process goes from
aqZJ ñQ>oQ> go
the running state to the
(A) Blocked state (A) ãbmŠS> ñQ>oQ>
(B) Ready state (B) aoS>r ñQ>oQ>
(C) Suspended state (C) gñn|S>oS> ñQ>oQ>
(D) Terminated state (D) Q>a{_ZoQ>oS> ñQ>oQ>

121. ________ is a type of diagram that 121. ________ EH$ àH$ma H$m AmaoI h¡ Omo
represents an algorithm or process {XemË_H$ Vram| go OwS>| EH$ {deof H«$_ _| MaUm|
by showing the steps as boxes H$mo {d{^Þ àH$ma Ho$ ~Šgo Ho$ ê$n _| {XImH$a EH$
of different types in a particular
sequence connected with directional à{H«$`m `m EëJmo[aX_ H$m à{V{Z{YËd H$aVm h¡ &
arrows.
(A) Pie chart (A) n¡ MmQ>©
(B) Flow chart (B) âbmo MmQ>©
(C) Algorithm (C) EëJmo[aX_
(D) Data chart (D) S>mQ>m MmQ>©

122. The efficiency of an algorithm in 122. _o_moar Amdí`H$VmAm| Ho$ g§X^© _| EH$ EëJmo[aX_
terms of its memory requirements is H$s XjVm __________ Ûmam _mnr OmVr h¡ &
measured by
(A) The maximum memory needed (A) EëJmo[aX_ Ûmam Amdí`H$ A{YH$V_ _o_moar
by the algorithm
(B) The minimum memory needed by
(B) EëJmo[aX_ Ûmam Amdí`H$ Ý`yZV_ _o_moar
the algorithm (C) EëJmo[aX_ bmoS> H$aZo Ho$ {bE Amdí`H$
(C) The primary memory needed to
àmW{_H$ _o_moar
load the algorithm
(D) The maximum disk space needed (D) EëJmo[aX_ Ûmam Amdí`H$ A{YH$V_ {S>ñH$
by the algorithm ñWmZ

-39- Set-A

123. For which of the purpose a parallelogram 123. âbmoMmQ>© _| {H$g CÔoí` Ho$ {bE g_m§Va MVw^©wO H$m
is used in a flowchart ?
Cn`moJ {H$`m OmVm h¡ ?
(A) To denote the step in which data (A) EH$ ñQ>on> H$mo {Zê${nV H$aZo Ho$ {bE {Og_|
is stored S>mQ>m ñQ>moa {H$`m OmVm h¡
(B) To denote a step in which decision (B) EH$ ñQ>on> H$mo {Zê${nV H$aZo Ho$ {bE {Og_|
has to be made {ZU©` {b`m OmZm h¡
(C) To show a step that cannot be (C) EH$ ñQ>on {XImZo Ho$ {bE {Ogo hb Zht {H$`m
resolved Om gH$Vm h¡
(D) To show input or output (D) BZnwQ> `m AmCQ>nwQ> {XImZo Ho$ {bE

124. While writing an algorithm, 124. EëJmo[aX_ {bIVo g_` {ZX}e __________ go
instructions are written from {bIo OmVo h¡ &
(A) Top to Bottom (A) D$na go ZrMo
(B) Left to Right (B) ~mE| go XmE|
(C) Right to Left (C) XmE| go ~mE|
(D) It depends on the author of the (D) `h EëJmo[aX_ Ho$ boIH$ na {Z^©a H$aVm h¡
algorithm

125. _________ control is used to provide 125. _________ H$ÝQ´>mob H$m Cn`moJ, Xygao H$ÝQ´>mob
an identifiable grouping for other Ho$ {bE nhMmZo OmZo `mo½` J«yn (g_yh) àXmZ
controls.
H$aVm h¡ &
(A) Frame (B) Label (A) \o«$_ (B) bo~b
(C) List box (D) Text box (C) {bñQ> ~m°Šg (D) Q>oŠñQ> ~m°Šg

126. A text box can hold as many as _____ 126. EH$ Q>oŠñQ> ~m°Šg _| _____ H¡$aoŠQ>a aI
characters. gH$Vo h§¡ &
(A) 2052 (B) 2048 (A) 2052 (B) 2048
(C) 1024 (D) 2058 (C) 1024 (D) 2058

-40- Set-A

127. In MICR, C stands for 127. E_.AmB©.gr.Ama. _| "gr' Š`m h¡ ?
(A) Code (B) Colour (A) H$moS> (B) H$ba
(C) Character (D) Computer (C) H¡$aoŠQ>a (D) H$åß`yQ>a

128. The given Boolean expression 128. Xr hþB© ~y{b`Z A{^ì`pŠV A.B + A.B + A.B
A.B + A.B + A.B is equivalent to g_Vwë` h¡
(A) A + B (B) A + B (A) A + B (B) A + B
(C) A + B (D) A + B (C) A + B (D) A + B

129. Which of the following disks can record 129. {ZåZ{b{IV _| go H$m¡Z-gr {S>ñH$ EH$ hr ~ma [aH$mS>©
only once ? H$a gH$Vr h¡ ?
(A) CD-R and DVD-R (A) CD-R Am¡a DVD-R
(B) CD-RW and DVD-RW (B) CD-RW Am¡a DVD-RW
(C) Blu-Ray Disk (C) Blu-Ray {S>ñH$
(D) CD-RAM and DVD-RAM (D) CD-RAM Am¡a DVD-RAM

130. Which of the following is designed to 130. {ZåZ{b{IV _| go {H$go H$åß`yQ>a Ho$ g§MmbZ H$mo
control the operations of a computer ? {Z`§{ÌV H$aZo Ho$ {b`o {S>µOmBZ {H$`m J`m h¡ ?
(A) Utility software (A) `y{Q>{bQ>r> gm°âQ>do`a
(B) Application software (B) EßbrHo$eZ gm°âQ>do`a
(C) System software (C) {gñQ>_ gm°âQ>do`a
(D) All of the above (D) CnamoŠV g^r

131. _____ keys are present on the top row 131. H$s~moS>© H$s erf© n§pŠV na _____ Hw§$Or _m¡OwX
of the keyboard. hmoVm h¡ &
(A) Function (A) \§$ŠeZ
(B) Type writer (B) Q>mBn amBQ>a
(C) Numeric (C) Ý`y_o[aH$
(D) Navigation (D) Zo{dJoeZ

-41- Set-A

132. A light sensitive device that converts 132. EH$ àH$me g§doXZerb CnH$aU Omo S´>mB§J, _w{ÐV
drawing, printed texts or other images nmR> `m AÝ` N>{d`m| H$mo {S>{OQ>b ê$n _| n[ad{V©V
into digital form is
H$aVm h¡, _____ H$hbmVm h¡ &
(A) Keyboard (B) Plotter (A) H$s~moS>© (B) ßbm°Q>a
(C) OMR (D) Scanner (C) Amo.E_.Ama. (D) ñH¡$Za

133. Binary language is also called 133. ~mBZar ^mfm H$mo _______ ^r H$hm OmVm h¡ &
(A) Assembly language (A) Eg|pãb ^mfm
(B) Machine language (B) _erZ ^mfm
(C) High level language (C) CÀM ñVar` ^mfm
(D) All of the above (D) CnamoŠV g^r

134. Which is high level language ? 134. CÀM ñVar` ^mfm H$m¡Z-gr h¡ ?
(A) BASIC (B) cobol (A) ~o{gH$ (B) H$mo~mb
(C) pascal (D) All of the above (C) nmñH$b (D) CnamoŠV g^r

135. Which programming languages are 135. {ZåZ ñVa H$s ^mfmAm| Ho$ ê$n _| H$m¡Z-gr àmoJ«mq_J
classified as low level languages ? ^mfmAm| H$mo dJuH¥$V {H$`m J`m h¡ ?
(A) fortran, pascal (A) \$moQ>´©Z, nmñH$b
(B) prolog, Expert System (B) àmobmoJ, EŠgnQ>© {gñQ>_
(C) Assembly Language (C) Eg|pãb b¢½doO
(D) Knowledge based system (D) Zm°boO ~ogS> {gñQ>_

136. Which of the following is machine 136. {ZåZ{b{IV _| go H$m¡Z-gm _erZ ñdV§Ì àmoJm« _ h¡ ?
independence program ?
(A) High level language (A) CÀM ñVar` ^mfm
(B) Assembly language (B) Eg|pãb ^mfm
(C) Machine language (C) _erZ ^mfm
(D) Low level language (D) {ZåZ ñVar` ^mfm

-42- Set-A

137. Floating point representation is used 137. µâbmoqQ>J nm°B§Q> [aàoO|Q>oeZ H$m Cn`moJ Š`m ñQ>moa H$aZo
to store Ho$ {bE {H$`m OmVm h¡ ?
(A) Boolean values (A) ~y{b`Z _mZ
(B) Rational numbers (B) n[a_o` g§»`m
(C) Real integers (C) dmñV{dH$ nyUmªH$
(D) Real numbers (D) dmñV{dH$ g§»`m

138. Cache memory acts between 138. H¡$e _o_moar {H$gHo$ ~rM H$m`© H$aVm h¡ ?
(A) CPU and RAM (A) gr.nr.`y. Am¡a a¡_
(B) RAM and ROM (B) a¡_ Am¡a amo_
(C) CPU and ROM (C) gr.nr.`y. Am¡a amo_
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

139. A communication between the 139. _mBH«$mo H§$ß`yQ>a Ho$ KQ>H$m| Ho$ ~rM g§Mma ES´>og Am¡a
components of a micro computer {ZåZ{b{IV _| go {H$gHo$ _mÜ`_ go hmoVm h¡ ?
take place via address and which of
the following ?
(A) Address bus (A) ES´>og ~g
(B) Register (B) a{OñQ>a
(C) Data bus (C) S>oQ>m ~g
(D) I/O bus (D) AmB©/Amo ~g

140. Which characteristics of RAM memory 140. a¡_ _o_moar H$s H$m¡Z-gr {deofVm Bgo ñWm`r ^§S>maU
make it not suitable for permanent Ho$ {bE Cn`wŠV Zht ~ZmVr h¡ ?
storage ?
(A) Its speed (A) BgH$s J{V
(B) Its volatility (B) BgH$s ApñWaVm
(C) Its reliability (C) BgH$s {dídgZr`Vm

(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-43- Set-A

PART – III
Analytical ability and Logical {díbofU Ed§ VH©$ epŠV
Reasoning

141. Find the wrong term in the given series. 141. {X`o J`o loUr _| JbV H«$_ H$mo kmV H$a| &
7, 3, 2, 2, 8, 16, 92, 640 7, 3, 2, 2, 8, 16, 92, 640

(A) 8 (A) 8

(B) 7 (B) 7

(C) 16 (C) 16

(D) 640 (D) 640

142. Six students A, B, C, D, E and F are 142. 6 N>mÌ A, B, C, D, E Am¡a F EH$ _¡XmZ _|
sitting in a field. A and B are from ~¡R>o h¢& A Am¡a B {~bmgnwa go h¡ Am¡a ~mH$s g^r
Bilaspur, while the rest are from
am`nwa go h¡& D Am¡a E bå~o h¢ O~{H$ AÝ` g^r
Raipur. D and E are tall while others
N>moQ>o H$X Ho$ h§¡& A, C Am¡a D b‹S>{H$`m± h¡§ O~{H$
are short. A, C and D are girls, while
other are boys. Which short boy are AÝ` b‹S>Ho$ h§¡& H$m¡Z-gm N>moQ>o H$X H$m b‹S>H$m
from Raipur ? am`nwa go h¡ ?

(A) C (A) C

(B) B (B) B

(C) E (C) E

(D) F (D) F

-44- Set-A

143. Which of the following diagrams 143. {ZåZ{b{IV _| go H$m¡g-gm {MÌ ñZmVH$, H${d> Ed§
represents the relationship between {ejH$ Ho$ _Ü` ghr g§~§Y H$mo Xem©Vm h¡ ?
graduate, poet and Teachers ?

(A) (A)

(B) (B)

(C) (C)

(D) (D)

144. Study the following figure carefully 144. {ZåZm§{H$V {MÌ H$m Ü`mZnyd©H$ AÜ``Z H$a|
Doctor {Ibm‹S>r S>mŠQ>a
Players 25 4
25 4 17 17
22 3
22 3 8
8
30
30

Artist
H$bmH$ma

How many player will also be Artist ? Eogo {H$VZo {Ibm‹S>r h¡ Omo H$bmH$ma ^r h¡ ?
(A) 25 (A) 25
(B) 22 (B) 22
(C) 30 (C) 30
(D) 33 (D) 33

-45- Set-A

145. If H = 10, HER = 37 then what is equal 145. `{X H = 10, HER = 37 Vmo heroic {H$gHo$
to heroic ? ~am~a hmoJm ?
(A) 58 (A) 58
(B) 64 (B) 64
(C) 70 (C) 70
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

146. Count the number of cubes in the given 146. {X`o J`o {MÌ _| KZ H$s g§»`m H$m JUZm H$a|
figure

(A) 8 (A) 8
(B) 9 (B) 9
(C) 12 (C) 12
(D) 15 (D) 15

147. A cube is painted blue on all faces and 147. EH$ KZ Ho$ g^r gVh na Zrbm a§J noÝQ> H$aVo h§¡
is then cut into 125 cubes of equal size. Am¡a BgHo$ ~mX Bgo 125 g_mZ KZm| _| H$mQ>Vo h¡§ &
How many cubes are painted on one ~VmBE {H$ {H$VZo KZm| Ho$ Ho$db EH$ gVh _| noÝQ>
face only ? hþAm h¢ ?
(A) 16 (B) 32 (A) 16 (B) 32
(C) 54 (D) 48 (C) 54 (D) 48

-46- Set-A

148. If the seed catalog is correct, then if 148. `{X ~rO gyMr ghr h¡, V~ `{X Aà¡b _| ~rO ~mo`m
the seeds are planted in April, then OmVm h¡, V~ OwbmB© _| \y$b {Ib|Jo & \y$b OwbmB© _|
the flowers bloom in July. The flowers {Ibo & Bg{bE, `{X ~rO gyMr ghr h¡, V~ ~rO
bloom in July. Therefore, if the seed Aà¡b _| ~mo`m OmVm h¡ & `h h¡
catalog is correct, then the seeds are
planted in April. It is
(A) Valid (B) Invalid (A) d¡Y (B) Ad¡Y
(C) Inconclusive (D) Absurd (C) A{ZUr©V (D) ~H$dmg
149. If “Hindustan” is written as 149. `{X “Hindustan” H$mo “134265984”
“134265984” how will “India” be {bIm OmVm h¡, Vmo “India” H$mo Cgr H$moS> _| Š`m
written in that code
{bIm Om`oJm ?
(A) 34328 (B) 34239 (A) 34328 (B) 34239
(C) 34325 (D) 34238 (C) 34325 (D) 34238

150. How many parallelograms are there in 150. ZrMo {XE {MÌ _| {H$VZo g_m§Va MVw^w©O h¢ ?
the following figure ?


(A) 16 (B) 18 (A) 16 (B) 18
(C) 20 (D) 17 (C) 20 (D) 17

151. Find the odd one out. 151. {^ÝZ H$mo kmV H$s{O`o &


(A) (1) (B) (2) (A) (1) (B) (2)
(C) (3) (D) (4) (C) (3) (D) (4)

-47- Set-A

152. The question image is embedded and 152. àíZ {MÌ {H$g EH$ CÎma {MÌ _| gpÝZ{hV Ed§ Nw>nm
hidden inside which one of the answer
hþAm h¡ ?
images ?


(A) 1 (B) 2 (A) 1 (B) 2
(C) 3 (D) 4 (C) 3 (D) 4

153. If machine is coded as 153. `{X machine H$m g§Ho$V
19 – 7 – 9 – 14 – 15 – 20 – 11 19 – 7 – 9 – 14 – 15 – 20 – 11
What is the code for danger ? h¡ Vmo danger H$m H$moS> Š`m hmoJm ?
(A) 13 – 7 – 20 – 10 – 11 – 25 (A) 13 – 7 – 20 – 10 – 11 – 25
(B) 13 – 7 – 20 – 9 – 11 – 25 (B) 13 – 7 – 20 – 9 – 11 – 25
(C) 10 – 7 – 20 – 13 – 11 – 24 (C) 10 – 7 – 20 – 13 – 11 – 24
(D) None of the above (D) Cn`w©ŠV _| go H$moB© Zht

154. How many triangles are there in the 154. {ZåZm§{H$V AmH¥${V _| {H$VZo {Ì^wO h¡ ?
following figure ?

(A) 6 (B) 7 (A) 6 (B) 7
(C) 8 (D) 9 (C) 8 (D) 9

-48- Set-A

155. Find the missing term. 155. {dbwßV nX H$mo kmV H$a| &
25 25 49 25 25 49
16 5 49 25 4 25 36 ? 9 16 5 49 25 4 25 36 ? 9
81 25 16 81 25 16

(A) 3 (A) 3
(B) 2 (B) 2
(C) 4 (C) 4
(D) 5 (D) 5

156. Choose the correct answer figure in 156. {ZåZ{b{IV àíZ {MÌ loUr Ho$ {bE {X`o J`o CÎma
the sequence of the following problem {MÌ {dH$ënm| _| go ghr {dH$ën N>m±{Q>E
figure


Problem figure àíZ {MÌ

(A) (A)

(B) (B)

(C) (C)

(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-49- Set-A

157. Which of the following is the missing 157. {ZåZ _| go {dbwßV {MÌ H$m¡Z-gm h¡ ?
figure ?

2 E 3 5 B 7 ? 17 B 19 2 E 3 5 B 7 ? 17 B 19

(A) 10 V 12 (A) 10 V 12


(B) 11 X 13 (B) 11 X 13


(C) 11 B 13 (C) 11 B 13


(D) 10 B 12 (D) 10 B 12


158. Which of the following is not a 158. {ZåZ _| go H$m¡Z-gm H$WZ Zht h¡ ?
statement ?
(A) How are you ? (A) Vw_ H¡$go hmo ?
(B) Mars is inhabited. (B) _§Jb na Am~mXr h¡ &
(C) 3 + 5 = 7. (C) 3 + 5 = 7.

(D) All bats have lungs. (D) g^r M_JmX‹S>m| H$mo \o$\$‹S>m hmoVm h¡ &

-50- Set-A

159. If ‘given any x’ is symbolized by ‘(x)’, 159. `{X "{X`o hþE {H$gr x' H$mo ‘(x)’ Ûmam ì`ŠV {H$`m
‘there is atleast one x such that’ is OmVm h¡, "H$_ go H$_ EH$ x Bg àH$ma h¡ {H$' H$mo
symbolized by ‘(∃x)’, ‘or’ is symbolized ‘(∃x)’ Ûmam, "`m' H$mo ‘V’ Ûmam, VWm "_aUerb'
by ‘V’ and ‘mortal’ is symbolized by ‘M’, H$mo ‘M’ go ì`ŠV {H$`m OmVm h¡, V~ _whmdam "{X`o
then the paraphrase ‘given any x, x is
hþE {H$gr x Ho$ {bE, x _aUerb h¡ `m H$_ go H$_
mortal or there is at least one x such
EH$ x Bg àH$ma h¡ {H$ x _aUerb h¡' H$mo ì`ŠV
that x is mortal’ is symbolized as
{H$`m Om gH$Vm h¡ Bg àH$ma
(A) (x)MxV(xMx) (A) (x)MxV(xMx)
(B) (∃Mx)V(x)Mx (B) (∃Mx)V(x)Mx
(C) (x)MxV(∃x)Mx (C) (x)MxV(∃x)Mx
(D) (x)Mx(V∃x)Mx (D) (x)Mx(V∃x)Mx

160. If ‘H’ is an attribute symbol for human, 160. `{X ‘H’ _mZd Ho$ {bE EH$ EQ´>rã`yQ> g§Ho$V h¡, EH$
‘a’ and ‘r’ for an individual constant, ì`pŠVJV AMa Ho$ {bE ‘a’ VWm ‘r’ h¡, naÝVw Ho$ {bE
‘∧’ for but, and ‘~’ for negation, then ‘∧’, VWm {ZfoY Ho$ {bE ‘~’ h¡, V~ "AañVw _mZd
‘Aristotle is human but Raipur is not h¡ naÝVw am`nwa _mZd Zhr§ h¡' Vm{H©$H$V: Vwë` h¡
human’ is logically equivalent to
(A) Ha ∧ Hr (B) ~Ha ∧ Hr (A) Ha ∧ Hr (B) ~Ha ∧ Hr
(C) ~Ha ∧ ~Hr (D) Ha ∧ ~Hr (C) ~Ha ∧ ~Hr (D) Ha ∧ ~Hr

161.
The missing term of the following 161. {ZåZ loUr H$m {dbwßV nX h¡ :
series is :
6, 20, 42, _____, 156 6, 20, 42, _____, 156
(A) 72 (B) 70 (A) 72 (B) 70
(C) 110 (D) 75 (C) 110 (D) 75

162.
The next term of the following 162. {ZåZ loUr H$m AJbm nX h¡ :
series is :
10, 24, 50, 120, _______ 10, 24, 50, 120, _______
(A) 170 (B) 130 (A) 170 (B) 130
(C) 144 (D) 70 (C) 144 (D) 70

-51- Set-A

163. M and S are children of SH and ID. 163. M VWm S, SH VWm ID Ho$ ~ÀMo h¢ & R VWm D,
R and D are children of RJ but grand
RJ Ho$ ~ÀMo h¢ naÝVw SH H$s nmo{V`m± h¢ & RJ, RA
daughters of SH. RJ is husband of RA
and son of ID. RN is brother of ID but H$m n{V h¡ VWm ID H$m nwÌ h¡ & RN, ID H$m ^mB© h¡
brother-in-law of SH. Then which of naÝVw SH H$m gmbm h¡ & V~ {ZåZ _| go H$m¡Z gË`
the following is not correct ? Zht h¡ ?
(A) SH is father of RJ (A) SH, RJ H$m {nVm h¡
(B) RJ is son of RN (B) RJ, RN H$m ~oQ>m h¡
(C) D is not grand daughter of RN (C) D, RN H$s nmoVr Zht h¡
(D) RJ is brother of M and S (D) RJ, M VWm S H$m ^mB© h¡

164. A and B are sisters. C is son of D. E 164. A VWm B ~hZ h¢ & C, D H$m nwÌ h¡ & E, F H$m
is father-in-law of F. G is son of A. H ggwa h¡ & G, A H$m nwÌ h¡ & H, F H$s ~oQ>r VWm G
is daughter of F and sister of G. L is
daughter of B and sister of C. Then
H$s ~hZ h¡ & L, B H$s nwÌr VWm C H$s ~hZ h¡ &
V~
(A) A is mother of L (A) L H$s _m± A h¡
(B) D is mother of L (B) L H$s _m± D h¡
(C) D is son of E (C) D, E H$m ~oQ>m h¡
(D) A is wife of F (D) A, F H$s nËZr h¡

165.
If ‘+’ stands for ‘–’, ‘–’ for ‘×’, ‘×’ 165. `{X ‘+’ h¡ ‘–’ Ho$ {bE, ‘–’ h¡ ‘×’ Ho$ {bE, ‘×’ h¡ ‘÷’
for ‘÷’ and ‘÷’ for ‘+’, then in the Ho$ {bE VWm ‘÷’ h¡ ‘+’ Ho$ {bE, V~ {ZåZ AZwH«$_
following sequence, how many digits _| {H$VZo A§H$ h¢ {OZHo$ R>rH$ nhbo ‘×’ VWm ~mX _|
are immediately preceded by ‘×’ and
followed by ‘–’ ? ‘–’ h¡ ?
25 + 5 – 6 – 7 + 8 – 6 + 7 + 3 ÷ 4 × 5 ÷ 25 + 5 – 6 – 7 + 8 – 6 + 7 + 3 ÷ 4 × 5 ÷
3+2×5 3+2×5
(A) 1 (B) 2 (A) 1 (B) 2
(C) 3 (D) 0 (C) 3 (D) 0

-52- Set-A

166. Choose the cube that is similar to the 166. dh KZ Mw{ZE Omo {H$ {X`o J`o H$mJO Ho$ Qw>H$‹S>o (P)
box formed from the given sheet of go ~Zo ~m°Šg Ho$ g_ê$n hmo &
paper (P).


(A) 1 and 4 only (B) 1 and 2 only (A) Ho$db 1 VWm 4 (B) Ho$db 1 VWm 2
(C) 2 and 3 only (D) 3 and 4 only (C) Ho$db 2 VWm 3 (D) Ho$db 3 VWm 4

167. A cuboid of length 8 cm, breadth 6 cm, 167. 8 go_r bå~mB©, 6 go_r Mm¡‹S>mB© VWm 1 go_r D±$MmB©
and height 1 cm is painted by red on dmbo EH$ KZm^ Ho$ 8 go_r × 6 go_r dr_m dmbo
two faces of dimension 8 cm × 6 cm,
by green on two faces of dimension
Xmo gVhm| H$mo bmb go, 8 go_r × 1 go_r dr_m
8 cm × 1 cm and by yellow on two dmbo Xmo gVhm| H$mo hao go VWm 6 go_r × 1 go_r
faces of dimension 6 cm × 1 cm. Now dr_m dmbo Xmo gVhm| H$mo nrbo go n|Q> {H$`m OmVm
the cuboid is cut into cubes of edges h¡ & A~ KZm^ H$mo àË`oH$ 1 go_r H$moam| (EOog)
1 cm each. Then number of cubes with
two faces red and remaining faces with
dmbo KZm| _| H$mQ>m OmVm h¡ & V~ d¡go KZm| H$s g§»`m
no colour is {OgHo$ Xmo gVh bmb hm| VWm eof gVhm| na H$moB©
a§J Z hmo, h¡
(A) 24 (B) 32 (A) 24 (B) 32
(C) 48 (D) None of these (C) 48 (D) BZ_| go H$moB© Zht

168.
The next letter in the following 168. {ZåZ loUr _| AJbm Aja h¡ :
series is :
C, A, F, D, M, K, Q, ____ C, A, F, D, M, K, Q, ____
(A) O (A) O
(B) P (B) P
(C) S (C) S
(D) Q (D) Q

-53- Set-A

169. Which number replaces the question 169. H$m¡Z-gr g§»`m àíZ {M• "?' H$mo à{VñWm{nV
mark ? H$aVr h¡ ?



(A) 13 (A) 13

(B) 10 (B) 10

(C) 18 (C) 18

(D) 17 (D) 17

170. Which number replaces the question 170. H$m¡Z-gr g§»`m àíZ {M• "?' H$mo à{VñWm{nV
mark ? H$aVr h¡ ?



(A) 5 (B) 7 (A) 5 (B) 7
(C) 9 (D) 15 (C) 9 (D) 15

-54- Set-A

171. Which number replaces the question 171. H$m¡Z-gr g§»`m àíZ {M• "?' H$mo à{VñWm{nV
mark ?
H$aVr h¡ ?


(A) 1 (B) 5 (A) 1 (B) 5
(C) 10 (D) 4 (C) 10 (D) 4

172.
If ‘∆x’ stands for ‘3 times of x’, ‘∇x’ 172. `{X ‘∆x’ h¡ "x Ho$ VrZ JwUm' Ho$ {bE, ‘∇x’ h¡
stands for ‘one third of x’, and ‘ x’ "x Ho$ EH$ {VhmB©' Ho$ {bE, VWm ‘ x’ h¡ ‘x2’ Ho$
stands for ‘x2’, then
(20 + ∆5 – ∇36 ÷ 2 × ∆2) = ?
{bE, V~
(20 + ∆5 – ∇36 ÷ 2 × ∆2) = ?
(A) 1 (A) 1
(B) 6 (B) 6
(C) 4 (C) 4
(D) None of these (D) BZ_| go H$moB© Zht

173. Find the number which comes next in 173. {X`o J`o loUr _| AmZo dmbm AJbm g§»`m kmV
the given series. H$a| &
2, 7, 14, 23, 36, ? 2, 7, 14, 23, 36, ?
(A) 47 (A) 47
(B) 49 (B) 49
(C) 51 (C) 51
(D) None of the above (D) CnamoŠV _| go H$moB© Zht

-55- Set-A

174. Find the odd one out. 174. {ZåZ{b{IV _| go {^Þ H$mo kmV H$a| &
(A) 87 – 111 (A) 87 – 111
(B) 67 – 101 (B) 67 – 101
(C) 128 – 162 (C) 128 – 162
(D) 56 – 88 (D) 56 – 88

175. Find the odd one out. 175. {ZåZ{b{IV _| {^Þ H$mo kmV H$a| &
(A) RAT (B) OUT (A) RAT (B) OUT
(C) BED (D) LET (C) BED (D) LET

176.
Find the missing term in given 176. {X`o J`o Aja loUr _| {dbwßV nX kmV H$a| &
alphabet series.
AZB, CYD, EXF, ? , IVJ AZB, CYD, EXF, ? , IVJ
(A) GUH (A) GUH
(B) GHU (B) GHU
(C) GWH (C) GWH
(D) GHW (D) GHW

177. Choose odd figure from given alternative. 177. {X`o J`o {dH$ën _| go {df_ {MÌ H$mo Mw{ZE &
(A) (A)

(B) (B)

(C) (C)

(D) (D)

-56- Set-A

178. If ‘÷’ means ‘+’, ‘–’ means ‘÷’, ‘×’ means 178. `{X ‘÷’ H$m AW© ‘+’, ‘–’ H$m AW© ‘÷’, ‘×’ H$m AW©
‘–’, and ‘+’ means ‘×’ then ‘–’, Am¡a ‘+’ H$m AW© ‘×’ hmo, Vmo
[(36 × 4) − 8 × 4] = ? [(36 × 4) − 8 × 4] = ?
[4 + 8 × 2 + 16] [4 + 8 × 2 + 16]

(A) 8 (A) 8
(B) 16 (B) 16
(C) 1 (C) 1
(D) 0 (D) 0

179. Find the next two numbers of the 179. {ZåZ Am±{H$H$ loUr H$s Xmo AJbr g§»`mE± kmV
following number series H$s{OE
3, 5, 5, 7, 11, 13, 17, 19, 41, 43, _ , _ 3, 5, 5, 7, 11, 13, 17, 19, 41, 43, _ , _

(A) 59, 61 (A) 59, 61

(B) 43, 47 (B) 43, 47

(C) 67, 69 (C) 67, 69

(D) None of these (D) BZ_| go H$moB© Zht

180. If following sequence is written in the 180. `{X {ZåZ AZwH«$_ H$mo CëQ>o H«$_ _| {bIm Om`, V~
reverse order, then which element will H$m¡Z-gm Ad`d ~m`t N>mo‹S> go 12 d| Ad`d go
be 4th to the right of 12th element from Xm{hZr Amoa 4 Wm hmoJm ?
left end ?
5q∇÷ef∆zr∗t lp87cn# 5q∇÷ef∆zr∗t lp87cn#
(A) ÷ (A) ÷
(B) (B)
(C) E (C) E
(D) 8 (D) 8

-57- Set-A

PART – IV
General Awareness gm_mÝ` AÜ``Z
181. E-Saathi app is associated with which 181. B©-gmWr En H$m g§~§Y {ZåZ ‘| go {H$ggo h¡ ?
of the following ?
(A) Health (A) ñdmñ϶
(B) Education (B) {ejm
(C) Police (C) nw{bg
(D) Environment (D) n¶m©daU

182. SERO survey is associated with which 182. goamo gd}jU {ZåZ ‘| go {H$ggo g§~§{YV h¡ ?
of the following ?
(A) Maleria (A) ‘bo[a¶m
(B) Dengu (B) S>|Jy
(C) Plague (C) ßboJ
(D) Covid (D) H$mo{dS>

183. Baul folk song is associated with which 183. ~mCb bmoH$JrV {H$g amÁ¶ go g§~§{YV h¡ ?
State ?
(A) Meghalaya (A) ‘oKmb¶
(B) Odisha (B) Amo{‹S>em
(C) West Bengal (C) npíM‘ ~§Jmb
(D) Haryana (D) h[a¶mUm

184. Which of the following is India’s 184. 2019 H$s BO Am°’$ Sy>BªJ {~OZog [anmoQ>© ‘|
ranking in 2019’s report of Ease of ^maV H$m ñWmZ {ZåZ ‘| go H$m¡Z-gm h¡ ?
Doing Business ?
(A) 77th (A) 77 dm±
(B) 63rd (B) 63 dm±
(C) 100th (C) 100 dm±
(D) 102nd (D) 102 dm±

-58- Set-A

185. The aim of Mission Sagar is to help 185. {‘eZ gmJa H$m bú¶ {ZåZ ‘| go {H$g Xoe H$mo
which of the following countries ? gh¶moJ nhþ±MmZm h¡ ?
(A) Maldives (A) ‘mbXrd
(B) Madagascar (B) ‘oS>mJmñH$a
(C) Seychelles (C) goeoëg
(D) All the above (D) Cnamo³V g^r

186. Vande Bharat Express runs between 186. d§Xo ^maV E³gàog H$m n[aMmbZ {H$Z Xmo eham|
which of the two cities ? Ho$ ‘ܶ hmoVm h¡ ?
(A) Varanasi to Delhi (A) dmamUgr go {Xëbr
(B) Jaipur to Delhi (B) O¶nwa go {Xëbr
(C) Bhopal to Delhi (C) ^monmb go {Xëbr
(D) Dehradun to Delhi (D) XohamXÿZ go {Xëbr

187. What is the rank of India in Economic 187. Am{W©H$ ñdV§ÌVm gyMH§$mH$ 2020 ‘| ^maV H$m
Freedom Index 2020 ? ñWmZ ³¶m h¡ ?
(A) 143rd (A) 143 dm±
(B) 130th (B) 130 dm±
(C) 120th (C) 120 dm±
(D) 105th (D) 105 dm±

188. Which Financial service is available in 188. ^maV ‘| H$m¡Z-gr {dÎmr¶ godm¶| CnbãY h¢ ?
India ?
(A) Banking (A) ~¢qH$J
(B) Stock Market (B) ñQ>mH$ ‘mH}$Q>
(C) Insurance (C) ~r‘m
(D) All the above (D) Cnamo³V g^r

-59- Set-A

189. Aurora name is given to which of the 189. Am°amoam Zm‘ {ZåZ ‘| go {H$go {X¶m J¶m h¡ ?
following ?
(A) Satellite (A) CnJ«h
(B) Missile (B) {‘gmBb
(C) Super computer (C) gwna H§$߶yQ>a
(D) Tank (D) Q>|H$

190. Who amongst the following has 190. ~mobmo En {ZåZ ‘| go {H$gZo àma§^ {H$¶m h¡ ?
launched the BOLO app ?
(A) Apple (A) Enb
(B) Google (B) JyJb
(C) Microsoft (C) ‘mBH«$mogmâQ>
(D) Samsung (D) go‘g§J

191. Which of the following is a Maharatna 191. {ZåZ ‘| go H$m¡Z-gm ^maV H$s ‘hmaËZ
Company of India ? H$ånZr h¡ ?
(A) H. A. L. (A) EM. E. Eb.
(B) B. S. N. L. (B) ~r. Eg. EZ. Eb.
(C) B. H. E. L. (C) ~r. EM. B©. Eb.
(D) N. M. D. C. (D) EZ. E‘. S>r. gr.

192. Which of the following is a largest 192. ^maV H$m g~go ~‹S>m CÚmoJ {ZåZ ‘| go
industry of India ? H$m¡Z-gm h¡ ?
(A) Coal (A) H$mo¶bm
(B) Steel (B) BñnmV
(C) Textile (C) H$n‹S>m
(D) Petroleum (D) noQ´>mo{b¶‘

-60- Set-A

193. Hubble telescope was made by which 193. h~b Xÿa~rZ H$m {Z‘m©U {ZåZ ‘| go {H$gZo {H$¶m h¡ ?
of the following ?
(A) B. A. R. C. (A) ~mH©$
(B) I. S. R. O. (B) Bgamo
(C) N. A. S. A. (C) Zmgm
(D) J. A. X. A. (D) Om³gm

194. Which State got first rank in good 194. CÎm‘ A{^emgZ gyMH$m§H$ ‘§o df© 2021 ‘| àW‘
governance Index in 2021 ? ñWmZ {H$g àXoe H$mo àmßV hþAm h¡ ?
(A) Karnataka (A) H$Zm©Q>H$m
(B) Maharashtra (B) ‘hmamîQ´>
(C) Tamil nadu (C) V{‘bZmSy>
(D) Chhattisgarh (D) N>ÎmrgJT>

195. Sattriya dance is associated with which 195. gm{̶m Z¥Ë¶ H$m g§~§Y {ZåZ ‘| go {H$g àXoe
State ? go h¡ ?
(A) Manipur (A) ‘Urnwa
(B) Assam (B) Ag‘
(C) Kerala (C) Ho$abm
(D) Uttar Pradesh (D) CÎma àXoe

196. Which State is famous for Kalamkari 196. H$b‘H$mar {MÌH$bm Ho$ {bE H$m¡Z-gm amÁ¶
painting ? à{gÕ h¡ ?
(A) Madhya Pradesh (A) ‘ܶàXoe
(B) Andhra Pradesh (B) AmÝY«àXoe
(C) Maharashtra (C) ‘hmamîQ´>
(D) Tamil Nadu (D) V{‘bZmSy>

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197. Which is a largest public sector bank 197. df© 2020 ‘| gmd©O{ZH$ joÌ H$m g~go ~‹S>m
in year 2020 ? ~¢H$ H$m¡Z-gm h¡ ?$
(A) H. D. F. C. (A) EM. S>r. E’$. gr.
(B) S. B. I. (B) Eg. ~r. AmB©.
(C) I. C. I. C. I. (C) AmB©. gr. AmB©. gr. AmB©.
(D) I. D. B. I. (D) AmB©. S>r. ~r. AmB©.

198. Which of the following is not a 198. {ZåZ ‘| go H$m¡Z-gm ^maV ‘| AZwg{y MV ~¢H$
scheduled bank of India ? Zht h¡ ?
(A) Central Co-operative Bank (A) Ho$ÝÐr¶ ghH$mar ~¢H$
(B) State Co-operative Bank (B) amÁ¶ ghH$mar ~¢H$
(C) Public Sector Bank (C) gmd©O{ZH$ joÌ Ho$ ~¢H$
(D) Private Sector Bank (D) {ZOr joÌ Ho$ ~¢H$
199. Aarogya setu app gives information of 199. Amamo½¶ goVy En {ZåZ ‘| go {H$g g§H«$‘U H$s
infection of which of the following ? OmZH$mar XoVm h¡ ?
(A) Plague (A) ßboJ

(B) Swine flue (B) ñdmBZ âby
(C) Ebola (C) B~mobm
(D) Covid (D) H$mo{dS>

200. What is ‘Jeevan’ made for the treatment 200. H$mo{dS Ho$ CnMma hoVw {Z‘uV ‘OrdZ’ ³¶m h¡ ?
of Covid ?

(A) Test kit (B) Ventilator (A) Om°§M H$sQ> (B) d|{Q>boQ>a
(C) Vaccine (D) Mask (C) Q>rH$m (D) ‘mñH$



-62- Set-A

-63- Set-A

SET – A

CÎma A§{H$V H$aZo H$m g_` : 3 K§Q>o A{YH$V_ A§H$ : 200
Time for making answers : 3 Hours Maximum Marks : 200

ZmoQ> :
1. Bg àíZ nwpñVH$m _| Mma ^mJ hm|Jo & BZ ^mJm| _| A§H$m| H$m {ddaU {ZåZmZwgma h¡ -
(i) ^mJ I – J{UV 1 – 100 100 A§H$
(ii) ^mJ II – H$åß`yQ>a OmJê$H$Vm 101 – 140 40 A§H$
(iii) ^mJ III – {díbofU Ed§ VH©$ epŠV 141 – 180 40 A§H$
(iv) ^mJ IV – gm_mÝ` AÜ``Z 181 – 200 20 A§H$
àË`oH$ àíZ 1 A§H$ H$m h¡ & g^r àíZ hb H$aZm A{Zdm`© h¡ &
2. àíZm| Ho$ CÎma Xr JB© OMR CÎma-erQ> (Am§ga erQ>) na A§{H$V H$s{OE Ÿ&
3. F$UmË_H$ _yë`m§H$Z Zht {H$`m OmdoJm Ÿ&
4. {H$gr ^r Vah Ho$ H¡$bHw$boQ>a `m bm°J Q>o~b Ed§ _mo~mBb \$moZ H$m à`moJ d{O©V h¡ Ÿ&
5. OMR CÎma-erQ> (Am§ga erQ>) H$m à`moJ H$aVo g_` Eogr H$moB© AgmdYmZr Z ~aV| {Oggo `h \$Q> Om`o `m Cg_|
_mo‹S> `m {gbdQ> Am{X n‹S> Om`o {OgHo$ \$bñdê$n dh Iam~ hmo Om`o Ÿ&

Note :
1. This question booklet contains Four Parts. The distribution of marks in these parts are
as follows –
(i) Part I – Mathematics 1 – 100 100 Marks
(ii) Part II – Computer Awareness 101 – 140 40 Marks
(iii) Part III – Analytical Ability and 141 – 180 40 Marks
Logical Reasoning
(iv) Part IV – General Awareness 181 – 200 20 Marks
Each question contains 1 mark. All questions are compulsory.
2. Indicate your answers on the OMR Answer-Sheet provided.
3. No negative marking will be done.
4. Use of any type of calculator or log table and mobile phone is prohibited.
5. While using OMR Answer-Sheet care should be taken so that the Answer-Sheet does
not get torn or spoiled due to folds and wrinkles.

-64- Set-A