VIJAYA BANK CLERK EXAM, 7-2-2010

ANSWERS 1.(4)  2.(4)  3.(3)  4.(2) 5. (2)  6. (2)  7.(4)   8. (3) 9.(1)  10. (5)  11. (3)  12. (3) 13. (3)  14. (5)  15. (2)  16. (3) 17. (5)  18. (4)  19. (5)  20. (4) 21. (3)  22. (2)  23. (3)  24.(2) 25.(2)  26. (4)  27. (4)  28.(1) 29. (2)  30. (4)  31. (4)  32. (I) 33. (3)  34.(4)  35. (2)  36. (3) 37. (5)  38. (4)  39. (4)  40.(1) 41. (2)  42. (5)  43. (4)  44.(4) 45. (3)  46.(1)  47. (3)  48.(1) 49. (2)  50. (5)  51. (2)  52. (1) 53. (1)   54.(2)  55. (3)  56.(5) 57. (1)  58. (4)  59.(1)  60.(5) 61. (4)  62. (3)  63. (4)  64.(2) 65. (5)  66. (2)  67. (4)  68.(4) 69. (3)  70. (2)  71. (4)  72. (2) 73. (2)  74.(1)  75. (3)  76; (3) 77. (2)  78. (3)  79. (2)  80.(5) 81. (3)  82. (5)  83. (2)  84.(5) 85. (3)  86.(5)  87. (3)  88.(4) 89. (3)  90.(1)  91. (1)  92.(2) 93. (4)  94. (2)  95. (3)  96. (1) 97.(2)  98. (1)  99. (2)  100. (5) 101. (3)  102. (2)  103. (1)  104. (4) 105. (1)  106. (4)  107. (3)  108. (2) 109. (3)  110. (4)  111. (2)  112. (1) 113. (4)  114. (1)  115. (4)  116. (3) 117. (2)  118. (3)  119. (3)  120. (4) 121. (2)  122. (3)  123. (4)  124. (1) 125. (5)  126. (1)  127. (3)  128. (4) 129. (2)  130. (5)  131. (4)  132. (1) 133. (2)  134. (3)  135. (1)  136. (3) 137. (2)  138. (1)  139. (4)  140. (2) 141. (5)  142. (1)  143. (3)  144. (2) 145. (4)  146. (1)  147. (4)  148. (5) 149. (2)  150. (3)  151. (2)  152. (2) 153. (4)  154. (4)  155. (3)  156. (2) 

157. (3)  158. (4)  159. (5)  160. (5) 161. (1)  162. (3)  163. (4)  164. (4) 165. (2)  166. (2)  167. (3)  168. (1) 169. (3)  170. (3)  171. (3)  172. (2) 173. (4)  174. (5)  175. (4)  176. (2) 177. (1)  178. (3)  179. (3)  180. (4J. 181. (5)  182. (4)  183. (2)  184. (1) 185. (5)  186. (2)  187. (1)  188. (3) 189. (5)  190. (4)  191. (3)  192. (2) 193. (4)  194. (3)  195. (4)  196. (1) 197. (1)  198. (5)  199. (4)  200. (5) 

EXPLANATIONS

 

 

(23 - 28) : {i) All coins are glasses —» Universal Affirmative (A-type) (ii) Some glasses are cups —» Particular Affirmative (I-type) (iii) No man is tiger —> Universal Negative (E-type) (iv) Some men are not tigers —> Particular Negative (O-type)

23. (3) Some cups are boxes.

All boxes are pins. I + A => I‐type of Conclusion "Some cups are pins" This is Conclusion III.

24. (2) Some pens are pencils.  

All Pencils are caps.  

I + A.=> I-type of Conclusion "Some pens are caps".

All pencils are caps.

All caps are buses. A+A=> A‐type of Conclusion "All pencils are buses" Some Pens are Caps.  All caps are buses. I + A =» I‐type of Conclusion "Some pens are buses". This is Conclusion II.  

25. (2) All shirts are skirts.  

All skirts are banks. A+ A =» A‐type of Conclusion "All shirts are banks".  All shirts are Banks. All banks are roads. A + A => A-type of Conclusion "All skirts are roads". All banks are roads.

All roads are brushes. A+A => A-type of Conclusion “All banks are brushes”.

              All Shirts are banks. 

All banks are roads. 

A+A => A-type of Conclusion “All shirts are roads.

All shirts are roads.

All roads are brushes”

A+A => A-type of Conclusion “All shirts are Brushes “.

Conclusion III is Converse of it.

26. (4) Some spoons are plants.

All Plants are crows.

I+A=> I-type of Conclusion “Some spoons are crows”.

Conclusion II is Converse of it.

Conclusion III is Converse of the third Premise.

27. (4) Some hens are ducks.

All ducks are pigeons.

I+A=>I-type Conclusion "Some hens are pigeons".  

All ducks are pigeons.

All pigeons are sparrows.

A+A=> A-type of Conclusion All ducks are sparrows”.

This is Conclusion I

Some hens are pigeons.

All pigeons are sparrows.

A+A =>A-type of Conclusion Some hens are sparrows”.

Some hens are pigeons .

All piggeons are spaarrows.

I+A => I-type of Coonclusion Soome hens aree Sparrows

Conclussion III is converse of it.

28.((1) No tige

29. (2) KL %O @ M * TheCon./1. NII. MIII. KIV. L

30. (4) A B $ C % D ©TherConI. DII. BIII. EIV. A

31.(4) Fp @ R ©S %TherConI. R II. SIII. PIV. S

32. ( 1 )

H ★ I $J @ KTher

Some cat

er is cat. 

ts are lions. 

E+1 =>01,type of Con

K © L => K ≥ L % O => L < O @ M => O = M

* N = * M ≤Nerefore, K ≥ L <clusions:

N © O => N ≥ OM $ L => M > LK * N => K ≤ NL @ N => L =

A * B => A ≤ B $ C => B > C % D => C < D © E => D ≥ E refore, A ≤B >clusions: $ A = > D > A

B $ D => B > DE % C => E < CA @ E => A =

F $ P => F > P R => P = R S => R ≥ S T S < T refore, F > P =clusions: % F => R < F

S * P => S,≤ P : TP © T => P ≥ TS % F => S < F

) G % H =» G <

★ I => H ≤ I $ J=> I > J K => J = K refore, G < H ≤

clusion 'Somee lions are  noot tigers “ 

 

< Q = M ≤N

O : True L : True N : Not true N : Not true

B

> C < D ≥ E

: Not true D : Not true

C : Not true E : Not true

= R ≥ S < T

: True True

T : Not true F : True

< H

≤ I > J = K

ConI. G

clusions:

II. G% I => G < I : True

I I IG % J G < J : N

IV. H. K $ I = > K >

Not true I : Not true J : Not true H ★ J => H ≤

33. (3) VV @ W => V = W

W %X * Y

% X => W < XX

Y $ Y => x ≤ y

TherZ => Y > Z

Conrefore, V = W << X ≤ Y > Z

I. Z $clusions:

II. Y$ X = > Z > X : Not true

III. WY © V => Y ≥

IV. YW % Y => W

≥ V : Not true

Y @ W => Y < Y : True = W : Not truee

34. (4) 35. (2) 36. (3) 37. (5) 38. (4) 39. (4) 40. "(1

41. (2)

Except in A E

B is between D

G and F are sit

) A is third to th

) G is to the im

) D is to the im

1) C is to the im

The following

(l)to(2)

F, in all others

D and E.

tting between A

he right of E.

mmediate right o

mmediate left of

mmediate left o

g changes occu

(2) to (3)

s the arrangemment is anticlockkwise.

A and D.

of A.

f B.

of H.

ur in the subseqquent figures :

42.(5) T half

The design 'P' d

diago

43. (4) Probtinue

44. (4)

pairsdesigrowsposicont

45. (3) F

inter 46. (1) F

(3) t 47. (3) F

of eaand

48. (1) F

and from

49. (2) F

(3) t 50. (5) F

clocdesig

51. (2) 26

=> 1 => ?

52.(1)

=>

f . step in anticldescends stepw

onally downwa

From Problemblem Figure (2ed in the subse

From Problems of adjacent dgn of the midds interchange ption while the tinued in the sa

From Problem rchange positio

From Problem Fthe entire desig

From Problem Fach column infrom Problem

From Problem the left design

m Problem Figu

From Problem to (4) and from

From Problem Fkwise directiongns (A) and [Z

6 x 451 - ? = 611726-7 = 610? = 11726 - 61

=963-6=95

> 957* =478

= =5=5*5=25

lockwise direcwise and ascend

ard and upward

m Figure (1) to 2) to (3) the lefequent figures a

m Figure (1) todesigns of the mdle row movespositions. Fromtwo pairs of a

ame order in th

Figure (1) to ons. Similar ch

Figure (1) to (2gn rotates throu

Figure (1) to (2nterchange posi

Figure (5) to'A

Figure (1) to (ns move to theure (3) to* (4) a

Figure (1) to m Problem Figu

Figure (1) to (2n. From Proble

Z] interchange p

6103 03 03 = 5623

57

85

5

tion and rotateds in one step a

d respectively

(2) the two paft most designsalternately.

o (2) the right mmiddle and lows to the left mom Problem Figadjacent designhe subsequent f

(2) the line shanges occur fr

2) the left and ugh 90° anticlo

2) the two coluitions so as the

Answer Figure.

(2) the rightmoe right side andarid from Prob

(2) the designsure (5) to Answ

2) the central dem Figure (2) tpositions. Thes

es through 90° after being rota

and is replaced

airs of designss of both the r

most design ower rows intercost position Wgure (3) to (4)ns of the upperfigures.

segments moverom Problem F

the right desigockwise. These

umns of designe two middle d.

ost designs movd all the three blem Figure (5)

s are arranged wer Figure.

design is replacto (3) the lowese two steps ar

anticlockwise ated through 90

d with a new d

s, situated alonows move to t

f the upper rowchange position

While the two p) the rightmostr and middle r

e to the opposigure (3) to (4)

gns are replacee two steps are

ns interchange designs. Simila

ve to the left sidesigns are re

) to Answer Fig

in reverse ord

ced with a newer left design mre continued alt

after every two0° clockwise. T

esign in each s

ng the two diagthe right most

w moves to thns. From Problpairs of adjacet design of therows interchang

site side diago) and from Pro

ed with new decontinued alter

positions. Simar changes occ

ide after beingeplaced with negure.

der. Similar- ch

w design and allmoves diagonal

ternately in the

o figures. The tThe arrow mov

subsequent figuthird design m

ves moves

ure.

gonals interchaposition. Thes

ange positions.se two steps ar

From re con-

he left most polem Figure (2) ent designs of e lower row mge positions. T

osition while thto (3) the righthe upper and

moves to the leThese three ste

he two ht most

lower ftmost

eps are

onally and the blem Figure (5

two middle d5) to Answer F

designs Figure.

esigns. From Prnately in the s

roblem Figuresubsequent figu

e (2) to ures.

milarly, th^ uppecur from Probl

er and lower dlem Figure (3)

designs to (4)

g rotated througew designs. Si

gh 90° anticlocimilar changes

ckwise occur

hanges occur ffrom Problem FFigure

l the four desigly to the uppere subsequent fi

gns move one sr right corner aigures.

step in and the

53. ( 1 ) ? = 11797 - 3695 = 8102

54.(2) (12.25 x 4.02 - 14.26) x ? = 699.7

=> (49.245 - 14.26) x ? = 699.7 => 34.985 x? = 699.7 => 34.985*? =699.7

>? = .. =20

 =

55. (2) (12.25 * 4.02-14.26)*? = 699.7

(49.245-14.26)* ? = 699.7 34.985 * ?= 699.7

. ? .

= = 20  

55. (3) +? 2 = 61.28

=> 25.28 +? 2 = 61.28 =>? 2 = 61.28 -25.28 = 36 =>? = √ = 6

56 ?.(5) = 311.88

? = . = 69

.