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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
.