r , 2)FLO · . 1. 3) The series consist of vowels A, E,I, 0, U 5. 2) The series isobtained as...
5
3) F, E, D 4) D. C, B various terms of a letter series and given with one term 8. Z, Y, X, T, S, R, N, M, L, ? Directions (1-35): In each of the following question, 1) H, E, F 2) H, G, F 2) Q, U 4) R, V 1) Q, V 3} R, U H, M,?, T,? 3) L 2) N 4) K 6. U,T,R, Q, 0, ? I} M 2) N 4) P 5. B,D,G, I, L, ? 1) M 3) 0 2) N 4) R 1) M 3)0 Y, U, R, P,? 4. 2} u 4) S B, D, H, N,? 1) V 3) R 3. 2) M 4) N 1) L 3} K B, C, E, H,? 2. 2) M 4)S 1. A,E,I, r, U 1) R 3)0 missing as shown by (?j Choose the correct answer out of the given alternatives. Clearly, the first letters of the terms are alternate. The sequence followed by the numbers is +2., +3, +4, +5 ...... The last letter of each term is 7. three steps forward of the last letter of the preceeding term. Thus, the next term would be K18R. Solution: (2) 4) K18S 3) L18R 2. Choose the term which will continue the folfowing series: C4F, E61, G9L, 1130? 1) K17R 2} K18R Example - 2: -.-;~)o K L __ .3---» 0 Hence the answer is (I). Clearly, the first, second and third letters of each term are respectively moved one, two and three steps forward to obtain the corresponding Ipttefs of the next term. So, the missing term is FKO. E ~ F Solution : (1) 4) FOK 3)GKO 2) FLO What will be the next term in ; CEH, DGJ, ElL? 1)FKO 1. Example - 1: ALPHABETIC SERIES , TYPE 1 SERIES www.tnpscquestionpapers.com www.tnpscquestionpapers.com
Transcript of r , 2)FLO · . 1. 3) The series consist of vowels A, E,I, 0, U 5. 2) The series isobtained as...
3) F, E, D 4) D. C, Bvarious terms of a letter series and given with one term
8. Z, Y, X, T, S, R, N, M, L, ?Directions (1-35): In each of the following question, 1) H, E, F 2) H, G, F
2) Q, U
4) R, V1) Q, V
3} R, U
H, M,?, T,?
3) L2) N4) K
6. U,T,R,Q, 0, ?I} M
2) N
4) P
5. B,D,G,I, L, ?1) M
3) 0
2) N4) R
1) M3) 0
Y, U, R, P,?4.
2} u4) S
B, D, H, N,?1) V3) R
3.
2) M4) N
1) L
3} K
B, C, E, H,?2.
2) M4)S
1. A, E,I, r , U1) R3) 0
missing as shown by (?j Choose the correct answerout of the given alternatives.
Clearly, the first letters of the terms are alternate.The sequence followed by the numbers is +2.,+3, +4, +5 ...... The last letter of each term is 7.three steps forward of the last letter of thepreceeding term. Thus, the next term would beK18R.
Solution: (2)
4) K18S3) L18R
2. Choose the term which will continue thefolfowing series:C4F, E61, G9L, 1130?
1) K17R 2} K18R
Example - 2 :
-.-;~)o KL __ .3---» 0
Hence the answer is (I).
Clearly, the first, second and third letters of eachterm are respectively moved one, two and threesteps forward to obtain the corresponding Ipttefsof the next term. So, the missing term is FKO.
E ~ F
Solution : (1)
4) FOK3)GKO2) FLO
What will be the next term in ; CEH, DGJ,ElL?1)FKO
1.
Example - 1 :
ALPHABETIC SERIES, TYPE 1
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9, c e" b j, t I, o?? 19. FKN, JOR, OIL, O??1) n, q 2) m, q 1) 5, W 2) T, W3) q, m 4) q, n 3) W, T 4) W, Y
10. DGI, KNP, PSU, N ? ? 20. Win, OJL, S??1) Q, T 2) Q, 5 1) NP 2) MQ3) R, T 4) M, N 3) Nf 4) WY
,11. JMK, VYW,PSQ, B? ? 21. alf, ene, i?d, mr?1) C, E 2) D,E 1) p, c 2) c. p3) E,C 4) M, N 3) a, C 4), q, C
12. GEC, MKI, TRP, N? ? 22. cr, gw, ka, oe, si, w?1) R, T 2) 5, T 1) n 2) m
3) T, R 4) S, U 3) q 4) p
13. . FKN. J??, DIL, OTW 23. acd, dca, mpq, qpm, fgh, ???1) R,5 2) 0, R 1) ghf 2) hfg
3) 0, M 4) P, S 3) hgf . 4) fhj
14. WRT, OJL, SN?, L?I 24. AB, DEF, IJKL, PQR?T, ??1) P, .G 2) Q, H 1) 5, X, Y 2) 5, Y, Z3) R,H 4)R,G 3) T, Y, Z 4) X, Y, Z
15, HF, KC, NK, QH, SU. ?? 25. AC, EG, HJ,??1) VO 2;VR 1) M. N 2) K, L3)VM 4) WR 3) J,L 4) J, I
16. OGI, KNp, PSU, ??? 26. BAT,DCV, FEX, ?1) MQS 2) NQT 1) HGI 2) HGZ
3)NQS 4) NRS 3) HIJ 4) HGY
17. JMK, VYW.?S? BEC 27. MN, LO, KP, ?1} MN 2)PQ 1) RS 2) JQ3) PR 4) ST 3) IJ 4) QR
18. GEe, MKI, TRP, ??? 28. C, G, L, R, ?1} VTQ 2) MTR 1) U 2) S
3) VTR 4) MQS . 3) Y 4} Z
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1. 3) The series consist of vowels A, E, I, 0, U 5. 2) The series is obtained as below:written in the order as above. B ~ D
2. 1) The series is of the form D ~ G
G +2B ~ C
C +2 E ~ L
L .2 NE +3 H
H +4 L 6. 2) The leiters are alternately moved one and twosteps backward to obtain the successive terms.
3. 1) The letters are obtained as below :
B +2 D 7. 1) Letters are in reverse order in which term theD +4' H last 1, 2, 3 and 4 letters are missing between
H +6 N two consecutive letters.
N +8 V8. 2) Beginning from Z in the alphabets 3
consecutive letters are written and then 34. 3) The letters are obtained as below :
consecutive letters are 'deleted and so on.Y -4 U
U --3 R 9. 4) The series may be divided into groups asR -2 P shown ceb / jli / oqn.
P ~ ° In each group, first letter is moved one step
ANSWERS
29. QR, PSJ OT, ?? MV 1) EU 2) EV..,1) NT 2) NU 3~ EF 4) FV
3) NW 4) SU33. AD, EI, IN, PS ?
30. C, H, M, R, ? 1) WY 2) XX
1) X 2} W 3) WX 4) WW
3) Y 4)V34. BDF, CFI. DHL. ?
,...,." 1) ElM 2) EJO31. JAZ, LEX, NIV, POT 3) CJM 4) EMI
1} TOC 2) PQR-
3) RSR 4) ACM 35. RGS, SRG, GSR, ?i) GRG 2) SRS
32. BY,ex, OW,? 3) RGS 4) SRG
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backward to obtain the third letter while the 16. 3) The arrangement of alphabets are as fr ..third letter is moved. Three steps forward to D ~ G ~->obtain the second letter. K -~> N ----_-:?_._)- P
p ~ S ·2 U10. 2) Here the letters are obtained as follows : Therefore the order must be
0 .3 G ~ N __ +3_7 Q .~ s--~K +3 N .2 P--~P ·3 S ~ U 17, 2) The letters are arranged as follows .
J ~ M 2 KN ~3 Q ~ s --~v ~ y ~ WP ~ S z Q11. 3) The letters are arranged as follows :B .3 E 2 CJ +3 M -2 K -.-~
V +3 Y -2 W The arrangement of letters are as follows :18. 3)p ~ S ~ Q G ~ E __ 2_> C
.' B .3 E ~ C M ~ ·K __ 2_~ I'T ~ R --~-> P
12. 3) The letters are obtained as follows : V ~ T ~ 1\G ·2 E -2 C
M -2 K -2 19. 2) Here the first letter the first series il!cn~c:,:.'d
T -2 R ~ P by 5 to obtain second letter is increas- by :~
N T Rto obtain the third letter.-2 ~ 0 ~ T _._'."._.:> W
13. 2) Here th", letters are arranged as follows: 20. 1) The first letter of the first series decreased L;,;F +5 K .3 N 5 to obtain the second letter of the first seriesJ +5 0 +3 R and then the second letter increases by ~ toD +5 +3 L obtain the third letter.
0 +5 T ~ W S -~ N -~-> P
14. 1) Here the arrangement is as follows: 21. I} The first letters of each series is increased by4 and the second letter of all series is increased
W -5 R ~ Tby 2 and third letter of each series is
0 --5 J ~ L decreased by one.S 5 N .2 P -.~ A .4 IIIa e .. _.- -- -->L --5 G .2
-_+~ n +2 p ~-}
f .__ l_~ e 1 d _~l __ )-----)-
15. 2) The first letter of each term is moved threesteps forward and the second letter is moved 22. 2) The _lettersin each term are moved four si~i;~three steps backward to obtain the forward to obtain. The corresponding lettercorresponding letters of the next letter. of the next term.
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A .--~* E ~ 30. 2) C ~ H
H ·1 J H ,5 M--~C .3 G .• 2 M +5 RJ ~ L R ~ W
26. 2) The first two letters in each words are 31. 3) J ~ L +2
alphabets starting A from the second N ~ P 2 Rposition to first position only. Therefore the
A 14 E .4first two ietters of the next word is HG. The
I .4 0 ,4 Sthird letter in each word starting from firstword exceeds by 2. Z ~ X ··2
T ·,2 V V ~ T -2 1..-.~ --)
V ,-2 XX .2 Z
32. 2) B ~ C ~ D ~ E
So, the answer is HGZY ~ X ~ W ~ V
27. 2) The second letters each sequence 33. 3) A +4 E .5 J +-6 Pis a
continuing series. +7 W
N 0 D +5 ,5 N .5 S~
0 P+5
~P ~ Q Therefore the required term is X.
Similarly the first letters34. 2) C EB ~ ~ D ~
In each sequence is a decreasing seriesD +2 F .2 H ~ J
M ~ LF ~ ,3 L ~ 0
L ~ K
K ~ T 35. 3) The last letter becomes the first letter is eachSo, the answer is JQ. subsequent series.
;.~... 3} The series tan be splitted as follows : 28. 3} C ,4 G---'----7
acd, dca G +S L
The reciprocal of first series is the second. L .6 RSimilarly mpq ~ qpm R -7 Y---7
Similarly fgh ~ hgf So, the answer .isY
24. i) The series consists of letters 2, 3, 4, 5, etc.and one letter is between first two series, 2
29. 2) Q ~ p
letters between second and third and so on.p ~ 0
s, y, z is the correct one. 0 ~ NN ~ M
25. 3) The series goes on increasing as follows:
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