TYPE-1 :- Alphabet series

a) Increasing by a definite number
e.g i) IJKL? ( each letter increases by 1)
ii) AGMSY? ( each letter increases by 6 place to its right position)

b) Decreasing by a definite number
e.g. i) ZXVTRP ? ( each letter decreases by 2 places to its left )

c) Increasing successively
e.g. DEGJNS? ( +1,+2,+3,+4,+5)

d) Decreasing successively
i) ZYWTP ( -1,-2,-3,-4 ..)
ii) ZTOKHFE ( -6,-5,-4,-3,-2,-1)

e) Decreasing and Increasing by a constant value.
e.g. i) DFCEBDACZ (+2,-3,+2,-3,...)


EX-1: Z1A, X2D,V6G,T21J,R88M, P445P,?

First letter: ZXVTRP (-2,-2,-2,.....)
Second letter: ADGJMP ( +3, +3,+3,...)
Series of numerals: 1,2,6,21,88,445 ( x1+1, x2+2, x3+3...)
So next term is N2676S.

EX.2:- 2Z5,7Y7,14X9,23W11,34V13,?

First numeral- 2,7,14,23,34 (+5,+7,+9,+11..)
Second letter- ZYXWV ( decreases by 1 each time)
Third numeral- 5,7,9,11,13 ( increases by 2 each time)

EX-3 :- W-144 , U-121, S-100, Q-81,?
First letter- decreases by 2 each time
Second numeral- square of 12,11,10,9,8..

Type-III :- Continuous patterns series

Ex-1 : ab_ _ baa_ _ ab_
options i) aaaaa ii) aabaa iii) caabab iv) baabb

solution: our answer is ii) . Here series aba is repeated

Ex-2 :ab_aa_bbb_aaa_bbba
options i) abba ii) baab iii) aabb iv) abab

Solution- our answer is ii) . The series is abb/aaabbb/aaaabbbb/a. Thus the letter are repeated twice , then thrice , then four times and so on .

Ex.3 - _bc_ca_aba_c_ca
Options i)abcbb ii)bbbcc iii)bacba iv)abbcc
Solutions- our answer is i) . The series is abc/bca/cab/abc/bca. Thus the letter change in cyclic order .

Ex.4- _c_bd_cbcda_a_db_a
Options i) adabcd ii) bdbcba iii) cdbbca iv)daabbc
Solutions- our answer is i). The series is acdb/dacb/cdab/acdb/da. Each group of four letters contains the letters of the previous group in the order - third , first , second and fourth.

Ex.5:- a_bb_baa_bbb_aa_

Options i) aabba ii) bbaab iii)abaaa iv)baabb
Solutions:- our answer is iii). The series is aabbbb/aaabbb/aaaa. At each step , the number of a's increases by one while the number of b's decrease by one.

Ex.6- _aba_cabc_dcba_bab _a

Options i) abdca ii) bcadc iii) abcdb iv) cbdaa
Solutions- Our answer is i) . The series is aababcabcd/dcbacbabaa. The letters equidistant from the beginning and the end of the series is same .

Ex.7- mnonopqopqrs_ _ _ _ _
Options- i) mnopq ii)oqrst iii)pqrst iv) qrstu
Solutions- our answer is iii) . The series is mno/nopq/opqrs


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