logical reasoning questions and answers paper 11 - skillgunNote: Paper virtual numbers may be different from actual paper numbers . In the page numbers section website displaying virtual numbers .
The middle letters are static, so concentrate on the first and third letters.
The series involves an alphabetical order with a reversal of the letters.
The first letters are in alphabetical order: F, G, H, I , J.
The second and fourth segments are reversals of the first and third segments.
The missing segment begins with a new letter.
There are thee alphabetic series present in each set .
First is ABCD,Second is JKLM and Third is UVWX.
There are three character set of characters present in the given series .
Every sets starting char incremented by one char .
K__,L__,M__,N__ hence in the third set first char must be M . In the second position of missing set F must come hence below sequence is valid for second position characters .
_B_,_D_,_F_,_H_ same way in the third position __H,__J,__L,__N grouping all series together the actual series must be written as followed.
This series having 3 characters in the sequence of alphabetic order , leaving next 3 sequence of alphabetic characters .
Hence EFG is ignored and HIJ must be taken , KLM must be ignored and NOP must be considered so on ....
There are two alphabetical series here. The first series is with the first letters only: ABCD.
The second series involves the remaining letters: JK,LM,NO,PQ.
There are two alphabetical series here.
The first series is with the first two letters : ABCDE.
The second series involves the remaining letter: D, E,F,G.
There are two alphabetical series here.First is DE,FG,HI,JK and second is LM,NO,PQ,RS hence the series third set will become HIPQ .
All first positioned characters set having series of characters with a missing intermediate character . It means ACEGI characters must come in first position second position character series must be NPRTV , the third position series of characters must be BDFHJ and fourth position characters must be OQSUW . taking all last positioned characters in the explanation will leads to IVJW
Every second word is reversal of immediate previous word hence FAR is becoming RAF in the next sequence , GAS is becoming SAG , same way HAT must become TAH .
In the given series next word is having characters incremented by one character and removing first and last characters . Hence in DEFGHIJ if first and last characters are removed remaining characters will be EFGHI , if each char is adjusted to next char the set becomes FGHIJ. From FGHIJ again remove first and last character resultant is GHI , now adjust characters to next position will become HIJ . Now remove First and last char from HIJ will results I , then increment I by one char results J .
Because the letters are the same, concentrate on the number series, which is a simple 2, 3, 4, 5, 6 series, and follows each letter in order.
Every time number in the series is incremented by one and moved to next position hence E3FG must become EF4G , next sequence must be EFG5 so on.....
There are two series to look for here.
The first letters are alphabetical in reverse: V,U,T,S,R.
The second letters are in alphabetical order, beginning with D.
There are three series to look for here.
The last letters are alphabetical in reverse: K,J,I,H,G.
The first and second letters are in alphabetical QR,ST,UV,WX,YZ.
There are three series to look for here.
The first letters are alphabets in reverse: Z, Y, X, W, V.
The second letters are in alphabets beginning with A in ascending order .
The number series is as follows: 10,20,30,40,50
There is a relation between first and last characters . That is between M_N,K_J,___,E_D and the relation is series of characters separated with an intermediate characters . In the last set DE are coming in reverse order with an intermediate character . In the 2nd set KJ are coming in reverse order with an intermediate letter . In the first set MN are coming in ascending order with an intermediate character . Hence in the third set GH must come in regular order . that is G_H . Now let us evaluate logic for intermediate letter . In the letter series every fifth letter is placed in the series as an intermediate letter ( that is V , Q , L , G )
AB CDEF GHIJK LMNOP QRSTU VWXYZ
Hence answer is GQH .
Upon careful observation first letters are decremented by tow letters . It means S , U ,W coming as first characters in the right to left order . Hence next letter must be Y by skipping X . Second letters are coming in series by removing one intermediate character in the left to right order . That is _D__,_F__,_H__ . hence B must come as second character in first set that is _B__ . The first character logic is valid for third character and second character logic is valid for fourth letter .
BDF, GEC2, DF2H,____, 2FHJ
In the given series first set BDF is reversed in the second set and promoted each character by one and adding 2 in the end . hence second set becoming GEC which is reversed and upgraded form of FDB. In the third set 2 is promoted to third position again remaining characters are upgraded by one character after reversing . Reversal of GEC is CEG, when each letter upgraded to next letter it becomes DFH and 2 comes before H. Following the same logic yields to I2GE as answer .
Consider A to Z as numbers from 1 to 26 . If you add value of Y and M leads to 25+13=38 same way add numbers corresponding to ABCDEFGJ leads to 38 . hence KMN is correct whose corresponding numbers addition result is also 38 .
None of the above .
Every First character is in ascending order and also second and third letters .
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