logical reasoning questions and answers paper 158 - skillgunNote: Paper virtual numbers may be different from actual paper numbers . In the page numbers section website displaying virtual numbers .
All the above
All letters in the series is having 10,11,12,13 so on as their total positional values . Hence All options are correct . Since N , JD and FH is also having 14 as its positional value .
The series is having 5,4,3,2,1 as difference in letters positions from left to right . That means Position of J is 10 , E's position is 5 . Hence difference between positions is 10-5=5, same way for T and P which leads to 20-16=4 . Using same logic you will get Positional difference between P and M as 3 , Z and Y as 1 . Hence the missing letters positional difference must be two . The best fit in the given options is CA and whose positional difference is 2 .
Between B and F 3 letters are skipped , Between F and L 5 letters are skipped . Hence between L and Next letter 7 letters must be skipped hence next letter must be T .
MY & YM
LZ & ZL
MZ & ZM
None of the above
In the series first ,third and fifth words are reversed to get second fourth and 6th words . Now let us observe the relation between First third and fifth sets of elements . G's position in alphabetical series is 7 , N's position is 14 . I's position is 9 and R's position is 18, K's position is 11 and V's position is 22 . First letters are getting incremented by 2 in the first second and third series . Next letters position is doubled . Using the same logic M must come as next sets first letter since its position is 13 and doubled positioned letter is Z . Hence we will get M and Z , Z and M as next elements .
None of the above .
The series is having 5 multiplication table . that is 5,10,15,20,25,30. If you add positions of individual elements present A through Z series you will get the above derived positional series .
Explanation : B's position is 2 and C's position is 3 . 2+3=5 , H's position is 8 , A's position is 1 which leads to 8+1+1=10, same way O's position is 15 . Using same logic we must get the resulting letters total positions value as 30 . In the given options D and Z is suitable . Since D's position is 4 and Z's position is 26 whose some is 26+4=30 .
The series is Fibonacci series , only difference is letters are placed in place of numbers . The letters are positional equivalent of corresponding numbers . Fibonacci series can be written as 1,1,2,3,5,8,13,21,.... so on . A's position is 1 B's position is 2 C's position is 3 ,E's position is 5 . Same way 21st letter must be U .
The given series is having letters arrangement as per the below given virtual logical series .
(1^1)+1 , (2^2)+1,(3^3)+1,(4^4)+1,(5^5)+1 . 1 square 1 plus one gives 2 . two square two plus one gives 5 , same way last letter must be Z since 5 square 5 plus one gives 26 . Since 26th positioned letter is Z .
b _ cdbba _ cc _ dbb _ aaa _ ccddd
The series is bacd/bbaaccdd/bbbaaacccddd.
Thus each letter of first sequence is repeated two times in the second sequence and three times in the third sequence.
b _ baaa _ ccccd _ ddccc _ aa _ ab
The series is bbb/aaaa/cccc/dddd/cccc/aaaa/b.
The series is having letters positioned as per the below given logical number series . The series is 1^2+0,2^2+1,3^2+2,4^2+3,5^2+4 so on . Hence YZD is correct answer since , the letters cumulative value gives 55 .
This series is having 5 multiplication series . That is 5,1,5,5,2,10,5,3,15,5,4,20 so on .
Position of E is 5, J's position is 10 and whose one digit equivalent is 1+0=1 , E's position is 5 . Same way E's position is 5, next multiplication factor is 2 and K's position is 11 and the one digit equivalent is 1+1=2 , and 5X2=10 . DF's cumulative position value is 10 . Using same logic BC's total single digit position value is 5 , multiplication factor is 3 , AAA's total positional value is 3. hence Next number must be 15 .
The given series is having position/sum of positions ( when reduced to single digit number ) of individual elements as 3,2,6,4,3,12,5,4,20. In the given series K's position is 11, when the individual numbers are added together produces 1+1=2 , F's value is 6. Same way V's single digit equivalent is 4 ( 22 --> 2+2=4 ) , JB's single digit equivalent is 1+0+2=3 and L's position is 12 . Using same logic 20th position element is T .
Count the total positional values of individual letters present in each set which leads to a series having 27,28,29_. The next number must be 30 . Now look at WEB , whose total positional equivalent is 30 .
Explanation: W's position in A through Z series is 23, E's position is 5 , B's position is 2 ( 23+5+2=30 ) .
For reference :
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Total positional equivalent of all letters in BULL is 47 (2+21+12+12) , COW Value is 41 (3+15+23) same way DOG value is 26 .
Total positional value of all letters in first set is 5 (ABB=1+2+2) , BAD 's total value of all letters position is 7 (2+1+4) , same way DAD's position is (4+1+4=9) . same way next letters sum must be 11 . Hence matched word is BED ( 2+5+4 ) .
Difference between elements position in each set is following Fibonacci series that is 1,1,2,3,5,8 so on . B's position is 2 and A's position is 1 . K's position is 11 and J's position is 10 and the difference is 1 . E's position is 5 and C's position is 3 and the difference is 2 . same way last letters position difference must be 8 .
The total letters position for J,T,ZD,TJ,FD is 10,20,30,40,50 ( ZD's letters positional sum is 26+4=30 ) . Same way TJFD's total value is 20+10+6+4=40 . Hence next letters positions sum must be equal to 50 there fore YY is correct answer .
Geometric sum of letters position in each set leads to 6,60,600.... so on series that is (BC=2X3=6 , FJ=6x10=60 EBFJ=5x2x6x10=600, ...) . Hence next set's Geometric sum must be 6000. TCJJ is having geometric sum of letters as 6000 ( 20X3X10X10=6000 ) .
All the above .
Positional difference between first and next letter present in each set is -2 . Position of O - Position of Q= 15-17 , Position of F -Position of H =6-8 =-2 , Hence all the given options are correct .
In the series first and second letters positional geometric sum and addition of third letters position ( If any ) leads to a series of numerical or number series which is represented as 13,14,15,16,17 so on . That means FBA's total value is (FXB)+A=(6X2)+1 , DCB's total value is (DXC)+B=(4X3)+2, so on . Hence correct answer is CEB ( (CXE)+B=(3X5)+2 =17 ) .
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