Logical Reasoning - Letter and Symbol Series - Discussion

In the above question they are asking for sequential series ok. Just concentrate the series in first term B₂CD = B+B+C+D (4 letter) in third term i.e., BCD₄ = B+C+D+D+D+D (total 6 letter) similarly 4th term contain 7 letter and last term contain 8 letter therefore according to series there must be contain 5 letter in 2nd term.

Now see the option i.e., how many option you have that contain 5 letter only two option are there i.e., option A & opt B so we go for option B only because no one term contain two word simultaneously more than one times except option B i.e., in option A (it contain 5 letter but here two word i.e., B, B and C, C simultaneously) thats why we go for option B. This is the right answer.

Hi,

In the above quest they are asking for sequential series ok...just concentrate the series in first term b2cd = b+b+c+d(4 letter) in third term i.e bcd4 = b+c+d+d+d+d(total 6 letter)similarly 4th term contain 7 letter and last term contain 8 letter therefore according to series there must be contain 5 letter in 2nd term. now see the option i.e how many option u have that contain 5 letter only two option are there i.e option A & opt B so we go for option B only because no one term contain two word simultaneously more than one times except option B i.e in option A(it contain 5 letter but here two word i.e B,B and C,C simultaneously)thats y we go for option B. this is the right ans.

Because all numbers are moving in circular path.

First number 2 on B (i.e B2Cd) , then it get incremented by one i.e. 3 moves to next alphabet (i.e C) form BC3D (Ans). Then again number get incremented by one i.e 4 moves to next alphabet (i.e D) form BCD4, Now Here Alphabet are Three only (B, C, D) which are in circular form (means After D here B will come) , again numbers moves to 1st alphabet i.e 'B' Hence next incremented number get moved to B (i.e B5CD) by this way After BC6D number will come BCD7 (In Advanced).

Steps.

1. Give number to all alphabets as 1, 2, 3, 4 (a-1, b-2, c-3).

2. You will notice that number series 234 is repeated.

3. Observe that in the first set, the first alphabet is repeated 2 times, in the second set should repeat. Times, in third series third alphabet is repeated four time. The circle is repeated with increasing number of repeating alphabet. So at the blank space second alphabet should repeat 3 times.

First number B2CD = B + B + C + D = 4 "B2". First series start with 2.

Second number BC3D = B + C + C + C +D = 5 "C3". Second series Sequence 3.

Third Numbr BCD4 = B + C + D + D + D + D = 6 "D4" Third Series sequence 4.

Fourth Number B5CD = B + B + B + B + B + C + D =7 "B5" Alternator continue with number 5.

Fifth number BC6D =B +C + C + C + C + C + C + D =8 "C6" This the sequence with 6.

In the question, there are only three alphabets in sequence B,C,D....The Numbers are also in sequence where the 3rd sequence/number is missing hence BC3D is the answer.

Replace the question with the given answer:
B2CD, (BC3D), BCD4, B5CD, BC6D

Now the numbers are in sequence 2,3,4,5,6...

Hope you got it now guys!

B2CD, _BC3D_, BCD4 = in B2CD B has number 2, in BCD4 D has a number 4 C has no number so answer is BC3D. The series of the number is 2, 3, 4.

B2 has 2 in first set of words, C has missing number so we have entered 3 for C in second set of words i.e., C3, D4 in third set of words.

On this letter series the number will take b to c, and c to d, and again d to b like triangle wise. So first, b series they had been mentioned the substring of 2. So, what will happen, next to b, c series will get the next process, after that d, next c.

It is very simple thing letter is same its didn't change but number is in sequence the sequence is 2 3 4 5 6 and in the options b and d contains 3, but in d option it also had 2 so it is not correct option then b is the correct.

The letters are the same for each case only the no.s 2,3,4....... are moving in cyclic order i.e. from b to c & from c to d and and when one cycle is finished we again go back to the previous one i.e. to B.