## Logical Reasoning Questions To Prepare For SBI PO Exams

Below are four reasoning questions based on some conditions.

Directions to solve :

In each question below is given a group of numbers/symbols followed by five combinations of letter codes labeled (a), (b), (c) and (d). You have to find out which of the combinations correctly represents the group of numbers/symbols based on the following coding system and the conditions and mark the correct options as your answer. (Conditions override the coding system whenever applicable).

NO. /Symbols : 2 # 7 \$ 1 % 9 & 3 8 * 4 + 6
Letter Codes : M V O I H E L R W Y S D N A G

Conditions :

I. If the third element is an even number and the last element is a symbol then that even number is to be coded as the code for the symbol.
II. If an odd number is immediately preceded as well as immediately followed by symbols then the odd number is to be coded as ‘X’.
III. If there are no even numbers then the codes of the first and the third elements have to be interchanged.

Question 1

8%9+24

a) SLRAVN
b) SLXAVN
c) ALRAVN
d) SLXRAN

Solution :

Check 8%9+24 with given three conditions as follows:

Third element 9 is not an even number; therefore condition I is not applicable.
And, 8%9+24 contains even numbers 8, 2 and 4; so we cannot apply condition III.

But, the odd number 9 in 8%9+24 is immediately preceded by the symbol % and immediately followed by the symbol +.
Therefore, we can apply condition II.

So that the corresponding code for 9 is X.

Now, let us replace each of the symbols other than 9 by their corresponding code as per coding system and replace 9 with X based on conditions. (Remember conditions override coding system. Though 9 has a code of R as per coding system, the code X based on conditions will override and will take priority over R.)

Elements: 8 % 9 + 2 4
Codes : S L X A V N

Question 2

\$371%2

a) HYILVE
b) HIELVY
c) HYXELV
d) HYIELV

Solution :

Check \$371%2 with given three conditions as follows:

Third element 7 is not an even number; therefore, we cannot apply condition I.
Each of the odd numbers 3 and 1 in \$371%2 is either immediately preceded by the symbol or immediately followed by the symbol; and 7 immediately preceded by and followed by numbers; so, the condition II is not applicable.

The given combination in question contains an even number 2; so that, we cannot apply condition III as well.

Therefore, \$371%2 is coded as follows:

Elements: \$ 3 7 1 % 2
Codes : H Y I E L V

Question 3

913\$7#

a) YERHXO
b) REYHIO
c) REYHXO
d) YEOHXR

Solution :

Check 913\$7# with given three conditions as follows:

Third element 3 is not an even number; condition I is not applicable.
The odd number 7 in 913\$7# is immediately followed and preceded by symbols.
Therefore, we can apply condition II.
That is, 7 is coded as X.
There are no even numbers in 913\$7#.
So that the condition III also applicable.
That is, the codes of 9 and 3 will be interchanged.
Therefore the code for 9 = Y
And, the code for 3 = R

Elements : 9 1 3 \$ 7 #
Codes : Y E R H X O

Question 4

3+82*&

a) YASVDW
b) SAWVDW
c) SASVDW
d) YAWVDW

Solution :

Check 3+82*& with given three conditions as follows:

Third element 8 is an even number and the last element is a symbol; condition I is applicable.

That is, the code for 8 = code for & = W.

The odd number 3 in 3+82*& is immediately followed by + but not preceded by any symbol.
Therefore, we cannot apply condition II.

And, there are even numbers 2 and 8 in 3+82*&.
Therefore, the condition III is also not applicable.

Elements : 3 + 8 2 * &
Codes : Y A W V D W