M4maths Previous Puzzles

23 Nov 2009 xvYBUz ?

If AxyBzC = ZcbYaX
And BbyCDw = YybXWd
Then xvYBUz =

OPtion

1) acBYFe
2) ecBYFa
3) cfBYEa
4) ebCYFa
5) yeBCFa
6) ecBYDf
7) ceBYFa
8) caYBFe
9) eaFBYc
10) aeDYBc

Solution

Rules for making these codes are
1. Capital letter should be capital.
2. Small letter should be small.
3. Letters are changed according to the sequence given below
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
z y x w v u t s r q p o n m l k j i h g f e d c b a
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
Z Y X W V U T S R Q P O N M L K J I H G F E D C B A

Correct Option: 7 Best Solution (4)

susmit   15 years AGO

matching of first letter tp last letter, second to second last and so on. also capital and small letters matched

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Dr.Dipin Singh   15 years AGO

using
z/Z in place of a/A
y/Y ........... b/B
x/X in place of c/C
.
.
c/C in place of x/X
b/B in place of y/Y
a/A in place of z/Z

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SURAJ   15 years AGO

Here A is repalced by Z, B with Y & so on...
SO code for xvYBUz = ceBYFa

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sujit   15 years AGO

count the letters from forward and the corresponding letter from back wards will replace it
A=Z
B=Y
C=X and so on
So xvYBUz = CebyFa

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