Elitmus
Exam
Logical Reasoning
Cryptography
FMA
AXX
-----------
KXFV
KXFV
VZML
------------
MKXQWV
Read Solution (Total 7)
-
- Question is wrong right question is
FMA
AXX
-----------
KXFV
KXFV
VZML
------------
MKXQZV
Now solution as we can see that
v+1=m
k+z=k this is only possible if z=0 or z=9 but we are getting a carry in v+1=m it means that z is greater no and hence it is 9
now
FMA
AXX
-----------
KXFV
KXFV
V9ML
------------
MKXQ9V
now u can see that
f+v=9 it is possible with values like (1,8),(2,7),(3,6),(4,5)
by hit and trail (1,8) not possible
go for now (2,7)
7ma
a
-------
29ml
--------
now by hit and trial a=4
7m4
4
--------
29m6
--------
again by hit and trail
m=3
734
4
-------
2936
-------
now writing whole equation
734
4XX
---------
KX72
KX72
2936
------------
3KXQ92
------------
now X can be 3 or 8 but 3 is occupied by M so X=8
734
488
------
5872
5872
2936
_________
358192
- 8 years agoHelpfull: Yes(10) No(2)
- 734
488
------
5872
5872
2936
_______
358192
- 8 years agoHelpfull: Yes(1) No(1)
- As I remember, there is Z in place of W.
734
488
-------
5872
5872
2936
_______
358192 - 8 years agoHelpfull: Yes(1) No(2)
- 734
488
-------
5872
5872
2936
--------
358192 - 8 years agoHelpfull: Yes(0) No(3)
- 734
488
-------
5872
5872
2936
-----------
358192 - 8 years agoHelpfull: Yes(0) No(1)
- Thank you, Ravi for your solution
- 7 years agoHelpfull: Yes(0) No(0)
- I gotvboth z and w values as 9 which is not accord to rule. Will any one explain it and correct answer
- 7 years agoHelpfull: Yes(0) No(0)
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