Latest Elitmus Aptitude Question SOLUTION: ABM PMV --------- XMSW XFMB BXSA ---------- XFFWPW

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(#M40152761) ELITMUS Logical Reasoning question Cryptography Keep an EYE

ABM
PMV
---------
XMSW
XFMB
BXSA
----------
XFFWPW

Asked In Elitmus chayan paria (8 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

Sourabh Prasad  (8 years ago)

648
*789
------------
5832
5184*
4536**
-----------
511272

Is this solution Helpfull? Yes (4) | No (1)  

Ankur Singh Panharia  (8 years ago)

648
*786
--------
5832
5184*
4536**
-----------
511272

Is this solution Helpfull? Yes (3) | No (2)  

JUGAL KHAJURIYA  (8 years ago)

Ans 648
*789
-----------

Is this solution Helpfull? Yes | No   

mohit mehra  (8 years ago)

Please explain your answer properly to make it approve?
hint for this comes by ABM * M = XFMB
so, first of all we have to find the M
by putting M as 1,2,3,4,7,8,9
by checking all possibilities we get M as 8 which full fill our need,
after that we get A as 6, B as 4, M as 8
as see that as we have another hint in last product
B + 1 = X,so we just find the B then we get X as 5
In 648 * V = XMSW(58SW)
by putting V as 9 becoz we use atleast all no. thats y I use 9 by using it we get our solution as below :
648
*786
--------
5832
5184*
4536**
-------
511272

Is this solution Helpfull? Yes | No (1)  

chandan kumar  (8 years ago)

here m*m = b at unit digit
so m may be 2,3,4,7,8,9 and corresponding b is , 4,9,6,9,4,1 further

b+1=x so x may be 5,7,5,2 at corresponding m = 2,4,8,9

now x is at 1st place in 1st and 2nd multiplication and x+x generets a carry =1 so x may be 5, or 7

in 2nd multiplication m*a= 5f it only possible if m=8

so we have m=8,b=4,x=5

now x+x +carry=f => 10+1=11 => f=1 ,=>s=3 ,and so that


after some further calculation

solution is 648*789=511272

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rakesh kumar  (8 years ago)

648
789
-------
5832
5184
4536
---------
511272

Is this solution Helpfull? Yes | No   

Mayank Sharma  (8 years ago)

1) As you can see that,
M*M=B, M*B= M
is only possible when M=8 and B=4
2) M*V can be one of 0,2,6
(B=4 and M=8 is already used)
3) Now put different values of
V from (2,3,5,6,7,9) such that
W is 0 or 2 or 6, and S should not be 4
or 8.
4) As putting V=9, we got S=3 and W=2.
5) Now,
S+B=P
or, 3+4=7
so, P is 7
6) Now you can see that,
M+M+A=W is only possible when
A is 6
7) We got A=6, B=4, M=8, V=9, P=7
and W=2
8) By putting all the values we got,

6 4 8
7 8 9
_____________
5 8 3 2
5 1 8 4
4 5 3 6
_______________
5 1 1 2 7 2

Is this solution Helpfull? Yes | No (1)  

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