Latest Infosys Aptitude Question SOLUTION: Q. If AFTER+TWO=YEARS then what is the value of Y+E+A+R+S?

A F T E R
- - T W O
---------
Y E A R S

4 9 7 0 3
- - 7 2 8
----------
5 0 4 3 1

guys if u notice the question then u'll get that "F = E" it means the value of 'F' is changed to 'E' and it is possible when the previous iteration carry 1.
but in the nxt iteration, value of 'A' is also changed to 'Y' it means the previous iteration i.e 'F' carry 1, and the previous iteration have only one alphabet i.e 'F' and its carry 1 then the value of 'F = 9' and whatever value will come from previous iteration i.e 'T + T' must carry 1 then it will satisfy the values "F=9,E=0"now put these value in the respective alphabet.

now check the last column i.e "R+O=S" also check the nxt column too i.e "E+W=R" and u know the value of 'E=0' from here u've to find 'W=R' and if u notice then u'll identify that the value if 'W' is changed to 'R' it means its carry 1 from its previous iteration i.e "R+O=S" then the value of 'S' must be greater than 10 then it will carry 1.
now check what are the possible value will give more than 10 and remember 9 and 0 already assign to F and E respectively. So '8+2=10' is not possible,
'8+3=11' it is possible and it carry 1 then assign the value of "R=3,O=8 and S=1"
now check 'E+W=R' u've "E=0,R=3" from here u'll get the value of 'W=2'.

Now check "T+T=A" no matter whatever value of 'A' will come.u've to remember that 'T+T=A' must carry 1.
if u put '5+5=10' i.e '0' not possible 'E=0' already assign,
'6+6=12' i.e '2' not possible 'W=2' already assign,
'7+7=14' i.e '4' it is possible.
then take the value of "T=7,A=4 and Y=5"

if still u've any doubt then ask me... :)

PLEASE TELL ME HOW U GUYS SOLVE THIS TYPE OF QUESTIONS QUICKLY?IS THERE ANE SHORT TRICK?PLEASE...

A F T E R = 4 9 7 0 3
T W O = 7 2 8
Y E A R S =5 0 4 3 1

49703 + 728 = 50431

So, Y+E+A+R+S = 5+0+4+3+1 = 13

google "Verbal_arithmetic wikipwdia". there is a complete example. after understand that try simple question where correcter will be repeat more. then go through cryptography puzzles of elitmus on m4maths. Sorry but there is no short trick. Even I learned this thing 3 days before.

after = 29608
+too = +677
years = 30285
so Y+E+A+R+S = 3+0+2+8+5 = 18

i cant really understand this probs... guys pls tech me how 2 solve these step by step...its my request. !!!

AFETR+TWO=YEARS
49703+728=50431
Y+E+A+R+S=13

EXPLANATION:: (MY APPROACH)
OKK when you'll write after+two=years in general addition convention(first write 'after' then write 'two' below it and draw two parallel lines then write 'years' ("i hope this much is clear"))then you can notice that there is nothing below 'AF' but in summation part they get changed into 'YE' respectively that means they are changed and only possible way is by carry.(carry can be 1 only in such type okk i hope u r getting it)
'F' will receive carry from adding 'T'+'T' but what about 'A'. just observe it and there is only one possible number is there between 0-9 to which when u add 1, that will generate carry. yess that is 9 so 'F' is 9; 9+carry(ie 1) will make 'E' as 0.
now you know values of total 3 places in question.
notice that 'E'+'W'='R' and 'E' is 0 here again there is case of carry because in summation part there is 'R'. this sum can not generate any carry because 0+(any number of range 0-9 will be that number only "simple na ??")

## this is my approach. now make some kind of cases and you will surely get the result.

hey plz tell the very first step to last step to solve these type of questions?

It seems both answer of Ganesh and Upendra are correct so which one i have to select in exam?

@ganesh u told that 8+3 = 11 which is greater than 10. so we can also take 7+4 = 11 which is also correct.????

Solving a cryptarithm by hand usually involves a mix of deductions and exhaustive tests of possibilities. For instance, the following sequence of deductions solves Dudeney's SEND + MORE = MONEY puzzle above (columns are numbered from right to left):
S E N D
+ M O R E
============
M O N E Y
1>From column 5, M = 1 since it is the only carry-over possible from the sum of two single digit numbers in column 4.
2>Since there is a carry in column 5, O must be less than or equal to M (from column 4). But O cannot be equal to M, so O is less than M. Therefore O = 0.
3>Since O is 1 less than M, S is either 8 or 9 depending on whether there is a carry in column 4. But if there were a carry in column 4, N would be less than or equal to O (from column 3). This is impossible since O = 0. Therefore there is no carry in column 4 and S = 9.
4>If there were no carry in column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1.
5>If there were no carry in column 2, then ( N + R ) mod 10 = E, and N = E + 1, so ( E + 1 + R ) mod 10 = E which means ( 1 + R ) mod 10 = 0, so R = 9. But S = 9, so there must be a carry in column 2 so R = 8.
6>To produce a carry in column 2, we must have D + E = 10 + Y.
7>Y is at least 2 so D + E is at least 12.
8>The only two pairs of available numbers that sum to at least 12 are (5,7) and (6,7) so either E = 7 or D = 7.
9>Since N = E + 1, E can't be 7 because then N = 8 = R so D = 7.
10>E can't be 6 because then N = 7 = D so E = 5 and N = 6.
11> D + E = 12 so Y = 2.

can u please explain clearly

please explain the logic relating to these type of q.s

@GOURAV yeah u can take '7+4=11' but after assigning these value u'll get two alphabet having same value which is not possible. So, '7+4=11' is not possible.

@ganesh ,man do u have solution for this,world+trade=center.value of c+e+n+t+e+r=??.

@ NANDISH

W O R L D + T R A D E = C E N T E R
5 3 6 8 4 + 7 6 0 4 2 = 1 2 9 7 2 6

C+E+N+T+E+R = 1+2+9+7+2+6 = 27

29607
+ 674
30281

therefore 3+0+2+8+1=14

29604(AFTER)+647(TWO)=30251(YEARS) is also correct.
So, 3+0+2+5+1=11