ROckHIsT wrote:
In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals
Attachment:
findginsum_figure.png
A. 2
B. 3
C. 4
D. 5
E. 6
How to tackle such questions? Do we have more question like this?
Sol: Here we are adding 2 nos and get a 3 digit number
When you look at the given nos we see that E has to be 1 only
Now let's do back calculation and see where end up
Consider G= 2
then we can have possible values of B&D (1,1 or0,2 or 2,0) But then each digit can have only 1 value So ignore G
Consider G= 3 then we have B&D (1,2 or 2,1 or 3,0) Now 1,2 and 2,1 can be ruled out cause we said E has to be 1 then we have a case 3,0 but then A=3 and B=0 then G=3 which is not possible
Consider G=4 then we have B&D (2,2 or 1,3 or 3,1 or 4,0) Consider 4,0 again if A=4 then B=0 and G=4 which is not possible as each Alphabet represent only single digit
Consider G=5 then we have possible values of B&D (2,3 or 5,0 or 4,1 or1,4) Only case is 2,3 Keep G=5 and the only possible nos
62+43= 105
We can stop here but just to check
Consider G=6 then we have (3,3 or 4,2 or 1,5 or 6,0) Now 1,5 and 6,0 can be ruled out. Consider 4,2
X4+Y2= 1A6 Which is impossible because we will need bigger nos let's say 7 8 and 9 as possible values of X and Y but these values are restricted from Q stem.
So ans has to be D
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