this seems weird:i added all the given ones: A+B+D+F+H+2(C+E+G+I) = 13*4 = 52and A+B+C+D+E+F+G+H+I = 1+2+..9 = 45diffing these 2, C+E+G+I = 52-45 = 7and there are no 4 integers in the set {1,2,...9} such that their sum is 7.....am i going wrong somewhere...?
Directions for Qs. 1 to 2: Refer to the following data and answer the following questions.A B C D E F G H IEach of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 is represented by a different letter in the figure.A + B + C, C + D + E, E + F + G and G + H + I is equal to 13.1. Which of the digits does E represent?(1) 9 (2) 4 (3) 7 (4) 12. Which of the digits does D represent?(1) 8 (2) 7 (3) 3 (4) 7 or 8
c+e+g=7.... so ecombinations of 3 digits whose sum is 13 are.... 1,3,9; 1,4,8; 1,5,7 2,3,8; 2,4,7; 2,5,6 3,4,6 c, e, g has to be from 1,2,4 we know that 'e' forms combination with both c and g (C + D + E, E + F + G) from the combinations we can see that only 4 forms combination with 1 and 2.....so e=4
for the second problem we just need to substitute e=4 and check whether there is any inconsistency.....i got that even if d=7 or 8, there is no inconsistency in the equations....i am doubtful...
c+e+g=7.... so ecombinations of 3 digits whose sum is 13 are....1,3,9; 1,4,8; 1,5,72,3,8; 2,4,7; 2,5,63,4,6c, e, g has to be from 1,2,4we know that 'e' forms combination with both c and g (C + D + E, E + F + G)from the combinations we can see that only 4 forms combination with 1 and 2.....so e=4for the second problem we just need to substitute e=4 and check whether there is any inconsistency.....i got that even if d=7 or 8, there is no inconsistency in the equations....i am doubtful...
Directions for Qs. 1 to 2: Refer to the following data and answer the following questions.A B C D E F G H IEach of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 is represented by a different letter in the figure.A + B + C, C + D + E, E + F + G and G + H + I is equal to 13.1. Which of the digits does E represent?(1) 9 (2) 4 (3) 7 (4) 12. Which of the digits does D represent?(1) 8 (2) 7 (3) 3 (4) 7 or 8
A+B+C+D+E+F+G+H+I = 45 ----- Eqn 1
A+B+D+F+H+I + 2 ( G + E + C ) = 52 ---- Eqn 2
Subtracting we get G + E + C = 7 --------- Eqn 3
Now C +D = F + G & E + F = H + I ------- Eqn 4 & 5
Also (C + D + E) + (E + F + G) = 26
( D + C ) + 2 E + ( F + G ) = 26 ---- Eqn 6
LET D + C = F +G = $
now 2 $ + 2 E = 26 = > $ + E = 13 ----- eqn 7
We can see that E will always have EVEN VALUE since $ + E = odd ( 13 )
i.e E can be either 2 or 4 ( since E has to be less than equal to 4 )
going by the options E = 4....
since E = 4 ; D + C = 9 ; D can be 7 or 8.......since 3 is ruled out....
bhai... approach batao....mujhe toh Question hi samaj nahi aaya
Even i could not make heads/tails of it initially
Will have to recheck the approach
When you rotate the inner wheel 2 times to the left..you get 8 instead of A
So the 1st answer starts with 8..the right wheel is rotated 2 steps to the left.
So instead of A you get C..for the word APPLE i used the system where i went from the inner wheel to the outer till the letter APP and then back again in the inner wheel
For the second question..i have no clue
For the last one..we have HOCJ9
I wrote the letters before and after each of the letters in HOCJ9
So i got 2 series GNBIA and IPDK8
From these 2 series we have to make a meaningful word such that the inner has to be moved left/right and the outer wheel has to be moved left/right
