.banana+guava=orange.what is the value of o+r+a+n+g+e??

• BANANA
  +GUAVA
  ---------------
  ORANGE
  --------------
  
  means A+A=E, N+V=G, A+A=N, N+U=A, A+G=R, B+_=O
  
  So just start from last
  A+A=E
  "A" can have d values ranging {0,1,2,3,4,5,6,7,8,9}
  first lets take A=0... A+A= 0+0=0... Again d result is 0... means E=0... So not possible for 2 alphabets to have same number...
  
  Lets take A=1... A+A= 1+1=2... A=1, E=2,... v have to find d value for "N"...
  Look A+A=N (3rd term)... A+A is already "E" (ie E=2A)... So definitely N should not be 2A... Therefore N=2A+1... There should be remainder from the 2nd term...
  so N=2A+1... N=2(1)+1= 3... Assign a value for V to get a remainder... Take V=7
  3+7=0 carry 1... 1+A+A=3...
  next N+U=A...result should be 1... 3+8=1 carry 1...
  next A+G=R... 1+1+7=9... There is no carry which means B nd O ll have d same value... so not possible... So try for different values...
  Note here A+G=R should produce a carry... then only v can yield B+carry=O
  
  By trial nd error method, Again start from last... Try A=2... Its also not possible... Agan A=3... Not possible...
  Now A=4... E=2A=8... N=2A+1=9...
  For getting a carry If i take V=1, A+G ll not produce carry... So take V=2... N+V=G... 9+2=1 carry 1... 1+A+A=9... N+U should produce 4... 9+5=4 so U=5... next A+G = 4+1=5 not producing carry... Here 4+(>5) only yields a carry...
  So v should get G value >5...
  So again trail nd error method, N+V=G... Lets take V=7 ..N+V=G... 9+7=6 carry 1...
  1+A+A=N... N+U=A... 9+5=4 carry 1... 1+A+G=R... 1+4+6=1 carry 1... 1+B+_=O...
  Values remaining is {0,2,3}
  1+0+_=1 not possible
  1+2+_=3
  So
  B=2, A=4, N=9, A=4, N=9, A=4
  G=6, U=5, A=4, V=7, A=4
  O=3, R=1, A=4, N=9, G=6, E=8
  
  O + R + A + N + G + E
  3 + 1 + 4 + 9 + 6 + 8 = 31
  
  Ans : 31
  
• 10 years agoHelpfull: Yes(36) No(1)
• make me understand ,, how you are assigning values to alphabets... how this will work.. please tell.. gerrally BANANA means 21141121 becoz B stands on 2nd pos.,A on 1 and N on 14....how can we assign other values to alphabets except positioning....
• 10 years agoHelpfull: Yes(7) No(3)
• please explain me how this type of questions can be solved...
• 10 years agoHelpfull: Yes(2) No(0)
• Ans is 31:
  By trial and error method::
  &
  banana+guava=orange
  placing a=4,we get 4+4=8,i.e e=8,
  taking n=9,v=7,g=6,u=5,b=2 and adding
  249494+65474=314968,adding this digits:3+1+4+9+6+8=31
• 10 years agoHelpfull: Yes(1) No(0)
• B A N A N A
  1 2 5 2 5 2
  __G U A V A
  + 3 7 2 8 2
  **************
  O R A N G E
  1 6 2 5 3 4
  
  THEN O+R+A+N+G+E=1+6+2+5+3+4=21
• 10 years agoHelpfull: Yes(1) No(6)
• B A N A N A
  G U A V A
  ---------------------
  O R A N G E
  
  as you can see in one sight O - B = 1 and N and E also consecutive. N - E = 1
  We have to hit and try using A={1,2,3,,4,5,6,7,8,9} all possible values.
  and A also can't be zero.
  
  At first sight it seems very tough to test for every value of A. BUt as there are a lot of interrelations between digit. wrong assumptions can be caught very easily.
  I am demoing here for A=1.
  
  
  B 1 3 1 3 1
  G U 1 V 1
  ---------------------
  O R 1 3 G 2
  
  From the above, 1 + G or 1 + 1 + G should be
  >10. So G can be either 8 or 9. But 1 + 1 = 3 is not possible. So A ≠≠ 1.
  Similarly A ≠≠ 2 as well as 3. For A = 4
  
  B 4 9 4 9 4
  G U 4 V 4
  ---------------------
  O R 4 9 G 8
  
  From the above, U = 5 and G is 6 for V = 7.
  
  2 4 9 4 9 4
  6 5 4 7 4
  ---------------------
  3 1 4 9 6 8
  
  So O+R+A+N+G+E=3+1+4+9+6+8=31
• 8 years agoHelpfull: Yes(0) No(0)