A+B+C+D+E=FG, FG =10F+G
Ato G are different letters , if FG is as large as possible , what is value of G.

• a=9,b=8,c=7,d=6,e=5 total sum of fg=35
  
  take f=3 and g=5;
  fg=10*3+5;
  
  so it will be 5 for g value........
• 12 years agoHelpfull: Yes(24) No(17)
• 9+8+4+6+5=32, hence FG-32, => G=2
• 12 years agoHelpfull: Yes(16) No(2)
• vikas can u plz explain more clearly
• 12 years agoHelpfull: Yes(5) No(5)
• G=0;
  
  POSSBLE LETTERS ARE 0,1,2,3,4,5,6,7,8,9.
  SINCE "FG" IS MAXIMUM POSSBLE NO. IS MAY BE 98. BUT IF WE ADJUST REMAINING LETTERS FOR EXPRESSION A+B+C+D+E with 3+4+5+6+7; it will satisfy.
  So, "FG" to become larger no. try 4+5+6+7+8 = 30.
  10*3+0=10*3+0=30.
  so, value of G=0.
• 12 years agoHelpfull: Yes(4) No(16)
• FG=24
  FG=10F+G;
  HENCE
  FG=10*2+4
  FG=24
  ans G=4
• 12 years agoHelpfull: Yes(3) No(29)
• is the ans 1
• 12 years agoHelpfull: Yes(1) No(1)
• Vikas how you come to say that FG=24?
  
• 12 years agoHelpfull: Yes(0) No(1)
• a=5,b=6,c=7,d=8.
  so, 5+6+7+8=30.
  here then f=3, g=0.
  proof: given, fg=10f+g.
  30=10x3+g.
  hence, g=0.
• 12 years agoHelpfull: Yes(0) No(1)
• let a=4,b=5,c=6,d=7,e=8.
  so, 4+5+6+7+8=30
  here then f=3, g=0.
  proof: given fg=10f+g
  so, 30=10X3+g.
  hence g=0.
• 12 years agoHelpfull: Yes(0) No(1)