1. If the equation A+B+C+D+E=FG is two digit number, whose value is 10F+G and the letters A,B,C,D,E,F and G each represents different digits.If FG is as large as possible, what is the value of G?
(a)4 (b)5 (c)1 (d)3

• 9+8+7+6+5=35 is not possible..
  so now we can go for 34 which is not possible for any coimbination..
  and next for33 is not possible..
  and nnow we can go for 32 =9+8+6+5+4 is possible..therfore, 2 is the answer..
• 12 years agoHelpfull: Yes(22) No(16)
• 9+8+7+5+2=31 then anser is 1
• 12 years agoHelpfull: Yes(16) No(5)
• 9+8+7+5+2=31..
  ans=1
• 12 years agoHelpfull: Yes(4) No(2)
• Given That A,B,C,D,E,F are different digits, so assume the values of them as 4,5,6,7,9 on added them as 4+5+6+7+9=31 which represents 3 for F and 1 for G..and the values are not same of the others....so the ans is 1
• 12 years agoHelpfull: Yes(3) No(3)
• ans is 2
  9+8+4+6+5=32
  
• 12 years agoHelpfull: Yes(3) No(3)
• the ans for g is G=5
  for simple just go through option if we consider 4 it will not equal when u sub no. for a=0,b=2,c=3,d=4,e=5 which is not equal to FG. Befor proceeding this fix the values for f=1 and A=0 and if we consider option (b)5 sub in equ 10(1)+5=15, and sub values for a=0,b=2,c=3,d=4,e=6 and the sum is 15 and FG is 15 so ans is 5...
• 12 years agoHelpfull: Yes(3) No(6)
• 9+8+7+6+5=35
  10(f)+g
  10(3)+5=35
  as,FG should be as large as possible,so 5 ans
• 12 years agoHelpfull: Yes(2) No(13)
• guys wer it is mentionened u tak values less than 10.
  4+6+7+8+10=10(3)+5=35 so 5 ans
• 12 years agoHelpfull: Yes(2) No(1)
• a=9,b=8,c=7,d=6,e=5 so fg=35
  10f+g=35 so f=3 and g=5.. i guess so
• 12 years agoHelpfull: Yes(1) No(16)
• 9+8+1+5+2=31 where FG=31 i.e; f=3,g=1.
• 12 years agoHelpfull: Yes(1) No(3)
• 9+8+6+5+4=32
  ie g=2
• 12 years agoHelpfull: Yes(1) No(4)
• FG is as large as possible and all the 7 numbers should be different.
  By trial and Error method,
  9 + 8 + 7 + 6 + 5 = 35…5 is getting repeated twice.
  9 + 8 + 7 + 6 + 4 = 34…4 is getting repeated
  9 + 8 + 7 + 5 + 4 = 33…3 repeats
  9 + 8 + 6 + 5 + 4 = 32
  None of the numbers repeat in the above case and 32 is the maximum number FG can have. The value of G is 2.
• 10 years agoHelpfull: Yes(1) No(0)
• 3+4+5+6+9=27 ans
• 12 years agoHelpfull: Yes(0) No(4)
• 9+8+7+5+4=33 so the ans is 3
• 12 years agoHelpfull: Yes(0) No(5)
• largest possible is 2..
  but as per the option 1
• 12 years agoHelpfull: Yes(0) No(1)
• ans will be 31.bcoz
  9+8+7+6+5=35 now we need to take f and g as different digit like (1,2,3,4),but by using it we can not get 35 value.similarly for 32,33,34
  but for 31 =9+8+7+5+2=10*3+1. which is true .so ans will be 1 for g.
  
• 12 years agoHelpfull: Yes(0) No(0)
• the sum should b maximum ,n digits r diffrerent.
  thr is posibility wth 32 instead of 31,so g=1 is not correct
  9+8+6+5+4=32=10(3)+2,
  g=2,
• 12 years agoHelpfull: Yes(0) No(0)
• 10+9+11+13+8=51
  10*5+1=51
  
• 10 years agoHelpfull: Yes(0) No(0)
• 9+8+7+5+4 =33
  The ans is 3
• 9 years agoHelpfull: Yes(0) No(0)