Que. Given A+B+C+D+E= FG. If FG is Maximum as possible and A,B,C,D,E,F,G all are distinct digits(0-9) then what is the value of G.
a) 1
b) 2
c) 3
d) 4
e) 5

• G=2
  atq, all are distinct digitd(0-9)
  9+8+7+6+5=35, not possible coz E=G=5
  9+8+7+6+4=34 not possible coz E=G=4
  9+8+7+4+5=33 not possible coz F=G=3
  9+8+4+6+5=32, possible coz all are distinct
  so, ans is 2
  
• 8 years agoHelpfull: Yes(23) No(2)
• A+B+C+D+E=FG
  9+8+7+6+5=35 Here E=G, So it is not possible.
  again 9+8+7+6+4=34 Here also E=G, so it is also not possible
  again 9+8+7+5+4=33 Here F=G
  Finally 9+8+6+5+4=32 In this case condition is satisfies, so all are distinct and value of G=2.
• 8 years agoHelpfull: Yes(12) No(2)
• 4+6+5+8+9=32 because all distinct and maximum . G=2
• 8 years agoHelpfull: Yes(3) No(0)
• 9+8+7+6+5+4=35
  G=5
• 8 years agoHelpfull: Yes(1) No(12)
• e)2
  9+8+6+5+4=32
• 8 years agoHelpfull: Yes(1) No(1)
• 9+8+7+6+5+2 = 31
  Since ABCDEFG all are distinct.
  G = 1
• 8 years agoHelpfull: Yes(1) No(4)
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  AS FG IS MAXIMUM AS POSSIBLE ,SO VALUE OF A+B+C+D+E=9+9+9+9+9=45=9*5
  SO VALUE OF G IS 5 AS 9 IS NOT GIVEN IN OPTION...
  
• 8 years agoHelpfull: Yes(1) No(19)
• 0+1+3+4+6=7*2
  G=2
• 8 years agoHelpfull: Yes(1) No(0)
• 3 a=8,b=7,c=1,d=6,e=5 f=9,g=3
• 8 years agoHelpfull: Yes(1) No(0)
• according to the question maximum value of A+B+C+D+E=9+9+9+9+9=45
  given A+B+C+D+E=FG
  now if FG=45 then the value of G will be 5 or 9
  but 9 is not given in the option so value of G is 5
• 8 years agoHelpfull: Yes(0) No(8)
• for max 5+6+7+8+9=35
  ans=5
• 8 years agoHelpfull: Yes(0) No(5)
• JHILIK SIL= A,B,C,D,E,F,G all are distinct so how it can possible 9+9+....
  
  HIMANSHU and VISWANADH : A,B,C,D,E,F,G all are distinct and your solution you have 5 for 2 times
  all of three can't be right...
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+5+2+0=6*4
  so 4 will be the value of g
• 8 years agoHelpfull: Yes(0) No(1)
• as all are distinct and we have to maximize so 1st solution should be
  9+8+7+6+5=35 (but here 5 exist twice,at E and at G position)
  so next target is to find 34(9+8+7+6+4) here also 4 exist twice
  so next target is to find 33(9+8+7+6+3) here 3 exist thrice but 33 can be found by summing 9+8+7+4+5
  so G is 3
• 8 years agoHelpfull: Yes(0) No(1)
• KAVITA i your answer is correct.
  
  Supratim Das : 9+8+7+4+5=33 so G=3 but here 3 is 2 times.
• 8 years agoHelpfull: Yes(0) No(1)
• 0+2+3+4+6=15
  then g=5
• 8 years agoHelpfull: Yes(0) No(0)
• 0+1+7+8+9=25=fg
  g=5
• 8 years agoHelpfull: Yes(0) No(2)
• ans is 4
  
• 8 years agoHelpfull: Yes(0) No(2)
• ans= e)5
  A=9
  B=8
  C=7
  D=6
  E=5
  A+B+C+D+E=35
• 8 years agoHelpfull: Yes(0) No(3)
• 9+8+6+5+4+0=32
  2=so,ans
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+6+5+4=32
  ans=2
• 8 years agoHelpfull: Yes(0) No(0)
• 3bcz root of 9
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+7+4+2=30 ie
  6*5 hence ang is 5
  here all digits are different
• 8 years agoHelpfull: Yes(0) No(0)
• 3 is answer
  
• 8 years agoHelpfull: Yes(0) No(0)
• Answer is b:
  Use trail and error method..by giving the last number(G) in the options in the last as one..and giving the nearest possible sum.of their 10th digit place number(assume 3) then by filling the values of E D C B A and adding the by assigning the biggest number to the last one then we get 3 repeated..then we keep 2 in the last position and follow the same procedure..we get the answer by second trail itself..9+8+6+5+4=32.
• 8 years agoHelpfull: Yes(0) No(0)
• here A+B+C+D+E= 5+ 6+ 4+ 8+ 9=32
  NOW G=2
• 8 years agoHelpfull: Yes(0) No(0)
• A,B,C,D,E=9,8,7,6,5soFG=35thenG=5
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+7+6+5=35 wrong
  9+8+7+6+4=34 wrong
  9+8+7+4+5=33 wrong
  9+8+4+6+5=32 right
  ans is 2
• 8 years agoHelpfull: Yes(0) No(0)
• 5 AS MAXIMUM IS NEEDED SO 9+8+7+6+5=35.... 35=5*7 SO ACCORDING TO THE OPTION ..... 5 IS ANSWER..
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+7+4+5=32
• 8 years agoHelpfull: Yes(0) No(0)
• max 5+6+7+8+9=35
  ans=5
• 8 years agoHelpfull: Yes(0) No(0)
• 5: 5+6+7+8+9
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+6+5+4=32
  => G=2
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+7+6+5=35
  f=3
  g=5
• 8 years agoHelpfull: Yes(0) No(0)
• distinct and max as possible so 9+8+7+6+5=35 so g=5
• 8 years agoHelpfull: Yes(0) No(6)
• 1+5+6+7+8=9*3
  ans=3(c)
• 8 years agoHelpfull: Yes(0) No(5)
• 9+8+7+6+5=35...... E=G
  9+8+7+6+4=34.......E=G
  9+8+7+6+3=33.......F=G
  9+8+7+6+2=32.... G=2 (b)
• 8 years agoHelpfull: Yes(0) No(1)
• 9+8+6+3+2=28
  28=7*4
  hence, 9+8+6+3+2=7*4
  According to the question all the digits should be distinct and must lie between 0-9
  Therefore F=7, G=4
  Ans: Option (D)=4.
• 8 years agoHelpfull: Yes(0) No(0)
• option (e)- i.e. 5
• 8 years agoHelpfull: Yes(0) No(0)
• 9+8+7+5+2=31 so according to me answer should be a) 1.
• 8 years agoHelpfull: Yes(0) No(1)