The sum of 3 consecutive numbers of the four numbers A, B, C, D are 4613,4961,5010,5099 then what is the largest number among A,B,C,D ?
a) 1948 b) 1463 c) 1601 d) 1550

• Let, S1,S2,S3,S4 be the sums of A,B,C,D taking 3 of them at a time.
  So, by the given data in the problem,it follows that :
  S1+S2+S3+S4 = 4613+4961+5010+5099
  =>S1+S2+S3+S4 = 19683....[eqn 1]
  Now, Without the loss of generality we can assume
  S1= A+B+C
  S2= B+C+D
  S3= C+D+A
  S4= D+A+B
  -------------------------------------------------------
  adding the above terms we get,
  S1+S2+S3+S4 = 3(A+B+C+D)
  =>19683 = 3(A+B+C+D); [From [eqn 1] ]
  =>(A+B+C+D)=(19683/3)
  =>(A+B+C+D)=6561
  
  So,the greatest term among them =sum of all four - sum of lowest three terms
  So,the greatest term among them=6561-4613 = 1948.
  
  So,the greatest term among them = 1948
  
  So, the correct answer is : OPTION a)1948
• 11 years agoHelpfull: Yes(145) No(3)
• Let, S1,S2,S3,S4 be the sums of A,B,C,D taking 3 of them at a time.
  So,
  S1+S2+S3+S4 = 4613+4961+5010+5099
  =>S1+S2+S3+S4 = 19683
  Now,
  S1= A+B+C
  S2= B+C+D
  S3= C+D+A
  S4= D+A+B
  adding the above terms we get,
  S1+S2+S3+S4 = 3(A+B+C+D)
  =>19683 = 3(A+B+C+D); [From [eqn 1] ]
  =>(A+B+C+D)=(19683/3)
  =>(A+B+C+D)=6561
  =>the gretest term among them =sum of all four -lowest no of the series
  => 6561-1948
  =>1948 is the answer
• 11 years agoHelpfull: Yes(16) No(4)
• thanks, i understood the answer.
  but i have one question, from where can i find this kind of question (i.e. source). i am preparing for TCS , ans for that i am solving their papers. i did not come across to any such question.
  
  from where did you find this question!? let me know such question bank for new pattern of TCS aptitude.
  
  thanks.
• 11 years agoHelpfull: Yes(15) No(11)
• D value is largest 1948
• 11 years agoHelpfull: Yes(2) No(11)