there is a list of the 7 positive integer,such that (n+1)th number is always greater or equal to the 6th number.the average of the first five number is 15, and the average of the last five number is 22.What can be the maximum value of 5th number in the sequence
(a)22 (b)23
(c)28 (d)29

• a1,a2,a3,a4,a5,a6,a7
  a1+a2+a3+a4+a5=5*15=75
  a3+a4+a5+a6+a7=5*22=110
  (n+1)>=n
  a1=a2=a3=a4=a......(1)
  a6=a7=a5............(2)
  from 1 and 2
  2a+3a5=110
  4a+a5=75
  a5=29 (ans)
• 10 years agoHelpfull: Yes(15) No(7)
• let the no. be : a1 a2 a3 a4 a5 a6 a7
  Given a1+a2+a3+a4+a5=75--(i); a3+a4+a5+a6+a7=110;--(ii) we have to find the max. value of a5.
  Now, as it is given in question term(n+1)>=term(n).
  So, to find max value of a5 we have to set a5,a6,a7 equal.
  when u will put 29, then should satisfy (a3+a4)=23,
  So, at the max values for that would be (12+11).Now it must also satisfy(a1+a2)=24. But the max values for a1 & a2 can not be greater than 11. so a1=a2=11. This does not make total 75.
  
  now when we will check that for 28, It will satisfy the condition.
  Thus answer is : 28.
  
• 10 years agoHelpfull: Yes(9) No(5)
• Ans. c)28
  Suppose Numbers are A,B,C,D,E,F,G
  Now we have
  A+B+C+D+E=15*5=75...............(1)
  C+D+E+F+G=22*5=110..............(2)
  From 1&2 we can have (F+G)-(A+B)=35.........(3)
  Now hit n try for options starting from maximum
  Case1: if E=29
  then min(F+G)=58 and C+D=23
  A+B=23(satisfying eq.1 & eq.3)
  but for the given condition in question B
• 10 years agoHelpfull: Yes(6) No(0)
• guys those who have given elitmus exam pls forward me the study material watever u have on anjalilathiya@gmail.com...pls..i have test this sunday...n also tell me wat is the probability of these same questions geting repeated.
• 10 years agoHelpfull: Yes(1) No(22)
• n
• 10 years agoHelpfull: Yes(1) No(1)
• Ans is 23..
  a1,a2,a3,a4,a5,a6,a7(all are +ve integers)
  
  a1+a2+a3+a4+a5=5*15=75------(1)
  
  a3+a4+a5+a6+a7=5*22=110-----(2)
  
  n
• 10 years agoHelpfull: Yes(0) No(1)