Q. A, B, C, D, E and F are six positive integers such that
B + C + D + E = 4A
C + F = 3A
C + D + E = 2F
F = 2D
E + F = 2C + 1
If A is a prime number between 12 and 20, then

1. The value of F is
(A) 14
(B) 16
(C) 20
(D) 24
(E) 28

2. Which of the following must be true?
(A) D is the lowest integer and D = 14
(B) C is the greatest integer and C = 23
(C) B is the lowest integer and B = 12
(D) F is the greatest integer and F = 24
(E) A is the lowest integer and A = 13

• 0 down vote
  
  
  B + C + D + E = 4A ------1
  
  C + F = 3A ----------------2
  
  C + D + E = 2F -----------3
  
  F = 2D --------------------4
  
  E + F = 2C + 1-----------5
  
  Also A can 13,17 or 19
  
  Eliminating D from all the equations we get
  
  B+2F = 4A ------6
  
  C+F = 3A -------7
  
  C+E = 3F/2 -----8
  
  E+F = 2C+1 ----9
  
  Adding 7 and 8
  
  2C+E+F = 3(A+F/2)
  
  4C+1 = 3(A+F/2)
  
  Substituting value of C from 7
  
  4(3A - F) + 1 = 3(A+F/2)
  
  => A = (11F - 2)/18 => Only for A = 17, F is an integer
  
  Therefore A = 17, F = 28, C = 23 , D = 14, B = 12, E = 19
  
  1.) F = 28
  
  2.) B = 12 and the lowest
  
• 11 years agoHelpfull: Yes(25) No(2)
• 1. Solve through options..it would be easy and would take less time. Thus,answer is (E)28.
• 11 years agoHelpfull: Yes(5) No(3)