Required solution to an aptitude question

Question

The following is an aptitude problem (question no: 29-32), I am trying to solve:-

Questions 29 - 32:

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

1. 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

Now there are 5 equations to solve 6 variables. So I am at a loss on how should I start solving the problem? Any help from anybody is appreciated.

Answer 1

Keep substituting and work with the equations to end up having equations for the other variables in terms of A, i.e. something like B = 10A+3. There are only certain values of A that are a prime number between 12 and 20. For those values of A, using your equation for F, you can compute the possible values of F. Once the noninteger or nonpositive values are removed there should be only one value for F. Likewise the second question can be answered.

Answer 2

F=2D implies F must be even.
E+F=2C+1 implies E+F must be odd, hence E must be odd (as F is even).
Let E=2q+1, hence 2q+1+F=2C+1, hence 2q+2D=2C, hence C=q+D
C+D+E=2F, hence q+D+D+2q+1=2(2D), hence 3q+1=2D, hence q must be odd.
Let q=2r+1, hence 6r+4=2D, hence D=3r+2.

Summarising so far, we get:
D=3r+2
F=2D=6r+4
E=2q+1=4r+3
C=q+D=5r+3

C+F=3A, hence 5r+3+6r+4=3A, hence 11r+7=3A, hence 11(r-1)+18=3A, hence r-1 must be a multiple of 3.
Let r-1=3s, hence 11(3s+1)+7=3A, hence 33s+18=3A, hence A=11s+6.

Summarising so far, we get:
D=3r+2=9s+5
F=6r+4=18s+10
E=4r+3=12s+7
C=5r+3=15s+8
A=11s+6

Finally B+C+D+E=4A, hence B+15s+8+9s+5+12s+7=4(11s+6)

Hopefully you can finish off from here.

Answer 3

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