The average (arithmetic mean) of four different positive integers is 9. If the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer, what is the highest possible value of the fourth integer?

(A+B+c+D)/4=9
A+B+c+D=36
A=3B, B=C-2 therefore C=B+2
putting in equation
3B+B+B+2 +D=36
5B+D=34
D=34- 5B
To maximize D we need to minimize 5B
Now we know all are positive so Minimum B = 1
therefore Maximum D =29 (34-5(1)) but in this case A and C will have same value
so lets PUT B=2 so
D=24

StrugglingGmat2910 Archit3110

At b=1, a=c=3.

It's mentioned in the question that 4 numbers are distinct.

At b=2, a=6; c=4; and d=24 (this case satisfies all the conditions)

You can't have a case in which D greater than 24 and satisfies all the conditions mentioned in the questions. Anyways, you can wait till the OA gonna revealed.

I think nick1816 is right.
There is restriction on the selection of numbers.. i.e first the numbers should be different & secondly they should be +ve . The maximum value of D is to be arrived keeping these two restrictions in mind. You are ignoring "different numbers" restriction. Similarly, if we ignore +ve numbers restrictions, we may get even higher value of D like considering b as -ve
d= 34-5b
d= 34-5(-1) = 34+5 = 39.. even higher value of D.

So, i think both restrictions are to be followed--- DIFFERENT & +VE..

a,b,c,d=9*4=36 (distinct positive integers)
a=3b, c=b+2;
3b+b+2+b+d=36
5b+d=34
b=34-d/5
d=4,9,…24,29

d=29: b=34-29/5=1, a=3b=3, c=b+2=3, a=c, invalid;
d=24: b=34-24/5=2, a=6, c=4, valid.

Ans (E)