Age(time) problem

Fokin wrote:the first act was 18 minutes shorter than the third act and half as long as
the second act.

I have decided in the following way:

46*3=138
1 act =x-18
2 act= (x-18)/2
3 act= x

If third act = x mins long, first act = x - 18 mins
First one is half as long as second, meaning second one is double that of first.
So second act = 2(x-18). This is where u went wrong.

If you use this, you will get D.

n a certain performance of a 3-act play, the first act was 18 minutes shorter than the third act and half as long as
the second act. If the average (arithmetic mean) length of the 3 acts was 46 minutes, how many minutes long was the
third act?

(A) 30 (B) 39 (C) 46 (D) 48 (E) 66

soln:
let the first act be of x mins:
so third act will be of x+18 mins
second act 2x
sum=x+x+18+2x=4x+18
now arithmetic mean=sum/no of terms
46=(4x+18)/3
138=4x+18
120=4x
x=30
so third act was x+18=30+18=48

46*3=138
1 act =x-18
2 act= (x-18)/2
3 act= x
So, x-18+x/2 -9 +x= 138
5x/2= 165
5x=330
x=66
in your solution you are going wrong at the bold part
2 act =2(x-18) as first act is half the length of second act