Problem Solving

in 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-act was 46 minutes, how many minutes long was the third act?

A. 30
B. 39
C. 46
D. 48
E. 66

The OA is D.

Please, can anyone explain this PS question for me? I can't get the correct answer. I need your help. Thanks.

swerve wrote:in 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-act was 46 minutes, how many minutes long was the third act?

A. 30
B. 39
C. 46
D. 48
E. 66

Total number of minutes = (number of acts)(average number of minutes per act) = 3*46 = 138.
We can PLUG IN THE ANSWERS, which represent the time for the third act.
When the correct answer is plugged in, the total number of minutes = 138.

B: 39 minutes
Since the first act is 18 minutes shorter than the 39-minute third act, the first act = 39-18 = 21 minutes.
Since the 21-minute first act is 1/2 as long as the second act, the second act = 42 minutes.
Total time = 39+21+42 = 102 minutes.
The total number of minutes is TOO SMALL.
Eliminate A and B.

B: 48 minutes
Since the first act is 18 minutes shorter than the 48-minute third act, the first act = 48-18 = 30 minutes.
Since the 30-minute first act is 1/2 as long as the second act, the second act = 60 minutes.
Total time = 48+30+60 = 138 minutes.
Success!

The correct answer is D.

Algebra:

Let t = the third act.
Since the first act is 18 minutes shorter than the third act, the first act = t-18.
Since the first act is 1/2 as long as the second act, the second act is twice as long as the first act:
2(t-18).
Since the total number of minutes for all 3 acts = 138, we get:
t + (t-18) + 2(t-18) = 138
4t - 54 = 138
4t = 192
t = 48.