In a certain three-act play, the first act was 18 minutes shorter than the third act and half as long as the second act. If the average length of the three acts was 46 minutes, how many minutes long was the third act?16 Mark For ReviewA) 30B) 39C) 46D) 48
Question
In a certain three-act play, the first act was 18 minutes shorter than the third act and half as long as the second act. If the average length of the three acts was 46 minutes, how many minutes long was the third act?16 Mark For ReviewA) 30B) 39C) 46D) 48
Solution
Let's denote the length of the first act as F, the second act as S, and the third act as T.
From the problem, we know that:
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The first act was 18 minutes shorter than the third act, so F = T - 18.
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The first act was half as long as the second act, so F = S/2.
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The average length of the three acts was 46 minutes, so (F + S + T)/3 = 46.
Now we can solve these equations step by step.
First, substitute F from the first equation into the second equation: T - 18 = S/2. Multiply both sides by 2 to get S = 2T - 36.
Next, substitute F and S from the first and second equations into the third equation:
(T - 18 + 2T - 36 + T)/3 = 46 Simplify to get 4T - 54 = 138 Add 54 to both sides to get 4T = 192 Divide both sides by 4 to get T = 48.
So, the third act was 48 minutes long. The answer is D) 48.
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