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Chapter 3: Problem 30

Seven airline passengers in economy class on the same flight paid an average of \(\$ 361\) per ticket. Because the tickets were purchased at different times and from different sources, the prices varied. The first five passengers paid \(\$ 420, \$ 210, \$ 333, \$ 695\), and \(\$ 485\). The sixth and seventh tickets were purchased by a couple who paid identical fares. What price did each of them pay?

Short Answer

Expert verified
Each of the last two airline passengers paid $192

Step by step solution

01

Calculate the total cost for all tickets

This can be done by multiplying the average ticket price by the total number of tickets. Using the provided average ticket price of $361 and the total number of tickets being 7, the total cost for all tickets is therefore \( 7 \times 361 = 2527$ \).
02

Calculate the total cost for the first five tickets

Simply add together the costs of the first five tickets, which are $420, $210, $333, $695, and $485. This gives us a total of \( 420 + 210 + 333 + 695 + 485 = 2143$ \).
03

Find the cost of the last two tickets

Subtract the total cost of the first five tickets from the total cost for all tickets. In other words, subtract the value from step 2 from the value from step 1. \( 2527 - 2143 = 384$ \). Since the sixth and seventh tickets cost the same, each ticket is therefore \( 384/2 = 192$ \).

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