Question

Ninety-three passengers rode in a train from City A to City B. Tickets for regular coach seats cost $115. Tickets for sleeper cars seats cost $290. The receipts for the trip totaled $20,320. How many passengers purchased each type of ticket?

Answer 1

To solve this, we must set up a Systems of Equations problem 
Let's say C stands for Coach seats bought and S stands for sleeper cars bought
Here are our equations we have enough information to make. 
c + s = 93 
115c + 290s = $20,320 
Lets solve for Coach seats first 
To do that we need a value for s 
In the first equation, s = 93 - c
Lets plug that in for S in the second equation 
115c + 290(93 - c) = $20,320 
Distribute and simplify 
115c + 26,970 - 290c = $20,320 
Combine like terms 
-175c + 26,970 = $20,320 
-175c = -6650
divide 
c = $\frac{-6650}{-175}$
c = 38
We just found out 38 coach seats were bought. 
Plug in 38 for C in this equation ⇒ 115c + 290s = $20,320 to find S
115(38) + 290s = $20,320 
Distribute and Simplify 
4,370 + 290s = $20,320 
290s = $15,950 
Divide 
S = $\frac{15950}{290}$
S = 55
We just completed the problem!
38 Coach seats were bought and 
55 Sleeper car seats were bought.