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Tickets to a show cost $10 in advance and $15 at the door. If 120 tickets are sold for a total of $1390, how many tickets were bought in advance?

 Apr 5, 2020
 #1
avatar+252 
+2

Here is how I solve it:

Say all the tix were bought in advance: that would mean 1200 dollars were sold. For each ticket we turn into bought at the door, we gain 5 dollars. The difference of 1390 and 1200 is 190. So 190/5 38 were bought at the door. So 82 were bought in advance.

 

Check

38*15=570

82*10=820

820+570=1390

 

smileysmileysmiley

 Apr 5, 2020
edited by MooMooooMooM  Apr 5, 2020
 #2
avatar+631 
+1

That's a unique way to solve it!

 #3
avatar+34 
+2

We can solve this using a system of equations. 

Let \(x=\)tickets bought at the door

Let \(y=\)tickets bought in advance

 

Our equations are 

\(10y+15x=1390\)

\(y+x=120\)

 

We can solve elimination (Feel free to use a different strategy):

 

Multiply the bottom equation by \(10\)

 

\(10y+10x=1200\)

 

Subtract the bottom equation

 

\(5x=190\)

 

Divide both sides by \(5\)

 

\(x=38\)

 

Plug that value into the bottom equation (or the top if you want)

 

\(y+38=120\)

 

Subtract \(38\) from both sides

 

\(y=82\)

 

\(82\) tickets were bought at the door.

 

Hope this helped!!!

-Ako

 Apr 5, 2020

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