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# help

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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
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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   Apr 5, 2020
edited by MooMooooMooM  Apr 5, 2020
#2
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That's a unique way to solve it!

#3
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We can solve this using a system of equations.

Let \(x=\)tickets bought at the door

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