A girl scout troop sells 62 tickets to their mother and daughter dinner, for a total of $216.If the tickets cost $4.00 for mothers and $3.00 for daughters how many of each ticket did they sell?


3 Answers

Mandy Clark Profile
Mandy Clark answered
Ok so you need 2 equations. I like to use variables that go with the problem
Ok so the first equation is the quantity equation. How many of each ticket was sold for a total of how many tickets? We don't know how many of each ticket was sold, so that's where the variables come in.
the second equation is the value equation. How much is each ticket worth? Then it equals the total cost of all the tickets
Ok so you have
You have to get one variable to eliminate it. This is appropriately named the elimination method:) personally, I like to multiply. I don't know why it just works easier for me. So I need to get one variable the same, so I'll solve for M, so I'll multiply equation 1 by 4.
4M+4D=248 I multiplied the entire thing by 4
4M+3D=216 now that you have a variable the same (4M) you can subtract the second equation from the first
4M-4M is 0, so that's gone and you're left with
D=32 plug 32 back into the very first equation
M+32=62 subtract 32 from 62
so 32 daughter tickets were sold and 30 mother tickets were sold:D
Oddman Profile
Oddman answered
If all 62 tickets are sold for daughters, the revenue is $186. The revenue is actually $30 more. We know that selling a mother ticket instead of a daughter ticket increases revenue by $1, so there must have been 30 mother tickets sold. The remaining 32 tickets were daughter tickets.
John McCann Profile
John McCann answered
A systems problem. Change all money to pennies for convenience of operation.

M = moms

D = daughters

- 300 (M + D = 62)
400M + 300D = 21600

- 300M - 300D = - 18600
400M + 300D = 21600

100M = 3000

M (moms) = 30 tickets * $4.00 = $120.00

D(daughters) = 32 tickets * $3.00 = $96.00
------add check

62 tickets and $216.00
------all checks, so Moms bought 30 tickets and daughters bought 32

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