# Thread: Can this be solved with this info?

1. ## Can this be solved with this info?

Hi!

I read this problem in yesterday and I bothers me that I cant put up an equation on how to solve it.

A bunch of people were going to ride a rollercoaster.
A childrens ticket was $15, and an adult ticket was$22.
Tickets were sold for a total of $349. How many children and adults were there? 2. Let$\displaystyle x$be the number of children and$\displaystyle y$be the number of adults. Then$\displaystyle 15x+22y=349$Since$\displaystyle 349\equiv19\mod22$, we must require that$\displaystyle 15x\equiv19\mod22$. Note that$\displaystyle 15\times3\equiv1\mod22$, so$\displaystyle x\equiv3\times19\equiv13\mod22$. Since$\displaystyle 0\leq x\leq23$, we must require$\displaystyle x=13$. So$\displaystyle x=13$and$\displaystyle y=7$. 3. Hello, Twig! This can be solved if you're familiar with Modulo Arithmetic. But I'll show a primitive algebraic approach. A bunch of people were going to ride a rollercoaster. A childrens ticket was$15, and an adult ticket was $22. Tickets were sold for a total of$349.
How many children and adults were there?
Let $\displaystyle C$ = number of children.
Let $\displaystyle A$ = number of adults.
Note that $\displaystyle A\text{ and }C$ are nonnegative integers.

The $\displaystyle C$ children's tickets were $15 each; they cost a total of$\displaystyle 15C$dollars. The$\displaystyle A$adult tickets were$22 each; they cost a total of $\displaystyle 22A$ dollars.

The total ticket sales was $349: .$\displaystyle 15C + 22A \:=\:349$Solve for$\displaystyle A\!:\;\;A \:=\:\frac{349-15C}{22}$Then: .$\displaystyle A \;=\;\frac{(330 + 19) + (-22C + 7C)}{22} \;=\;\frac{(330-22c) + (19 + 7C)}{22}$. .$\displaystyle A \;=\;\frac{330-22C}{22} + \frac{19C + 7C}{22} \quad\Rightarrow\quad A\;=\;15 - C + \frac{19+7C}{22}$.[1] Since$\displaystyle A$is an integer,$\displaystyle 19+7C$must be a multiple of 22. . .$\displaystyle 19 + 7C \:=\:22b\;\text{ for some integer }b.$Solve for$\displaystyle C\!:\;\;C \:=\:\frac{22b-19}{7} \;=\;\frac{(21b + b) +(-21 + 2)}{7}$. .$\displaystyle C \;=\;\frac{(21b - 21) + (b + 2)}{7}\quad\Rightarrow\quad C \;=\;3b - 3 + \frac{b+2}{7}$.[2] Since$\displaystyle C$is an integer,$\displaystyle b+2$must be a multiple of 7. The least value of$\displaystyle b$is: .$\displaystyle b = 5$Substitute into [2]: .$\displaystyle C \;=\;3(5) - 3 \frac{5+2}{7}\quad\Rightarrow\quad\boxed{ C \,=\,13}$Substitute into [1]: .$\displaystyle A \;=\;15 - 13 + \frac{19+7(13)}{22} \quad\Rightarrow\quad\boxed{ A \,=\,7}\$

There were 13 children and 7 adults.

4. ## hey

Hi!

I am not familiar with the Modulo Arithemtic.

I didnt not understand your explanation. I followed up till the "Solve for A".
Then I dont what you are doing.

PS: I took this problem from my little brothers math book and he is 15....I cant imagine that they would need this complex math.

5. I learned a method of solving these types of problems today.
Called Diofantic eqauations.

There seem to be many different approaches.

6. Originally Posted by Twig
I learned a method of solving these types of problems today.
Called Diofantic eqauations.

There seem to be many different approaches.
Diophantine, in English anyway.

RonL

7. hi

Can anyone please explain how "mathsucks" solution works?

Thanks