# Can this be solved with this info?

• March 21st 2008, 04:03 PM
Twig
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? • March 21st 2008, 04:41 PM math sucks Let $x$ be the number of children and $y$ be the number of adults. Then $15x+22y=349$ Since $349\equiv19\mod22$, we must require that $15x\equiv19\mod22$. Note that $15\times3\equiv1\mod22$, so $x\equiv3\times19\equiv13\mod22$. Since $0\leq x\leq23$, we must require $x=13$. So $x=13$ and $y=7$. • March 21st 2008, 05:00 PM Soroban Hello, Twig! This can be solved if you're familiar with Modulo Arithmetic. But I'll show a primitive algebraic approach. Quote: 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 $C$ = number of children.
Let $A$ = number of adults.
Note that $A\text{ and }C$ are nonnegative integers.

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

The total ticket sales was \$349: . $15C + 22A \:=\:349$

Solve for $A\!:\;\;A \:=\:\frac{349-15C}{22}$

Then: . $A \;=\;\frac{(330 + 19) + (-22C + 7C)}{22} \;=\;\frac{(330-22c) + (19 + 7C)}{22}$

. . $A \;=\;\frac{330-22C}{22} + \frac{19C + 7C}{22} \quad\Rightarrow\quad A\;=\;15 - C + \frac{19+7C}{22}$ .[1]

Since $A$ is an integer, $19+7C$ must be a multiple of 22.
. . $19 + 7C \:=\:22b\;\text{ for some integer }b.$

Solve for $C\!:\;\;C \:=\:\frac{22b-19}{7} \;=\;\frac{(21b + b) +(-21 + 2)}{7}$

. . $C \;=\;\frac{(21b - 21) + (b + 2)}{7}\quad\Rightarrow\quad C \;=\;3b - 3 + \frac{b+2}{7}$ .[2]

Since $C$ is an integer, $b+2$ must be a multiple of 7.
The least value of $b$ is: . $b = 5$

Substitute into [2]: . $C \;=\;3(5) - 3 \frac{5+2}{7}\quad\Rightarrow\quad\boxed{ C \,=\,13}$

Substitute into [1]: . $A \;=\;15 - 13 + \frac{19+7(13)}{22} \quad\Rightarrow\quad\boxed{ A \,=\,7}$

There were 13 children and 7 adults.

• March 22nd 2008, 09:57 AM
Twig
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.
• August 5th 2008, 11:51 AM
Twig
I learned a method of solving these types of problems today.
Called Diofantic eqauations.

There seem to be many different approaches.
• August 6th 2008, 02:41 AM
CaptainBlack
Quote:

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
• August 6th 2008, 06:07 AM
Twig
hi

Can anyone please explain how "mathsucks" solution works?

Thanks