Can this be solved with this info?

**Twig** · March 21st 2008, 04:03 PM

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?

**math sucks** · March 21st 2008, 04:41 PM

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 $y=7$ .

**Soroban** · March 21st 2008, 05:00 PM

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 $C$ = 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.

**Twig** · March 22nd 2008, 09:57 AM

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.

**Twig** · August 5th 2008, 11:51 AM

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

There seem to be many different approaches.

**CaptainBlack** · August 6th 2008, 02:41 AM

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

**Twig** · August 6th 2008, 06:07 AM

hi

Can anyone please explain how "mathsucks" solution works?

Thanks

Thread: Can this be solved with this info?

LinkBack

Thread Tools

Display

Can this be solved with this info?

hey

Similar Math Help Forum Discussions

Getting Less info same stat as more info

Is there enough info to do this problem?

Graph Info

Limit...using given info

info help for matlab/ODE

Search Tags