# algebra (combinations)

• Feb 1st 2010, 01:51 PM
sri340
algebra (combinations)
If Julie has $1.25 in dimes and nickels, how many different combinations of the coins could she have? • Feb 1st 2010, 05:39 PM TheEmptySet Quote: Originally Posted by sri340 If Julie has$1.25 in dimes and nickels, how many different combinations of the coins could she have?

Setting up the relationship we get

$\displaystyle .25Q+.05N=1.25 \iff 25Q+5N=125$

Now lets isolate N to get

$\displaystyle 5N=125-25Q \iff N =25-5Q \iff N=5(5-Q)$
Now just plug in in $\displaystyle Q=0,1,2,3,4,5$ to get all of the ordered pairs.
• Feb 1st 2010, 05:42 PM
Soroban
Hello, sri340!

Quote:

Julie has $1.25 in dimes and nickels. How many different combinations of the coins could she have? Let: .$\displaystyle \begin{array}{ccc} d &=& \text{no. of dimes} \\ n&=&\text{no. of nickels} \end{array}$Then: .$\displaystyle \begin{array}{c}d\text{ dimes have a value of }10d\text{ cents} \\ n\text{ nickels have a value of }5n\text{ cents}\end{array}$Their total value is 125 cents: .$\displaystyle 10d + 5n \:=\:125 \quad\Rightarrow\quad n \:=\:25 - 2d$So we have: . . .$\displaystyle \begin{array}{|c|c|} \hline
\text{Dimes} & \text{Nickels} \\ \hline
0 & 25 \\ 1 & 23 \\ 2 & 21 \\ 3 & 19 \\ \vdots & \vdots \\ 10 & 5 \\ 11 & 3 \\ 12 & 1 \\ \hline\end{array}\$

There are 13 combinations.