Isabella has some dimes and some quarters. She has at most 25 coins worth a minimum of $4.45 combined. If Isabella has 17 dimes, determine all possible values for the number of quarters that she could have.

No Solutions

Step-by-step explanation:

Define Variables:

May choose any letters.

\text{Let }d=

Let d=

\,\,\text{the number of dimes}

the number of dimes

\text{Let }q=

Let q=

\,\,\text{the number of quarters}

the number of quarters

\text{\textquotedblleft at most 25 coins"}\rightarrow \text{25 or fewer coins}

“at most 25 coins"→25 or fewer coins

Use a \le≤ symbol

Therefore the total number of coins, d+qd+q, must be less than or equal to 25:25:

d+q\le 25

d+q≤25

\text{\textquotedblleft a minimum of \$4.45"}\rightarrow \text{\$4.45 or more}

“a minimum of $4.45"→$4.45 or more

Use a \ge≥ symbol

One dime is worth $0.10, so dd dimes are worth 0.10d.0.10d. One quarter is worth $0.25, so qq quarters are worth 0.25q.0.25q. The total 0.10d+0.25q0.10d+0.25q must be greater than or equal to \$4.45:$4.45:

0.10d+0.25q\ge 4.45

0.10d+0.25q≥4.45

\text{Plug in }\color{green}{17}\text{ for }d\text{ and solve each inequality:}

Plug in 17 for d and solve each inequality:

Isabella has 17 dimes

\begin{aligned}d+q\le 25\hspace{10px}\text{and}\hspace{10px}&0.10d+0.25q\ge 4.45 \\ \color{green}{17}+q\le 25\hspace{10px}\text{and}\hspace{10px}&0.10\left(\color{green}{17}\right)+0.25q\ge 4.45 \\ q\le 8\hspace{10px}\text{and}\hspace{10px}&1.70+0.25q\ge 4.45 \\ \hspace{10px}&0.25q\ge 2.75 \\ \hspace{10px}&q\ge 11 \\ \end{aligned}

d+q≤25and

17+q≤25and

q≤8and

0.10d+0.25q≥4.45

0.10(17)+0.25q≥4.45

1.70+0.25q≥4.45

0.25q≥2.75

q≥11

\text{It is not possible to have }q\le 8\text{ AND to have }q\ge 11\text{.}

It is not possible to have q≤8 AND to have q≥11.

\text{Therefore there is NO SOLUTION}

Therefore there is NO SOLUTION