1. ## quarters and dimes

This is hard to visualize because quarters is 25 cents and dimes is 10 cents. I can't figure this out. Please help.

A student has $6.80 in dimes and quarters. The number of dimes is two more than half the number of quarters. How many of each type of coin does the student have? Answer key: 22 quarters, 13 dimes Thanks. 2. Originally Posted by shenton This is hard to visualize because quarters is 25 cents and dimes is 10 cents. I can't figure this out. Please help. A student has$6.80 in dimes and quarters. The number of dimes is two more than half the number of quarters. How many of each type of coin does the student have?

Answer key: 22 quarters, 13 dimes

Thanks.
Translate into an equation. let $\displaystyle x$ be the number of quarters, and the number of dimes is $\displaystyle 2+x/2$ then:

$\displaystyle 680=25\,x+10\,(2+x/2)$.

Now solve this for $\displaystyle x$ which tells you the number of Q's, and from that the number of D's can be found.

RonL

3. Hello, shenton!

If you're familiar with Systems of Equations, here's another approach.

A student has $6.80 in dimes and quarters. The number of dimes is two more than half the number of quarters. How many of each type of coin does the student have? Answer key: 22 quarters, 13 dimes Let:$\displaystyle \begin{array}{cc}D \:= & \text{number of dimes} \\ Q\:= & \text{number of quarters}\end{array}$The$\displaystyle D$dimes are worth 10¢ each. . . Their value is:$\displaystyle 10D$cents. The$\displaystyle Q$quarters are worth 25¢ each. . . Their value is:$\displaystyle 25Q$cents. The total value of the coins is: .$\displaystyle 10D + 25Q\:=\:680$(cents) [1]$\displaystyle \underbrace{\text{The number of dimes}}\;\underbrace{\text{is}}\;\underbrace{\text {two more than}}\;\underbrace{\text{half the number of quarters}}$. . . . . .$\displaystyle D\qquad\qquad\quad =\qquad\quad2\quad+\qquad\qquad\qquad\quad\frac{1} {2}Q $We have: .$\displaystyle D\:=\:2 + \frac{1}{2}Q\quad\Rightarrow\quad 2D - Q\:=\:4$[2] Divide [1] by 5: .$\displaystyle 2D + 5Q\:=\:136$. . Subtract [2]: .$\displaystyle 2D \,- \;Q \:= \;\;4$And we have: .$\displaystyle 6Q\,=\,132\quad\Rightarrow\quad\boxed{Q \,=\,22}$Substitute into [2]: .$\displaystyle 2D - 22 \:=\:4\quad\Rightarrow\quad\boxed{D\,=\,13}$Therefore, there are$\displaystyle 22$quarters and$\displaystyle 13\$ dimes.