quarters and dimes

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December 22nd 2006, 08:29 PM
shenton

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.
December 22nd 2006, 10:06 PM
CaptainBlack

Quote:

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 $x$ be the number of quarters, and the number of dimes is $2+x/2$ then:

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

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

RonL
December 23rd 2006, 04:06 AM
Soroban

Hello, shenton!

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

Quote:

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: $\begin{array}{cc}D \:= & \text{number of dimes} \\ Q\:= & \text{number of quarters}\end{array}$

The $D$ dimes are worth 10¢ each.
. . Their value is: $10D$ cents.
The $Q$ quarters are worth 25¢ each.
. . Their value is: $25Q$ cents.

The total value of the coins is: . $10D + 25Q\:=\:680$ (cents) [1]

$\underbrace{\text{The number of dimes}}\;\underbrace{\text{is}}\;\underbrace{\text {two more than}}\;\underbrace{\text{half the number of quarters}}$
. . . . . . $D\qquad\qquad\quad =\qquad\quad2\quad+\qquad\qquad\qquad\quad\frac{1} {2}Q$
We have: . $D\:=\:2 + \frac{1}{2}Q\quad\Rightarrow\quad 2D - Q\:=\:4$ [2]

Divide [1] by 5: . $2D + 5Q\:=\:136$
. . Subtract [2]: . $2D \,- \;Q \:= \;\;4$

And we have: . $6Q\,=\,132\quad\Rightarrow\quad\boxed{Q \,=\,22}$

Substitute into [2]: . $2D - 22 \:=\:4\quad\Rightarrow\quad\boxed{D\,=\,13}$

Therefore, there are $22$ quarters and $13$ dimes.