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Thread: quarters and dimes

  1. #1
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    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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shenton View Post
    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
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  3. #3
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    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.

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