SOLUTION: A child is collecting state quarters and new $1 coins. If she has a total of 21 coins, and the number of quarters is 9 more than the number of dollar coins, how many of each type o

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Question 1175437: A child is collecting state quarters and new $1 coins. If she has a total of 21 coins, and the number of quarters is 9 more than the number of dollar coins, how many of each type of coin does she have?

Found 2 solutions by ikleyn, ankor@dixie-net.com:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

You can solve it in this way, using one single equation


    D + (D + 9) = 21  coins


        2D      = 21 - 9 = 12

         D               = 12/2 = 6


obtaining the answer  6  one-dollar coins  and  6+9 = 15 quarters.



Or, alternatively, you can write a system of two equations


    Q + D = 21

    Q - D =  9


and then add the equations, obtaining

    2Q = 21 + 9 = 30

     Q          = 30/2 = 15  quarters and  15-9 = 6 one-dollar coins.



You can choose any of these two way - they lead you to the same answer.

Solved, explained and completed.


--------------

To see a million other similar solved problems, look into the lesson
    - Word problems that lead to a simple system of two equations in two unknowns
in this site.



Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
A child is collecting state quarters and new $1 coins.
If she has a total of 21 coins, and the number of quarters is 9 more than the number of dollar coins, how many of each type of coin does she have?
let x = no. of dollar coins
She has 9 more quarter than dollar coins, therefore
(x+9) = no. of quarters
:
x + (x+9) = 21
2x = 21 - 9
2x = 12
x = 12/2
x = 6 dollar coins
:
I'll let you find the number of quarters, see that they add up to 21

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