SOLUTION: Kevin and Randy Muise have a jar containing 28 coins, all of which are either quarters or nickels. The total value of the coins in the jar is 4.00. How many of each type of c

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Kevin and Randy Muise have a jar containing 28 coins, all of which are either quarters or nickels. The total value of the coins in the jar is 4.00. How many of each type of c      Log On


   

Question 1035410: Kevin and Randy Muise have a jar containing
28 coins, all of which are either quarters or nickels.
The total value of the coins in the jar is 4.00.
How many of each type of coin do they have?
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Kevin and Randy Muise have a jar containing
28 coins, all of which are either quarters or nickels.
The total value of the coins in the jar is $4.00.
How many of each type of coin do they have?
Let the number of quarters be x
Let the number of nickels be y


                      Value      Value
Type       Number       of         of
 of          of        EACH       ALL
coin        coins      coin      coins
-------------------------------------------
quarters      x      $0.25       $0.25x
nickels       y      $0.05       $0.05y
-------------------------------------------
TOTALS       28      -----       $4.00

 The first equation comes from the "Number of coins" column.

  %28matrix%283%2C1%2CNumber%2Cof%2Cquarters%29%29%22%22%2B%22%22%28matrix%283%2C1%2CNumber%2Cof%2Cnickels%29%29%22%22=%22%22%28matrix%284%2C1%2Ctotal%2Cnumber%2Cof%2Ccoins%29%29
                 x + y = 28

 The second equation comes from the "Value of all coins" column.

  %28matrix%284%2C1%2CValue%2Cof%2CALL%2Cquarters%29%29%22%22%2B%22%22%28matrix%284%2C1%2CValue%2Cof%2CALL%2Cnickels%29%29%22%22=%22%22%28matrix%285%2C1%2CTotal%2Cvalue%2Cof%2CALL%2Ccoins%29%29

           0.25x + 0.05y = 4

Get rid of decimals by multiplying every term by 100:

                25x + 5y = 400

 So we have the system of equations:
           system%28x+%2B+y+=+28%2C25x+%2B+5y+=+400%29.

We solve by substitution.  Solve the first equation for y:

           x + y = 28
               y = 28 - x

Substitute (28 - x) for y in 25x + 5y = 400

    25x + 5(28 - x) = 400
     25x + 140 - 5x = 400
          20x + 140 = 400
                20x = 260
                  x = 13 = the number of quarters.

Substitute in y = 28 - x
              y = 28 - (13)
              y = 15 nickels.

Checking:  13 quarters is $3.25 and 15 nickels is $0.75
            That's 28 coins.
            And indeed $3.25 + $0.75 = $4.00
Edwin

Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.
Kevin and Randy Muise have a jar containing
28 coins, all of which are either quarters or nickels.
The total value of the coins in the jar is 4.00.
How many of each type of coin do they have?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let n = the number of nickels.
Then the number quarters is 28-n

Nickels value is 5n.
Quarters value is 25*(28-n).

The "total value equation" is

5n + 25*(28-n) = 400  (written in cents).

5n + 700 - 25n = 400,

-20n = 400 - 700,

-20n = -300,

n = 15.  15 nickels.

And 28-15 = 13 quarters.

Answer.  15 nickels and 13 quarters.

Check. 5*15 + 25*13 = 400.   OK!