Matthew R.

asked • 10/28/16

a cashier has 25 bills consisiting of twice as amny ones as tens, two fewer fives than ones, and the rest twenties. if the total value is $140findhowmanyshe has

stumped and confused

Elizabeth H.

x = number of ones
y = number of fives
z = number of tens
n = number of twenties
 
Total number of bills is 25, so
x + y + z + n = 25
 
Total value is $140, so
$1(x) + $5(y) + $10(z) + $20(n) = $140
 
You know twice as many ones as tens, so
x = 2z  OR  z = x/2
 
You know two fewer fives than ones, so
x = y + 2
 
Putting these two equations together, you get
z = (y + 2) / 2  OR  z = y/2 + 1
 
Now substitute those values for x and z into the first equation above
   x     + y +       z      +  n = 25
(y+2) + y + (y/2 + 1) + n = 25
2y + y/2 + 3 + n = 25
n = 22 - 2y - y/2
 
Now substitute the values for x, z, and n into the second equation above
   1x      + 5y +         10z     +          20n            = 140
1(y + 2) + 5y + 10(y/2 + 1) + 20(22 - 2y - y/2) = 140
  y + 2   + 5y +    5y + 10    + 440 - 40y - 10y   = 140
452 - 39y = 140
-39y = -312
y = 8
 
So to answer the question:
y (number of fives) = 8
x (number of ones) = y + 2 = 10
z (number of tens) = (y/2) + 1 = 4 + 1 = 5
n (number of twenties) = 22 - 2y - y/2 = 22 - 16 - 4 = 2
 
So there are 10 ones, 8 fives, 5 tens and 2 twenties
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10/28/16

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